2014 Topic 1: Moving About

Preliminary Physics 2014

Topic 1: Moving About Initially compiled by CM and AD Nov 2007 Amended AD Jan 2010/2012/2014

Icons •  Prac*cal Ac*vity

•  Teacher Demonstra*on

•  Text ques*ons to do

•  Another document to reference.

Pg:

Q:

Prac

Demo

Doc

Context -‐ Transport

•  Physics is all about being able to apply general rules, theories and laws to a wide variety of situa*ons.

•  Imagine the number of situa*ons you could think of that would involve transport… that’s what you need to be able to understand!!!

Things we will learn about in this topic:

•  1. Describing Mo*on •  2. Forces and Newton’s Laws •  3. Work and Energy •  4. Collisions •  5. Safety devices in cars – Assessment here! •  6. Forces in 2-‐dimensions

The assessment this term….

…is not un*l week 9! •  So don’t worry about it yet. •  It is going to be a project on vehicle safety. You will do some of it in class and some at home.

•  You will undertake an experiment on crashing cars in the lab, do some research and also get a chance to sit in and look at some very cool (and some very old) cars…

So let’s start, shall we?

Transport and accidents

The pracDcal use of physics -‐ safety devices in cars

Why has this occurred? What sort of things will have increased the likelihood of fatali*es, What would have decreased the likelihood of road fatali*es?

The physics of vehicles Part 1: Describing MoDon ‘Vehicles do not typically travel at a constant speed’

Read Chapter 9

P173-188

Scalar and Vector QuanDDes •  Scalar quan**es are those that only have a

magnitude (value indica*ng a measurement). •  Eg.

–  A distance measurement is given in metres -‐ I walked 2km to school.

–  A mass measurement is given in kilograms – The car has a mass of 1500kg.

•  Think of some more example of scalar quan**es…

Pg:189

Q:1

Vectors •  Vector quan**es have both a magnitude and a DIRECTION.

•  Examples of Vector QuanDDes: –  Displacement (Δr) – think of a change in posi*on. This must have both magnitude and direc*on.

–  Velocity (v) – same units as speed but also has a direc*on. Uses displacement rather than distance to calculate velocity.

Speed and Velocity •  Speed and velocity both measure how fast something is

going. So what’s the difference? •  To calculate average velocity we need:

d(1) the displacement, , of an object. It is a straight line distance from a reference position in a given direction. - IT HAS DIRECTION Distance does not have a direction.

(2) the time taken, t, for this displacement to occur. The AVERAGE VELOCITY is the displacement undergone by the object per unit time

ie it is the rate of change of displacement.

So.. What do you think the difference is between speed and velocity?

takentimentdisplacemevelocityAverage =

tdvave =

!

Direction is important when we talk about velocity.

takentimedistancespeedAverage =

Direction is not important when we talk about speed. Speed does not have a direction. Think of an example of when the difference between speed and velocity would be important.

tdvave =

Expt: Speed of

cars

Average and instantaneous speed

•  Watch the video of the 400m sprint from the Olympics. Calculate the Cathy Freeman’s average speed over the distance.

•  Ques6on – can your maximum running speed exceed her average speed over the distance?????

http://www.youtube.com/watch?v=JElsXXQqaUU

There is a difference between DISTANCE & DISPLACEMENT

•  Distance is how far you travelled. •  Displacement is how far you are from some reference posi*on and in what direc4on.

•  Speed is the distance travelled per unit *me. •  Velocity is the displacement travelled per unit *me.

DIRECTION IS IMPORTANT Displacement and Velocity Versus Distance and Speed

A car travels 500 m to the right, turns around and travels another 1000 m to the left. The car travelled with a uniform speed and the time taken was 150s. Find

(a) the total distance travelled (b) the total displacement (c) the average speed (d) the average velocity

500m 1000m

500m

Back to Cathy Freeman’s 400m sprint…

•  Cathy Freeman runs the 400m in about 56 seconds.

•  The distance is 400m around the track. However between the start and finish line

there is no displacement (they are at the same place). Why is it so?

•  Because velocity needs direc4on as well as magnitude, on any part of the track her velocity will have an opposite pair with the same magnitude but opposite direc4on. Although the average magnitude of all her instantaneous veloci4es is 7.1 ms-‐1 in magnitude, the direc4ons cancel each other out. Therefore the average velocity is zero.

Start/Finish 11.7

56400

−=

=

=

mssm

tdSpeedav

10560

−=

=

Δ

Δ=

mssm

trVav

In some problems we will need to CONVERT units., usually m/s to km/h or the other way. How do we do this?

Question: is 100km/h faster or slower than 100m/s? think about it logically and then make a decision.

Answer: a football field is almost 100m long. If 100km/h was more than 100m/s, think about how far you would be travelling every second on the freeway! WOW it would be fast!

So… 100km/h must be LESS than 100m/s

Converting Units

So… how do we go from km/h to m/s?

Example: Change 20 km/hr to a value in m/s.

== m/hrkm/hr110002020 x

=m/min60100020x

m/s6060100020xx

(a) The Concorde flies at an average speed of 1440 km/hr. How long will it take to fly 100 km(in secs)?

(b) How far will the Concorde fly in 1 minute?

Converting Units Pg:189

Q:2-6

Displacement and Velocity

Quantity Vector or Scalar?

