VCE PHYSICS

1
VCE PHYSICS

• Unit 3
• Topic 1
• Motion in 1 2 Dimensions

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Unit Outline
Unit Outline

• To achieve this outcome the student should
  demonstrate the knowledge and skills to
• apply Newtons laws of motion to situations
  involving two or more forces acting along a
  straight line and in two dimensions
• analyse the uniform circular motion of an object
  moving in a horizontal plane (FNET mv2/R) such
  as a vehicle moving around a circular road a
  vehicle moving around a banked track an object
  on the end of a string.
• Apply Newtons 2nd Law to circular motion in a
  vertical plane consider forces at the highest
  and lowest positions only
• investigate and analyse the motion of
  projectiles near the Earths surface including a
  qualitative description of the effects of air
  resistance
• apply laws of energy and momentum conservation
  in isolated systems
• analyse impulse (momentum transfer) in an
  isolated system, for collisions between objects
  moving along a straight line (F?t m?t)
• apply the concept of work done by a constant
  force
• work done constant force x distance moved in
  the direction of the force
• work done area under force distance graph
• analyse relative velocity of objects along a
  straight line and in two dimensions
• analyse transformations of energy between
  kinetic energy strain potential energy
  gravitational potential energy and energy
  dissipated to the environment considered as a
  combination of heat, sound and deformation of
  material
• kinetic energy i.e. ½ mv2 elastic and
  inelastic collisions in terms of conservation of
  kinetic energy
• strain potential energy i.e. area under
  force-distance graph including ideal springs
  obeying Hookes Law ½ kx2
• gravitational potential energy i.e. mg?h or
  from area under force distance graph and area
  under field distance graph multiplied by mass
• apply gravitational field and gravitational force
  concepts g GM/r2 and F GM1M2/r2
• apply the concepts of weight (W mg), apparent
  weight (reaction force, N) , weightlessness (W
  0) and apparent weightlessness (N 0)
• model satellite motion (artificial, moon,
  planet) as uniform circular orbital motion (a
  v2/r 4p2r/T2)
• identify and apply safe and responsible
  practices when working with moving objects and
  equipment in investigations of motion.

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Chapter 1

• Topics covered
• The S.I. System.
• Position.
• Scalars Vectors.
• Vector Addition Components.

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1.0 The S. I. System
The system of units used in Physics is the
Systeme Internationale dUnits or more simply
the S. I. System. The system has two important
characteristics Different units for the same
physical quantity are related by factors of
10.(eg. mm cm km)
The system is based on 7 Fundamental Units, each
of which is strictly defined.
All other units, so called DERIVED UNITS, are
simply combinations of 2 or more of the
Fundamental Units.
5
1.1 Position
To specify the POSITION of an object, a point of
ORIGIN needs to be defined. It is from this point
all measurements can be taken.
For example on the number line below the point
labelled 0 is the origin and all measurements are
related to that point.

• Thus a number called -15 is 15 units to the left
  of 0 on the number line.

A number called 30 is 30 units to the right of 0.
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1.2 Scalars Vectors

• Before proceeding, it is important to define two
  general classes of quantities.
• 1. SCALAR QUANTITIES
• These are COMPLETELY specified by
• A MAGNITUDE (ie a NUMBER)
• and A UNIT
• Examples of Scalar Quantities would be
  Temperature (17oC), Age (16 years), Mass (2.5
  kg), Distance (150 m).

• 2. VECTOR QUANTITIES
• These are COMPLETELY specified by
• A MAGNITUDE (ie. A NUMBER)
• and A UNIT
• and A DIRECTION
• Examples of Vector Quantities would be
  Displacement (2.7 km, West), Force (15 N,
  Downward), Acceleration (1.5 ms-2, S.E.)

VECTORS ARE GENERALLY REPRESENTED BY ARROWS The
length of the arrow represents the magnitude of
the vector. The orientation of the arrow
represents the direction of the vector.
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1.3 Vector Addition
VECTOR ADDITION
Two Forces act at the Centre of Mass of a body.
The first of 4N East
and the second of 3N South
Which way will the body move ?
SINGLE VECTOR DIAGRAM
In a direction, and with a force, that is the
sum of the 2 vectors
Direction Sin ? 3/5 ? ? Sin-1 3/5
36.90
A Vector of Magnitude 5 units Direction NE or
N45E or 45T
THE RESULTANT FORCE HAS A MAGNITUDE OF 5 N
DIRECTED AT E 36.90 S
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1.4 Vector Subtraction
An object moving East at 8.0 ms-1
changes its velocity to 8.0 ms-1 South
The velocity change (?v) is given by vf - vi
(- vI ) is a negative vector. It can be converted
to a positive one by reversing its
direction. Then, by performing a vector addition,
the velocity change ?v can be obtained.
What is the objects change in velocity ?
Direction Tan ? 8/8 1.0 ? ? Tan-1
1.0 450
THE CHANGE IN VELOCITY 11.3 ms-1 AT S 450W
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1.5 Vector Components
A Jump Jet is launched from a 150 ramp at a
velocity of 40 ms-1
What are the vertical
and horizontal components of its velocity ?
Vertical Component V VERTICAL 40 Sin 150
10.4 ms-1
Horizontal Component VHORIZONTAL 40 Cos 150
38.6 ms-1
V VERTICAL and VHORIZONTAL are the COMPONENTS of
the planes velocity.
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Motion - Revision Questions Question type
Vectors

• Adam is testing a trampoline. The diagrams show
  Adam at successive stages of his downward motion.
• Figure C shows Adam at a time when he is
  travelling DOWNWARDS and SLOWING DOWN.

A Acc is UPWARD. In order to meet the
requirements set - travelling downward BUT
slowing down, he must be decelerating ie.
Accelerating in a direction opposite to his
velocity. Thus acc is upward.
Q1 What is the direction of Adams acceleration
at the time shown in Figure C ? Explain your
answer.
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Chapter 2

• Topics covered
• Distance versus Displacement.
• Speed versus Velocity.
• Acceleration.
• Graphical Representations.

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2.0 Distance vs Displacement
JOURNEY No 1. Distance Displacement

• Distance is a Scalar Quantity having a magnitude
  and a unit. The S.I. unit for Distance is the
  metre (m)
• Distance is best thought of as How far you have
  travelled in your journey.
• Displacement is a Vector Quantity having a
  magnitude, a unit and a direction. The S.I. unit
  for Displacement is the metre (m), plus a
  direction
• Displacement is best thought of as How far from
  your starting point you are at the end of your
  journey.
• Distance and Displacement may or may not be
  numerically equal, depending on the nature of the
  journey.

