Module 2 UNDERSTANDING MOTION 2
Module 2 – Understanding motion
Module
2
UNDERSTANDING MOTION
2
Written by
Dr Janet Taylor
Revised by Lucy George 2001
Module 2 – Understanding motion
2.1
Introduction
Module 1 – The Nature of Physics
Inside us, blood flows. Inside plants, fluids circulate. In the heavens, stars and planets rotate and orbit. Within the atom, electrons are constantly on the move.
Even things that appear to be at rest are moving, relative to another object. Motion is observed everywhere. As easy as motion is to recognise it is hard to describe. This module is concerned with our understanding of motion – what it is, how it is measured and represented and how different forces and energies relate to different motions.
Before starting this module consider the following scenario.
Many of you, as drivers, would have been faced with the dilemma of whether or not you should speed through an orange light or stop? Lateness for appointments, stress levels, red light cameras and intuition aside, how can we decide if we should risk it? (There are certain calculations we can make to judge more accurately if we will make it through.) We will come back to this challenge at the end of this module. Viewing the physics video will also help you with this challenge.
2.1 Linear motion
2.1.1 How to measure motion
For an object to be in motion it must be changing position relative to another object. Speed is a measure of how fast something is moving or changing position. It is described as a rate or a quantity divided by time. In this case the quantity is the distance moved. So if you are late for work and running to catch a bus, you might run 100 metres in 20 seconds.
Your average speed would be calculated as follows:
Average speed = time interval
=
100 metres

20 seconds
= 5 metres/second
This is written as 5 m/s or 5 ms
–1
If you miss the bus and catch a taxi to the next town 40 kilometres away, it might take
35 minutes. Your average speed would then be:
2.2
TPP7160 – Preparatory physics
Average speed =
35 minutes
=
40 km

0.58 hour
= 68.6 km/hour (rounded to one decimal place)
This is written as 68.6 km/h or 68.6 km h
–1
. In SI units this would be converted to ms
–1
by the following calculation
68.6
1000
60 60
= 19.0 ms
–1
(rounded to 1 decimal place).
(Note that if you have a number of calculations keep them as accurate as possible before finally rounding.)
However if you glance at the speedometer of the taxi during the above trip, you would notice that it changes. On a straight run it could read 100 km/hr, while on the corners it could read 20 km/hr.
The speedometer measures the speed of a car at any one instant. This is termed instantaneous
speed.
Distance and speed are quantities that tell you something about the size or magnitude of the change in position, and how fast that position has changed.
The problem is that distance and speed don’t tell you anything about the direction of that change in position. To take this into account two further measures were developed:
displacement and velocity.
Displacement is a measure which incorporates the magnitude and direction of the change in position.
Velocity is a measure which incorporates how fast the change in position takes place and the direction of that change.
2.1.2 How to represent motion
You can represent velocity diagrammatically by drawing a line whose length represents the speed (or magnitude of velocity) and its direction the direction of the velocity. Such lines are called vectors.
Scalars are quantities like speed or distance which have no direction i.e.
Scalar
(magnitude only)
distance
(e.g. 10 m) speed
(e.g. 10 ms
–1
)
Vector
(magnitude and direction)
displacement
(e.g. 10 m east) velocity
(e.g. 10 ms
–1
north)
Module 2 – Understanding motion
2.3
Later you might like to add to this table as we investigate other scalar and vector quantities.
Example:
Let’s consider the situation discussed previously. Imagine you are running for the bus at
6.7 metres/second with a wind blowing at 2.7 metres/second in the same direction. Such a situation could be represented by scaled vectors so that 1 cm represents 1 metre per second.
Scale: 1cm : 1ms
–1
6.7 m/s
6.7 cm
2.7 m/s
2.7cm
To find your velocity:
•
Draw an arrow to represent each individual velocity. The arrow points in the direction of the velocity and its length represents the size of the velocity.
•
Redraw the second arrow so that its tail is at the tip of the first.
•
Draw a vector from the tail of the first arrow to the tip of the second arrow.
6.7
2.7
If you write:
V y
to be the vector representing your velocity.
V w
to the vector representing the velocity of the wind.
V wy
to be the vector representing your final velocity.
thus
V y
+ V w
= V wy
Consider if you were running into the wind. It is harder running when the wind is against you because
V wy
=
V y
– V
w
This can be represented using vectors.
V y
V w
In the olympics wind ‘assistance’ is considered when deciding whether new records are allowed.
2.4
TPP7160 – Preparatory physics
Example:
Consider the situation of running for the bus with the wind blowing perpendicular to your direction of running (as shown) what would be your resultant speed relative to the ground and what would be your direction (relative to the ground means as observed by someone who is stationary, since the earth itself rotates on its own axis and also revolves around the sun).
You could represent it using the following vectors.
V w
= wind velocity your result ant velocity
V y
= your velocity
To calculate the resultant velocity we need to calculate two things:
(i) magnitude
(ii) direction.
Firstly draw a vector diagram by completing the parallelogram below
(which in this case is a rectangle).
r
2.7
σ
6.7 m/s
Then using Pythagoras’ Theorem
2.7
2
+ 6.7
2
=
r
2
r
2
r
2
= 2.7
2
+ 6.7
2
= 52.18
r
= 7.2 ms
–1
!"The resultant velocity has a magnitude of 7.2ms
–1
.
2.7 m/s
Module 2 – Understanding motion
2.5
To calculate the direction of the velocity, use the vector triangle and the tan ratio. tan # =
2.7

6.7
# = 21.9$
= 0.4030
So you are running at 7.2 metres per second at an angle of 21.9$ to your original direction.
This velocity is called the resultant velocity.
The wind has increased your speed but at the same time changed your direction.
Vectors are a very useful tool to help understand and predict behaviours of moving objects.
They can be used to represent many different quantities, if these quantities have magnitude and direction components. You will return to the study of these later.
Activity 2.1
Question 1
A dog searching for a bone walks 3.5 m south, then 8.2 m at an angle of 30$"north of east and finally 15 m west. Find the dog’s resultant displacement vector by drawing the vectors accurately to scale.
Question 2:
A motor boat is travelling due east at 14 ms
–1
across a river that is flowing north at 6 ms
–1
.
(i) Use both graphical (by scaled drawing) and mathematical methods to find the resultant velocity of the boat.
(ii) If the river is 100 m wide, how long does it take the boat to reach the other side?
(iii) How far downstream is the boat when it reaches the other shore?
Question 3:
A plane is headed directly east at 340 km/hr when the wind is from the south at 45 km/hr. What is the plane’s velocity with respect to the ground?
2.6
TPP7160 – Preparatory physics
2.1.3 Acceleration
Previously we examined instantaneous velocity where, while driving along in a taxi, the speedometer variously changed from 60 to 80 then back to 60 km/hr (for example). If you calculate how fast this velocity is changing then you are calculating acceleration.
Acceleration = time interval (change in time)
For instance, if you were driving in the above taxi along a straight road and increased your velocity from 60 to 80 km/hr in a twosecond period then
Acceleration =
80 km/hr – 60 km/hr

2 second
'
% change in velocity
change in time
(
&
= 10 km/hour/second
(10 km per hour per second) or (2.8 ms
–2
in SI units)
This is written as 10 km/h/s or in SI units as 2.8 ms
–2
On the other hand, if you decrease your velocity from 80 to 60 km/hr:
Acceleration =
60 –
2
80
= –10 km/hour/second (or –2.8 ms
–2
)
Acceleration can be considered as the rate at which velocity changes. Just as velocity is dependent on the magnitude and direction of the change in position change, so acceleration is dependent on the magnitude and direction of the velocity change.
If you ride on a merrygoround you can feel the changes that are taking place as you swing around. The speed (or the magnitude of the velocity) of the merrygoround will be constant but the velocity will be changing because of the changing direction, so therefore there is an acceleration. (As long as one component is changing, be it the magnitude or the direction of the velocity.)
Suppose that a taxi does not change its speed (the magnitude of the velocity) but turns a corner. Is this acceleration?
Yes, because although the speed is not changing the direction is – thus a change in velocity has resulted – this is defined as acceleration.
Acceleration has magnitude and direction components and can be represented by a vector. It is a vector quantity.
Module 2 – Understanding motion
2.7
Activity 2.2
Question 1:
A car’s velocity increases from 3.0 ms
–1
to 3.5 ms
–1
over a
4 second period. What is its acceleration?
Question 2:
The handbrake gives way on your car and it rolls down the hill reaching a speed of 10 ms
–1
in the first 3 seconds.
What is its acceleration?
Question 3:
A road train slows down from 50 ms
–1
to 10 ms
–1
in 15 seconds in order to round a corner. What is its acceleration?
2.1.4 Motion in a straight line
Let us consider a simple form of motion – motion in a straight line where the object accelerates uniformly (i.e. acceleration is constant). There are many examples of this in real life.
• a train constantly accelerating along a straight track
• a dragster speeding along a drag strip
• an athlete running the 100 m (as long as we assume that the acceleration is constant)
• fruit falling from a tree.
In all of these situations an object is falling or travelling in a straight line with constant acceleration (at least for part of the time).
It would be handy for us, as well as scientists in general, to have a simple set of equations that describe how far and how fast an object is moving or falling.
To derive such equations the first step is to define the variables.
Let a = constant acceleration
u
= the initial velocity of an object moving in a straight line
v
= the velocity after time, t
t
= time
s = the distance travelled in time, t
We know from section 2.1.3
acceleration = change in time
a
=
v
–
t u
(using SI units this would be ms
–2
)
2.8
TPP7160 – Preparatory physics
Making v the subject of the formula above,
v = u + at
(Equation 1)
This is often referred to as the 1st equation of motion.
Also from section 2.1.1
average velocity = change in time
(average velocity is used when velocity changes over the period of time)
u
+
2
v
=
t
Making s the subject of the formula above,
s
=
s
=
'
%
u
+
v

2
(
&
t
'
%
u
+
u
2
+
at
*
(
&
t substituting for v = u + at (from equation 1)
s
=
ut
+
ut
+
at
2
2
!" s =
ut
+
2
at
2
(Equation 2)
This is referred to as the 2nd equation of motion.
Finally consider the equations
s
=
)
u
+
2
v
*t and
a
=
v
–
t u
we rearrange both equations to make t the subject of the formula
t
=
2s

u
+
v
and
t
=
v
–
u

a
equating for t we have,
2s

u
+
v
=
v
–
u

a
(v + u) (v – u) = 2as(crossmultiply)
v
2
– u
2
= 2as
!
v
2
=
u
2
+ 2as (Equation 3)
This is referred to as the 3rd equation of motion.
Module 2 – Understanding motion
2.9
The three equations for motion in a straight line for uniform or constant acceleration are
v
= u + at
s
= ut +
at
2
2
v
2
=
u
2
+ 2as
In situations of bodies falling on or near the surface of the earth, the acceleration due to gravity is called g with an approximate magnitude of 9.8 ms
–2
, and directed towards the centre of the earth.
Example:
A hammer is dropped from the roof of a building.
(a) What would its velocity be after 4s?
(b) How far does it fall during this time.
Solution:
(a) Firstly set out your given information
u
(initial velocity) = 0 (dropped)
a
(acceleration due to gravity) = 9.8 ms
–2
t (time) = 4 s
Since u, a, t are known, and we want to find its final velocity v after 4 s, we need to think about the concept involved here.
Basically a hammer is dropped.
As it drops it gains speed because of gravity accelerating it.
u = 0
g = 9.8 ms

–2
t = 4 s
An image of what’s going on is always helpful.
As we visualise the event, the actual concept here is a cceleration.
v = ?
2.10
TPP7160 – Preparatory physics
We know that acceleration or
= change in velocity

time
!
a
=
9.8 =
v
–
t u v
–
4
0
9.8 × 4 = v
(a = g in this case) or v = 39.2 ms
–1 in the direction of the acceleration.
We could also just refer to the first equation of motion where
v = u + at
(relating the variables we are interested in.)
v
= 0 + 9.8 " 4
= 39.2 ms
–1
A word of advice: Students tend to look at physics problems by firstly asking themselves
‘which formula do I use?’ Instead I would advise you to firstly:
•
•
•
•
Set out your given information
Draw a picture or vector diagram labelled with the given data
Think about what basic concept(s) is involved here
Use a formula(s) that describes that concept(s) in a nutshell (that’s what a formula does)
As you can see I chose to use
a
=
v
–
t u
in the example above because it describes our central concept of acceleration.
Remember the 3 equations of motion are simplified versions describing the concepts of linear motion. The first equation of motion v = u + at came from simplifying
a
=
v
–
t u
.
So I hope you will not make the mistake of solving physics problems by asking ‘which formula do I use?’ but instead think about the concepts involved first. With experience and practice, you will learn to use simple basic formulas with ease. Remember formulas are concepts in a nutshell.
(b) This question asks how far the hammer falls during this time. We want to find its displacement or change in position.
Module 2 – Understanding motion
2.11
Since there is a change in velocity due to acceleration, we know that
u = 0
s = ?
t = 4 s
v = 39.2 ms
–1
g = 9.8 ms
–2
average velocity
=
displacement
=
time average velocity
"
time s
=
%
#
u
+
2
v
&
$
"
time
=
%
#
0 + 39.2