Displacement

Speed

Velocity

Distance

Complete the table below:

AcceleraDon •  Another Vector! •  Accelera*on is the rate of change of velocity. (how much the velocity changes every second).

–  a = accelera*on (ms-‐2) –  Δv = change in velocity (ms-‐1) –  Δt = change in *me (s)Units = ms-‐2

–  u = ini*al velocity (ms-‐1) –  v = final velocity (ms-‐1)

tuv

tvaav Δ

−=

Δ

Δ=

Describe the acceleration of

this ball….

Pg:190

Q:8-11

AcceleraDon quesDon: •  A car starts at rest and increases its speed to 100 kmh-‐1 in only 4.5s when

driving straight in a northerly direc*on. What is its accelera*on? •  Data: •  Ini*al velocity (vi) = 0 ms-‐1 North •  Final velocity (vf)= 100kmh-‐1North = 28 ms-‐1 (do conversion first!) •  Δv = vf – vi = 28 – 0 ms-‐1 North •  Δt = 4.5 s

2

1

2.65.4

28

−

−

=

=

ΔΔ

=

msas

msa

tva

av

av

av

In other words, every second, the car increases its velocity by 6.2ms-1

P195 acceleration

How do I show direcDon? 1D

•  When dealing with objects that are travelling in straight lines, we use posi*ve and nega*ve signs to signify the two opposite direc*ons.

•  E.g. Jess and Mark run towards each other, Jess at 4ms-‐1 Mark at 5ms-‐1. Because their direc*ons are opposite we can define one direc*on as posi*ve.

•  They take 0.5 seconds to stop when they meet each other. •  Calculate the accelera*on of both people. •  Discuss the meaning of the signs in the answers.

How do I show direcDon? 2D

•  Bearing = Clockwise from North (E.g. Bearing 250°T)

•  Rela*ve Direc*on = direc*on with respect to another. (E.g. S32°W – meaning from South, go 32 degrees toward the West.

Bearing Relative 325 °

N25°W 167 °

!  Complete the table using the compass above.

QuesDon -‐ Wenona to Thredbo •  Total distance by car = 500km •  Total *me by car = 5 hours 40 mins •  Total magnitude of displacement =

300km •  Total *me by Helicopter = 1hour 30

mins (assume helicopter travels in a straight line)

!  Draw in a vector that shows the displacement between the start and finish.

!  Use a protractor to determine the bearing of this vector. !  Calculate the average speed of the car. !  Calculate the average velocity of the car. !  Determine the average speed of the helicopter. !  Determine the average velocity of the helicopter.

Expt

p 194 Going Home

EquaDons of uniformly accelerated moDon from the HSC Data Sheet

•  When can these be used?

1.  When the accelera*on is constant

2.  When you know 3 things – each of these equa*ons has 4 unknown quan**es

22

22

2

221

xx

xx

x

yyy

yy

yyy

uv

uvtux

yauv

tatuy

tauv

=

=

=Δ

Δ+=

+=Δ

+=

Uniform acceleraDon quesDon:

•  A ball rolls from rest down an incline with a uniform accelera*on of 4.0m/s/s.

a)  What is its speed aier 8 seconds? b)  How long will it take to reach a speed of 36m/s? c)  How long does it take to travel a distance of 200m,

and how fast is it going at that *me? d)  How far does it travel in its third second of mo*on? A: 32m/s B: 9.0s C: 10s; 40m/s D: 10m

Gravity •  We already know a likle bit about this…

•  The value of gravita*onal accelera*on is 9.8ms-‐2 on the earth’s surface but this does vary depending on your distance from the centre of the earth.

•  As you get further from the centre of the Earth, the accelera*on decreases.

Expt: heavy vs light

Problems with gravity – what happens in these situaDons with ‘a’ and ‘v’?

1. Toward the trampoline.

2. Just before reaching highest point.

3. In contact with the trampoline moving downwards.

1. Just after jumping out of the plane.

2. After reaching terminal velocity.

3. As the parachute is opened.

1. Just leaving the ground.

2. Just before clearing the bar.

3. On the way down.

4. In contact with the mat, still moving downwards.

Expt – measuring ‘g’

•  Method 1 – using a pendulum

•  Method 2 – using a mo*on sensor or *cker *mer

Prac: measuring

‘g’

So ‘g’ is 9.8m/s/s…

•  How fast is that? •  Here’s something to think about: •  hkps://www.youtube.com/watch?v=9YUtFpLpGn

Uniform slow motion

Uniform faster motion

Ticker Dmer tapes The ticker timer is used to measure displacement and time for objects moving in a straight line. A dot is made every 1/50th of a second. The distance between dots tells you how far the object moved in that time interval. It is often best to use a group of adjacent spaces between dots to give the distance and time, for example 10 spaces equates to the distance moved in 1/10th of a second.

Air resistance and falling objects Prac: air

resistance

Hammer and

feather

Graphs of MoDon •  Graphs are really useful things •  They allow us to see trends in data much more easily than data in tables

•  These trends tell us how the x and y axes are related

Graphs of MoDon •  Review of Cartesian Graphs

–  For a graph with an x and y axis, the

–  Therefore if I had a value that was given by the formula P = xy, this could be obtained from the graph by finding the area under the graph (P = area under the graph).