At the end of the run Distance 100
m. Displacement 100 m
JOURNEY No 2. Distance ? Displacement
At the end of the one lap run Distance 400
m. Displacement 0 m
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2.1 Speed vs Velocity

• INSTANTANEOUS vs AVERAGE VELOCITY
• The term velocity can be misleading unless a
  specific label is attached.
• The label indicates whether the velocity is an
  Average value calculated over a long period of
  time OR an Instantaneous value calculated at any
  instant of time.
• A simple example illustrates
• A journey of 40 km across the suburbs takes 1
  hour
• VAV 40/1 40 kmh-1
• BUT VINST could be anything from 0 kmh-1 (stopped
  at traffic lights) to VINST 100 kmh-1
  (travelling along the freeway).

• Speed is defined as the Time Rate of Change of
  Distance.
• Speed is a Scalar Quantity.
• Mathematically Speed
  Distance/Time
• The S.I. unit for Speed is metres/sec (ms-1)
• Velocity is defined as the Time Rate of Change of
  Displacement.
• Velocity is a Vector Quantity.
• Mathematically Velocity
  Displacement/Time
• The S.I. unit for Velocity is metres/sec (ms-1),
  plus a direction

IN ALL CALCULATIONS AND EQUATIONS USED IN THE
COURSE, ASSUME INSTANTANEOUS VALUES ARE REQUIRED
UNLESS OTHERWISE STATED.
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2.2 Some Common Speeds  
 
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2.3 Acceleration

• Acceleration is defined as the Time Rate of
  Change of Velocity.
• Acceleration is a Vector Quantity.
• Mathematically Acceleration
  Velocity/Time
• The S.I. unit for acceleration is metres/sec/sec
  (ms-2)
• Since acceleration is a vector quantity, a body
  travelling with a constant speed but in a
  constantly changing direction must be
  accelerating.
• So a cyclist travelling around a corner at
  constant speed is, in fact, accelerating ! (More
  of this later).

ACCELERATING VEHICLE
The velocity and acceleration are in the same
direction
DECELERATING VEHICLE
The velocity and acceleration are in opposite
directions.
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2.4 Graphical Representations

• Much of the information delivered in this Physics
  course is presented graphically.
• Generally, graphs tell a story and you need to
  develop the ability to read the story the graph
  is telling.
• There are two basic families of graphs you should
  be familiar with
• (a) Sketch Graphs, paint a broad brush, general
  picture of the relationship between the
  quantities graphed.
• (b) Numerical Graphs from which exact
  relationships may be deduced and/or exact values
  may be calculated.

SKETCH GRAPHS
The Story As time passes, the distance of the
object from its starting point is increasing in a
uniform manner (the slope is constant). This is
the graph an object moving at constant speed.
The Story As time passes, the velocity of the
object is increasing in a uniform manner (the
slope is constant). This is the graph a
constantly accelerating object
The Story As time passes, the displacement of
the object is increasing more quickly (the slope
is increasing at a constant rate).
The Story As time passes, the distance of the
object from its starting point does not change.
This is the graph a stationary object.
This is the graph a constantly accelerating
object
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Motion - Revision Questions Question type
Sketch Graphs
In a road test, a car was uniformally accelerated
from rest over a distance of 400 m in 19 sec. The
driver then applied the brakes, stopping in 5.1
sec with constant deceleration.
The graphs A to F below should be used to answer
the questions below. The horizontal axis
represents time and the vertical axis could be
velocity or distance.
Q2 Which of the graphs, A to F, represents the
velocity time graph for the entire journey ?
A Graph B
Q3 Which of the graphs, A to F, best represents
the distance time graph of the car for the entire
journey ?
A Graph E
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2.5 Exact Graphical Relationships

• You are required to be familiar with graphs of
• Distance or Displacement Versus Time
• Speed or Velocity Versus Time and
• Acceleration Versus Time
• These graph types and the exact information
  obtainable from them can be summarised in the
  table given below. Put this table on your cheat
  sheet.

Distance or Displacement versus Time
Distance or Displacement
Speed or Velocity
No Useful Information
Speed or Velocity
Distance or Displacement
Speed or Velocity versus Time
Acceleration
Acceleration versus Time
No Useful Information
Velocity
Acceleration
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2.6 The Equations of Motion

• HINTS FOR EQUATIONS USE

u v a x t

• These are a series of equations linking velocity,
  acceleration, displacement and time.
• THE EQUATIONS CAN ONLY BE USED IN SITUATIONS
  WHERE THE ACCELERATION IS CONSTANT.
• The 3 most important of these equations are
• 1. v u at
• 2. v2 u2 2ax
• 3. x ut ½at2
• where u Initial Velocity (ms-1) v Final
  Velocity (ms-1) a Acceleration (ms-2)
  x Displacement (m) t Time
  (s)

When using these equations always list out the
information supplied in the question and what you
are required to calculate then choose the
appropriate equation to use.
Sometimes it is necessary to choose a positive
direction ie. up or down for vertical motion
questions or left or right for horizontal motion
questions.
Questions are often asked which require a 2 step
process to get to the answer, ie. A value for
acceleration may be needed before the final
velocity can be found.
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Motion - Revision Questions Question type
Equations of Motion
In a road test, a car was uniformally accelerated
from rest over a distance of 400 m in 19 sec.
The driver then applied the brakes, stopping in
5.1 sec with constant deceleration.
Q4 Calculate the acceleration of the car for the
first 400 m.
A Firstly, list information u 0 v ? a
? x 400 m t 19 s
Choose the appropriate equation x ut
½at2 400 0 ½a(19)2 a 2.22 ms-2
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Motion - Revision Questions Question type
Average Speed
In a road test, a car was uniformally accelerated
from rest over a distance of 400 m in 19 sec. The
driver then applied the brakes, stopping in 5.1
sec with constant deceleration.
Q5 Calculate the average speed for the entire
journey, covering both the accelerating and
braking sections.
Need to know u, the initial speed for the braking
section which equals the final speed for the
accelerating section.
A Average Speed Total Distance
Total Time For the accelerated part of
journey Distance 400 m Time 19
sec For the braking part of journey Distance
needs to be calculated Time 5.1 sec
For accelerating section u 0 v ? a 2.2
ms-2 x 400 m t 19 s
v u at 0 (2.2)(19) 41.8 ms-1
Now can calc s
x ut ½at2 (41.8)(5.1) ½(-8.2)(5.1)2
213.2 - 106.6 106.6 m
Need to calc acc v u at 0 41.8 a(5.1) a
- 8.2 ms-2
To get braking distance, use Eqns of Motion u
? v 0 a ? x ? t 5.1s
Braking list becomes u 41.8 ms-1 v 0 a ? x
? t 5.1s
List does not contain enough info the calculate s
Still not enough info
Total Distance 400 106.6 506.6 m
Total Time 19 5.1 24.1 s So, Average Speed
506.6/24.1 21 ms-1
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Chapter 3

• Topics covered
• Newtons Laws.
• Force in Two Dimensions.
• Momentum and Impulse.
• Conservation of Momentum.

