MOMENTUM AND IMPULSE

Chapter 7

Linear Momentum, p

ntotal pp

vmp

Units of momentum: kg(m/s) or (N)(S)

Since velocity, v, is a vector, momentum, p, is a vector.

p is in the same direction as v

Momentum is a measure of how hard it is to stop or turn a moving object. It is moving inertia.

(single particle)

(system of particles)

Linear Momentum, p

vmp

Which car has more momentum? A or B

A

B

The faster car, A.

Linear Momentum, p

vmp

Which car has more momentum? A or B

A

B

The more massive vehicle, B.

Newton’s 2nd Law

t

pF

t

mvmv

t

vvm

t

vmmaF 00 )(

The rate of change of momentum of a body is equal to the net force applied to it.

Example - Washing a car: momentum change and force. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it (that is we ignore any splashing back). What is the force exerted by the water on the car?

Nt

pp

t

pF ifx

x 301

300

smv /20In each second, 1.5 kg of water leaves hose and has v=20 m/s.

0

/30)20(5.1

ff

ii

mvp

smkgmvp

By Newton’s 3rd law, force exerted by water on the car is +30 N

Example - Washing a car: momentum change and force. What if the water splashes back from the car? Would the force on the car be more or less?

ipsmkgmvp ii /30

0 ff mvp

If the water splashes back

In each second, 1.5 kg of water leaves hose and has v=20 m/s.

Change in momentum would be greater and so the force should be greater.The car exerts a force on the water not only to stop it, but an extra force to give it momentum in the opposite direction

fp

vi

t

vm

t

pFnet

pJtFnet Fnet

vNewtons 2nd Law

Slope ~ Fnet

Impulse, J

area = J=Dp

Contact begins

Contact ends

vf

ptFJt

pF

net

net

Impulse-Momentum Theorem

Impulse is the product of a net external force and time which results in a change in momentum

Impulsive forces are generally of high magnitude and short duration.

Units are N s or kg m/s

ptFJ net

Impulse-Momentum Theorem

0 2 4 6 8 100

1

2

3

4

5

t (sec)

F (

N)

“riding the punch”

Impulse is the area under an F-t graph.

DO NOW – Force to stop a car: momentum change, force and impulse. A 2200 kg vehicle traveling at 26 m/s can be stopped in 21 s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22s if it hits a concrete wall. What impulse is exerted on the vehicle in eachof these stops? What net force is exerted in each case? smv /26

ixfxxxx ppptFJ

For all three

x

ixfxx

Jsmkg

vvmJ

/200,57

)260(2200)(

DO NOW – Force to stop a car: momentum change, force and impulse. What net force is exerted on the vehicle in each of these stops?

t

JF x

x

smv /26

Nt

JFnet 2734

21

200,57

Nt

JFnet 053,15

8.3

200,57

Nt

JFnet 000,260

22.0

200,57

Gentle brake

Slam brake

Concrete wall

tFJ netxx

STOP BY: 0.13

Gs

0.70 Gs

12 Gs

Problem: This force acts on a 1.2kg object moving at 120.0m/s. The direction of force is aligned with velocity. What is the new velocity of the object?

0

1000

2000

3000

0 0.2 0.4 0.6 0.8 1

Fnet

(N)

t (s)

vf = 328 m/s

0 0.2 0.4 0.6 0.8 1

-3000

-2000

-1000

0

t (s)

F (

N)

Problem: This force acts on a 1.2kg object moving at 120.0m/s. The direction of force is aligned with velocity. What is the new velocity of the object?

vf = -88.3 m/s

Impulse

Impulse

pB

pC

2m

1.5m

Dp=?Fnet=?

pA=0

pD=0

EXAMPLE: A 100 g ball is dropped from a height of h = 2.00 m above the floor. It rebounds vertically to a height of h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on the ball. Assume that the time interval of the collision is 0.01 sec.

ExampleA 100 g ball is dropped from a height of h = 2.00 m above the floor. It rebounds vertically to a height of h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on the ball. Assume that the time interval of the collision is 0.01 seconds.

smghv

mvmgh

KU

EE

oB

BA

BGA

BA

/26.62*8.9*22

21 2

smghv

UK

EE

C

gDC

DC

/4.55.1*8.9*22

smkgp

smkgp

vmp

after

before

/*54.0)4.5(100.0

/*626.0)26.6(100.0

NF

F

vvmvmFt o

6.116

))26.6(4.5(100.0)01.0(

)(

Problem: A 150-g baseball moving at 40 m/s 15o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35o above the horizontal. What is the impulse exerted by the bat on the ball? If the collision took 2.3 ms, what was the average force of the bat on the ball?