–  If I had a value that was give the formula Q = Δy/Δx, I could calculate this using the graph by finding the gradient of the line (Q = gradient of the line)

Cartesian Graph

0246810121416

0 10 20 30 40 50

x (units)y

(uni

ts)

ΔxΔy

runriseGradient ==

xycurve under the area =

DISPLACEMENT TIME GRAPHS

•  Plots the displacement of an object against *me.

•  The slope of a displacement-‐*me graph is equal to the velocity.

SlopeRiseRun

DisplacementTime

Velocity= = =Δ

Δ

Motion game using

dat.log.

Displacement

Time(s)

5

10

15

20

25

30

35

40

1 2 4 5 6 7

(m)

Rise displacement= Δ

Run = timeΔ

smtsv /5

420

+=+

=Δ

Δ=!

!

Displacement (m)

Time (s)

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Displacement/*me graphs for accelerated mo*on

Displacement

Time (s)

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

ACCELERATED MOTION

Displacement

Time (s)

The instantaneous velocity of an

object is the slope of the tangent to the displacement time curve at that

time

VELOCITY TIME GRAPHS

•  Plots the velocity of an object at each instant of *me

•  The slope of a velocity *me graph represents the accelera*on. Some*mes these graphs are called speed/*me graphs

!!

avt

riserun

= = = −Δ

Δslope of v t graph

• The area between the graph and the time-axis represents the displacement of the object

Velocity Time (s)

+4

+2

-2

-4

+6

2 4 6 8 10 12 14

+8 Velocity/time graph

Describe what is happening to the object in this time interval. Calculate the accelerations of each section

Velocity Time (s)

+4

+2

-2

-4

+6

2 4 6 8 10 12 14

+8 The area above the t axis is

positive displacement

The area below the t axis is

negative displacement

The slope of a v-t graph gives the acceleration

Velocity Time (s)

+4

+2

-2

-4

+6

2 4 6 8 10 12 14

+8

2/5.146 sm

runrisea +=

+==

+1.5

0

-2.0

0

+4.0

ariserun

m s= = =03

0 2/ariserun

m s= =−

= −63

2 2/ariserun

m s= = =02

0 2/ariserun

m s= =+

= +41

4 2/

A bh=

= =

12

12 4 6 12.

A lb=

= =3 6 18.

A bh=

= =

12

12 3 6 9.

A ==

12 2 4

4.A ==

2 48.

A = 2

Total distance=12+18+9+4+8+2=53 m

Total displacement=12+18+9-4-8-2=+25 m

Accelera*on *me graphs

tvaΔΔ

= tav Δ=Δ

!  Plots the acceleration of an object relative to a stationary reference point over a period of time.

!  For straight line motion, a negative acceleration would be when the object is experiencing a change in velocity in the opposite direction to that which was defined as positive.

y-axis = acceleration, x-axis = time Remember therefore . By finding the area under a velocity vs time graph, we can find the change in velocity. Describe the motion of the object in the graph below.

t (s)

a (ms-2)

Check your understanding -‐ Graphs of constant accelera*on

Pg:189

Q:12-15

Pg:191

Q:23-24

Review: Graphing MoDon of a Ball Thrown into the Air

t (s)

r (m)

t (s)

v (ms-1)

t (s)

a (ms-2) Use these axis to sketch a graph of the motion of a ball thrown into the air vertically and returns to the original position.

Accel. using

dat.log.

Accel. Down slope

tick.timer

Graphs Summary

Displacement vs Time

Velocity vs Time

Acceleration vs Time

Pg:190-192

Q:16-22

Moving About Part 2: Forces and Newton’s Laws of Mo6on

Read Chapter 10 p197-217

(2d vectors covered

later)

Newton’s First Law -‐ Iner*a •  Iner*a is defined as “the tendency of an object to resist a change in mo*on.” –  Imagine throwing a ball…… –  Imagine throwing an elephant… – Which one do you think has greater iner*a? –  The greater a mass, the greater its iner*a, therefore the harder it is to change its mo*on (larger force needed).

•  Newton’s First Law states: –  “An object will remain at rest or at constant velocity un*l acted upon by a non-‐zero net force.”

Transla4onal Equilibrium •  In a transla6onal equilibrium all the forces on an object

cancel each other out, in the sense that the (vector) sum of all forces is zero, so there is no net force ac*ng on it. All the forces are balanced. So the state of transla*onal equilibrium is equivalent to the condi*on for no accelera*on.

•  Compare to Newton’s first law of mo4on: First Law An object will stay at rest or move at a constant speed in a straight line unless acted upon by an unbalanced force.

Newton 1: an applica*on

•  While travelling in a bus standing up, you are asked to write observa*ons for the following scenarios based on Newton’s first law of mo*on.

!  Bus accelerates

!  Bus stops suddenly

!  Bus turns to the right

Pg:219

Q:11-13

•  Net force (Fnet) – When all the forces ac*ng on an object are added together, the resultant force is the Net Force. –  The accelera*on of an object is propor*onal to the NET force on the object and inversely propor*onal to the mass of the object.

•  m is the MASS of the SYSTEM in kg. this is the mass of everything which the force acts on.

•  The accelera*on is how fast the system is gesng faster (or slower) in metres per second per second.

Newton’s Second Law Fnet=ma –  Eg. Bill pushes his trolley with a force of 20N, there is a resistance

force of 5N ac*ng in the opposite direc*on. Determine the net force ac*ng on the trolley.