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3.0 Newtonian Motion

• Sir Isaac Newton (1642 - 1727) was unique for a
  number of reasons, but mostly because he
  developed a set of laws describing the motion of
  objects in the universe.
• Prior to Newton, scientists believed that a set
  of laws existed which explained motion on Earth
  and these laws had to be modified to describe
  motions in all other parts of the universe.
• Newton was the first scientist to realise that
  all motion anywhere in the universe could be
  described by a single set of laws which then had
  to be modified for use in the friction riddled
  confines of the Earth.

Isaac Newton Aged 26
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3.1 Newtons Laws
INERTIA That property possessed by all bodies
with mass whereby they tend to resist changes to
their motion. It is associated with an objects
mass more mass, more inertia. Inertia is NOT a
force.
Newton developed 3 laws covering motion in the
universe, they are
LAW 1. THE LAW OF INERTIA. A body
will remain at rest, or in a state of uniform
motion, unless acted upon by a net external force.
LAW 2. The acceleration of a body is directly
proportional to the net force applied and
inversely proportional to its mass. (a ?F/m)

• LAW 3.
• For every action there is an equal and opposite
  reaction

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Motion - Revision Questions Question type
Newtons Laws

• A seaplane of mass 2200 kg takes off from a
  smooth lake as shown. It starts from rest, and is
  driven by a CONSTANT force generated by the
  propeller. After travelling a distance of 500 m,
  the seaplane is travelling at a constant speed,
  and then it lifts off after travelling a further
  100 m.

Total Opposing
Force (N)
10000
8000
6000
4000
2000
Distance (m)
0
100
200
300
400
500
600
The total force opposing the motion of the
seaplane is not constant. The graph shows the
TOTAL FORCE OPPOSING THE MOTION of the seaplane
as a function of the distance travelled.
Q6 What is the magnitude of the net force acting
on the seaplane after it has travelled a distance
of 500 m from the start ?
A At d 500 m the plane is travelling at
CONSTANT VELOCITY,
So SF 0
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Motion - Revision Questions Question type
Newtons 2nd Law
Q7 What is the magnitude of the seaplanes
acceleration at the 200 m mark ?

• A At d 500 m the seaplane is subject to 0 net
  force (see previous question).
• Thus, Driving Force Opposing Force 10,000 N
  (read from graph).
• At d 200 m the total opposing force 2000 N
  (read from graph)
• So SF 10,000 - 2000 8000 N
• Now, we know that SF ma
• So, a SF/m 8000/2200 3.64 ms-2

Total Opposing
Force (N)
10000
8000
6000
4000
2000
Distance (m)
0
100
200
300
400
500
600
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Motion - Revision Questions Question type
Work

• Q8 Estimate the work done by the seaplane
  against the opposing forces in travelling for a
  distance of 500 m.

A Work Force x Distance Area under F vs d
graph.
Area needs to be calculated by counting
squares. Each square has area 2000 x 100 2 x
105 J
Total number of squares (up to d 500 m) 9
whole squares (x)
6 part squares (p)
12 whole squares.
So Work done (12) x (2 x 105) 2.4 x 106 J
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Motion - Revision Questions Question type
Newtons 2nd Law
In Figure 2, a car of mass 1000 kg is being towed
on a level road by a van of mass 2000 kg. There
is a constant retarding force, due to air
resistance and friction, of 500 N on the van, and
300 N on the car. The vehicles are travelling at
a constant speed.
Q9 What is the magnitude of the force driving
the van?
As the vehicles are travelling at constant speed,
a 0 and thus SF 0 Thus driving force total
retarding force Thus driving Force 800 N
Q10 What is the value of the tension, T, in the
towbar?
Looking at the towed car alone the forces acting
are tension in the towbar and the retarding
force. Since the car is travelling at constant
velocity, SF 0, so tension retarding force
300 N
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Motion - Revision Questions Question type
Newtons 3rd Law
The figure shows a cyclist with the bicycle
wheels in CONTACT with the road surface. The
cyclist is about to start accelerating forward.
Q11 Explain, with the aid of a clear force
diagram, how the rotation of the wheels result in
the cyclist accelerating forwards.
A The wheels rotate in the direction shown.
The force labelled FTR is the force the tyre
exerts on the road. This force is directed in the
opposite direction to the acceleration and thus
cannot be the force producing that acceleration.
The force labelled FRT is the Newton 3 reaction
force arising from the action of FTR. It is this
force (directed in the same direction as the
acceleration) that actually produces the
acceleration of the bike and rider.
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3.2 Newtons Laws Restrictions Consequences
RESTRICTIONS The laws only apply at speeds much,
much less than the speed of light. The laws apply
equally in ALL inertial frames of reference.
CONSEQUENCES
As far as Newtons 1st law is concerned rest
and uniform motion are the same state. You
cannot perform any test which can show whether
you are stationary or moving at constant velocity.
FRAME OF REFERENCE ? A frame of reference is best
described as your point of observation. An
inertial frame of reference is one that is either
stationary or moving with constant velocity. A
non inertial frame of reference is
accelerating. Humans in non inertial frames tend
to invent fictitious forces to explain their
experiences
The action and reaction law requires there to be
TWO bodies interacting, the ACTION force acting
on one body and the REACTION acting on the
other. Deciding when an Action / Reaction
situation exists can be done by answering the
question Does the second (reaction) force
disappear immediately the first (action) force
disappears ? If the answer is yes, you have an
action reaction pair.
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Motion - Revision Questions Question type
Relative Motion
A train is travelling at a constant velocity on a
level track. Lee is standing in the train, facing
the front, and throws a ball vertically up in the
air, and observes its motion. Q12 Describe the
motion of the ball as seen by Lee.
Lee sees the ball move straight up and down.
Sam, who is standing at a level crossing, sees
Lee throw the ball into the air. Q13 Describe
and explain the motion of the ball as seen by
Sam.
From Sams point of view, the ball follows a
parabolic path made up of the vertical motion
imparted by Lee and the horizontal motion due to
the train.
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3.3 Force in Two Dimensions

• A Force is either a Push or a Pull.
• A Force is either a CONTACT type force or a FIELD
  type force.
• Force is a Vector quantity having both a
  magnitude, a unit and a direction.
• Force is NOT one of the 7 fundamental units of
  the S.I. System and thus it is a Derived
  quantity.
• The unit for Force is kgms-2. This was assigned
  the name the NEWTON (N), in honour of Sir Isaac.
• Forces can act in any direction and the TOTAL,
  NET or RESULTANT force is the vector sum of all
  forces acting on a body.
• The body will then ACCELERATE in the direction of
  the RESULTANT FORCE, according to Newton 2.