vi = 40

vf = 55

)(54.12

))6.38(45(15.0

SN

mvmvpJ ixfxxx

35o

15o

)(17.3

)4.105.31(15.0

sN

mvmvpJ iyfyyy

022 2.14,9.12 sNIIIyx

ptFJ net

onet NF 2.14,5624

ptFI net

Impulse-Momentum

Theorem

Impulse on a graph: area under the F-t curve

KEdFW netnet

Work on a graph: area under the F-x curve

Work-EnergyTheorem

VECTOR SCALAR

Problem: a tennis player receives a shot with the ball (0.6 kg) travelling horizontally at 50.0 m/s and returns the shot with the ball travelling horizontally at 40.0m/s in the opposite direction. A) what is the impulse delivered to the ball by the racket? B) what work does the racquet do on the ball?

vi=50

vf=40Ns

mvmvpI ifxx

54)5040(6.0

J

mvmvKEWifnet

270)5040(3.0 22

2212

21

t

p

1

2

3

4

The figure gives the momentum vs time for a particle moving along an axis. A force directed along the axis acts on the particle.a) Rank the four indicated regions according

to the magnitude of the force, greatest first.

b) In which region is the particle slowing?

1 > 3 > 2 =4

3

Conservation of Momentum

1m1v1 2

m2v2

2m2v’2

1m1v’1

1 2F21F12 0extF

'' 22112211 vmvmvmvm momentum before = momentum after

as long as NO EXTERNAL FORCE ACTS on the SYSTEM

Conservation of Momentum

1m1v1 2m2v2

2m2v’21

m1v’1

1 2F21F12

tFvmvmp 1211111 '

tFtFvmvmp 122122222 '

''''

''

2121

22112211

22221111

ppppvmvmvmvmvmvmvmvm

tFpt

pF

net

BALL 1

BALL 2

Newtons 2nd Law

Newtons 3rd Law

Newtons 2nd Law

0

0

system

system

systemsystem

p

Ft

pF

If there are NO external forces

Conservation of Momentum can be extended to include any number of interacting bodies

Total momentum of system (vector sum of momenta of all objects)

LAW OF CONSERVATION OF MOMENTUM – The total momentum of an isolated system of bodies remains constant

0 systemp

LAW OF CONSERVATION OF MOMENTUM

The total momentum of an isolated system of bodies remains constant.

A system is a set of objects that interacts with each other.

An isolated system in one in which the only forces present are those between the objects of the system and those will be zero because of Newtons 3rd law. ( )

0systemisolatedF

Momentum in space

Example. Railroad cars collide: momentum conserved. A 10,000 kg railroad car traveling at a speed of 24 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward?v1 = 24 v2= 0

v’

smv

pp

vvmp

smkg

vmp

beforebefore

after

before

/12'

'20000'2

)/(000,240

)24(10000

1

11

DO NOW Rifle recoil. Calculate the recoil velocity of a 5.0 kg rifle that shoots a 0.050 kg bullet at a speed of 120 m/s.

v’B

smv

v

pp

vvvmvmp

p

R

R

beforebefore

RRBBRRafter

before

/2.1'

6'50

6'5)120(05.0'5''

0

v’R

before shooting

after shooting

Types of Collisions In all collisions where ΣFext = 0, momentum is conserved

Elastic Collisions No deformation occurs.Kinetic energy is also conserved.

ExplosionsReverse of perfectly inelastic collision, kinetic energy is gained.

Inelastic Collisions: Deformation occurs.Kinetic energy is lost.

Perfectly Inelastic CollisionsObjects stick together, kinetic energy is lost.

22

1 mvK

Elastic Collisions

Inelastic Collisions:

http://www.science-animations.com/support-files/collisions.swf

Example. Railroad cars. A railroad car of mass 3000 kg, moving at 20 m/s eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s westward. The railroad cars stick together after the impact. a) What is the magnitude and direction of the velocity of the combined

trains after the collision? b) What is the impulse exerted on the smaller train by the larger train?

On the larger train by the smaller?c) Prove that the collision is inelastic by kinetic energy analysis.

v1 = 20

v’

v2 = 20

smv

v

pp

vvvmmp

sNvmvmp

afterbefore

after

before

/0.10'

'400040000

'4000')10003000(')(

)(40000)20(1000)20(3000

21

2211

Example. Railroad cars. A railroad car of mass 3000 kg, moving at 20 m/s eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s westward. The railroad cars stick together after the impact. a) What is the magnitude and direction of the velocity of the combined

trains after the collision? b) What is the impulse exerted on the smaller train by the larger train?