20N

5N

Taking right as positive

Fnet = ∑Forces

Fnet = 20N + -5N

Fnet = 15N

Because answer is +ve, direction is to the right.

Newton’s Second Law

•  Fnet = ma –  The accelera*on of an object is propor*onal to the NET force on the

object and inversely propor*onal to the mass of the object. •  Eg. If Bill’s trolley had a mass of 50kg, what would be the accelera*on of

the trolley?

•  If Bill wanted the trolley to remain at a constant velocity, what forward force would he have to apply? (assume resistance force is always 5N)

•  Show that the accelera*on would be zero when this force is applied.

5N

Newton 2 Applica*ons •  While sisng in a car, it accelerates at 5ms-‐2. Calculate the net force ac*ng

on your body assuming you have a mass of 60kg.

•  Calculate the force required to throw a ball 50g ball at a velocity of 20ms-‐1 if it takes 0.2s from rest to release of the ball.

Pg:220

Q:22-24 +28

¨  E.g. A medicine ball of mass 6kg is dropped off of a very tall building.

¨  Aier some *me, the air resistance provides an upwards force of 10 Newtons.

¨  What is the net force and the accelera*on of the ball at this instant?

Net force and accelera*on

Weight force

Air resistance

¨  The weight force can be calculated from F=mg

¨  ‘g’ is the gravita*onal field strength.

¨  On the surface of the earth, this is 9.8m/s/s or 9.8N/kg

¨  So the weight is 6x9.8 = 58.8N


58.8N

10N

¨  The net force is the effec*ve force ac*ng on the system which accelerates it.

¨  In this case, the net force is: 58.8 -‐ 10 = 48.8 N

We can work out the accelera*on too: a = F/m = 48.8/6 = 8.1m/s2


58.8N

10N

Pg:219

Q:9-10

Newton 2 Applica*on

•  A train locomo*ve has a mass of 100 tonnes. The carriages it pulls has a mass of 40 tonnes each. A resistance force exists when the train is moving of 3000N on each carriage and 5000N on the locomo*ve.

•  The train accelerates at 1ms-‐2. –  A) determine the net force ac*ng on the whole train. –  B) determine the thrust force applied by the locomo*ve. –  C) determine the force the coupling is exer*ng on the last carriage.

Pg:221

Q:31-32

Experiment: forces in systems

We can measure: The mass of the system The weight force which accelerates the mass The accelera*on of the system So we can measure all of the things in Newton's second law equa*on. We can then compare the results to confirm the Law!

!  We will set up a system like this.

!  The mass which hangs off the edge of the table accelerates the mass and the trolley.

Forces in systems

•  The applied force = weight force ac*ng on m1 –  Fweight = m1a –  Fweight = m1 x 9.8ms-‐2

•  This applied force must accelerate the en*re mass of the system. •  Fnet = mtotal atrolley •  mtotal = m1 + mtrolley

•  Therefore atrolley = Fnet/(m1 + mtrolley)

!  The mass m1 is under the influence of gravity and will therefore have a force acting on it.

!  The trolley will be accelerated by the applied force. M1

M2

Weight force= m1g

P 224 Newton 2

Mass vs Weight

•  One force that we have referred to is a weight force. •  We know this is the force ac*ng on an object due a

gravita*onal field. •  The gravita*onal field strength and the mass of an object

determines the weight exerted on the object. •  The gravita*onal field strength is given in the units Nkg-‐1 (the

force exerted on each kilogram of mass). –  Force is measured in Newtons –  F=ma therefore units for Newtons is kgms-‐2 –  Gravita*onal field units = Nkg-‐1 = kgms-‐2/kg = ms-‐2

–  Gravita*onal field strength = accelera*on due to gravity!!

Mass vs Weight

•  Mass does not change! •  Weight depends on accelera*on due to gravity. –  F=ma –  Fweight = mg (where g = accelera*on due to gravity)

•  Calculate the force that the Earth exerts on you (g = 9.8Nkg-‐1)

Pg:219

Q:3-6 + 9

Weight on different planets

•  Complete the following table.

Planet Gravitational field strength

Your Weight

Mercury 0.36 Nkg-1

Jupiter 26.04 Nkg-1

Mars 3.75 Nkg-1

Pluto 0.61Nkg-1

Newton’s Third Law

•  For every ac*on there is an equal and opposite reac*on. – When you push a wall, the wall pushes back on you.

– When you stand on the ground, the ground pushes back on you.

– When you push a trolley, the trolley pushes back on you.

– You can only hit something hard that can exert a large force back on you.

Kicking different

balls

‘Magic Statement’! •  These pairs of forces are called ac*on-‐reac*on pairs. To always

get these correct, use the following formula and never fail. –  A exerts a force on B –  B exerts an equal and opposite force on A

•  Ac*on-‐reac*on pairs must ALWAYS act on different objects.

Label the following ac6on-‐reac6on pairs Pg: 221

Q: 29

IdenDfy 6 pairs in the following:

Newton 3 QuesDon: •  These forces are equal and in opposite direc*ons but are they an ac*on-‐reac*on pair?

Normal-reaction force of ground acting on person

Weight force acting on the

person !  Applying the formula

–  Weight exerts a force on person –  Normal reaction exerts equal and opposite

force on person

!  This is not an action-reaction pair as both forces are acting on the same object!