AN OBJECT UNDER THE ACTION OF 4 FORCES
Perform a vector addition of the forces.
The Resultant Force (FRES ) will give the
direction of the acc.
In which direction will the object accelerate ?
The object will be subject to FRES and
accelerate in that direction
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Motion - Revision Questions Question type
Force in 1 2 Dimensions
A cyclist is towing a small trailer along a level
bike track (Figure 1). The cyclist and bike have
a mass of 90 kg, and the trailer has a mass of 40
kg. There are opposing constant forces of 190 N
on the rider and bike, and 70 N on the trailer.
These opposing forces do not depend on the speed
of the bike.
The bike and trailer are initially travelling at
a constant speed of 6.0 m s-1.
Q14. What driving force is being exerted on the
road by the rear tyre of the bicycle?
A Constant speed implies SF 0 So driving
force total retarding force Total retarding
force 190 70 260 N driving force
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3.4 Momentum Impulse

• Newton called Momentum the quality of motion
  and it is a measure of a bodys translational
  motion - its tendency to continue moving in a
  particular direction.
• Roughly speaking, a bodys momentum indicates
  which way the body is heading and just how
  difficult it was to get the body moving with its
  current velocity.
• Momentum is a Vector Quantity.
• Mathematically Momentum (p)
  m.v where, m mass (kg),
  v velocity (ms-1)
  p momentum (kgms-1)

• Impulse is the transfer mechanism for momentum.
• In order to change the momentum of a body you
  need to apply a force for a certain length of
  time to produce the change.
• Impulse is a Vector Quantity.
• Mathematically Impulse (I)
  F.t where F Force (N)
  t time (s)
  I Impulse (Ns)
• From Newton 2 (F ma) and the definition of
  acceleration, (a v/t), we get
• F mv/t
• ? Ft mv
• Thus Impulse Momentum

35
Motion - Revision Questions Question type
Momentum
A small truck of mass 3.0 tonne collides with a
stationary car of mass 1.0 tonne. They remain
locked together as they move off. The speed
immediately after the collision was known to be
7.0 ms-1 from the jammed reading on the car
speedometer. Robin, one of the police
investigating the crash, uses conservation of
momentum to estimate the speed of the truck
before the collision.
Q15 What value did Robin obtain?
A PBEFORE PAFTER
PBEFORE (3000)(x)
PAFTER (3000 1000)(7.0)
So, 3000x 28,000 x 9.3 ms-1
The calculated value is questioned by the other
investigator, Chris, who believes that
conservation of momentum only applies in elastic
collisions. Q16 Explain why Chriss comment is
wrong.
A Momentum is conserved in ALL types of
collisions whether they be elastic or inelastic.
KE is not conserved in this type (inelastic)
collision.
36
Motion - Revision Questions Question type
Momentum
A car of mass 1000 kg travelling on a smooth road
at 5.0 ms1 collides with a truck that is
stationary at a set of traffic lights. After the
collision they are stuck together and move off
with a speed of 2.0 ms1 Q17 How much momentum
did the car transfer to the truck?
A Mom is ALWAYS conserved. Mom of car
before (1000)(5) 5000 kgms-1 Mom of car
after (1000)(2) 2000 kgms-1 Since
mom is conserved Mom loss by car Mom gain
by truck So, Mom transferred to truck
3000 kgms-1
Q18 What is the mass of the truck?
A p mv so, m p/v 3000/2
1500 kg
37
Motion - Revision Questions Question type
Impulse
A car of mass 1000 kg travelling on a smooth road
at 5.0 ms1 collides with a truck that is
stationary at a set of traffic lights. After the
collision they are stuck together and move off
with a speed of 2.0 ms1
Q19 If the collision took place over a period of
0.3 s, what was the average force exerted by the
car on the truck?
For the truck Impulse Change in momentum
Ft mv F(0.3) 3000 F
10,000 N
38
Motion - Revision Questions Question type
Momentum
A railway truck (X) of mass 10 tonnes, moving at
6.0 ms-1, collides with a stationary railway
truck (Y), of mass 5.0 tonnes. After the
collision they are joined together and move off
as one.
Q20 Calculate the final speed of the joined
railway trucks after collision.
A In ALL collisions Momentum is conserved. So
Mom before collision Mom after collision Mom
before (10 x 103)(6.0) (5 x 103)(0) Mom after
(15x 103)(v) So v
60,000/15,000 4.0 ms-1
39
Motion - Revision Questions Question type
Impulse
A railway truck (X) of mass 10 tonnes, moving at
6.0 ms-1, collides with a stationary railway
truck (Y), of mass 5.0 tonnes. After the
collision they are joined together and move off
as one at a speed of 4.0 ms-1.
Q21 Calculate the magnitude of the total impulse
that truck Y exerts on truck X
A Impulse Change in Momentum Truck Xs change
in momentum Final momentum Initial
Momentum (10 x 103)(4.0) (10 x
103)(6.0) -2.0 x 104 kgms-1 The
mechanism for this change in momentum is the
impulse supplied by Truck Y So, I 2.0 x 104
Ns
40
3.5 Elastic Inelastic Collisions
All collisions (eg, cars with trees, cyclists
with the footpath, neutrons with uranium atoms,
bowling balls with pins etc.) fall into one of 2
categories
(b) INELASTIC COLLISIONS, where Momentum is
conserved BUT Kinetic Energy is NOT conserved.
(Most collisions are of this type). The lost
Kinetic Energy has been converted to other forms
of energy eg, heat, sound, light.

• (a) ELASTIC COLLISIONS, where BOTH Momentum AND
  Kinetic Energy are conserved. (Very few
  collisions are of this type). If anywhere, these
  will most likely occur on the atomic or subatomic
  level.