On the larger train by the smaller?c) Prove that the collision is inelastic by kinetic energy analysis.v1 = 20

v’

v2 = 20

NsJJ

NsvmvmpJ

smv

3000

3000)]20(10[1000'

/0.10'

21

22222

Newton’s 3rd Law

Impulse on smaller train is equal to its change in momentum

J2J1

Example. Railroad cars, inelastic collision.

v1 = 20

v’

v2 = 20

smv /0.10'

JvmmKKK

JvmvmKKK

after

before

000,200')(''

000,8002

2121

21

2222

12112

121

K is reduced so collision is inelastic

Example. Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon?

v’B = ?

smv

v

pp

vvmvmp

p

B

B

beforebefore

BBBCCafter

before

/5.22'

'102250

)'(10)5.1)(150(''

0

v’c = 1.5

J

vmvmKE

KE

BBCCafter

before

2700

''

02

212

21

Example. Pool or billiards. A billiard ball of mass 0.5 kg moving with a velocity of 3 m/s collides head on in an elastic collision with a second ball of equal mass at rest (v2 = 0). What are the speeds of the 2 balls after the collision?v2= 0

v1 = 3

21

21

212211

2211

''3

)''(5.05.1

'5.0'5.0''

)(5.10)3(5.0

vv

vv

pp

vvvmvmp

sNvmvmp

beforebefore

after

before

Example. Pool or billiards. v2= 0

v1 = 3

21 ''3 vv

22

22

22

21

22

21

22

21

2222

12112

121

2222

12112

121

')'3(''9)''(25.025.2

)''(25.0''''

25.2

vvvvvv

KKvvvmvmKKK

JvmvmKKK

afterbefore

after

before

from conservation of momentum:

Since collision is elastic, kinetic energy is also conserved:

3'2 v0'1 v

In a head-on collision:Which truck will experience the greatest force?Which truck will experience the greatest impulse?Which truck will experience the greatest change in momentum?Which truck will experience the greatest change in velocity?Which truck will experience the greatest acceleration?Which truck would you rather be in during the collision?

Truck Collision

In a head-on collision:

Which truck would you rather be in during the collision?

Truck Collision

same

same

same

Example. A nuclear collision. A proton of mass 1.0 u (unified atomic mass units) traveling with a speed of 20,000m/s has an elastic, head-on collision with a Helium (He) nucleus (mHe = 4.00 u) at initially rest. What are the velocities of the proton and Helium nucleus after the collision?

vHe= 0vP = 20,000

HeP

beforebefore

HePHeHePPafter

HeHePPbefore

vvpp

vvvmvmp

vmvmp

'4'1000,20

'4'1''

000,200)20000(1

Example. A nuclear collision.

vHe= 0vP = 20,000

HeP vv '4'1000,20

22

228

2222

122

1

822

122

1

'2)'420000(5.0'2'5.0102

'2'5.0''

102

HeHe

HeP

afterbefore

HePHeHePPafter

HeHePPbefore

vvvvx

KKvvvmvmK

xvmvmK

from conservation of momentum:

Since collision is elastic, kinetic energy is also conserved:

8000' Hev000,12' Pv

Example. Propulsion in space: explosion.

Ion Thruster

tFJ

Nt

pF fuel

5000

Rocket

Rocket thrust: LARGE force over short time

tFJ

Nt

pF ion

092.0

Ion Thruster: small force over LONG time

Example. Propulsion in space: explosion. An astronaut at rest in space fires a thruster pistol that expels 35 g of hot gas at 875 m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol?

v’A = ?

smv

v

pp

v

vmvmp

p

A

A

beforebefore

A

GGAAafter

before

/36.0'

63.30'840

)875(035.0'84

''

0

J

vmvmK

K

GGAAafter

before

404,13

''

02

212

21

v’G = 875

Conservation of Momentum can also be applied in 2 or 3

dimensionsFor 2-dimensional collisions• Set coordinate system up with x-

direction the same as one of the initial velocities

• Label and resolve velocity vectors into x and y components in a sketch

• Resolve momentum vectors into x and y components when working the problem

• Use conservation of momentum independently for x and y dimensions.

v1=0.8

v’2=0.3

v’1

q1

q2=350

2 Dimensional Problem: A pool player hits the 14- ball in the x-direction at 0.80 m/s. The 14-ball knocks strikes the 8-ball, initially at rest, which moves at a speed of 0.30 m/s at an angle of 35o angle below the x-axis. Determine the angle of deflection of the 14-ball.

q1= 17.20

v’1= 0.58 m/s

A 2.5 g bullet is fired into a 215 g wooden block of a ballistic pendulum. After the collision, the block and embedded bullet rise to a maximum height of 1.20m. Find the speed of the bullet.

'870025.0

'2175.0

')(')(0

vv

v

m

vmmv

vmmvmppp

pp

b

b

Pbb

Pbbb

PbPb

afterbefore

85.42'

'221

ghv

ghmvmEE

BB

TB

smvvb /422'87

The diagram depicts the before- and after-collision speeds of a car that undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car crumples up and sticks to the wall.

a. In which case is the change in velocity the greatest?

b. In which case is the change in momentum the greatest?

c. In which case is the impulse the greatest?

d. In which case is the force that acts upon the car the greatest (assume same contact times)?