Momentum

•  Momentum (P) -‐ The mass of an object mul*plied by its velocity.

•  If an object has a large momentum, it is more difficult to change its velocity.

p=mv –  p = momentum (kgms-‐1) – m = mass (kg) –  v = velocity (ms-‐1)


Momentum

•  Momentum (P) -‐ The mass of an object mul*plied by its velocity.

•  If an object has a large momentum, it is more difficult to change its velocity.

p=mv –  p = momentum (kgms-‐1) –  m = mass (kg) –  v = velocity (ms-‐1)

The units of momentum are: Ns or kgm/s


Momentum magic! •  Because the net force is equal to the change in momentum (Newton2) and forces are ac*on-‐reac*on pairs of equal magnitude (Newton3), the total change in momentum of the system is equal to zero!!!!

m1u1+m2u2=m1v1+m2v2

•  This is known as the CONSERVATION OF MOMENTUM

•  For this to be true, we need to consider the TOTAL system, i.e. everything interac*ng

Expt: collision

ConservaDon of Linear Momentum quesDon

•  A billiard ball, mass 40g, is ini*ally rolling to the lei at 20cm/s. It hits a second iden*cal sta*onary ball. The ini*al ball slows to 5cm/s lei immediately aier the collision. What is the final speed of the other ball?

Linear Momentum quesDon 2

•  A toy train, mass 420g, moving at 30cm/s south collides with a carriage moving at 32cm/s north. If they become coupled together, what is their common velocity aier collision?

Linear Momentum quesDon 3 •  A 2.1-‐kg cart is rolling along a fric*onless, horizontal track

towards a 1.4-‐kg cart that is held ini*ally at rest. •  The carts are loaded with strong magnets that cause them

to akract one another. Thus, the speed of each cart increases.

•  At a certain instant before the carts collide, the first cart's velocity is +5.9 m/s, and the second cart's velocity is -‐2.0 m/s.

(a)  What is the total momentum of the system of the two carts at this instant?

(b)  What was the velocity of the first cart when the second cart was s*ll at rest?

Pg: 244

Q: 20-25

Impulse (Δp) !  Impulse is the change in momentum !  But how can that be if momentum is

conserved? !  Momentum is NOT conserved if we consider only

one thing. !  Think about playing snooker…when the white ball

hits a coloured ball, the coloured ball moves… the P of the ball changes…

Impulse (Δp)

Force vs Time - Soccer Ball

0

50

100

150

200

0 0.1 0.2 0.3 0.4 0.5

Time (s)

Forc

e (N

)

!  Impulse is the change in momentum of an object: – Δp=pf – pi

– Δp=mfvf – mivi !  Usually the mass of an

object remains the same, therefore: – Δp=mfvf – mivi

– Δp=mfv – miu – Δp=m(v – u)

Impulse is also equal to : force x time

Impulse quesDon

•  A ball with a mass of 0.10 kg has a velocity of 7 m/s. It strikes a concrete wall perpendicularly and bounces straight back with velocity of 4 m/s. what is the impulse on the wall?

•  If the ball is in contact with the wall for 0.1seconds, what is the average force on the wall?

•  If the wall is softer, the contact will take place over a

longer time. What will this do to the peak force? •  Can you think of an example to do with vehicles where this

is important?

Newton’s 2nd and Momentum

•  Newton actually defined his second law in terms of momentum!

•  Net Force = rate of change in momentum.

matuvmtpFnet

=Δ

−=

Δ

Δ=

)(

Impulse quesDon 2

•  Determine the impulse of the ball aier being struck for the 0.4s

•  Calculate the final velocity of the 500g ball once it leaves the boot.

•  Calculate the average net force ac*ng on the ball when kicked.

Force vs Time - Soccer Ball

0

50

100

150

200

0 0.1 0.2 0.3 0.4 0.5

Time (s)

Forc

e (N

)

Pg:242

Q:9-19

Impulse quesDon 3

•  In a game of snooker a player hits a 0.20kg ball with the cue, exer*ng an average force of 40N south on the ball for 12 milliseconds.

•  A. What is the impulse exerted on the ball? •  B. What is the change in momentum of the ball?

•  C. With what velocity does the ball leave the cue?

A) 0.48Ns; B) 0.48Ns; C) 2.4m/s south

So do we understand momentum?

•  We know that •  IMPULSE = Change in Momentum

•  And… •  In collisions, momentum is conserved.

Read both the above statements. Considered together they don’t make sense. Why not? How can you explain that they are actually correct?

Collisions simulaDon

Review – Newton’s 3 laws

•  Cut and paste the following link into your browser –

hkp://www.grc.nasa.gov/WWW/K-‐12/airplane/newton.html

Read the slides on Newton’s 3 laws, and do the 11/12 ac*vity for Newton’s 3rd law. Also, have a look at the ‘movie’ of the Wright

Brothers


Moving About Part 3: Work and

Energy

Work – what is it? •  Work is the product of the magnitude of the displacement and the component of the force ac4ng in the direc4on of the displacement

•  The units are JOULES

θ

θ

cos)cos(

FdWxFW

=

Δ=!

θ

Work •  Most of the *me F is in the direc*on of d so θ = 0° and cos 0° = 1 so…

FdW =max

Work •  Work generally falls into 2 categories:

1.  When you force something to move against the influence of an opposing force –  (push-‐ups & gravita*onal force, walking

& fric*on, bow & elas*c force) 2.  When you change the speed of something

–  A net force is present

How are Work and Energy related?