41
Motion - Revision Questions Question type
Elastic/Inelastic Collisions
A small truck of mass 3.0 tonne collides with a
stationary car of mass 1.0 tonne. They remain
locked together as they move off. The speed
immediately after the collision was known to be
7.0 ms-1 from the jammed reading on the car
speedometer. Robin, one of the police
investigating the crash, uses conservation of
momentum to estimate the speed of the truck
before the collision at 9.3 ms-1
Q22 Use a calculation to show whether the
collision was elastic or inelastic.
A Total KE before collision ½ mv2 ½
(3000)(9.3)2
129735 J
Total KE after collision ½ mv2 ½
(4000)(7.0)2
98,000 J
KE is NOT conserved. So collision is INELASTIC
42
Motion - Revision Questions Question type
Inelastic Collisions
A railway truck (X) of mass 10 tonnes, moving at
6.0 ms-1, collides with a stationary railway
truck (Y), of mass 5.0 tonnes. After the
collision they are joined together and move off
as one at 4.0 ms-1.
Q23 Explain why this collision is an example of
an inelastic collision. Calculate specific
numerical values to justify your answer.
A In Inelastic collisions Momentum is conserved
BUT Kinetic Energy is not. For this
collision pBEFORE (10 x 103)(6.0) (5 x
103)(0.0) 6 x 104 kgms-1 pAFTER (15 x
103)(4.0) 6 x 104
kgms-1 KEBEFORE ½mv2 ½(10 x 103)(6.0)2
1.8 x 105 J KEAFTER ½mv2 ½(15 x
103)(4.0)2 1.2 x 105 J Thus Mom IS conserved
but KE is NOT conserved, therefore this is an
inelastic collision
43
3.5 Conservation of Momentum

• The Law of Conservation of Momentum states that
  in an isolated system, (one subject to no outside
  influences), the total momentum is conserved.
• So in a collision (say a car hitting a tree and
  coming to a stop), the total momentum before the
  collision total momentum after the collision.

Mass of car plus passengers m
Momentum p mv
THE CONSERVATION LAW DOES NOT ALLOW MOMENTUM TO
DISAPPEAR. THE APPARENTLY LOST MOMENTUM HAS, IF
FACT, BEEN TRANSFERRED THROUGH THE TREE TO THE
EARTH. Since the Earth has a huge mass (6.0 x
1024 kg) the change in its velocity is negligible.
44
3.6 The Physics of Crumple Zones Air Bags
A car crashes into a concrete barrier. The change
in momentum suffered by the car (and passengers)
is a fixed quantity. So, Impulse, (the product of
F and t), is also fixed. However, individual
values of F and t can vary as long as their
product is always the same. So if t is made
longer, consequently F must be smaller. Crumple
Zones increase the time (t) of the collision. So,
F is reduced and the passengers are less likely
to be injured. The same logic can also be applied
to Air Bags
The air bag increases the time it takes for the
person to stop. So the force they must absorb is
lessened. So they are less likely to be
seriously injured.
A further benefit is this lesser force is
distributed over a larger area
45
Motion - Revision Questions Question type
Momentum/Impulse

• In a car the drivers head is moving
  horizontally at 8.0 ms-1 and collides with an air
  bag as shown. The time taken for the drivers
  head to come to a complete stop is 1.6 x 10-1 s.
  This collision may be modelled as a simple
  horizontal collision between the head of mass 7.0
  kg and the air bag.

Q24 Calculate the magnitude of the average
contact force that the air bag exerts on the
drivers head during this collision.
A Impulse Change in Momentum F?t
?(mv) So F ?(mv)/?t
(7.0)(8.0)/(1.6 x 10-1) 350 N
46
Motion - Revision Questions Question type
Air Bags/Crumple Zones
In a car the drivers head is moving horizontally
at 8.0 ms-1 and collides with an air bag as
shown. The time taken for the drivers head to
come to a complete stop is 1.6 x 10-1 s. This
collision may be modelled as a simple horizontal
collision between the head of mass 7.0 kg and the
air bag.

• Q25 Explain why the driver is less likely to
  suffer a head injury in a collision with the air
  bag than if his head collided with the car
  dashboard, or other hard surface.

• A The change in momentum suffered by the
  drivers head is a FIXED quantity no matter how
  his head is brought to rest.
• Therefore the product of F and t (ie Impulse) is
  also a fixed quantity.
• However the individual values of F and t may be
  varied as long as their product always remains
  the same.
• The air bag increases the time over which the
  collision occurs, therefore reducing the size of
  the force the head must absorb so reducing the
  risk of injury.
• The air bag also spreads the force over a larger
  area, reducing injury risk.
• Without the air bag the drivers head may hit a
  hard surface decreasing the time to stop his head
  and necessarily increasing the force experienced
  and thus the likelihood of injury.
• In addition the force will be applied over a much
  smaller area increasing the likelihood of severe
  injury

47
Chapter 4

• Topics covered
• Centre of Mass.
• Weight.
• Reaction Force.
• Bouncing Balls.
• Friction.
• Various Force Applications

48
4.0 Centre of Mass

• In dealing with large objects it is useful to
  think of all the objects mass being concentrated
  at one point, called the Centre of Mass of the
  object.
• The C of M of the object is the point around
  which it will spin if a torque or turning force
  is applied to the object.

Centre of Mass
For oddly shaped objects eg. a boomerang, the C
of M may fall outside the perimeter of the object.
For regularly shaped objects eg. squares or
rectangles, cubes or spheres the Centre of Mass
of the object is in the geometric centre of the
object
49
4.1 Centre of Mass - Systems

• For a system of 2 or more bodies, the
• position of the C of M may be determined
• from the formula

XC of M is the position of the Centre of Mass of
the System as measured from a CHOSEN REFERENCE
POINT. m1,m2,m3, etc are the masses of the
individual components of the System x1, x2, x3
etc are the distances measured from the CHOSEN
REFERENCE POINT to the centres of mass of the
systems components.
WHAT IS THE CENTRE OF MASS OF THIS SYSTEM?
CHOSEN REFERENCE POINT IS A
30 kg mass
50 kg beam
2.875 m from A
50
4.2 Weight

• The effect of a Gravitational Field on a Mass is
  called its WEIGHT.
• Weight is a FORCE and therefore a Vector
  quantity.
• Mathematically W mg
  where W
  Weight (N) m mass (kg)
  g Gravitational Field
  Strength (Nkg-1)
• Weight acts through the Centre of Mass of the
  body and is directed along the line joining the
  centres of the the two bodies between which the
  Gravitational Field is generated.
• On Earth, the Gravitational Field of Strength
  9.8 Nkg-1 gives any mass under its influence
  alone an acceleration of 9.8 ms-2

Object of Mass (m)
Weight Force acts through the Centre of Mass and
is directed toward the Centre of the Earth
51
4.3 Reaction Force

• Any stationary object which is under the
  influence of the Earths Gravitational Field must
  be subject to a force equal in size, but opposite
  in direction to, its weight.
• This equal but opposite force is called the
  NORMAL REACTION FORCE.
• The Normal Reaction Force only arises from the
  action of the weight force and DOES NOT EXIST AS
  AN ISOLATED FORCE IN ITS OWN RIGHT.