2
&
$
"
4
= 78.4 m in the direction of the acceleration.
OR we could have used the equation of motion
s = o
ut
+
1

2
gt
s
=
1

2
× 9.8 " 4
2
2 where u = 0,
= 78.4 m or
!
v
2
=
u
o
2
+ 2as
(39.2)
2
= 0 + (2 × 9.8 × s)
s = 78.4m
or
2as = v
2
2
s =
2a
=
39.2
2

2
"
9.8
= 78.4m
Home experiment
Try this simple experiment for yourself.
You will need a small ball, a stopwatch or a watch from which you can count seconds, and possibly an assistant.
Throw a ball straight up into the air to an approximate known height (e.g. the height of your house). Measure the time it takes for the ball to reach the ground after it has reached its maximum height. Using the equations of motion, assuming that the acceleration due to gravity is 9.8 ms
–2
, calculate the velocity of the ball when it reaches the ground.
2.12
TPP7160 – Preparatory physics
•
What would you expect the velocity to be?
_______________________________________________________________________
•
Detail accurately what methods you used to make the measurements in this experiment and how you calculated the velocity.
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
•
What did you calculate the velocity to be?
_______________________________________________________________________
•
Were your predicted and observed velocities different?
_______________________________________________________________________
•
How can you explain this, and could the experiment be improved?
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
•
Did you refer to any books or other sources of information to complete this experiment. If so list them here.
_______________________________________________________________________
_______________________________________________________________________
Well, how did you go with this first home experiment? Before starting the experiment you could have calculated, using Newton’s Equations of Motion, what you might expect the time a ball was in the air if it was thrown straight up.
Module 2 – Understanding motion
2.13
t t
Height of ball
10 m
s
20 m
s
C
Time of flight = 2t
(time to reach maximum height = time to fall down to starting point)
Final velocity (v)
u = 0
t s
=
ut
+
1
at
2
2
,
u
= 0
a
= 9.8 ms
–2
v
= u + at,
u
= 0,
a
= 9.8 ms
–2
t
= 1.42 s
t t
= time it takes to drop through a displacement of 10m
10 =
1

2
"
9.8
"
t
2
v
= 9.8 "!1.42
v
= 13.9 ms
–1
v = ?
t
2
= 2.04
t
= 1.43 s time of flight = 2t
= 2.86 s
!After falling back to its starting point, its final velocity = 13.9 ms
–1
v
= u + at
s
=
ut
+
1
at
2
2
u = 0
v =
9.8 "!'()'
20 =
1

2
"
9.8
"
t
2
= 19.8 ms
–1
t
= 2.02 s time of flight = 4.04 s
C
v = ?
Then you could have conducted the experiment calculating the velocity of the ball by measuring the maximum height of the throw and the time the ball takes to reach this maximum height
%
# velocity = time
&
$
.
If your calculations in the experiment do not match your theoretical prediction, then you have to look at your methods and conditions on the day to determine what other factors could cause your results to differ from the theoretical. Maybe it was a windy day, or maybe you could not estimate the height accurately enough. A range of factors could be involved but you were there so only you know what they could be.
2.14
TPP7160 – Preparatory physics
Activity 2.3
Question 1: A bus moving at 15 ms
–1
slows down at a rate of 2 ms
–2
. How far will it go before it stops?
Question 2: A driver of a car travelling at 60 km/hour sees a child crossing the road 100 m away. Calculate the acceleration required by the car to stop before hitting the child (in SI units). Is your answer possible?
Explain.
Question 3: What velocity would a bungi jumper reach two seconds after jumping from a platform? How far would she have fallen during this time?
Question 4: A cat falls from a branch of a tree and is in the air for 0.5 seconds before touching the ground. How high was the branch from the ground?
Question 5: What is the acceleration of a car that reaches a speed at 18 ms
–1 from rest after travelling 240 m? The motorist having reached this speed maintains it through a built up area where the maximum speed is 60 km h
–1
. Is the motorist above or below this speed limit and by how much?
Question 6: A lift is travelling downward at 4.4 ms
–1 and is accelerated upward at 1.5 ms
–2
for two seconds. What is its final velocity? (remember that velocity includes direction.)
Section review
In this section we have examined concepts of motion including velocity and acceleration. We have also discussed how these concepts can be represented using vectors and how problems involving motion can be solved using Newton’s equations of motion. You should now assess yourself as to whether you have met the module objectives.
You should now be able to:
• explain the difference between speed and velocity
• use vectors and scalars to represent velocity and speed
• explain the meaning of acceleration and its relationship with velocity
• use equations of motion.
In your study of this section you should have been making notes of the main points and listing those concepts that were difficult to understand.
Have you been able to revise those concepts that were difficult to understand?
Have you sought help?
Use the following posttest to see how you are going so far.
Module 2 – Understanding motion
2.15
Posttest 2.1
Question 1: Complete this summary of the previous section.
The rate of change of displacement with time is called __________ .
Velocity is a ____________________ quantity i.e. its ___________________ and
______________________ must be specified.
The magnitude of the velocity of a body is called its ___________________ .
Speed is a __________________ quantity.
If we divide total displacement by time taken the result is the ____________________ velocity in the direction of displacement.
The velocity of a moving object at any particular time is called ____________________ velocity.
The SI unit of velocity is ___________________ .
If velocity is changing with time, the rate at which it changes is called the
____________________ of the body.
The SI unit of acceleration is ____________________ .
If the velocity changes by a constant amount every second, the acceleration is said to be
____________________ .
Gravitational acceleration at any particular point on the earth’s surface is an example of
_____________________ acceleration.
If an object undergoes a displacement s is time t, starting with an initial velocity, u, and acceleration a in the direction of the displacement and after time t the velocity is v, three equations of motion can be written
1.
2.
3.
If the body starts from rest these equations become
1.
2.
3.
2.16
TPP7160 – Preparatory physics
Question 2: Can a body have a constant speed and a changing velocity? Can a body have a constant velocity and a changing speed? Explain your answers.
Question 3: If a ball is thrown straight up into the air, describe the changes in its velocity and acceleration.
Module 2 – Understanding motion
2.17
2.2 Motion and forces
In our world pigs don’t fly or if they did we would probably look for supporting wires or some special effects. Throughout our lives we have been observing nature and have accepted how objects behave on our planet. In the past many observers have watched and recorded how objects move. Isaac Newton (1642–1727), the famous scientist and mathematician, synthesised the work of many others to formulate these observations into three laws:
Newton’s First Law of Motion (Law of Inertia)
Newton’s Second Law of Motion
Newton’s Third Law of Motion
2.2.1 Newton’s First Law of Motion (Law of Inertia)
Have you ever slipped on a wet floor, or worse still, tried to stop when ice skating? Or have you ever tried the magicians’s trick of pulling a tablecloth out from under a set of cups and saucers without dropping the crockery? These are all examples of the operation of Newton’s
First Law of Motion, often called the Law of Inertia.
Newton’s First Law says:
An object will continue moving in a straight line at a constant speed and a stationary object will remain at rest unless acted upon by an unbalanced force.
Home experiment
Try this experiment at home.
You will need a glass of water, a dollar coin and a square of shiny cardboard bigger than the top of the glass.
Place the cardboard over the top of the glass and the coin in the centre of the cardboard and then try to remove the cardboard.
Hint: Try flicking the cardboard out with your finger or ruler.
•
What do you think will happen to the coin?
•
Describe in detail the methods you used to perform this experiment.
•
Describe what did happen to the coin.
•
What is your explanation of this?
•
Could this experiment be improved?
•
Did you refer to any books or other people to help you in your explanations? If so list what or who you referred to.
2.18
TPP7160 – Preparatory physics
There are many other examples of the operation of Newton’s first law in our everyday world around us.
•
You don’t slip on a dry floor, but can very easily slip on a wet floor, because the force of friction is much less on the wet floor than the dry floor.
•
When driving your car, if you take your foot off the accelerator it does not automatically come to a stop. It coasts along for a while. If you want it to stop you have to apply the brakes, that is, apply another force.
The force then is what we have to apply to cause a change in motion or a change in velocity. It is a vector quantity because it has a magnitude and a direction and is measured in a unit called a Newton. A force of magnitude of 1 Newton (N) causes a mass of one kilogram to accelerate at a rate of one metre per second per second.
The concept of a force is important throughout physics and will be returned to a number of times throughout these modules.
What is Inertia?
Newton’s first law of motion is often called the Law of Inertia – but what is inertia? Have you ever tried to push a loaded shopping trolley uphill or to push a bogged car out of a hole.
These objects are said to possess ‘inertia’ – a tendency to remain in the same state of motion
(in this case still). Their inertia is dependent on their mass, so that mass is often used as a measure of inertia.
Mass is often confused with weight. Mass is the measure of the amount of matter in an object; it depends only on the type and number of atoms in it. In SI units it is measured in kilograms.
Weight on the other hand is not a fixed property of an object but depends on the amount of gravitational force acting on it, hence it varies with location.
Weight is directly proportional to the acceleration due to gravity. In SI units weight = mass " acceleration due to gravity.
Note that the term ‘weight’ is often used incorrectly by many of us, when we get a measure on the scales that is really our mass, not weight, as weight is a force and is measured in Newtons.
Note mass is a scalar quantity, while weight is a vector.
Activity 2.4
Question 1: What is your weight in Newtons?
Question 2: What would it be on the moon which has a gravitational force
1
/
6
that of earth?
Question 3: What would it be on Mars where the acceleration due to gravity is
8.3 ms
–2
?
Question 4: What would it be in outer space?
Module 2 – Understanding motion
2.19
2.2.2 Newton’s Second Law of Motion
We all know it takes a greater force to push a full shopping trolley than an empty one.
This is an illustration of Newton’s Second Law which says that:
The acceleration of a body is directly proportional to the force acting on it and inversely proportional to the mass of the body.
In mathematical terms F = ma where F is the force in Newtons
m is mass in kilograms
a is acceleration in ms
–2
Weight is measured in Newtons and mass is measured in kilograms in the SI System.
What forces act on a body?
Consider your study book sitting on your desk. What forces are acting on it? Gravity would be one, but if it were the only force then the book would be accelerating downward through the desk. If this is not happening then the desk must also be pushing upward on the book.
There are often many forces acting on an object, but the object will stay still if the sum of all those forces is zero. Remember that force is a vector quantity with magnitude and direction.
So it follows that if an object is made to move then it must have a set of unbalanced forces acting on it. What we see is the affect of resultant force. For example if we push our study book across the desk there are another two forces acting on it; the force we exert and the friction of the desk’s surface. If we represent this diagrammatically we have:
Table’s surface
Actual diagram
Upward force exerted by desk
Backward force exerted by friction between book a nd desk by hand
Downward force exerted by gravity
A ‘free body diagram’
The first law tells us when a force is acting.
The second law tells what the force does when it acts.
So if we know the forces acting on a system of known mass we can predict its future motion.
This is intuitively accepted. For example the harder you throw a ball the further it goes and the greater the force the greater the acceleration. Newton’s second law defines the balance between resultant force and mass in producing an acceleration.
2.20
TPP7160 – Preparatory physics
However Newton’s second law does not imply that every time a force acts motion results. For example you can push against a wall without moving it; your study book on a desk is under the influence of gravity yet does not move down through the desk towards the centre of the earth.
In both cases the wall and the desk are exerting their own forces that balance the one that acts on them. It is only the net or unbalanced forces that actually give rise to acceleration.
Before we consider how these unbalanced and balanced forces act on an object, it is important to examine Newton’s third law.
2.2.3 Newton’s Third Law of Motion
Previously you looked at the situation of pushing a wall and it pushing back, or the study book on the desk not falling down towards the centre of the earth. Similarly if you touch a person that person cannot help but touch you back. These are all examples of Newton’s third law.
For every action there is an equal and opposite reaction.
Newton’s third law tells us that whenever a force is applied to an object, that object simultaneously exerts an equal and opposite force – forces always act simultaneously in pairs.
How does this tie in with Newton’s other laws? If you catch a ball, the first law says you have to exert a force so that the ball will stop, the third law says the ball will exert an equal and opposite force on our hand, and the second law says that a resultant force on your hand will cause your hand to move.
So how can Newton’s three laws help us to understand and analyse the forces working on moving and nonmoving bodies?
2.2.4 Forces acting horizontally and vertically
Consider again your study book sitting on your desk.
Table’s surface
Actual diagram
Upward force exerted by desk
Backward force exerted by friction between book a nd desk by hand
Downward force exerted by gravity
A ‘free body diagram’
Module 2 – Understanding motion
2.21
The forces acting on it would be as shown – downward force (F
w
weight = mg) and an upward force (F
n
) due to gravity (known as
) termed the normal force. As the book is stationary (in the vertical direction) these forces are in equilibrium so that sum of forces in the vertical direction = 0 i.e.
!F = 0 ( is the symbol for summation) or
mg + F n
= 0
Remembering that forces are vectors with direction, F
n
opposite directions i.e.
F n
and weight have equal magnitude but
= –mg (the negative sign here indicates direction only).
Now what happens if you push the book across the desk. If the desk is highly polished it will move easily, if on the other hand it has a rough surface it will be more difficult to move. Two more forces are involved – the force you exert by pushing the book (F
p
frictional force (F
f
acting on the book.
) and the frictional
) between the book and the surface. This means that there are four forces
F n
F
p
table’s surface
F
f
F
w
Since the book is being pushed forward, there must be a net force in the forward direction.
Summing horizontal forces,
F =
F net
F p
+ F f
= F net
(in this case, F p
> F f
for an acceleration to occur)
Experimentally it has been found that the frictional force increases as you first try to move an object. This force is called the static frictional force. Once the object begins to move, the frictional force acting is called the kinetic frictional force. This is much smaller than static friction. Friction is
• dependent on the nature of the two surfaces
• independent of the size (measured as contact area) of the two surfaces in contact.
The frictional force is actually proportional to the normal force pushing the two surfaces in contact with each other.
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TPP7160 – Preparatory physics
The result can be summarised by the equation:
F f
= "
s
F n
where
F f
is the magnitude of the frictional force
F n
is the magnitude of the normal force, exerted by the desk on the book
"
s
is coefficient of static friction which depends on the nature of the surfaces in contact.
Once the book moves, F
k
of kinetic friction.
= " k
F n
where F
k
is the force of kinetic friction " k is the coefficient
The coefficient of friction being a ratio of the magnitude of two forces is a unitless number. It depends on the types of materials in contact and the condition of the surfaces (polished, rough) and often other variables such as temperature. Typically the coefficient of static friction ("
s
ranges from 0.01 for smooth surfaces to 1.5 for rough surfaces while the values of "
k
smaller.
) are much
Note that if F
p
= F constant velocity.
k
i.e. net force = 0 there is no acceleration, and the book moves forward with
Example:
A person presses a 1.6 kg book against a vertical brick wall. If the static coefficient of friction between the book and the wall is 0.35, how hard would the person have to push to allow the book to slide down at a constant speed.
mg
F f
becomes
F n
F
push
F p
F w
Actual
F
frictional
Free body diagram
mg
Solution:
If the book is moving down at a constant speed it has no acceleration. There is no net force and the vertical forces are in equilibrium, with the gravitational force dragging it down equal to the frictional force holding it up.
i.e. Gravitational force =
mg
= 1.6 # 9.8
= 15.68 N
Frictional force, F
f
= 15.68 N.
Module 2 – Understanding motion
2.23
The force exerted on the book by the wall is the normal force, and is calculated from the equation:
F f
= "
s
F n
where
F f
= 15.68 N
F n
=
F

