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2

AP Physics 1 Momentum

2015shy12shy02

wwwnjctlorg

3

Table of Contents

bull Momentum

bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects

bull Types of Collisions

Click on the topic to go to that section

bull Conservation of Momentum

bull Collisions in Two Dimensions

4

Return toTable ofContents

Momentum

5

Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)

In our experience

bull When objects of different mass travel with the same velocity the one with more mass is harder to stop

bull When objects of equal mass travel with different velocities the faster one is harder to stop

Momentum Defined

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

2

AP Physics 1 Momentum

2015shy12shy02

wwwnjctlorg

3

Table of Contents

bull Momentum

bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects

bull Types of Collisions

Click on the topic to go to that section

bull Conservation of Momentum

bull Collisions in Two Dimensions

4

Return toTable ofContents

Momentum

5

Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)

In our experience

bull When objects of different mass travel with the same velocity the one with more mass is harder to stop

bull When objects of equal mass travel with different velocities the faster one is harder to stop

Momentum Defined

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

3

Table of Contents

bull Momentum

bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects

bull Types of Collisions

Click on the topic to go to that section

bull Conservation of Momentum

bull Collisions in Two Dimensions

4

Return toTable ofContents

Momentum

5

Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)

In our experience

bull When objects of different mass travel with the same velocity the one with more mass is harder to stop

bull When objects of equal mass travel with different velocities the faster one is harder to stop

Momentum Defined

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

4

Return toTable ofContents

Momentum

5

Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)

In our experience

bull When objects of different mass travel with the same velocity the one with more mass is harder to stop

bull When objects of equal mass travel with different velocities the faster one is harder to stop

Momentum Defined

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

5

Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)

In our experience

bull When objects of different mass travel with the same velocity the one with more mass is harder to stop

bull When objects of equal mass travel with different velocities the faster one is harder to stop

Momentum Defined

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

6

Define a new quantity momentum (p) that takes these observations into account

momentum = mass times velocity

Momentum Defined

p =mv

click here for a introductory video on momentum from Bill Nye

7

Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector

andbull momentum = mass times velocity

therefore

Momentum is a vector quantity shy it has magnitude and direction

Momentum is a Vector Quantity

8

There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist

We use the product of the units of mass and velocity

mass x velocity

kgsdotms

SI Unit for Momentum

9

1 Which has more momentum

A A large truck moving at 30 ms

B A small car moving at 30 ms

C Both have the same momentum Answ

er

10

2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms

Answ

er

11

3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left

Answ

er

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

8

There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist

We use the product of the units of mass and velocity

mass x velocity

kgsdotms

SI Unit for Momentum

9

1 Which has more momentum

A A large truck moving at 30 ms

B A small car moving at 30 ms

C Both have the same momentum Answ

er

10

2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms

Answ

er

11

3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left

Answ

er

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

9

1 Which has more momentum

A A large truck moving at 30 ms

B A small car moving at 30 ms

C Both have the same momentum Answ

er

10

2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms

Answ

er

11

3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left

Answ

er

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

10

2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms

Answ

er

11

3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left

Answ

er

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

11

3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left

Answ

er

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

12

4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms

Answ

er

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

13

5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms

Answ

er

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

14

ImpulseshyMomentum Equation

Return toTable ofContents

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

15

Suppose that there is an event that changes an objects momentum

bull from p0 shy the initial momentum (just before the event)

bull by Δp shy the change in momentum

bull to pf shy the final momentum (just after the event)

The equation for momentum change is

Change in Momentum

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

16

Momentum change equation

Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force

Momentum Change = Impulse

Look at the above equation Can you relate Newtons First Law to the Δp term

Δp would represent the external force

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

17

Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner

Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time

Momentum Change = Impulse

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

18

The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation

Impulse Momentum Equation

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

19

There no special unit for impulse We use the product of the units of force and time

force x time

Nsdots

Recall that N=kgsdotms2 so

Nsdots=kgsdotms 2 x s

= kgsdotms

SI Unit for Impulse

This is also the unit for momentum which is a good thing since Impulse is the change in momentum