Object Mass m
Table
R acts upwards from the boundary between the
table and the mass through the Centre of Mass
W acts downwards from the Centre of Mass toward
the centre of the Earth
R and W are NOT an ACTION - REACTION PAIR WHY ?
52
4.4 Bouncing Balls.

• When a ball, falling under the action of
  gravity, arrives at a hard surface, it is the
  Normal Reaction Force (R), which provides
• (a) The Force required to decelerate the ball to
  a stop and then,
• (b) The Force needed to accelerate the ball away
  from the surface.

53
4.5 Friction

• Friction is the most unusual of all forces as it
  cannot start an object moving.
• Friction can, however, slow or stop an object
  once it is moving or prevent it from starting to
  move.
• Slowing a moving object is the result of DYNAMIC
  FRICTION, while stopping an object from starting
  to move is the result of STATIC FRICTION.
• Generally speaking, for the same pair of
  surfaces, Static Friction ? Dynamic Friction.
• The size of the frictional force depends upon
• (a) The Roughness of the surfaces measured by the
  Coefficient of Friction (?)
• (b) The Separation of the surfaces.
• Friction does NOT depend on the AREA in contact.

OBJECT HELD STATIONARY BY STATIC FRICTION
The size of the Frictional Force Fr is calculated
from Fr ?R
54
4.6 Friction Applications

• Friction is NOT always a hindrance to living on
  Earth, often it is vital for movement over the
  Earths surface.
• Consider the following
• A car is accelerating along a Rough (meaning
  having Friction) Road Surface.

The frictional force between the tyre and the
road (FTR) is directed backwards. This force
CANNOT provide the forward propulsion.
In order for the Car to accelerate in the
direction shown a force must exist in that
direction.
IT IS THE REACTION FORCE WHICH PRODUCES THE
ACCELERATION
Thus it is the Force of the Road on the Tyre
(FRT) which gives rise to the acceleration
55
4.7 Various Force Situations Stationary and
Falling Bodies
MASS - STATIONARY ON A TABLE
MASS - FALLING WITH AIR RESISTANCE
a 0
? F ma ? F W - FR So, W - FR ma and FR
ma - mg m(a - g)
From Newton 2 ? F ma ?F R - W ma but a
0 So, R - W 0 and R W
56
4.8 Various Force Situations Lifts
LIFT - ACCELERATING UPWARDS
LIFT - ACCELERATING DOWNWARDS
R is a measure of the apparent weight of the
mass. When accelerating upward the mass is
heavier than normal, by an amount (ma), and
when accelerating downward it is lighter than
normal, by (ma).
For the Mass (m) ?F ma ?F R W so, R - W
ma and R W ma mg ma Thus R
m(g a)
For the Mass (m) ?F ma ?F W - R so ,W - R
ma and R W - ma mg - ma Thus R
m(g - a)
57
4.9 Various Force Situations Inclined Planes
Body of mass ( m) on an inclined plane
Weight and Normal Reaction act on mass
Weight broken up into 2 components
Force triangle and inclined plane triangle are
similar, ??s are equal.
Parallel (to the plane) component mgSin ?
Parallel Component can be transferred to act
through the C of M.
IT IS THIS COMPONENT THAT ACCELERATES THE MASS
DOWN THE PLANE
Perpendicular Component (mg Cos ?) R
Notice the Component of the Weight parallel to
the plane (mg Sin?) increases as the plane gets
steeper thus increasing the acceleration of the
mass down the plane.
58
4.10 Various Force SituationsConnected Masses
For m1 T - m1g m1a
?F ma
For m2 m2g - T m2a
For m1 T m1a
For m2 W - T m2a
Solve simultaneous equations to get acceleration
59
Chapter 5

• Topics covered
• Work.
• Work Energy.
• Power.
• Energy Types.
• Conservation of Energy.

60
5.0 Work

• The term WORK has a very strict definition in the
  physics world.
• If a FORCE moves an object through a DISTANCE,
  WORK has been done on that object.
• Mathematically
  W F.d where W Work
  (Joules) F Force (N)
  d Distance (m)
• This formula can only be used if the Force
  remains constant through the course of doing the
  work.
• Work is a SCALAR quantity.
• No work has been done if the force has not
  caused the object to move.
• No work has been done if the object begins and
  ends its movement in the same place. ie travels
  in a circle.

WORK DONE BY A VARIABLE FORCE
If the Force varies in the course of doing the
Work (as in stretching a spring), the work done
can only be calculated from the area under the
Force versus distance graph.
61
5.1 Work Energy
An object is said to possess energy if it has the
ability or capacity to do work. Work is the
Transfer Mechanism for Energy meaning that if
some work has been done on an object the amount
of energy it possesses has been changed.
In the Physics world, Work and Energy are
intimately related. Energy is very difficult to
define. It is easy to say what energy can do but
not so easy to say what it is. Thus energy is
defined in terms of work.
WORK DONE ENERGY TRANSFERRED
62
Motion - Revision Questions Question type
Work and Energy
A model rocket of mass 0.20 kg is launched by
means of a spring, as shown in Figure 1. The
spring is initially compressed by 20 cm, and the
rocket leaves the spring as it reaches its
natural length. The force-compression
characteristic of the spring is shown in Figure 2.
Q26 How much energy is stored in the spring
when it is compressed?
A Compressing the spring requires work to be
done on it. (this work is stored as elastic
potential energy in the spring) Work done area
under graph up to a compression of 0.2 m
½ (0.2)(1000) 100 J
63
Motion - Revision Questions Question type
Energy Conversion
Q27 What is the speed of the rocket as it leaves
the spring?
A All the elastic potential energy stored at max
compression (100 J) will be converted to Kinetic
Energy at release. So 100 ½ mv2 v
v1000 31.6 ms-1
64
Motion - Revision Questions Question type
Equations of Motion
Q28 What is the maximum height, above the
spring, reached by the rocket? You should ignore
air resistance on the way up since the rocket is
very narrow.
v2 u2 2ax 0 (31.6)2 2(-10)x x 1000/20
50 m
u 31.6 ms-1 v 0 a -10 ms-2 x ? t ?
65
Motion - Revision Questions Question type
Newtons 2nd Law
When the rocket reaches its maximum height, the
parachute opens and the system begins to fall. In
the following questions you should still ignore
the effects of air resistance on the rocket, but
of course it is critical to the force on the
parachute. This retarding force due to the
parachute is shown as R in Figure 3, and its
variation as a function of time after the
parachute opened is shown in Figure 4.
Q29 What is the acceleration of the rocket at a
time 5 s after the parachute opens?
A At t 5.0 s, R 1.8 N Weight of Rocket W
mg (0.2)(10)
2.0 N
SF ma a SF/m (2.0 1.8)/ 0.2
1.0ms-2
66
Motion - Revision Questions Question type
Work and Energy
In a storeroom a small box of mass 30.0 kg is
loaded onto a slide from the second floor, and
slides from rest to the ground floor below, as
shown in Figure 4. The slide has a linear length
of 6.0 m, and is designed to provide a constant
friction force of 50 N on the box. The box
reaches the end of the slide with a speed of 8.0
m s1
Q30 What is the height, h, between the floors?
At the second floor the box has Potential Energy
mgh On reaching the ground floor this has been
converted to Kinetic Energy (½mv2) plus the work
done against friction in moving down the slope
thus, mgh ½ mv2 Fd
(30.0)(10)h ½ (30)(8.0)2 50(6.0)
h 4.2 m
67
5.2 Energy Types