"
f s
"
s
= 0.35
F n
=
F n
is force exerted by the wall and is unknown.
0.35
$ F
n
= 44.8 N
The force that the person exerts on the book to hold it against the wall has a magnitude of
44.8 N.
Activity 2.5
Question 1:
The coefficient of kinetic friction between rubber car tyres and a wet road is 0.50. If the driver of a 800 kg car travelling at 30 ms
–1 applies the brakes and skids to a stop:
(a) What is the size and direction of the force of friction that the road exerts on the car?
(b) What is the resulting acceleration of the car?
(c) How long will it take for the car to come to rest, and how far will it skid?
Question 2:
To drag a 50 N box along the footpath at a constant speed a horizontal force at 36 N is exerted. What is the coefficient of friction between the footpath and the box?
2.2.5 Forces at any angle
Have you ever taken a dog for a walk? If the dog sits down, then you exert a force on the lead to make the dog stand again. This force is made up of two components: a horizontal component that is pulling the dog forward and a vertical component pulling the dog up. Of course there are other forces acting on the dog but we will ignore these for the time being.
upward component forward component
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TPP7160 – Preparatory physics
It is possible to resolve the force on the lead into these two components. The process of transforming a vector into its horizontal and vertical components is called vector resolution.
If the force used to pull the dog is 10 N and is exerted at an angle at 35% to the horizontal, then this can be represented by the following vector diagram.
10 N
35°
Using trigonometry, to resolve the force in the lead, the components will be:
F sin q
F
F
cos q q
Vertical component of F=
Horizontal component of F=
F sin!&!'!10 sin 35% =
F cos &!'!10 cos 35%
5.73 N
= 8.19 N.
In general, any vector can be resolved into its horizontal and vertical components. Remember that the vector to be resolved is always the hypotenuse in a right angled triangle.
The technique of resolution of vectors into their components is useful we want to sum up vectors in a certain direction.
Consider the following examples. You will need to refresh your knowledge of basic coordinate geometry and trigonometry before progressing further. You will also need a pair of compass and protractor for your geometrical constructions.
Example:
Consider two forces acting on a body B, one of 30 N at 30% to the horizontal and the second of
20 N at 140%!to the horizontal. What would be the resultant force on the body?
Solution:
1. This problem could be solved graphically by the addition of vector components.
To solve the problem graphically, you need to accurately construct a parallelogram (figure 2.1) using the force vectors as the sides. Another way is to draw a vector triangle, (figure 2.2) by joining the vectors tip to tail and making sure that the directions are maintained.
The following diagrams illustrate the graphical solution, drawn correctly to a scale of
1 inch: 10 N
30 N
C
20 N
140º
30º
B
D
20 N
Resultant
30 N
A
Given
Figure 2.1:
B
69º
Module 2 – Understanding motion
2.25
20 N
The resultant force on B is 30 N and directed at 69° to the horizontal.
Resultant
B
69º
30º
30 N
Figure 2.2
2. The problem could also be solved by resolving each vector into its horizontal and vertical components and then adding each new component to give the resultant. This is shown as follows:
Force Horizontal Component (x) Vertical Component (y)
30 N 30 cos 30% = 25.98 N 30 sin 30% = 15 N
20 N 20 cos 40% = 15.32 N 20 sin 40% = 12.86 N
40º
30º
The horizontal component for the resultant vector, R
x
components, 25.98 + (–15.32) = 10.66
is the sum of all the horizontal
(Note: negative sign denotes that it is in
the opposite direction, i.e. to the left)
The vertical component for the resultant vector, R
y
is 15 + 12.86
= 27.86
We will now add the resultant components by using Pythagoras’ Theorem.
This can be represented diagrammatically as:
R
θ
R x
=
10.66
R y
=
27.86
Using Pythagoras’ rule, the magnitude of the resultant vector is:
R
=
R x
2
+
R y
2
= 10.66
!
2
+ 27.86
!
2
= 29.82
= 30
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TPP7160 – Preparatory physics
Using trigonometry, the direction of the resultant vector is given by: tan & =
R

R y x
=
27.86

10.66
= 2.61
& = 69.1"
Thus the resultant force would be 30 N at 69".
3. The problem could also be solved by straight calculation
Firstly from Figure 2.1,
#
Applying the Cosine Rule, where
C
a
2
=
b
2
+ c
2
–2bc Cos
= 20
2
+ 30
2
– 2(20)(30) cos 70° b = 20 N a
70º
A
= 400 + 900 – 410.42
= 889.57
c = 30 N
= 889.57
a = 29.8 N
B
# The resultant force is approximately 30 N
To find the direction of the resultant force, we can apply the cosine rule or the sine rule.
The sine rule is easier in this case.
= 110°
ˆ
= 180° – 110°
= 70°
Using sin
b
= sin
a b A a
=
20 sin 70"
29.8
# $ %39°
# The direction of the resultant is at an angle of (39° + 30°) = 69° to the horizontal.
As you can see, it is much easier to use the graphical method.
Module 2 – Understanding motion
2.27
Note: Some of you may already be familiar with the sine and cosine rules. If not, you could use the calculation shown earlier where Pythagoras’ Theorem is employed.
Example:
Four forces act on a body at point A as in the following diagram. Use algebra and trigonometry to find the resultant force acting on the body. (Hint: Use the direction of the 80 N force as the frame of reference, and measure other angles anticlockwise.)
100 N
110 N
150°
45°
200°
80 N
A
160 N
Solution:
Table of component vectors.
Force (N)
80
100
110
160
Resultant
Horizontal Component (x)
80 cos 0" = 80
= 70.7
100 cos 45"
110 cos 150" = –95.3
160 cos 200" =
R x
= –95 N
–150.4
Vertical Component (y)
80 sin 0" = 0
= 70.7
100 sin 45"
110 sin 150" = 55
160 sin 200" = –54.7
R y
= 71N
Note that R
x
is directly opposite in direction to the 80 N force since it is negative.
See figure 2.3.
Using figure 2.4, the magnitude of
R
=
R x
2
+
R y
2
=
–
95 !
2
+ 71 !
2
= 118.6 N
R x
= 
95 N
Figure 2.3
R y
= 71 N
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TPP7160 – Preparatory physics
Direction of R, tan & =
R