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

20

6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers

A Increase the applied force

B Increase the time that the force is applied

C Maintain the same applied force

D Decrease the time that the force is applied

Answ

er

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

21

7 From which law or principle is the ImpulseshyMomentum equation derived from

A Conservation of Energy

B Newtons First Law

C Newtons Second Law

D Conservation of Momentum Answ

er

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

22

8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer

Answ

er

Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

23

Effect of Collision Time on Force

∆t (seconds)

F (newtons) Since force is inversely

proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object

Impulse = F(∆t) = F(∆t)

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

24

Every Day Applications

Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as

bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

25

Every Day Applications

I=FΔt=Δp Lets analyze two specific cases from the previous list

bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

26

Car Air Bags

I=FΔt=Δp

In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window

They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

27

Car Air Bags

I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you

The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating

Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

28

Car Air Bags

I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags

Rearranging the equation we have

F represents the Force delivered to the passenger due to the accident

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

29

Car Air Bags

I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller

Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure

Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

30

9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers

A An absolutely rigid car body that doesnt deform

B Deployment of an air bag for each adult in the car

C Deployment of an air bag only for the driver

D The proper wearing of a seatbelt or child seat for each person in the car

Answ

er

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

31

10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest

A Force delivered to the car

B Interval of time for the collision

C Momentum change

D Acceleration of the car Answ

er

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

32

Hitting a Baseball

I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum

This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)

Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum

What is the goal of the batter

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

33

Hitting a Baseball

I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats

Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

34

Hitting a Baseball

I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing

The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

35

Every Day Applications

I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied

Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car

bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

36

11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system

Answ

er

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

37

12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system

Answ

er

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

38

13 The momentum change of an object is equal to the ______

A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it

Answ

er

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

39

14 Air bags are used in cars because they

B Answ

er

A increase the force with which a passenger hits the dashboard

B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

40

15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers

A change in momentum

B force on the car

C impact time

D final momentum

Answ

er

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

41

16 In order to increase the final momentum of a golf ball the golfer should Select two answers

A maintain the speed of the golf club after the impact (follow through)

B Hit the ball with a greater force

C Decrease the time of contact between the club and the ball

D Decrease the initial momentum of the golf club Answ

er

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

er

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

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Attachments

Eqn1pict

42

17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force

Answ

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43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

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44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

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45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

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49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

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50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

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51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

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The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

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75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

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76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

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77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

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92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

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93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

43

18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity

Answ

er

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

44

19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car

Answ

er

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

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59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

45

Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems

So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time

In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph

Start with representing a constant force over a time interval Δt and plot Force vs time on a graph

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

46

Graphical Interpretation of Impulse

Note the area under the ForceshyTime graph is

height times base = F ∆t = I (impulse) = ∆p (change in momentum)

F [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

47

Graphical Interpretation of ImpulseF [N]

t [s]

F

∆t

Area =

Impulse (Change in Momentum)

By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

48

20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s

Answ

er

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

49

21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s

Answ

er

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

50

22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s

Answ

er

51

The Momentum of a System of Objects

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52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

51

The Momentum of a System of Objects

Return toTable ofContents

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

52

If a system contains more than one object its total momentum is the vector sum of the momenta of those objects

psystem = sum p

psystem = p1 + p2 + p3 +

psystem = m1v1 + m2v2 + m3v3 +

The Momentum of a System of Objects

Its critically important to note that

momenta add as vectors

not as scalars

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

53

psystem = m1v1 + m2v2 + m3v3 +

In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction

The Momentum of a System of Objects

Then Add the momenta to get a total momentum of the system

+shy

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

54

Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms

6 kg 14 kg13 ms 7 ms East (+)

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

55

23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south

Answ

er

psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)

direction is North

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

56

24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west

Answ

er

psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

57

25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south

Answ

er

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)

= 14 kgsdotms (North)

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

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110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

58

Conservation of Momentum

Return toTable ofContents

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

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67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

59

Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary

Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event

Conservation Laws

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

60

In the last unit we learned that energy is conserved

Like energy momentum is a conserved property of nature This means

bull Momentum is not created or destroyed

bull The total momentum in a closed system is always the same

bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force

Momentum is Conserved

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

61

We will use the Conservation of Momentum to explain and predict the motion of a system of objects