•  Energy is the ability to do work OR

•  Work causes changes in energy OR

•  Work is a transfer of energy

Why is work important in car crashes?

Work = force x displacement •  In a car crash, work is one of the ways most of the kine*c energy that the car has is converted into other forms.

•  HOW this conversion occurs, and HOW FAST it occurs will determine the likelihood of injury to the passengers.

•  Can you explain why?

Work quesDons:

•  What work is done by a force of 12N west which pushes a box 6.0m west?

•  How much work is done by a bricklayer liiing a brick weighing 40N a ver*cal height of 1.5m?

Energy -‐ Types • Mechanical Energy: Energy due to posi*on in a field force or energy due to movement

•  Non-‐mechanical Energy: Energy that does not fall into the above category

Energy – Flow Chart Energy

Mechanical

Kinetic

Linear

Rotational

Potential

Gravitational

Elastic

Electric

Magnetic

Non-mechanical

Light Sound Heat

Mechanical Energy - Types

1.   Kine6c Energy, KE: Energy of a moving object

–  Linear, KE → center of mass moving

–  Rota4onal, KR → object rota*ng around center of mass (not required as part of HSC course)

Linear KineDc Energy EquaDon

2

21mvKE =

Mechanical Energy - Types

2.   Poten6al Energy, PE: Energy due to posi*on in a field force

–  Gravita4onal, PEg –  Elas4c, PEs –  Electric, PEE –  Magne4c, PEB

*You must choose a zero point for these*

PotenDal Energy EquaDons

mghPEg =2

21 kxPEs =

Work and Energy

Work – Energy Principle or work done by a net force or net work done on an

object

EnergydFW netnet Δ==This means that the amount of energy input or output of a system is equal to the work done on or by that system. Give an example of this:

Energy ConservaDon •  The total energy is neither increased nor decreased in any process.

•  Energy can, however, be transformed from one type to another AND transferred from one body to another, BUT, the total amount of energy in the process remains CONSTANT!

•  Using this informa4on, and considering the KE equa4on, explain why speeding in a car is very dangerous.

Work – Energy Principle Redefined

•  So if energy is conserved we can write it this way using mechanical and non-‐mechanical energies

WPEKE =Δ+Δ

A skater of mass 60 kg has an initial velocity of 12 m/s. He slides on ice where the frictional force is 36 N. How far will the skater slide before he stops?

Work-‐energy principle quesDon:

Work-‐energy principle quesDon 2:

•  A hammer head of mass 0.50 kg is moving with a speed of 6.0 m/s when it strikes the head of a nail s*cking out of a piece of wood. When the hammer head comes to rest, the nail has been driven a distance of 1.0 cm into the wood. Calculate the average fric*onal force exerted by the wood on the nail.

Work energy principle quesDon 3

•  A hill rises 1.0m ver*cally for every 14m of its inclined length. A truck of mass 3000kg travels a distance of 40m up the slope at a constant speed. Find the work done by the motor if the fric*onal resistance is 100N down the hill.

Work-‐energy principle quesDon 4:

•  A car (m = 1150 kg) experiences a force of 6.00 x 103 N over a distance of 125 m. If the car was ini*ally traveling at 2.25 m/s, what is its final velocity?

Mechanical Energy ConservaDon

•  If we ignore non-‐conserva*ve forces (fric*on and the such), the implica*on is that no non-‐mechanical energies are present (heat, sound, light, etc) therefore…

0=Δ+Δ PEKEThis is the conservation of mechanical energy. It is useful when considering things which are thrown in the air, where the gain in GPE is offset by the loss in KE, as well as many other situations.

Work energy principle simula*on

Ques*on

•  Georgina jumps off a 10m diving pla�orm. She has a mass of 60kg. –  a) calculate the gravita*onal poten*al energy Georgina has when

standing on the pla�orm. –  b) determine the work done by gravity as she falls. –  c) what is Georgina's kine*c energy just before she hits the water? –  d) what is Georgina's velocity as she enters the water?

Thinking – Total Mechanical Energy •  Think about the Energy in a system as a circle. •  This amount of energy can not change! It can only be converted into different

types of energy or transferred to a different object. •  When mechanical energy is conserved, the circle stays the same size, it is just the

propor*on of the circle allocated to the different forms of energy that changes. •  Applying this to our diving example:

Egp Ek

Ek

Egp

Roller Coaster

A diver of mass m drops from a board 10.0 m above the water surface, as in the Figure. Find his speed 5.00 m above the water surface. Neglect air resistance.

ConservaDon of mechanical energy quesDon:

Pg:242

Q:1-8

ConservaDon of mechanical energy quesDon 2:

•  A 200g tennis ball is thrown ver*cally upwards at 2m/s. What is the maximum height it reaches above the point from which it was thrown?

A skier slides down the frictionless slope as shown. What is the skier’s speed at the bottom?

H=40 m

L=250 m

start

finish


Two blocks, A and B (mA=50 kg and mB=100 kg), are connected by a string as shown. If the blocks begin at rest, what will their speeds be aier A has slid a distance s = 0.25 m? Assume the pulley and incline are fric*onless.

s


x

Now a harder quesDon!