R y x
Resultant, R
R y
tan & =
–
71
95 tan & = –0.747
& = 143"
R x
θ original reference line
Figure 2.4
since xcomponent is negative and ycomponent is positive the angle & must be in the
2nd quadrant, so & = 143"
Resultant force is 119 N at 143"
Try to solve this graphically and check your answer with the solution above. (Hint: It is much easier to use graph paper when solving by drawing).
Activity 2.6
Question 1: Two soccer players kick a ball at exactly the same time. One player’s foot exerts a force at 65 N north. The other players foot exerts a force of 83 N east. What is the magnitude and direction of the resultant force on the ball?
Question 2: Two 15 N forces act concurrently on point A. Find the magnitude of their resultant when the angle between them is 35".
Previously we examined the nature of the forces that acted on a book on your desk, with the technique of resolution of vectors into components we can consider further situations where several forces act on a body.
Example:
A 40 kg block of stone is being pulled at a steady speed by rope at an angle of 30" to the horizontal and the tension in the rope being 100 N. Find:
(a) the frictional force
(b) the force exerted by the earth on the block.
Module 2 – Understanding motion
2.29
Solution: Let F
f
represent the frictional force, and F
n
represent the force exerted by the earth normal to the plane
F n
100
30 o
N
F f
F n
F f mg mg
(a) In the horizontal direction
The frictional force must be equal to the horizontal component of the pulling force.
Therefore F
f
= 100 cos 30"
(b) In the vertical direction
= 86.6 N
The downward force (weight) must be equal to the sum of the upward forces.
Therefore
mg
=
F n
+ F sin 30"
40 ' 9.8
=
F n
+ 100 sin 30"
F n
= 342 N
# The normal reaction of the earth is 342 N
Example:
A sign over a bookshop weighs 229 N and the angle of the supporting wire is 51"%to the horizontal. What is the force in the rope?
F
51°
Bookshop
229 N
F
51 °
F cos 51 °
F sin 51 °
mg = 229 N
2.30
TPP7160 – Preparatory physics
Solution:
In the vertical direction, when forces are in equilibrium so
mg
= F sin &"
229 = F sin 51"
F
= sin 51"
= 294.7 N
# Force in the rope is 294.7 N
2.2.6 A special force – gravity
Gravity is the most obvious force in our daily lives and we have already talked about it a great deal. If a stone is released and falls to earth, then its motion is due to a force acting on it. The force acting on it is called its weight and is due to something called gravity. The true nature of gravity is still under investigation.
Let’s have a look at some further situations where gravity plays an important role. On a hill (an inclined plane) gravity will still pull straight down, not at an angle downhill. However a component of the weight will act parallel to the inclined surface.
mg sin θ
θ
B
θ
mg
mg cos θ
θ
α
α
( +
A
= 90" and ( +
B
= 90"
#
B
=
ˆ
= &
The component of weight pulling the body down the hill is mg sin &.
Example:
A truck weighing 562 N is resting on a plane inclined at 20" to the horizontal.
Draw a diagram indicating the relevant forces.
Find the magnitudes of the parallel and perpendicular components of the weight.
Module 2 – Understanding motion
2.31
Solution:
Force parallel to the inclined plane F
p
is mg sin &.
F p
θ 20°
20
°
F n mg
F p
= 562 sin 20"
= 192 N
Force perpendicular to the inclined plane F
n
is mg cos &.
F n
= 562 cos 20"
= 528 N
Activity 2.7
Question 1: A car weighing 1.2 ' 10
4
N is parked on a 35" slope.
(a) Find the force tending to cause the car to roll down the hill.
(b) What is the force the car exerts perpendicular to the hill?
Question 2: A welloiled, frictionless wagon with a mass of 75 kg is steadily pulled uphill, using a force of only 110 N. What is the slope of the hill?
Question 3: A force of 89 N is needed to push a trolley up an incline of 35" to the horizontal. Find the weight of the trolley along the slope if friction is neglected.
Question 4: A child of mass 20 kg is standing on a plank inclined at 20" to the horizontal. How much force does the child exert normally
(perpendicularly) against the plank?
2.32
TPP7160 – Preparatory physics
Home experiment
Measurement of coefficient of static friction.
In this experiment you will need some small objects e.g. a coin, ice block, small ball etc., a stiff flat surface that you can tilt up and down, and a protractor.
Put a coin on the cover of a book, then slowly tilt the cover of the book until the coin begins to slide down the cover of the book.
What angle did you think the coin would begin to move?
Measure your angle with a protractor and determine what angle the coin actually began to move at.
Draw a diagram indicating the actual situation.
When motion is just ready to start friction force is at a maximum and the forces parallel or perpendicular to the book cover are balanced.
What information would you use to determine the coefficient of static friction.
__________________________________________________________________________
__________________________________________________________________________
If you said we would use the relationship
F f
= ) F
n
where and F
n s
is the coefficient of static friction, F
f
is the frictional force which is equal to W sin !
is the normal force which is equal to W cos !, you would have been correct.
Calculate the coefficient of static friction between the book and the coin.
__________________________________________________________________________
__________________________________________________________________________
Determine the coefficient of static friction for at least 5 objects you have around the home.
(Describe the method and characteristics of each object).
What combination has the greatest friction?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
Module 2 – Understanding motion
2.33
What combination has the least friction?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
Are there any limitations to using this technique to measure the coefficient of static friction?
How would these limitations effect your answer?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
Activity 2.8
Question 1:
Suppose a large block is at rest on an inclined plane. The angle of the inclined plane is slowly increased until at 42" to the horizontal the block begins to slide. What is the coefficient of static friction between the block and the incline?
Question 2:
When a force of 500 N pushes a 25 kg box up an incline of 40" the resultant force is 18 N.
(a) Draw a diagram indicating the relevant forces.
(b) Write an equation to determine the frictional force
(hint: balance all forces parallel to inclined plane).
(c) Write an equation to determine the normal.
(hint: balance all forces perpendicular to the inclined plane).
(d) Calculate the coefficient of kinetic friction.
Question 3:
At a tourist resort in the Southern Alps of Australia a horse pulls a sled up a steep snowcovered slope of 8". The sled has a mass of
300 kg and the frictional force between the sled and the snow is
350 N. What is the coefficient of friction between the snow and the sled?
2.34
TPP7160 – Preparatory physics
2.2.7 Falling bodies
If you ever get a chance to go to one of the science centres situated in most Australian capital cities they traditionally have a display of a feather and some solid object, usually a coin, falling in a vacuum. Observation of this tells us that the feather and the coin will land at the same time. Gravity, the attraction between two objects, the feather and the earth, say, is the only force acting on the feather because in a vacuum, air resistance (a frictional force) does not exist. We say the feather would be in free fall. A heavier object, such as the coin, is attracted to the earth with more force than the lighter object (feather). But they do not travel with different accelerations because the acceleration of an object depends on force and mass
(See Newton’s second law and the relationship F = ma). So double the force working on double the mass results in the same acceleration as half the force on half the mass.The feather and coin will arrive at the bottom of the tube at the same time. The acceleration will always be a constant, g = 9.8 ms
–2
.
However, not many of us experience seeing objects falling in a vacuum. What happens in the real situation of an object falling through our atmosphere. In air, feather and coins behave very differently.
Home experiment
You will need a feather or a small piece of paper about the size of a feather and a coin.
Drop the feather and coin from the same height.
What would you expect to happen?
__________________________________________________________________________
__________________________________________________________________________
Describe your methods used.
__________________________________________________________________________
__________________________________________________________________________
Describe what happened.
__________________________________________________________________________
__________________________________________________________________________
Give an explanation for your result.
__________________________________________________________________________
__________________________________________________________________________
If you have referred to any books, list them.
__________________________________________________________________________
__________________________________________________________________________
Module 2 – Understanding motion
2.35
Are there any limitations to the design of this experiment and how could you improve the design.
__________________________________________________________________________
__________________________________________________________________________
Did you notice that the feather dropped in air, accelerates very briefly then floats to the ground at constant speed. This happens because of the effects on the feather of air resistance which quickly increases until it equals the weight of the feather, so that the feather then has net force of zero acting on it. If there is no net force acting on the feather then the feather must no longer be accelerating and the speed of the feather is constant. This speed is called the terminal speed or if direction is considered, terminal velocity. For a feather it is about 0.02 ms
–1
.
From your observations what factors affect the amount of air resistance on a feather or other object.
__________________________________________________________________________
__________________________________________________________________________
If you said the size (i.e. crosssectional area) and the speed of the falling object you would have been correct.
The size is important because it determines the amount of air the object must fall through in falling, while the speed is important because at greater speeds the object collides with more air molecules increasing the impact forces.
Consider the situation of a parachutist or sky diver. They can control their rate of descent by changing the crosssectional area of their parachute or the position of their limbs and body.
height
wt
Drag or a ir resistance time
Two parachutists of the same weight must land at the same time. The first parachutist will leave the plane and slow down his fall by extending his arms and legs. Terminal velocity is reached at about 32 ms
–1
and the parachutist opens his chute to slow down his descent further.
(Terminal velocity about 5 ms
–1
). The second parachutist must travel faster than the first in order to catch up and reduces her surface area by bringing arms and legs closer to the body.
The terminal velocity of the 2nd parachute is about 34 ms
–1
. Both parachutists can land at the same time despite the fact that they left the plane at different times.
2.36
TPP7160 – Preparatory physics
Activity 2.9
Question 1: A 0.50 kg ball falls from the window of a tall building.
(a) At one point air resistance acting on the ball is 0.8 N. What is the acceleration of the ball?
(b) At another stage of its journey the ball is falling with constant velocity. What is the magnitude of the air resistance now?
(c) If there were no air resistance what would be the acceleration of the ball?
Question 2: Two balls of equal size, but different mass, (one solid and one a shell) are dropped at the same time from a hot air balloon. Both experience air resistance as they fall.
(a) Which ball reaches terminal velocity first?
(b) Do both hit the ground at the same time? Explain your answer.
Question 3: Why is it that a tennis ball dropped from the top of a 50 storey building will hit the ground with no greater velocity than if it was dropped from a 20 storey building.
2.2.8 Newton’s Law of Universal Gravitation
If we drop a stone it falls to the earth. The ancient Greeks explained this by saying that every object had a ‘home’ and would try to return home. For a stone, its home was the earth, and that was where it returned.
As knowledge increased, it was realised that if an object was to move a force was necessary. If a stone was released and fell to the earth then it was due to a force acting on it, and the force was called weight (see section 2.2.1) and due to something called gravity.
It is reported that Newton was sitting in his garden when an apple fell from a tree. Newton asked himself what would happen if you took the apple 100 metres up and dropped it – would it fall back to earth because of gravity? What if you took the apple to 1 000 metres or 10 000 metres would it drop back? Newton reasoned that the apple would fall back to earth from these distances and then asked the ‘big’ question – what if you take the apple to 380 000 km (the moonearth distance) – would it still fall back to earth. He reasoned that the apple would fall back to earth and the force on the moon that kept it in orbit, was the same type of force that would act on the apple – earth’s gravity.
Twenty years later after much thought, and cross checking, Newton published his Law of
Universal Gravitation which states:
Between any two objects in the universe there is an attractive force (gravity) that is proportional to the masses of the objects and inversely proportional to the square of the distance between them.
Module 2 – Understanding motion
2.37
In summary, in mathematical terms, if F is the force acting between 2 masses then
F
=
G
# first mass # 2nd mass
distance
2 where G is the universal gravitational constant 6.67 # 10
–11
Nm
2 kg
–2
.
G is an extremely small number, which means that the gravitational forces between everyday objects (you and your dog etc.) will be very small. Gravity only becomes a significant force when relatively large masses and relatively small distances are involved, as with the earth and the moon.
For large masses the distance d is measured from their centre of mass as shown.
d
Moon
(Note: not drawn to scale.)
Earth
Example:
1. What is the gravitational force between two balls each weighing 5.3 kg and 1 metre apart?
Solution:
F
=
F
=
Gm

d
2
m
2 where
m
1
= mass 1 in kg = 5.3 kg
m
2
= mass 2 in kg = 5.3 kg
d
= distance in m = 1 m
G
= universal gravitational constant
6.67
# 10
– 11
# 5.3
# 5.3

1
2
= 0.000 000 001
= 1 # 10
–9
N (a very small force.)
2. Calculate the mass of the earth, assuming it is a sphere of radius 6370 km.
Solution:
Let m
1
be the mass of the earth, and m
2 force on that object will be m
2 g.
the mass of an object on its surface. The weight of
So given
F
=
Gm

d
2
m
2
m
2
object
d m
2
g
=
Gm

d
2
m
2
m
1
Earth
Simplifying
m
1
=
g
#
d
2

G
2.38
TPP7160 – Preparatory physics
=
9.8
# & 6370
6.67
# 10
#
–
10
11
3
'