As with energy it will only be necessary to compare the system at two times just before and just after an event

Conservation of Momentum

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

62

The momentum of an object changes when it experiences an impulse (I=FΔt)

This impulse arises when a nonshyzero external force acts on the object

This is exactly the same for a system of objects

Conservation of Momentum

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

63

Conservation of Momentum

If there is no net external force on the system then F = 0 and since I = FΔt I is also zero

Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

64

26 Why dont the internal forces on a system change the momentum of the system

Answ

er

The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

65

27 An external positive force acts on a system of objects Which of the following are true Select two answers

A The velocity of the system remains the same

B The velocity of the system increases

C The momentum of the system decreases

D The momentum of the system increases

Answ

er

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

66

Types of Collisions

Return toTable ofContents

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

67

Types of Collisions

Objects in an isolated system can interact with each other in two basic ways

bull They can collide

bull If they are stuck together they can explode (push apart)

In an isolated system both momentum and total energy are conserved But the energy can change from one form to another

Conservation of momentum and change in kinetic energy can help predict what will happen in these events

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

68

Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form

bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

69

Inelastic Collisions

There are two types of Inelastic Collisions

bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy

bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

70

Elastic Collisions

There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions

In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law

Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

71

Explosions

A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound

A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound

Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

72

Explosions

In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed

An inelastic collision transforms kinetic energy into other forms of energy such as potential energy

An explosion changes potential energy into kinetic energy

Thus the equations to predict their motion will be inverted

The next slide summarizes the four types of collisions andexplosions

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

73

Collisions and Explosions

Event Description Momentum Conserved

Kinetic Energy Conserved

General Inelastic Collision

Objects bounce

off each otherYes

No Kinetic energy is converted to other forms of energy

Perfect Inelastic Collision

Objects stick together Yes

No Kinetic energy is converted to other forms of energy

Elastic Collision

Objects bounce

off each otherYes Yes

ExplosionObject or

objects break apart

Yes

No Release of potential energy increases kinetic

energy

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

74

28 In _______ collisions momentum is conserved

A Elastic

B Inelastic

C All

Answ

er

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

75

29 In ______ collisions kinetic energy is conserved

A Elastic

B Inelastic

C All

Answ

er

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

76

30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers

A Nothing kinetic energy is conserved

B More kinetic energy

C Thermal energy

D Light energy Answ

er

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

77

Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating

A BmAvA mBvB

A B

A BmAvA mBvB

+xthe prime means after

If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

78

Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object

A B

After (moving together)

pA+pB=(mA+ mB)v

A B

pA=mAvA

Before (moving towards each other)

pB=mBvB

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

79

31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (13500kg)(45ms) (13500+25000)kg= 16 ms

in the same direction as the first cars initial velocity

m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

80

32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it

Answ

er

m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)

= (100kg)(250ms) (100+15000)kg= 17 ms

in the same direction as cannon balls initial velocity

m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

81

33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision

Answ

er

m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

82

Explosions

In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

83

Explosions

A B

Before (moving together)

pA+pB=(mA+ mB)v

A BpA=mAvA

After (moving apart)

pB=mBvB

Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

84

34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion

Answ

er

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

85

35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going

Answ

er

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

86

Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved

This will give us two simultaneous equations to solve to predict their motion after the collision

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

87

Elastic Collisions

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA=mAvA pB=mBvB

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

88

Elastic Collisions

Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

89

Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)

frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22

m1v12 + m2v22 = m1v12 +m2v22

m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)

m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)

v1 + v1 = v2 + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 shy v2 = shy(v1 shy v2)

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

90

Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared

Do you recognize the terms on the left and right of the above equation And what does it mean

v1 shy v2 = shy(v1 shy v2)