To what height h does the block rise when moving up the incline? Hint – use the formula for potential energy in a spring to work out how much energy you start with. Remember this is conservation of mechanical energy – it is all transformed!

A 0.50 kg block rests on a horizontal, frictionless surface as in the figure; it is pressed against a light spring having a spring constant of k = 800 N/m, with an initial compression ‘x’ of 2.0 cm.

Part 4: Collisions

ElasDc & InelasDc Collisions

119

When we consider collisions we need to think about momentum and energy

There are three types of collisions we will consider:

1.  Elastic 2.  Perfectly inelastic

3.  Inelastic Elastic and perfectly inelastic collisions are

limiting cases. Most collisions actually fall into a category between the two extremes.

Perfectly inelas*c collision is a collision in which two objects s*ck together and move with a common velocity aier colliding.

120

Perfectly Inelastic Collisions

fii vmmvmvm !!! )( 212211 +=+

121


Kine*c Energy is not constant in inelas*c collisions. Some kine*c energy is converted to sound and/or heat, or causes deforma*on.

122


if

fff

iii

KEKEKE

vmvmKE

vmvmKE

=Δ

+=

+=2

22212

1121

2222

12112

1

123

To calculate the amount of kine*c energy that is lost:


A 90.0 kg rugby player moving south with a speed of 5.0 m/s has a perfectly inelas*c collision with a 95.0 kg opponent running north at 3.0 m/s.

a. Calculate the velocity of the players just aier the tackle.

b. Calculate the decrease in total kine*c energy as a result of the collision.

124

Perfectly inelastic collisions question

Perfectly inelasDc collisions quesDon 2

•  A freight train is being assembled in a switching yard. Boxcar #1 has a mass of 6.5 x 104 kg, and Boxcar #2 has a mass of 9.2 x 104 kg. If car 1 is moving with a velocity of +0.80 m·∙s-‐1, and car 2 hits it from behind with a velocity of +1.2 m·∙s-‐1, with what velocity will the two cars move with together aier coupling?

In inelas*c collisions, colliding objects bounce and move separately aier the collision, but the total kine*c energy decreases in the collision. Most collisions where things do not s*ck together are like this.

126

Inelastic collisions in general

An Elas*c collision is a collision in which the total momentum and the total kine*c energy remain constant.

127

Elastic Collisions

2222

12112

12222

12112

1

22112211

Energy Kinetic

Momentum

ffii

ffii

vmvmvmvm

vmvmvmvm

!!!!

!!!!

+=+

+=+

128

Elastic Collisions

Two 0.40 kg soccer balls collide elasDcally in a head-‐on collision. The first ball starts at rest, and the second ball has a speed of 3.5 m/s. Aker the collision, the second ball is at rest.

a. What is the final speed of the first ball? b. What is the kineDc energy of the first ball before

the collision? c. What is the kineDc energy of the second ball

aker the collision?

129

Elastic Collisions question

Part 5: Safety devices in cars

Read pages 233-240

Safety devices in cars

Why has this occurred? What sort of things will have increased the likelihood of fatali*es, which would have decreased the likelihood of road fatali*es? Make a list.

Safety devices in cars

•  1988 Holden Commodore

•  2010 Holden Commodore

NRMA Crashed Car Showroom

•  hkp://saferchoices.nrma.com.au/crashed-‐car-‐showroom.aspx

Safety devices in cars – avoiding collisions

Name of Device FuncDon Basic physics principle (if applicable)

Safety devices in cars – reducing effects of collisions

Name of Device FuncDon Basic physics principle

Pg: 245

Q: 27

Speed and safety – low speed zones

•  What happens to the amount of KE when the speed of an object is doubled? calculate the rela*onship!

•  What will this mean in the event of a collision? Write a paragraph explaining how this will change the result of a collision, and why.

http://www.rta.nsw.gov.au/cgi-bin/player.cgi?crashlab_fs2

Part 6: Forces in 2 dimensions


(2d vectors sections)

Forces as Vectors Review-‐ Vector Addi*on

•  Vector addiDon is a method of adding any vectors of the same type in a parDcular problem.

•  This could be the addiDon of two forces, two velociDes, two displacements….etc.

•  The thing to remember is that vectors have magnitude and direcDon so we must take into consideraDon both of these whenever we do a problem involving the addiDon (or subtracDon) of vectors.

Adding vectors: method 1. Tip to tail addi*on.

•  Forces can be represented best by arrows. This can show magnitude (length) and direc*on (orienta*on) therefore sa*sfy the addi*on of both.

•  Basketball when first dropped Basketball aier falling for some *me –  Where –  R = air resistance –  W = weight (W = mg) –  N = Normal reac*on force (a force reac*ng to the force of gravity)

•  Net force is downward therefore accelera*on is downward No net force therefore no accelera*on.

W W

R

+ = Fnet + = Fnet = 0

R

Tip to tail 2D example: •  A boat travels across a river at 2ms-‐1 and the river flows at

3ms-‐1. Calculate the velocity of the boat as measured by a bystander on the shore.

2ms-1

3ms-1 N

Vector SubtracDon or ‘change in’, 1D •  For 1D examples, it is fairly straigh�orward: •  Maths

•  A – B = A + (-‐B) •  Using vectors this looks a likle different

B = A = !  -B has the same magnitude but opposite

direction (- B) =

!  A+(-B) =

+ =

A+(-B)

Vector subtracDon 2D Example -‐ AcceleraDon

•  A car turns a corner from East to North travelling at 15ms-‐1 and it takes 2s to make the turn. –  Calculate the change in velocity and the accelera*on of the car during this

*me. (a=v-‐u/t) –  Remember – change in means final value MINUS ini*al value!