2
= 6.0 # 10
24 kg.
$% Mass of the earth is 6.0 # 10
24 kg.
(where d is the radius of the earth, assuming the object is relatively small).
We will return to Newton’s Law of Universal Gravitation later when we examine rotational motion.
Activity 2.10
Question 1: The earth’s radius is about 6 370 km. Scientists take a piece of experimental equipment weighing 15 kg to a height of 149 km above the earth’s surface.
(a) What is the object’s mass at this height?
Question 2:
(b) How much does the object weigh at this height?
If the radius of earth is 6 370 km and its mass is 6.0 # 10
24 kg, while Mars has a radius of 3 440 km and a mass 0.11 that of earth:
(a) What would a 200 N object weigh on Mars?
(b) What would be its acceleration due to gravity?
Question 3: An astronaut of mass 70 kg is in a space capsule 1 000 km above the surface of the earth:
(a) Calculate the gravitational force that the earth exerts on the astronaut.
(b) Compare the above to the gravitational force the astronaut would feel at the surface of the earth.
Question 4: The planet Jupiter has a mass of 1.90 # 10
27
kg and a radius of
7.14 # 10
7 m.
(a) What would be the acceleration due to gravity on the surface of
Jupiter?
(b) By what factor would your weight on Jupiter be larger than your weight on earth?
Question 5: If you wanted to make a profit by buying then selling a precious metal at different altitudes, should you buy or sell at the higher altitude? Explain your answer.
Module 2 – Understanding motion
2.39
2.2.9 Momentum
Over the years the word momentum has slipped into everyday use ... we say ‘that car had a lot of momentum’ or ‘I had so much momentum I couldn’t stop running’. But what exactly does the word momentum mean?
When running to catch a bus we would never run out in front of the bus to stop it, because we know it would be impossible ... the bus has too much momentum. Previously we examined the concept of inertia which was the tendency of an object to remain in the same state of motion, and was dependent on the object’s mass. Inertia of a stationary object is just a property of its mass, whereas momentum is the inertia of a moving object and is the product of mass and velocity.
Momentum, P = mass # velocity
P
=
mv
Momentum is a vector quantity and in SI units is measured in kg ms
–1
or kg m s
–1
.
If the momentum of an object changes, then either the mass or the velocity, or both change. In most cases it is the velocity, not the mass, that changes. If the velocity changes, then remembering Newton’s Second Law, the object must be accelerating, or under the influence of a force.
Consider Newton’s Second Law –
F
= ma,
F
= m
(t
a is acceleration and equals,
(v
(t
, a change in velocity over a change in time.
(note: ( is used to symbolise a small change. It is the Greek letter pronounced as delta)
Multiplying both sides of the equation by (t
F (t = m(v
This side of the equation is equivalent to the impulse, the product of the force and time.
This side of the equation is equivalent to a change in momentum.
Thus the impulse of a system is equal to the change in momentum. Another way of writing
Newton’s Second Law is, Force = rate of change of momentum.
F = 
(
t
Imagine your brakes fail in your car and you have the choice of crashing into a brick wall or a haystack. You probably don’t need to know physics to make the choice but it will help you understand why the haystack is the better choice. When your car crashes the momentum of the car changes, it is decreased.
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TPP7160 – Preparatory physics
In a crash it is desirable to reduce the amount of force in the car as much as possible this can be achieved by prolonging the time of impact. Because of the relationship between change in momentum and impulse, the longer time is compensated by a lesser force. So whenever you wish the force of impact to be small, increase the time of contact.
There are many examples of this used in everyday life.
Low force, long time pa dding on the dashboard of a vehicle.
safety nets for tight rope walkers.
bending legs when landing from a jump.
The reverse of this situation also occurs when you want a large force you reduce the time of impact. Karate experts use this principle when trying to crack piles of bricks or deliver a deadly blow.
Note that in all these cases the change in momentum is the same but the relative magnitudes of force and time may vary.
Example:
1. A squash player hits a ball of mass 0.04 kg at the front wall of the court with an initial velocity of 40 ms
–1
. The ball hits the wall perpendicularly and rebounds straight back with the same speed.
(a) What is the momentum as the ball hits the wall?
(b) What is the momentum as the ball rebounds?
(c) What is the change in momentum?
Solution:
(a) Initial momentum,P
i
=
mu
= 0.04 # 40 = 1.6 kg ms
–1
(b) Final momentum,P
f
=
mv
= 0.04 # –40 = –1.6 kg ms
–1
Note that the ball is now moving in the opposite direction hence the final velocity is
–40 ms
–1
.
Module 2 – Understanding motion
2.41
(c) The change in momentum, (
P
=
P f
– P
i
= –1.6 – 1.6
= –3.2 kg ms
–1
Activity 2.11
Question 1:
What is the momentum of the following objects?
(a) The earth revolves around the sun. The earth’s linear velocity is 3 # 10
4 ms
–1
and its mass is 6 # 10
24
kg.
(b) A 100 g snail crawling at 1 # 10
–6 ms
–1
.
(c) A rifle bullet of mass 15 g fired at 600 ms
–1
.
(d) An electron of mass 9.1 # 10
–31 kg travelling at
2.0 # 10
5 ms
–1
.
Question 2:
A 0.14 kg cricket ball is bowled at a velocity of 38 ms
–1
. After it is hit by the bat, it moves at –38 ms
–1
(a) What impulse did the bat deliver to the ball?
(b) If the bat and ball were in contact for 0.6 s what was the average force the ball exerted on the bat?
(c) Find the average acceleration of the ball during its contact with the bat.
Question 3:
A car moving at 40 km/hr crashes into a tree and stops in 0.25 m.
(a) Find the time required to stop the car.
(b If a child (20 kg) is an unrestrained passenger, and also stops in the same time, what average force is acting on the child?
Question 4:
A 55 kg dancer leaps 0.32 m vertically into the air.
(a) Calculate the dancer’s momentum on reaching the ground.
(b) What impulse is needed for her to stop?
(c) The dancer’s knees bend on landing lengthening the stopping time to 0.055 seconds. Find the average force exerted on the body.
Question 5:
In terms of impulse and momentum explain why are padded dashboards safer in cars.
2.42
TPP7160 – Preparatory physics
Conservation of momentum
Have you ever seen a performance on ice by Torville and Dean or other renowned iceskaters.
When they come together there is a symmetry in their movement. As they revolve around each other they exert forces which are equal in magnitude but opposite in direction. They are operating as a unified system, with no external forces operating (if we assume friction to be minimal). If the forces they exert on each other are equal then the impulse of these forces must also be equal, as they are acting over the same time period.
However, the impulse of the net force acting on a system is equal to the change in momentum.
So if there is a change in momentum of one of the skaters there must be an equal but opposite change in momentum of the other dancer. Overall we say that the net momentum of the system is constant.
This principle is called the conservation of momentum and states that:
The momentum of any closed, isolated system does not change.
There are many examples of conservation of momentum in real life but perhaps the most quoted is the situation which occurs when a gun is fired.
Anyone who has watched the movies or fired a gun knows that when a bullet is fired from a gun the bullet moves forward and the gun moves backwards. Special shooting jackets have been designed with padded shoulders to protect the wearer from the gun recoil. The momentum before and after firing is zero because the momentum of the rifle is equal and opposite to the momentum of the bullet. The rifle and the bullet are both part of a closed isolated system.
Explain in your own words how the principle of conservation of momentum can be used to explain rocket propulsion.
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
Momentum is also conserved in collisions ... the total momentum of a system of colliding objects remains unchanged throughout the entire collision.
In any collision: the total momentum before the collision = total momentum after the collision.
In elastic collisions total kinetic energy is conserved but in inelastic collisions, kinetic energy is not conserved. (In inelastic collisions, objects coalesce and move together as one body.)
Module 2 – Understanding motion
2.43
v m
1
m
2
u
1
Before Collision
u
2
= 0 m
1
m
2
u
1
Before Collision
u
2
= 0 m
1
v
1
After collision
m
2
v
2
m
1
After collision
v
Inelastic
Collision
Elastic
Collision
Example:
A 6 g bullet is fired horizontally into a 10 kg block of wood and sticks in it. After impact the block slides along the surface with a velocity of 40 cms
–1
. What is the initial velocity of the bullet?
Solution:
In this case the system involved consists of the bullet and the wooden block.
sum of initial momentum = sum of final momentum momentum (bullet) + momentum (block) = momentum (block and bullet)
(note: this is an inelastic collision)
In SI units –
m
1
u
1
+ m
2
u
2
= (m
1
+ m
2
)v
0.006 u
1
+ 10 0 = 10.006 0.40
(where u
1
is the initial velocity of the bullet)
0.006 u
1
= 4.0024
u
1
= 667 ms
–1
! Initial velocity of the bullet is 667 ms
–1
.
2.44
TPP7160 – Preparatory physics
Activity 2.12
Question 1: A 20 000 kg freight wagon is travelling at 5 ms
–1
along a straight track when it strikes a second stationary wagon of the same mass and couples with it. What will be their combined velocity after impact?
Question 2: A car with mass 1245 kg moving at 28 ms
–1
strikes another car of mass 2163 kg at rest. If the two cars coalesce (combine to become one), what is their velocity after impact?
Question 3: Two teenagers on roller skates face each other. One teenager has a mass of 80 kg and the other a mass of 60 kg. If they push each other away:
(a) Find the ratio of their velocities.
(b) Which student has the greater speed?
Question 4: The nucleus of an atom has a mass of 3.8 10
–25
kg and is at rest.
If it becomes radioactive and ejects a particle of mass 6.6 10
–27 kg and speed 1.5 10
7
ms
–1
, what is the recoil speed of the nucleus?
Question 5: A footballer with mass 95 kg running at 2.8 ms
–1
tackles an opposition player of mass 128 kg who was moving in the opposite direction. They both end up still on the ground. Calculate the speed of the opposition player.
Section review
In this section we have examined Newton’s three laws and how they apply to motion. We have studied forces, and the concept of momentum. You should now assess yourself as to whether you have met the module objectives.
You should now be able to:
• state Newton’s first law
• define mass and weight
• state and explain Newton’s second law
• explain the term force
• use vectors to represent forces
• describe applications of Newton’s second law, in particular friction and gravity
• explain the concept of momentum and its effects on state of motion
• describe applications of the principle of conservation of momentum
• state and explain Newton’s third law
• describe applications of Newton’s third law, using action and reaction forces
• use vectors to represent action and reaction forces
• calculate the components of a force.
Module 2 – Understanding motion
2.45
In your study of this section you should have been making notes of the main points and listing those concepts that were difficult to understand.
Have you been able to revise those concepts that were difficult to understand?
Have you sought help?
Use the following post test to test out your learning.
2.46
TPP7160 – Preparatory physics
Posttest 2.2
Question 1: Complete the following summary.
Newton’s fist law of motion states:
Any object continues in a ______________________ or of __________________ motion in a straight line, unless acted upon by an ____________________ force.
Force is a ____________________ quantity, i.e. to specify a force we must give both
____________________ and ____________________. For acceleration of a body to occur there must be an ___________________ force acting.
The tendency of a body which causes it to resist any change in its state of rest or motion is called its
__________________ .
Newton’s second law of motion states:
The acceleration of a body is ___________________ proportional to the resultant force, and _____________________ proportional to the mass of the body.
In symbols, with suitable units Newton’s second law may be written as:
____________________
The unit of mass in the SI system of units is ___________________ .
The unit of force in the SI system of units is ___________________ .
One ________________ is the force which applied to a mass of ______________ , results in an acceleration of ________________ ms
–2
.
The weight of a body is _______________ .
In symbols W = _________________ .
Momentum is the product of ________________ and _______________________ of a body.
It is a _________________ quantity and is measured in _____________________
(in SI units).
Impulse is a product of force and ___________________ .
The change in momentum of an object is equal to the ___________________ applied to it.
Module 2 – Understanding motion
2.47
In any interaction between two or more isolated objects the total ____________________ does not change. This is referred to as the law of ______________________ of
__________________ .
Newton’s Law of Universal Gravitation states that everybody in the universe attracts every other body with a force which is ___________________ to the ______________________ of the masses of the two bodies, and _______________ to the ______________________ of the distance between the two bodies.
If m
1
, m
2
are the masses of the bodies and d is their distance apart, F = ______________ where G is a constant called __________________ .
Newton’s third law states that if two bodies A and B interact, the force exerted by the body B on body A is __________________ and ____________________ to that exerted by body A on body B.
Question 2: If an unbalanced force F, acts on an object of mass m, producing an acceleration
a, for a time t, these quantities are related by: (Choose one of the following.)
(a) F
=
a m
(b) F = ma
2
2
(c) F
=
m a t
(d) F
=
m a /t
Question 3: Which of the following statements are incorrect?
(a) The force of gravity acting on an object is called weight.
(b) The weight of an object has the SI units the kilogram.
(c) If an unbalanced force acts on a moving object, it produces a change in the velocity.
(d) If an object suffers an acceleration it has been acted upon by an unbalanced force.
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TPP7160 – Preparatory physics
Question 4: Which of the following is a true statement concerning friction?
(a) Frictional forces are independent of the surface areas of contact.
(b) Dry frictional forces depend on the speed of sliding.
(c) The frictional force is equal to the normal force holding the two surfaces together.
(d) If the area between two surfaces doubles the friction force would double.
Question 5: If an unbalanced force acts on an object for a short time then:
(Choose one of the following)
(a) The impulse of the force equals the change in momentum of the system.
(b) The change in the acceleration of the system equals the impulse of the force.
(c) The change in the momentum of the system equals the magnitude of the unbalanced force.
(d) The change in the acceleration of the system is proportional to the magnitude of the unbalanced force.
Question 6: What do we mean by the components of a force and why is it useful to resolve (or separate) a force into its components? Give an example to explain your answer.
Question 7: Can a truck have the same momentum as a bullet? Explain your answer.
Module 2 – Understanding motion
2.49
2.3 Nonlinear motion
2.3.1 Projectile motion
Previously we have considered motion in a straight line, however the next time you water the garden with a hose or throw a ball, the motion you observe is not linear but follows the path of a curve.
Projectile motion is the curved motion of an object that is projected into the air. The motion will follow a parabolic path and will have vertical and horizontal components. Anything projected into the air will follow such a path: throwing a stone, shooting a gun, kicking a football, water from a hose, the path of dancers or athletes when jumping.
Home experiment
The only equipment you will need is two tennis balls.
Place one ball on the corner of your desk and another on the palm of your hand. Move your hand quickly so that the ball in your palm drops straight down and you hit the ball of the table sideways.
What would you expect to happen?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
What actually happened?
__________________________________________________________________________
__________________________________________________________________________
Can you explain your observations.
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
Are there any limitations to this experiment that would affect your interpretation of your conclusion?
__________________________________________________________________________
Let us consider the experiment further. Both balls when in the air are under the influence of the same acceleration caused by gravity, hence they both would have hit the ground together: they had the same vertical velocity. This occurred even though one of the balls had been projected by a horizontal force. This means that the horizontal and vertical components of velocity operate independently of each other.
2.50
TPP7160 – Preparatory physics
This principle is called the Independence of Velocities.
This principle is useful in helping us understand about, and calculate component velocities of projectiles in motion.
Example 1:
A stone is thrown horizontally with an initial velocity of 20 ms
–1 from a height of 50m above the ground.
(a) How long does it take for the stone to reach the ground?
(b) How far from the base of the building is the stone when it reaches the ground?
(c) Sketch the pathway of the stone.
Solution:
Consider the situation
u = 20 ms
–1
Given:
Initial horizontal velocity, u = 20 ms
–1
Displacement, s = 50 m
(a) Consider the vertical motion, (or motion in the ydirection)
Using the equation of motion s = ut +
1
/
2
at
2 where
s =
1
/
2
gt
2
s = vertical displacement (50 m)
t =
2s

g
u = initial vertical velocity (zero)
=
2 50
9.8
a = acceleration due to gravity, (9.8 ms
= 3.18 s
OR substituting into the equation we get
–2
)
Module 2 – Understanding motion
2.51
50 = 0 +
1
/
2
9.8 t
2
t
2
=
2 50

9.8
t
= 10.20
seconds
= 3.19 secs.
! Time taken to reach the ground is 3.19 s.
(b) In the horizontal direction displacement = velocity time (since its motion is independent of gravity)
s
=
ut
= 20 3.19
= 63.88
= 64 m
Distance travelled horizontally is 64 metres.
(c) The pathway or trajectory of the stone.
2.52
TPP7160 – Preparatory physics
Example 2:
Consider the same situation as discussed in Example 1, but this time the stone is not thrown horizontally but at an angle of 40" to the horizontal.
u y u
=
20 ms
–1
u
(a) How long does the stone take to reach the ground?
(b) How far from the base will it be when it hits the ground?
Solution:
The initial horizontal component of velocity will be,
u x
=
u cos 40°
= 20 cos 40°
= 15.32 ms
–1
The initial vertical component of velocity will be,
u y
= –u sin 40°
(using negative sign for upward direction)
= –20 sin 40°
= –12.86 ms
–1
(a) To find the time of flight, t, we use the equation,
s
= ut +
1
/
2
at
2
50 = –20 sin 40°t #
1
/
2
50 = (–12.86t) + 4.9t
2
9.8 t
2 where s = 50m
u y
= –u sin 40°
a = 9.8 ms
–2
(using positive sign for downward direction)
!
4.9t
2
–12.86t – 50 = 0
(This equation can be solved by using the formula for solving a quadratic equation. See
Appendix 2.)
Module 2 – Understanding motion
2.53
t
=
– $ – % & $ – 12.86
%
2
– $ 4 4.9
– 50 %
2 4.9
t
= 4.75 or –2.14 secs
! Time to reach the ground is approximately 4.8 seconds.
(b) To find the horizontal distance of the stone from the base, we use the equation
s =
u x t
s = u cos 40° × t
= 15.32 × 4.8
= 73.5 m
(horizontal component of velocity is a constant since it is not affected by gravity)
(or 72.8m if you used 20 cos 40° × 4.75)
Horizontal distance travelled is 73.5 metres.
Activity 2.13
Question 1: A crop duster is moving at 15 ms
–1
parallel to the flat ground 100 m below. At what horizontal distance must the plane be from its target field when it releases its load? Assume air resistance is negligible.
Question 2: A golf ball is hit and moves into the air with a velocity of 3.45 ms
–1 at an angle of 75" to the horizontal.
(a) Calculate its time of flight.
(b) How high will the ball reach before it starts to fall?
Question 3: A person fishing throws a leadweighted line into the water with an initial velocity of 20 ms
–1 at 45" to the horizontal. How far will the cast be when it touches the water?
Question 4: A dart player aims at the target s metres away and throws a dart horizontally at a speed of 11.3 ms
–1
. The dart hits the board 0.25 m below its target. How far is the player from the board?
Question 5: You are the long jump coach at the Australian Institute of Sport.
What factors would you have to consider if you were aiming to increase the length of a competitor’s jump? Discuss your answer.
2.54
TPP7160 – Preparatory physics
2.3.2 Circular motion
If you were to spin a bucket of water horizontally about your head, with any luck the water would stay in the bucket. Both the water and the bucket are moving in a circle. A centripetal force, directed towards the centre of the circle is provided by the tension in your arm. It is this net force that maintains an acceleration towards the centre of the circle. An acceleration causes a change in velocity, the effect of which is constant speed in a circle.
We have developed symbols and terminology for motion in a straight line; let us develop and extend these ideas to motion in a circle.
If you run around a track with radius r then two forms of velocity can be considered – angular velocity and linear velocity.
A r
q
Linear velocity at A (v)
s
B
a ngular velocity (ω)
In moving from A to B along the arc, the linear displacement along the arc is s while the angular displacement is '($clockwise). If the radius is r, then the angular displacement is defined as '( =
r
" or
arc length