• KINETIC ENERGY
• The Energy of Motion. It is the energy possessed
  by moving objects
• Mathematically
  K.E. ½mv2 where, K.E.
  Kinetic Energy (J ) m mass (kg)
  v velocity (ms-1)

• GRAVITATIONAL POTENTIAL ENERGY
• The Energy of Position. The energy possessed by
  an object due to its position.
• Mathematically
  P.E. mgh where, P.E.
  Grav. P. E. (J ) m mass
  (kg) g
  gravitational field
  strength (N kg-1) h height above
  zero point (m)
• The zero point for measuring height is usually,
  but not always, the surface of the Earth.

• ELASTIC POTENTIAL ENERGY
• The energy stored in elastic materials (eg.
  Springs rubber bands)
• Mathematically
  Es ½ kx2 where, Es
  Elastic P. E. (J ) k Spring
  Constant (N kg-1) x Extension or
  Compression (m)

Energy is not a directional quantity so all these
forms of energy are Scalar Quantities.
68
Motion - Revision Questions Question type
Elastic Potential Energy
The box then slides along the frictionless floor,
and is momentarily stopped by a spring of
stiffness 30 000 N m1
Q31 How far has the spring compressed when the
box has come to rest?
KE of box is converted to Elastic Potential
Energy in the spring
Thus ½ mv2 ½ kx2
½ (30.0)(8.0)2 ½ (30,000)(x)2
x 0.25 m
69
5.3 Power

• Power is defined as the Time rate of doing Work.
• Mathematically
  P W/t
  where,
• P Power (W - Watts) W
  Work (J ) t
  time (s)
• Power is a Scalar Quantity
• SINCE WORK DONE
  ENERGY TRANSFERRED
• Power can also be defined as the Time rate of
  Energy Transfer.
• Mathematically
  P E/t where,
  P Power (W - Watts) E Energy (J
  ) t time (s)
• It is useful to note that since W F.d and v
  d/t the original power formula P W/t (F.d)/t
  F.v
• This allows the calculation of the power of a
  body moving at constant velocity

70
5.4 Conservation of Energy

• The law of conservation of energy states
• ENERGY CAN BE NEITHER CREATED NOR DESTROYED BUT
  ONLY TRANSFORMED FROM ONE FORM TO ANOTHER.
• Energy conversion processes convert useful,
  ORDERED energy (energy capable of doing useful
  work) into useless, DISORDERED energy. This
  disordered energy is sometimes called Thermal
  Energy, or Low Grade Unrecoverable Heat.

The percentage efficiency of energy transfers is
calculated from Eff EOUT/EIN x 100/1 No
energy transfer process can be 100 efficient in
our friction ridden world. If 100 efficiency
could be attained perpetual motion machines would
then be possible.
71
5.5 Seeing Energy (1)
With a little practice you can watch energy
flow through a system just as an accountant can
watch money flow through an economy.
As KE leaves the object it slows down as in a
car slowing down when you lift off the
accelerator pedal.
The most obvious form of energy is Kinetic
Energy, the energy of motion. It is easy to see
when kinetic energy is transferred to or from an
object.
A water polo ball speeds up when you do work on
it during the action of throwing, you are
transferring energy from your body into the ball,
where the energy shows up in the motion of the
ball.
72
5.6 Seeing Energy (2)
Potential Energy is more difficult to see. It
can take many different forms, as shown in the
table below. In each case nothing is moving but
because the objects still have a great potential
to do work, they possess Potential Energy.
At certain points all a body's energy may be
potential whilst at another all its energy may be
kinetic. Recognizing these points will enable
the solutions to most problems in this area.
Form of Potential Energy Example
1. Gravitational Potential Energy A person standing at the top of a building
2. Elastic Potential Energy A wound clock spring
3. Electrostatic Potential Energy A cloud in a thunderstorm
4. Chemical Potential Energy A firecracker
5. Nuclear Potential Energy Uranium
73
Chapter 6

• Topics covered
• The Experience of Acceleration.
• Circular Motion.
• Centripetal Force.
• Vertical Circles.
• Projectile Motion.
• Projection Angles
• Projectile Graphs
• Real Life Projectiles

74
6.0 The Experience of Acceleration
The backward force you experience when a car
accelerates is caused by your bodys inertia,
its tendency not to want to accelerate.
The car and your seat are accelerating, and since
the seat back acts to keep you from falling
through its surface, it provides a forward
support force which causes you to accelerate
forward. But the seat cant exert a force
uniformly throughout your body. Instead, it
pushes only on your back and your back then
pushes on your bones, internal organs and tissues
to make them accelerate forward.
Nothing is more central to the understanding of
the laws of motion than understanding the
relationship between force and acceleration. Up
to now we have looked at forces and noticed they
can produce accelerations (Newtons 2nd Law). Now
we will reverse the process looking at
accelerations and noticing that they require a
force. For you to accelerate, something must
push or pull on you.
The experience of the accelerating car seat is
very similar to the experience of gravity when
you stand still on the Earths surface. In this
situation you feel heavy ie. you experience
your weight
Just where and how that force is exerted on you
determines how you feel when you accelerate.
75
6.1 Fictitious Forces
However despite the convincing sensations, the
backward, heavy feeling in your gut as you
accelerate is not due to a real force. This
experience of acceleration is explained by the
supposed action of a Fictitious Force. This
Fictitious Force always points in the direction
opposite to the acceleration that causes it and
its strength is proportional to the acceleration.
When the car seat is causing you to accelerate
forward, you also feel heavy your body senses
all the internal forces needed to accelerate its
pieces forward, and you interpret these
sensations as weight. This time you experience
the weight directed toward the back of the car.
The gravity-like force that you experience as
you accelerate is truly indistinguishable from
the force of gravity. No laboratory instrument
can determine directly whether you are
experiencing gravity or simply accelerating.
76
6.2 Apparent Weight
With the car stationary the only force you are
subject to is your Weight
As the car accelerates you are pushed back in
your seat by the fictitious force
The vector sum of the Weight and Fictitious force
produces the Apparent Weight you experience
whilst accelerating.
The larger the Acceleration, the larger the
Fictitious force, the more backward your
Apparent Weight becomes.
77
6.3 Circular Motion