The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

91

36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision

Answ

er

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

92

37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

93

38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2

Answ

er

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

94

Properties of Elastic Collisions

Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

95

Properties of Elastic Collisions

Using these equations we will predict the motion of three specific cases and relate them to observation

bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

96

Collision of same mass particles

v1 = v2 v2 = v1

m mm

v1

m

v2

v1 = v2 and v2 = v1

For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

97

Collision of same mass particles

The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

98

Collision of same mass particles

The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

99

Collision of same mass particles

The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)

v1 = v2 and v2 = v1

v1 = v2 v2 = v1

m mm

v1

m

v2

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

100

Large mass striking lighter mass at rest

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

v1 = v1 and v2 = 2v1

For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

101

Large mass striking lighter mass at rest

The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest

v1 = v1 and v2 = 2v1

m1 m1

m2 m2

v1 v2 = 0 v1 asymp v1 v2 = 2v1

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

102

Small mass striking heavier mass at rest

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

v1 = shyv1 and v2 asymp 0

For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

103

Small mass striking heavier mass at rest

v1 = shyv1 and v2 asymp 0

The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball

m2

m1

m2

v2 = 0

m1 v1v1

v2 = 0

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

104

39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball

Answ

er

v1shyv2 = shy(v1shyv2)

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

105

40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit

Answ

er

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

106

41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision

Answ

er

When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms

v1 = +6msv2 = shy3msv1 = v2 =

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

107

42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision

Answ

er

v1 = +6msv2 = shy3msv1 = v2 =

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

108

43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with

Answ

er

When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

109

Collisions in Two Dimensions

Return toTable ofContents

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

110

Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame

This means that the momentum conservation equation p = p can be solved independently for each component

This of course also applies to three dimensions but well stick with two for this chapter

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

111

Example Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection

Is momentum conserved in this problem

m

m

θ

θ

p

p

pxpy

pxpy

The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

112

Example Collision with a Wall

Momentum is not conserved

An external force from the wall is being applied to the ball in order to reverse its direction in the x axis

However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal

m

m

θ

θ

p

p

pxpy

pxpy

Now its time to resolve momentum into components along the x and y axis

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

113

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Apply the Impulse Momentum Theorem in two dimensions

What does this tell us

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

114

Example Collision with a Wall

m

m

θ

θ

p

p

pxpy

pxpy

Momentum is conserved in the y directionbut it is not conserved in the x direction

The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

115

44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball

A shymv

B shy2mv

C shymv cosθ

D shy2mv cosθ m

m

θ

θ

p

p

pxpy

pxpy An

swer

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

116

45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction

A shymv

B 0

C mv

D 2mv m

m

θ

θ

p

p

pxpy

pxpy An

swer

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

117

General Two Dimensional Collisions

m1 m2

Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

118

General Two Dimensional Collisions

This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal

To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest

The problem now is to find the momentum of m2 after the collision

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

119

General Two Dimensional Collisions

m1 m2

Before

p2 = 0

After

m1

m2

p2 =

This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions

Since the momentum in the y direction is zero before the collision it must be zero after the collision

And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

120

General Two Dimensional Collisions

m1

m2

Here is the momentum vector breakdown of mass 1 after the collision

m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1

m2

and this is the final momentum for mass 2 by vectorially adding the final px and py

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

121

46 After the collision shown below which of the following is the most likely momentum vector for the blue ball

A

B

C

D

E

before after

m1

m1

m2

m2

Answ

er

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

122

General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision

There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

123

General Two Dimensional Collisions

Given

Find

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

124

General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions

x direction yshydirection

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

125

before after600deg

120 kgshyms

θ

200 kgshyms

m2 m2

m1 m1

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

126

47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision

before after

30deg18 kgshyms

531degpinball

Answ

er

yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

127

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθ

One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

128

Perfect Inelastic Collisions in Two Dimensions

m1

m2Before

p1

p2

Afterm1m2

pθpx

py

pshyconservation in x in y

final momentum

final velocity

final direction

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

129

48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision

Answ

er

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

130

Explosions in Two Dimensions

p3

p2

p1

The Black object explodes into 3 pieces (blue red and green)

We want to determine the momentum of the third piece

During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

131

Explosions in Two Dimensions

θ

p1

p2

p3

The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece

before px = py = 0

after p1x + p2x + p3x = 0

p1y + p2y + p3y = 0

In this case the blue and red pieces are moving perpendicularly to each other so

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

132

49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece

A

B

C

D

Answ

er

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

133

50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece

Answ

er

Attachments

Eqn1pict

Attachments

Eqn1pict