radius
#
!
the unit for angular displacement '(is radians
In moving through 360º, the arc length = circumference
Hence '( = = 2 radians
r
Angular velocity ) is the rate at which the angular displacement '(is changing with time.
) =
%

t
rads
–1
......................................equation 1
It is often difficult to measure the angle moved per second, but easier to measure the frequency
(f) in number of revolutions per second or the Period (T) i.e. time taken for a single revolution.
The distance travelled over the circumference of a circle is 2*r,
! angular velocity linear velocity
Equating T,
!
&
=
T
rad s
v
=
T
2$
&
=
2$r

v v
= r& ms
–1
–1 ........................................equation 2
.........................................equation 3
.........................................equation 4
This equation relates the linear velocity to the angular velocity.
Module 2 – Understanding motion
2.55
Since T is the time taken for 1 revolution, and frequency f is the number of revolutions per sec
'
f
=
T
1
Substituting for in ()we have
T
&
=
2$

T
= 2$f or
& = 2$f
........................................ equation 5
Remember the merrygorounds that were common in playgrounds once? If you rode on one of these the faster the equipment rotated the faster your tangential or linear velocity was when you tried to jump off.
Consider two merrygorounds – one for young children and one for older children. The linear velocity on exit than the smaller apparatus is usually slower than that on the larger apparatus. This is always true even if you are rotating at the same speed as rotational speed is not dependent on distance from the centre. The linear velocity, however, is dependent on r, as shown by v = r *
V R
θ
R
If the angular velocity changes, then an object must be accelerating along the circle.
2.56
TPP7160 – Preparatory physics
Angular acceleration = (measured in radians per second per second)
t
,
=
&
–
&
1
........................................ equation 6
t
The corresponding linear acceleration which is always directed towards the centre is called the centripetal acceleration, and is given by
or a c
=
r
2
= r&
2
a c
= r&
2
(using equation 4)
r a c v
Think about these questions:
•
Why do clothes in the rapidly spinning tub of a spin dryer ‘stick’ to the sides of the tub but falls to the bottom when the tub stops spinning?
•
Why does the water not fall out of a can when it is being spun?
Example:
A spin drier (of a washing machine), with a diameter 30 cm, revolves at 900 rpm.
(a) What is the angular velocity of the spin drier in radians per second?
(b) Calculate the linear velocity of the spin drier.
(c) What is the centripetal acceleration of the spin drier?
Solution:
(a) The angular velocity is 900 revolutions per minute.
900 rpm =
900

60
= 15 rev per second
= 15 + 2$ rad s
–1
(1 rev = 2$ radians)
= 94 rad s
–1
' Angular velocity is 94 rad s
–1
(b) Linear velocity,
v
= r
= 0.15 + 94
= 14.1 ms
–1
(c) Centripetal acceleration, a
c
=
v
2

r
=
14.1
2

0.15
= 1330 ms
–2
Module 2 – Understanding motion
2.57
Example:
A car negotiates a roundabout of radius 30 m. If the wheels of the car can withstand a linear acceleration of 6 ms
–2
without sliding, what is its maximum linear velocity?
Solution:
Using the equation
a c
=
r
2
v
2
=
a c r v
=
a c r
=
6
+
30
=
13.4 ms
–1
a c
= 6ms
–2
and r = 30m
For cars, speed is usually expressed in km/hr so 13.4 ms
–1
You could multiply by 3.6 to convert from ms
–1
to kmh
–1
= 48 km/hr.
and divide by 3.6 for the reverse.
Activity 2.14
Question 1: The earth moves around the sun in a circular path of radius
1.5 + 10
11 m at a uniform speed. What is the centripetal acceleration of the earth towards the sun?
Question 2: In a biogenetic laboratory an ultracentrifuge spins a test tube in a circle of radius 15 cm at 1500 revolutions per second.
(a) What is the angular velocity of the test tube in radians?
(b) Calculate the centripetal acceleration of the test tube.
Question 3: The disc of a circular sander has a diameter of 20 cm and rotates at
5 000 revolutions per minute.
(a) What is the linear velocity of a point on the edge of the disc?
(b) What is centripetal acceleration at this point?
Question 4: In a nuclear accelerator, protons travel (at close to the speed of light
(3.00 + 10
8
ms
–1
) in a circular path of radius 0.751 cm.
What is the centripetal acceleration of the protons?
2.58
TPP7160 – Preparatory physics
2.3.3 Forces and circular motion
Newton’s second law says that if an acceleration occurs then it is due to the action of a resultant (or net) force. When an acceleration has constant magnitude and is always kept perpendicular to linear velocity, circular motion occurs. The force that maintains this acceleration is called the centripetal force, and is directed towards the centre of the circle.
F c
= ma c
=
mv
2

r
= mr&
2
v
Example:
A child ties a 20 g rock to the end of a 1 m long string and swings it above her head in a circular motion with a frequency of 3 revolutions per sec.
What is the centripetal acceleration of the rock and what is the corresponding tension in the string?
Solution:
a c
The rock is undergoing a centripetal acceleration given by the equation, where
a c
=
&
2
r
& = 3 revs
–1
= 3 +)2$)rads
–1
)
= 18.8 rads
–1 therefore a
c
= (18.8)
2
+ 1 = 353.4 ms
–2
To cause the above acceleration the string must exert a force
(pull) on the rock of mass 0.02 kg.
F c
= ma c
 7 N
= 0.02 + 353.4 = 7.07
' The tension in the string T = F
c
= 7 N
Module 2 – Understanding motion
2.59
Activity 2.15
Question 1: In a cyclotron, protons of mass 1.657 + 10
–27 kg move in a circular path of diameter 2 m in a large electromagnet. If the velocity of protons is 2.0 + 10
6
ms
–1
find:
(a) the time it would take for the protons to complete one revolution.
(b) the force the magnet exerts on the protons.
Question 2: A cord of 0.9 m in length can withstand a force of 2 N without breaking. It is used to whirl a 0.8 kg lead weight in a horizontal circle. Find the minimum period (time for one revolution) the weight can be whirled without breaking the string.
Question 3: The distance from the Earth to the Sun is 1.5 + 10
11 m. Estimate the mass of the Sun if the Earth has a period of 365 days and
G
= 6.67 + 10
–11
Nm
2
kg
–2
.
2.3.4 Other forms of rotational motion
In previous sections we looked at Newton’s laws of motion as they applied to straight line motion. These same laws can be extended to rotational motion.
Torque
At a local school fete your job is to run the chocolate wheel.
axis
r r
axis
F
F
⊥ clockwise
The wheel spins about an axis of rotation. If there is no friction on the wheel it can keep spinning indefinitely. The force you have applied to the wheel produces a torque or turning effect and results in a change in the wheel’s angular velocity. The subscript). is the perpendicular sign, and F
.
means that the effective force is always perpendicular to the distance r from the axis.
2.60
TPP7160 – Preparatory physics
But not all forces applied to the wheel will cause a torque. If you push parallel to the axis
(Figure 2.3) the wheel will not spin. Only a force applied perpendicularly to the axis will produce rotational movement. If you want to spin the wheel as fast as possible, you would:
Figure 2.5:
• pull as hard as possible because the torque is proportional to the force applied
• pull close to the rim of the wheel, not near the centre because the torque is proportional to the
. distance from the axle
• apply the force tangentially to the rim.
These conditions are summarised by saying, torque is the product of the force and the
perpendicular distance from the axis to the line of the action of the force.
Torque, / =
F
.
r
If the force applied is at an angle % to r, then / = r × F sin %
(where % is the angle between r and F).
The units of torque are in Newtonmetres (Nm). If there are opposing torques you can assign (+) and (–) to them. For example positive, the body is moving in a clockwise direction, and negative if the body is moving in an anticlockwise direction. Torque is also called the moment of a force.
The symbol)/ is the Greek letter ‘tau’.
r
F
⊥
θ
F
/ /
F
.
= F sin %
F
//
= F cos %
Example:
The radius of a chocolate wheel is 1.5 m and the operator spins the wheel with a 30 N force applied tangentially. What is the resultant torque?
Solution:
Torque, / =
F
.
r
= 30 × 1.5
= 45 Nm
Example:
A person pushes perpendicularly against a door at a distance of 500 mm from the hinges with a force of 10 N. What torque was produced?
Solution:
Torque, / =
F
.
r
= 10 × 0.5
= 5 Nm
If an object is acted up on by more than one torque, the object may still be in equilibrium, as the different torques cancel each other out. Alternatively one torque might outweigh the others and produce a net torque.
Module 2 – Understanding motion
2.61
F
F
F
F
These two forces produce opposing torques and the wheel remains stationary.
These two forces produce a ‘couple’.
The result is a torque = 2Fr (clockwise).
For any object to remain in equilibrium two conditions should apply.
•
The sum of all the forces should be zero, or 0 F = 0
•
The sum of all the torques must be zero 0 / = 0
Example:
Two children and their father want to balance a 4 m seesaw pivoted at its centre. The children, weighing 20 and 45 kg sit at one end 0.4 m apart. Where would their father, who weighs 85 kg, have to sit in order to balance the seesaw?
Solution:
Let x be the distance of the father from the pivot as shown in the diagram.
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TPP7160 – Preparatory physics
0.4 m
F
1
25 g
F
2
45 g a nticlockwise torques
4 m
Pivot
x
F
3
85 g clockwise torque
No forces other than F
1
, F
2
and F
3
are acting on the system. Thus, for the system to be in equilibrium,)0/)= 0 or the torques created by the children must equal the torque created by the father.
!"#"(F
1
$"2) + (F
2
$ 1.6) + F
3
(x) = 0 or sum of anticlockwise torques = sum of clockwise torques torque of 1st child + torque of 2nd child = torque of father
(25g $ 2) + (45g $ 1.6) = –(85g.x)
(25 $ 9.8 $ 2) + (45 $ 9.8 $ 1.6) = –85 $ 9.8 $ x
%
x
= –1.4 m
% The father must sit 1.44 m from the pivot on the opposite side.
(Note: &'("or (–) signs are meant to denote opposite directions in vectors)
Example:
An advertising sign weighing 200 N in the shape of a uniform beam of length L, holds another sign weighing 450 N as shown. Find the magnitudes of the forces exerted on the sign by its two supports.
support
Beam
Shoes
4 support
F
1
A
L