• Newtons 1st law says, in part, an object
  experiencing no net force will travel in a
  straight line at constant speed.
• Thus, in order to make an object travel in a
  circle ie. constantly change the direction of its
  velocity, a force must be constantly applied to
  it. This means the object is constantly
  accelerating.
• The number of times an object spins around per
  second is called its Frequency (f).
• An object travelling around a circle completes
  one full spin in a time interval called a Period
  (T). Period and frequency (f) are the inverse of
  one another. Thus
• T 1/f
• T Period (sec)
• f Frequency (Hz or sec-1)

CIRCLE OF RADIUS R
Circumference of circle 2 ? R
Time for one revolution Period T
Speed Distance , So v 2? R 2? Rf
Time T
78
6.4 Centripetal Acceleration
Since the direction of the velocity at any point
is the tangent to the circle at that point, the
direction of the velocity is constantly changing,
thus an object travelling in a circular path is
subject to a constant acceleration.
The size of that acceleration (called the
CENTRIPETAL (or centre seeking) ACCELERATION) is
given by aC v2/R
where
aC centripetal
acc (ms-2) v linear
velocity (ms-1) R radius of
circle (m) Substituting for v 2?R/T
gives aC 4 ?2 R/T2
CIRCLE VIEWED FROM ABOVE
aC vf - vi
THE CENTRIPETAL ACCELERATION, aC ,IS ALWAYS
DIRECTED TOWARD THE CENTRE OF THE CIRCLE.
79
6.5 Centripetal Force
The fact that an acceleration exists, directed
toward the centre of the circle
requires there to be a force acting in the same
direction.
This force is called the CENTRIPETAL FORCE (FC)
and is defined in terms of Newtons 2nd law.
Mathematically FC maC mv2/R m4?2R/T2
Girl Riding a Playground Ride
Centripetal Force is NOT a REAL FORCE in its own
right, but is supplied by other real, measurable
forces.
In the case opposite, the girl requires a
Centripetal Force to travel her circular path
The Tension (T) of her muscular grip on the
pivot pole of the ride provides the Centripetal
Force.
She, of course, will use the idea of a Fictitious
Force, centrifugal force, to explain the
outward pull she feels while on the constantly
accelerating ride.
80
Motion - Revision Questions Question type
Centripetal Force

• Mark Webber and his Formula 1 racing car are
  taking a corner at the Australian Grand Prix. A
  camera views the racing car head on at point X on
  the bend where it is travelling at constant
  speed. At this point the radius of curvature is
  36.0 m.
• The total mass of the car and driver is 800 kg.

Q32 On the diagram showing the cameras view of
the racing car, draw an arrow to represent the
direction of the NET force acting on the racing
car at this instant.
The magnitude of the horizontal force on the car
is 6400 N.
Q33 Calculate the speed of the car.
A FC mv2/R So v vFCR/m
v(6400)(36.0)/(800) 17 ms-1
81
Motion - Revision Questions Question type
Centripetal Force
Q34 Referring to the racing car from the
previous slide, explain (a) Why the car needs a
horizontal force to turn the corner. (b) Where
this force comes from.

• A (a) Newton 1 states all objects will continue
  in a straight line unless acted upon by a net
  force.
• In order for the car to travel in a circular
  path a constant force directed at right angles to
  the direction of motion (toward the centre of the
  circle) must exist.
• (b) The horizontal force arises from the
  frictional force exerted BY THE ROAD ON THE TYRES
  and is directed toward the centre of the circle
  the car is travelling in.

82
Motion - Revision Questions Question type
Centripetal Force
The safe speed for a train taking a curve on
level ground is determined by the force that the
rails can take before they move sideways relative
to the ground. From time to time trains derail
because they take curves at speeds greater than
that recommended for safe travel. Figure 5 shows
a train at position P taking a curve on
horizontal ground, at a constant speed, in the
direction shown by the arrow.
Q35 At point P shown on the figure, draw an
arrow that shows the direction of the force
exerted by the rails on the wheels of the train.
83
Motion - Revision Questions Question type
Centripetal Force
The radius of curvature of a track that is safe
at 60 km/h is approximately 200 m. Q36 What is
the radius of curvature of a track that would be
safe at a speed of 120 km/h, assuming that the
track is constructed to the same strength as for
a 60 km/h curve?
This can be solved as a ratio question m(60)2
m(120)2 200 x
x 800 m
Q37 At point Q the driver applies the brakes to
slow down the train on the curve. Which of the
arrows (A to D) indicates the direction of the
net force exerted on the wheels by the rails?
A B
84
6.6 Centrifugal Force
Despite its fictitious nature, centrifugal force
creates a compelling sensation of gravity like
force. The girl on the ride feels as though
gravity is pulling her outward as well as down
and must hold the handle tight in order not to
fall off.
Fictitious forces, such as Centrifugal Force, do
NOT contribute to the net force experienced by an
object. Thus if the girl lets go she will fly
off the ride in the direction of the linear
velocity, NOT in the direction of the fictitious
force.
A stationary object on Earth experiences a weight
force of 1g. The fictitious centrifugal force
experienced by 5 kg of clothes during the spin
cycle in a washing machine, travelling in a 0.25
m radius circle at 20 ms-1 is about 163 gs and
they have an apparent weight of 815 kg.
Anyone who talks about centrifugal force as being
a real, measurable force is talking rubbish and
does not understand physics, be wary of ALL they
say !!!!!
85
6.7 Banked Corners
Race track and road designers often build their
tracks or roads with banking on the corners.
This design feature is used to enhance the safety
of the track or road allowing users to round
corners at higher speeds (and with a greater
margin of safety) than they could if the corner
was not banked.
Any vehicle travelling around a corner with
velocity v needs a centripetal force (FC) acting
toward the centre of the corner
In the case of the flat, non banked, corner this
centripetal force (FC) is supplied by the
friction of the road against the tyres (FRT)
With the banked cor