Vector subtracDon 2D example: Momentum

•  Adam Gilchrist strikes a 300g cricket ball that was bowled to him at 140kmh-‐1 and sends it through square leg at 200km-‐1. The ball was in contact with the bat for 0.02s.

•  Calculate the change in momentum of the ball and hence the average force ac*ng on the ball during the hit.

u

v

Components of Vectors •  Any vector can be broken down into a number of components (usually to

make solving a problem easier). •  Using the fact that any two vectors can be added to create a “resultant”

vector, we can take the resultant and create any two (or more) vectors that add to give this resultant vector.

•  Eg. Oien when things happen at an angle we want to know just the horizontal or ver*cal component of the mo*on or the force ac*ng on the object. If a force acts on an object at an angle, we can divide it up into different components such as horizontal and ver*cal to determine the net force.

Horizontal component of applied force

Vertical component of applied force

1.  What are all the forces acting on the trolley?

2.  What is the normal reaction force?

3.  What is the net force?

4.  What is the acceleration of the trolley?

4kg

30°

Applied force = 10N

Components of vectors: Parallelogram Method

•  With two forces ac*ng in different direc*ons one can use the parallelogram method.

•  Two people pull on a large rock to try to move it both with a force of 200N. One pulls toward the North, the other toward the East. What will the net force be? Which direc*on will the block move?

1.  Place forces tail to tail maintaining their magnitude and direc*on. 2.  Draw in the net force as though it would create the diagonal of the parallelogram

(in this case a square) 3.  The net force is given in terms of magnitude (length) and direc*on (orienta*on). 4.  Therefore the resultant net force will be 283 N to the North East. (using

trigonometry)

N

bac

cba

283200200 22

22

222

=

+=

+=

=+

200N

200N

Components of vectors quesDon: •  A plane takes off with an air speed of 300kmh-‐1. The angle of take off is 30°.

•  Calculate how fast a car would have to travel along the runway to remain directly below the plane as it takes off..

30°

Air speed

velocity relative to the ground

Vertical velocity

Force as a Vector

•  Force is a vector quan*ty. If the sum of forces on an object is zero, the object is moving at a constant speed.

•  If the net force at a point is zero, the components of the forces in any direc*on will be zero.

P223 force as a vector

Normal ReacDon Force •  The normal reac*on force acts perpendicular to the surface. •  If the object is res*ng or moving on a surface is flat, this force is directly

opposite in direc*on and equal in magnitude to the weight force. •  If the surface is sloped, the reac*on force is s*ll perpendicular to the

surface. This means that the direc*on and magnitude of the force are no longer equal and opposite to the weight force.

Forces on objects on inclines •  Using components of forces, we can simplify problems with objects on inclines. To do this we break all forces that are not ac*ng down or up the slope, either perpendicular or parallel to the slope.

θ

W = mg

N

R

θ

mg

R

Wperp

Wpara

θ

!  Here the weight force has been broken up into the components parallel (Wpara) and perpendicular (Wperp) to the inclined plane. You can see that Wperp = N but in opposite directions. Also R = Wpara but in opposite directions.

!  Trigonometry can be used to calculate Wpara = mg sin θ !  and Wperp = mg cos θ !  Note: The angle θ is the same between the horizontal and the incline

plane as it is between mg and Wperp

Forces on inclines example: •  A car accelerates up a hill of 10° at 1ms-‐2. The mass of the car is

1000kg. The resistance force ac*ng up the hill is 300N. •  Determine

•  Weight of car (W) •  Component perpendicular to slope (Wy) •  Component parallel to the slope (Wx) •  Normal reac*on force. •  Net force on car. •  Thrust force provided by the car.

θ

W = mg

N Thrust

R

Pg: 220

Q: 16-19, 25,26

Uniform Circular MoDon •  Objects do not perform

uniform circular motion unless they are subject to a centripetal force.

•  This is a force that is always perpendicular to the velocity of the object.

•  The centripetal force is always directed toward the centre of the circular motion.

Circular MoDon and Centripetal Force

Going around a corner

When a vehicle turns a corner and maintains a constant speed, the vehicle is accelerating. How do we know this? The force required to turn the corner (centripetal force) is given by he equation:

F = mv2 / r We often hear about ‘centrifugal force’. What is it? How can we explain it?

Pg: 222

Q: 34

Banked Turns If the car is on a banked turn, the normal force (which is always perpendicular to the road's surface) is no longer vertical. The normal force now has a horizontal component, and this component can act as the centripetal force on the car! This helps the car move around the corner without sliding as it can provide some of the centripetal force. Given just the right speed, a car could safely negotiate a banked curve even if the road is covered with perfectly smooth ice!

Pg: 222

Q: 35-36

Centripetal AcceleraDon

•  Centripetal acceleraDon is the rate of change of tangen*al velocity:

•  We can deduce the direc*on of this centripetal force for uniform circular mo*on…. How?

ac = v2 / r

Movie of hammer throw

2014 Topic 1: Moving About

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Transcript of 2014 Topic 1: Moving About