2
C
200 N
L

4
450 N
F
2
B
Module 2 – Understanding motion
2.63
Solution:
L
4
L

4

2
Since the object is in equilibrium "F = 0 and "!"= 0
In the vertical plane the forces in equilibrium are represented by the equation
F
1
+ F
2
– 200 N – 450 N = 0 ( (–) sign for downward forces)
F
1
+ F
2
= 650 N............................................ equation (1)
Before the torque equation is written we can choose an axis that will make the calculations easier. The position of the axis is arbitrary, for if the sum of the torques is zero about one axis it is zero about all other axes parallel to the first. The axis is usually chosen, so as to eliminate one of the unknown forces. Let us choose the axis to pass through A. The torque equation is then,
–
2
L
200 –
4
450 + LF
2
= 0 ( (–) sign for clockwise torques about an axis through A)
Solving for F
2
we have,
F
2
%
F
2
=
=
438
438 N
To find F
1
we substitute F
2
= 438N into the equation (1)
F
1
+ F
2
= 650
F
1
= 650 – F
2
%"
%"
F
1
= 212 N
The two supporting forces are 438 N and 212 N.
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TPP7160 – Preparatory physics
Activity 2.16
F
1
Rock
35º
Pivot
Question 1: Two adults hold the ends of a uniform plank weighing 400 N. One adult is taller than the other so that the plank is at an angle of 25) to the horizontal.
(a) What force (vertical) must each person use to carry the plank?
(b) If a child weighing 140N sat on the beam way from its
4
" plank and child?
Question 2: A crowbar 2.0 m long is used to lever a rock out of a hole. The crowbar is pivoted 0.4 m from the rock and makes an angle 35° with the horizontal ground. See Figure above.
You now apply a force F
1
vertically as shown. Find:
(a) the torque being applied to the crowbar by you if F
1
= 250 N.
(b) the force F
2 subsequently exerted on the rock.
(c) What is the weight of the rock?
(d) Comment on the magnitudes of F
1
and F
2
and the advantage of using a simple machine such as a crowbar.
Question 3: A window cleaner, weighing 550 N, is standing 1 m from the end of a scaffold (2.4 m long) which is supported at each end by ropes.
How much tension is in each rope?
Question 4: In your own words explain why it is easier to remove a stubborn tyre nut with a longhandled spanner than with a shorthandled spanner.
Module 2 – Understanding motion
2.65
Section review
In this section we have examined Newton’s three laws and how they apply to nonlinear motion. In particular we have studied projectile motion and circular motion. You should now assess yourself as to whether you have met the module objectives.
You should now be able to:
• describe the motion of projectiles in terms of component forces
• use vectors to solve problems on projectile motion
• describe circular motion
• define angular velocity and acceleration
• explain the meaning of torque in circular motion
• solve problems involving circular motion
• use appropriate units for the quantities involved in projectile and circular motion.
In your study of this section you should have been making notes of the main points and listing those concepts that were difficult to understand.
Have you been able to revise those concepts that were difficult to understand?
Have you sought help?
Use the following post test to test out your learning.
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TPP7160 – Preparatory physics
Posttest 2.3
Question 1:
Uniform circular motion occurs when an object moves in a circular path with uniform
__________________.
Angular velocity is defined as ___________________.
The unit for angular velocity is ___________________ (in SI system).
The uniform linear velocity v of a body moving in a circle of radius r, is related to the angular velocity * by the equation __________________
Since the speed along the circle is uniform, there is no ______________acceleration, but the velocity is changing in _________________ .
The acceleration of the moving object is directed towards the centre of the circle and is called __________________ acceleration.
A body projected with velocity v at an angle + to the horizontal has two component velocities: ___________________ horizontally and__________________ vertically. Each of these components can be considered independently of the other.
Question 2: A torque ! acting on an object can produce one of the following:
(a) an acceleration of the object
(b) a change in the angular velocity
(c) friction
(d) an unbalanced force
Question 3: Many science fiction stories are based on space stations which appear to look like rotating wheels around a central axis. Why would this design simulate gravity and which way would be down?
Question 4: A golfer hits two shots as shown. Which hit
(a) has the shorter flight time?
(b) has the larger flight time?
(c) has the larger horizontal velocity?
q
=
60°
30
°
A
B
Module 2 – Understanding motion
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TPP7160 – Preparatory physics
2.4 Energy
Energy is all around us. Whenever we take a breath, ride in a car, climb some stairs, boil a kettle we use energy. As you sit at your desk reading this book millions of cells are using energy changing yesterday’s food into today’s mental action. Energy affects everything in our universe and the laws which govern its behaviour are among the most important in science today.
All of the situations above have one thing in common – a force is being applied. When you breathe your lungs exert a force as they pull and push air, in a car the fuel’s energy is converted so that the wheels of the car move, if you boil a kettle electrical energy is converted to heat energy to make the water molecules move, if you climb a set of stairs your muscles exert a force that lifts your legs against gravity.
In our everyday speech we use the word ‘energy’ a lot.
‘I feel energetic today ...’
‘The chocolate bar will energise you ...’
‘Come on children, sing energetically ...’
But in physics and other fields of science the word energy has a precise meaning, which is associated with the concept of work. Now work is another word that has many everyday meanings. If you back your car out of the driveway and crash into the garbage bin you may not consider that you have done any work but a physicist would, because a force has deformed the car’s metal a measurable distance. The formal definition of work is as follows:
If we move an object a distance s against a force F, then the work done is the product of F and s.
Work = Force $ distance = Fs (where s is measured in the direction of the force)
Note that if the force is at an angle , to the displacement, then Work = Fs cos ,
In SI units the unit of work is the Joule (J) and is defined to be the amount of work done when you exert a force of one Newton through a distance of one metre. Work is a scalar quantity.
It is often convenient to have a measure of work that also incorporates how long it takes to do that work. Power is the term that is used for this purpose. It is the work done per unit time.
Module 2 – Understanding motion
2.69
Average Power = time interval
The unit of power is thus Joules per second (Js
–1
) which is also known as a Watt, after James
Watt the developer of the steam engine.
Example:
If a motor is required to lift an elevator weighing 1.50 $ 10
4
N up to the second floor of a building 10 m up, find:
(a) the work done by the motor in lifting the elevator if it is empty
(b) the minimum power of the motor required to lift the elevator if the usual length of time for such a trip is 15 s
(c) the maximum power of the motor if the lift can carry 10 passengers (assume each passenger has a mass of 85 kg)
Solution:
(a) Work done = Fs cos !
F
In this instance is zero
Work done = 1.5 " 10
4
" 10
= 1.5 " 10
5
Nm
# The work done is 1.5 " 10
5
Joules
(b) Power =
= time
1.5
" 10
5

15
= 1 " 10
4
Js
–1
or W
1.50
# The minimum power required is 10000 W or 10 kW
(c) The weight of ten passengers is 10 " 85 " 9.8 and equals 8330 N
The total weight of the lift and passengers = 8330 + 1.5 " 10
4
= 23330 N
× 10
4
N
Power = work
time
=
=
23330 " 10

15
= 15553 W force " distance
time
= 15.6 kW
#!$he maximum power should be greater than 15.6 kW
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TPP7160 – Preparatory physics
Activity 2.17
60º
Question 1: When moving house you have to carry a 210 N box of books up a flight of stairs with a vertical displacement of 4 m and a horizontal displacement of 4.5 m.
(a) How much work against gravity do you do on the box?
(b) If you rushed up the stairs in 15 s, how much power did you use if you weigh 75 kg?
Question 2: A dock worker pulls a boat along the dock with a rope at an angle of
60% to the horizontal. How much work does the worker do if a force of 250 N is exerted on the rope and to pull the boat 25 m to its mooring?
Question 3: A bush walker weighing 710 N climbs a cliff 9.3 m high in
30 seconds.
(a) How much work does the walker do?
(b) What is the average power of the climber?
Doing work on any sort of system is sure to change the system in some way. This change results in an increase in the energy of the system. Similarly, if a system does work then this will result in a decrease in the energy of the system. Energy is thus defined as the ability to do work and has the same units as work (Joules). Energy appears in many forms – such as mechanical energy, chemical energy, heat energy and nuclear energy to name a few. In this module we will discuss mechanical energy only. In later modules we will investigate other forms of energy.
2.4.1 Mechanical energy
Mechanical energy may be of the form of Potential Energy (the energy related to an object’s position) and Kinetic Energy (the energy related to an object’s velocity).
Potential energy
A book sitting on a shelf certainly doesn’t seem to have much energy in a colloquial sense.
However, if you accidentally knock it off the shelf and it lands on your toe it is immediately obvious that it has done some work on your toe. An object can store energy because of its position, the energy is not obvious but is stored and held in readiness. Petrol for your car has
Module 2 – Understanding motion
2.71
potential energy, as do all fossil fuels, electric batteries, food and wound clocks. Perhaps the most obvious type of potential energy is that described in the previous example – gravitational potential energy. This is the energy due to an elevated position such as your book on its shelf, water in a reservoir or a child on the top of a slippery dip. The work done in this instance is the product of the force required to move the object upwards to that position and the vertical distance it is moved.
Gravitational Potential Energy = weight " height = mgh
Example:
10 books each of mass 500 g are stored on a shelf 1.5 m above the floor. What is the gravitational potential energy stored in these books?
Solution:
The total mass of the books is 10 "!&''g or 5 kg.
Gravitational potential energy = mgh = 5 " 9.8 " 1.5
= 73.5 Joules
#!73.5 Joules of gravitational potential energy is stored in these books.
Example:
At a fitness centre you have to move a set of weights weighing a total of 80N
(a) What work is done if you lifted them 1 m above the floor?
(b) How much gravitational potential energy is now stored in these weights?
(c) Explain in your own words the relationship between the work you have done and the gravitational potential energy stored in the weights.
Solution:
(a) To lift the weights 1 m, work done = Force " distance
= mgh
= 80 " 1 = 80 Joules.
(b) The gravitational potential energy now stored in the weights is due to its position above the floor.
Remembering that mg is equivalent to the weight of the object in this case 80 N
G.P.E. = mgh
= 80 " 1 = 80 Joules.
(c) The work you have done to lift the weights to a height of 1 m is now stored as gravitational potential energy at that height, Fs = mgh, i.e. Work = gravitational potential energy.
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TPP7160 – Preparatory physics
Activity 2.18
Question 1: A balloonist drops her lunch of mass 1 kg from the basket of the balloon which is 400 m above the ground.
(a) How much work was done on the lunch by the gravitational force?
(b) How much gravitational potential energy did it lose?
Question 2: In a garage two cars are lifted on the hoist to examine their steering.
(a) If the cars are of the same mass, and one car is lifted twice as high as the other what is the relationship between their respective gravitational potential energies?
(b) If the heights are the same but the mass of one car is twice the mass of the other, what is the relationship between their gravitational potential energies?
Question 3: In a hydroelectric plant in Tasmania water is stored 300 m above the level of the river, how much water (in kilograms) must be pumped to store 2 " 10
13
J?
Kinetic energy
An ice skater skating on ice will keep going at a constant speed for a long period of time
(forever if the surface was frictionless). To speed up she would have to do work with her legs.
It must follow then that speed is a measure of energy. The energy an object has because of its velocity is called kinetic energy.
The relationship between kinetic energy and velocity is
KE
=
mv
2
2
Thus we can determine the kinetic energy of anything if we know its mass and velocity.
Example:
Which has the greater kinetic energy, a car travelling at 35 km/hr or another car which is half the mass as the first travelling at 60 km/hr?
Solution:
To convert 35km/h to ms
–1
1hr
=
( 60 " 60 ) sec
s
= 9.72 ms
–1
=
60 " 1000

3600
= 16.67 ms
–1
Module 2 – Understanding motion
2.73
Let m be the mass of the first car, the mass of the second car will be
m

2
KE (1st car) =
1

2 m (9.72)
2
= 612.5 " m
= 47.2m Joules
KE (2nd car) =
1
( ) (16.67)
2
2 2
= 900 " m
= 69.4m Joules
Hence the lighter car has the greater kinetic energy because of its greater velocity.
Activity 2.19
Question 1:
A skier attains a speed of 220 km/hr on a downslope run while a runner briefly attains a speed of 44 km/hr on a track. If they both have a mass of 72 kg,
(a) what is the kinetic energy of the skier?
(b) what is the kinetic energy of the runner?
(c) compare the kinetic energies of the two athletes.
Question 2:
If a moving car increases its speed by a multiple of 4:
(a) by how much does its kinetic energy increase?
(b) how much more work must the car’s brakes perform to stop the car at this new speed compared to the work done to stop the car at its original speed?
2.4.2 Conservation of energy
In a previous section we examined the Law of Conservation of Momentum. A crucial point in this discussion was that the law operated only in an isolated closed system; that is, a collection of objects of constant total mass with no external forces acting on it. During energy transactions the Law of Conservation of Energy operates in the same way as the Law for
Conservation of Momentum.
Within a closed, isolated system, energy can change form but the total amount of energy remains constant.
Let’s think about what happens when you throw a ball into the air (see the figures on next page). With a large upward force it is propelled into the air with an upward velocity ... it has considerable initial kinetic energy. As it goes higher into the air the force of gravity causes the ball to slow down, until at its maximum height it has zero velocity and zero kinetic energy.
However, when its kinetic energy is zero, its gravitational potential energy is at a maximum.
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TPP7160 – Preparatory physics
Finally as the ball falls back down to the ground the gravitational potential energy decreases as the kinetic energy increases. At the point of contact with the ground the kinetic energy is once again at a maximum and the gravitational potential energy is at a minimum or close to zero.This means that the decrease in the potential energy of the object is equal to the increase in its kinetic energy. The sum of the potential and kinetic energy remains unchanged and is a constant.
B v = 0
A C
v maximum
Energy
A
PE
KE
B C constant value
Position of ball
The same principle applies no matter what energy transformations take place as long as they take place within a closed isolated system.
Example:
1. An arrow is shot upward into the air with a speed of 20 ms
–1
. What height would it reach when its speed is reduced to 8 ms
–1
if you ignore air resistance?
Solution:
The Law of Conservation of Energy says that total energy is conserved.
Total KE + Total PE = constant
Loss in kinetic energy = gain in potential energy
Kinetic energy is given by the formula, KE =
1

2
mv
2
Loss in Kinetic energy = " m " (20
2
2
– 8
2
)
Gain in potential energy = (mg) (h
2
–h
1
) where h
2
– h
1
is the change in height.
# mg (h
2
–h
1
) = m 20
2
2
( – 8
2
)
h
2
–h
1
=
( 20
2
2
– 8
9.8
2
)
"
"
m m

"
h
2
h
1
V
V
2
1
= 8
=
20 ms
− 1 ms
− 1
=
19.6
= 17.14
The arrow has an increase in height of 17.14 m.
Module 2 – Understanding motion
2.75
Activity 2.20
Question 1: A large piece of hail with mass 1.5 kg falls from a roof 8.00 m above the ground.
(a) Find the kinetic energy of the ice when it reaches the ground.
(b) What is the velocity of the ice when it reaches the ground?
Question 2: A 5 kg test rocket is fired vertically from its launching bay. If the fuel gives it a kinetic energy of 2176 J before it leaves the bay, how high will the rocket rise?
Question 3: A ball at the end of a string swings like a pendulum. If the ball’s velocity is 400 c ms
–1
when it passes its lowest point, what height will it rise before stopping and reversing direction?
Question 4: At sea level a nitrogen molecule in the air has an average kinetic energy of 6.2 " 10
–21
J. If its mass is 4.7 " 10
–26 kg,
(a) what is the molecule’s average velocity?
(b) If all its KE is converted into gravitational potential energy, how high could the molecule rise above sea level.
Question 5: On a roller coaster, a carriage starts at the top of a 35 m hill and then rolls down into a trough then up to the next hill 25 m high.
(a) What is the velocity of the carriage at the bottom of the first hill?
(b) What is the velocity of the carriage at the top of the second
Question 6: A 2000 kg truck is coasting down a 25° slope with a speed of
12 ms
–1
. At this point the driver applies the brakes. What constant braking force parallel to the road must be applied if the stopping distance is to be in 100 m?
(Hint: Total energy of the truck is equal to the work done by the braking force.)
Question 7: If a car travelling at 50 km/h, skids linearly for 15 m when the brakes are applied, how far will the car skid if it is travelling at 150 km/h?
Question 8: Does an object have energy if it has no momentum (and vice versa)? Explain your answer.
Question 9: Use your knowledge of the conservation of energy to design a new side show game for a local fete.
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TPP7160 – Preparatory physics
Section review
In this section we have studied the relationship between force, work and energy. We have also looked at the conservation of energy and the relationship between the kinetic and potential energy of a moving object. You should now assess yourself as to whether you have met the module objectives.
You should now be able to:
• explain the relationship between force, work and energy and use these relationships to solve problems
• explain the terms potential and kinetic energy
• describe examples of potential and kinetic energy in terms of quantitative energy changes; and use these energy changes to solve problems
• explain the principle of conservation of energy and use this principle to solve related problems
• use appropriate units of measurement for energy calculations.
You should have also made some notes of the main points in this section and listed the concepts that were difficult to understand.
Have you been able to revise those concepts that were difficult to understand?
Have you sought help?
Use the following post test to test out your learning.
Module 2 – Understanding motion
2.77
Posttest 2.4
Question 1:
When work is done on a system the energy of the system is ______________________ .
The energy of an object determined by its position is termed _____________________ , while that which is determined by its motion is termed _______________________ .
Energy is a _________________ quantity and is measured in _____________________
(in SI units). When an object falls freely to the earth it gains _____________________ energy and loses ____________________ energy. In a closed isolated system energy can change form but the total amount of energy is ________________________ .
This principle is called __________________ _________________________ .
If an object is moved a distance s, when a force F is applied, then ___________ is done and is the product of _________________ and _____________________ . If the force is at an angle + to the horizontal, then the work is determined by _________________.
Work is measured in ______________ (SI units) and is a ______________ quantity.
Work done over a set time interval is called ___________________ and is measured in
__________________ (SI units).
Question 2: When work is done then: (Choose one of the following.)
(a) energy is transformed from one form to another.
(b) an object is accelerated.
(c) only can friction be overcome.
(d) certain forms of energy are converted to potential energy.
Question 3: People commonly speak of ‘energy consumption’. Is this term appropriate or inappropriate? Explain.
Question 4: If you drop a ‘superball’ from a height could it bounce higher than this height?
Explain.
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TPP7160 – Preparatory physics
Orange light dilemma
Now to decide whether or not we should speed through that orange light ...
Upon approaching an intersection whose light has just turned orange, you can stop at a maximum negative acceleration, race through at some maximum positive acceleration, or maintain your same speed.
Suppose we wish to cross successfully.
When the light goes orange, we have 2 seconds.
Let the car be x m from the intersection, which is 9m wide. The car is initially travelling at
9 ms
1
.
Firstly, try to visualise the situation with a drawing.
(The distance to be covered is (x + 9) m in 2 secs with an acceleration of 3 m s
–2
)
u = 9 ms
1
Using
s
= ut +
at
2
2
x + 9 = (2 " 9) + (
1
2
" 3 " 4)
x + 9 = 18 + 6
x
= 15 m
#
x
= 15 m is the maximum distance we can be from the intersection to successfully cross under the conditions given.
Suppose we wish to stop.
Module 2 – Understanding motion
2.79
We want v to be 0 at x metres from that instant, with an acceleration of –3 ms
–2
, and
u
= 9 ms
–1 using
v
2
=
u
2
+ 2as
0 = 81 + 2 " (–3) " x
0 = 81 –!6x
6x = 81
#!
x
= 15.2 m
# In order to race through successfully you would have to be less than 15 metres from the intersection but to stop successfully you have to be more than 15 metres from the intersection.
What would you do if
(i) x ,!15 m?
(ii) x  15.2 m?
Answer
(i) You can only keep going because you can’t stop in time.
(ii) You should aim to stop because you will still be crossing when the red light comes on
(and you might have your photographs taken!)
(Why not work this out for your own personal driving speed).
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TPP7160 – Preparatory physics
Module 2 – Understanding motion
2.81
Module 2 – Understanding motion
Solutions to activities
Activity 2.1
Question 1:
N finish
15 m resultant
85° start
8.2 m
3.5 m
30 °
Scale 1 cm
≡
1 m
The dog’s displacement is 7.9 m and 85 W of N or at a bearing of 280° (bearings are stated from North, clockwise using 3 digits)
Question 2:
A
N
67
θ result a nt
C
B
6 ms
–1
(i) (a) Graphical solution
Scale 0.5 cm
1 ms
1
By measurement AC (resultant)
= 7.6 cm
= 15.2 ms
1
!"Resultant velocity
= 15 ms
–1 at 067° (as measured)
14 ms
–1
(i) (b) Mathematical solution AC
2
=
AB
2
+ BC
2
= 14
2
+ 6
2
#
= 196 + 36 = 232
AC = 15.23
resultant velocity = 15 ms
–1
2.82
TPP7160 – Preparatory physics since tan
'
!
=
'
6

14
= 0.4286
= 23.2( and the direction of the resultant velocity is 23° North of East or 67°
(ii) Effectively the boat travels 100 m across the river at a constant velocity of 14 ms
–1
(the component of its velocity at right angles to the river) time = velocity
=
14
= 7.14 secs
15.2 ms
1
6 ms
1
(iii) During this time the river has carried the boat downstream at 6 ms
–1
!
Displacement = velocity # time
= 6 # 7.14
= 42.9 m
Alternatively you can use the angle to find the displacement tan 23.2° =
x

100
!
x = 42.9 m
Therefore the boat is 42.9 m downstream.
14 ms
23.2º
1
100 m
x
Question 3:
90 − θ
θ
45 km h
1
340 km h
1 scale: 1 cm
40 km h
–1
Using this scale the plane’s velocity of 340 kmh
–1
$ "
40 and the wind velocity of 45 kmh
–1
$ "
40
$" 8.5 cm
$" %&%cm
By measurement, the resultant = 8.60 cm and ' = 7.5°
!
The resultant velocity of the plane = 8.6 × 40
= 344 kmh
–1
at a bearing of 082.5°
Module 2 – Understanding motion
2.83
By calculation, resultant =
=
340
2
+
117625
45
= 343 kmh
–1
2
!
tan
'
=
340
'
= tan
–1
45
7.5(
340
=
!" The plane’s velocity with respect to the ground (the ground being taken as the point of reference with zero velocity) is 343kmh
–1
at N82.5°E.
A more accurate answer can be arrived at if we used a larger scale diagram.
Activity 2.2
Question 1:
initial velocity final velocity
Question 2:
initial velocity of car final velocity of car change in velocity time taken acceleration
u
= 3.0 ms
–1
v = 3.5 ms
–1 acceleration = change in velocity over time
a
=
v
–
t u
=
3.5
–
4
3.0
= 0.13 ms
–2 where t = 4 secs in the direction the car is moving
u
= 0
v
= 10 ms
–1
)
v
=
v – u
= 10 – 0
= 10 ms
–1
a t
= 3 secs
=
)
v/t
=
3
= 3.3 ms
–2 down the hill
2.84
TPP7160 – Preparatory physics
Question 3:
initial velocity of train final velocity of train change in velocity
Time taken acceleration
u
= 50 ms
–1
v
= 10 ms
–1
)
v = 10 – 50 = –40 ms
–1
t = 15 secs
a
=
–40

15
= –2.7 ms
–2
Note: Negative acceleration because the roadtrain is decelerating or slowing down.
Activity 2.3
Question 1:
initial velocity of the bus, u = 15 ms
–1 final velocity
v
= 0 ms
–1 acceleration a = –2 ms
–2
(negative because bus is slowing)
To find distance travelled s, use the equation i.e.
v
2
=
u
2
+ 2as (since t is not involved)
0 = (15)
2
+ 2 × (–2) # s
2
!"
s
=
4
= 56.25 m
!"The bus will travel for 56.3 m before it stops.
Question 2:
60 km/hr =
60
#
1000
3600
ms
–1
(convert to SI units)
=
36
=
6
= 16.7 ms
–1
The car has to stop from this speed in 100 m if it is not to hit the child.
Module 2 – Understanding motion
2.85
i.e.
u
= 16.7ms
–1
s
= 100 m
v
= 0
To find acceleration use
v
2
=
u
2
+ 2as
0 = (16.7)
2
+ (2a # 100)
a
= –
2
16.76

200
= –1.39 ms
–2 or
–
5 kmh
–1
s
–1
66
To convert ms
–2
to kmh
–1 s
–1
(i.e. km per hour per sec)
=
1s
#
1s
=
=

+
1000
#

+

60
#
60
.
,

1000
.
,
#
1s
× 3600 h
–1
s
–1
= 5 kmh
–1
s
–1
This would be possible, provided the driver can do so in 12 s flat i.e. at a deceleration of –5 km h
–1 every second. It also depends on the driver’s reaction time which varies with age, state of concentration, and muscle reflexes.
Question 3:
Assuming she jumps horizontally, initial downward velocity
Acceleration in freefall g = 9.8 ms
–2
After two seconds, velocity
at)
v
= 0 + 2 # 9.8
u
= 0
(using v = u +
Distance travelled
at
2
)
= 19.6 ms
–1
s
= 0 # 2 + 
2
# 9.8 # 2
2
(using s = ut +
2
= 19.6 m
The jumper would have fallen 19.6 m in 2 secs.
Question 4:
Time of flight of cat
Acceleration using
= 0.5 secs
g = 9.8 ms
–2
s = ut +
1
at
2
2
s = 0 + #"9.8 # *0.5)
2
2
= 1.225 m
The branch is 1.2 m above the ground.
2.86
TPP7160 – Preparatory physics
Question 5:
Since the car starts from rest initial velocity u = 0
Distance travelled s = 240 m
Velocity v at end of this distance = 18 ms
–1 using
v
2
=
u
2
+ 2as
18
2
= 0 + 2 # a # 240
Speed limit, 60 kmh
–1
a =
2
480
=
480
= 0.68ms
–2
=
60
#
1000metres
1
#
60 60 sec
onds
=
60
#
10
3600
3
ms
–1
=
100

6
= 16.6 ms
–1
Since the motorist is travelling at 18 ms
–1
his speed is greater than the speed limit of 16.6 ms
–1
, by approximately 1.4 ms
–1 or 5 kmh
–1
.
Notice that to convert ms
–1
to kmh
–1 multiply by
1000 or a factor of 3.6
ms
–1
× 3.6 = kmh
–1 kmh
–1
/"3.6
= ms
–1
Question 6:
If we assign downwards as the positive direction, then initial velocity,
u
= 4.4 ms
–1 and acceleration,
a
= –1.5 ms
–2
After two seconds, v (final velocity) is given by
v
= u + at
= 4.4 + (–1.5 # 2)
= 4.4 – 3
= 1.4 ms
–1
Since v is positive the lift is still moving downward at 1.4 s
–1
.
The acceleration upward is the same as a deceleration downwards.
Module 2 – Understanding motion
2.87
Activity 2.4
Question 1:
To find my weight in Newtons I must first find my mass in kg.
Mass = 72 kg (say)
My weight is the force due to the attraction of the earth exerted on my mass.
Using the equation where
F
=
ma a
= acceleration due to gravity = 9.8 ms
–2
F = 72 # 9.8 N
= 705.6 N
Note: we often use W = mg to calculate the weight of an object on Earth.
Question 2:
On the moon my mass is the same, 72 kg

6
So
F m
= N
6
! My weight = 117.6 N on the moon
Question 3:
On Mars the gravitational force
F mars
= mass # acceleration
= 72 # 8.3
= 597.6 N
! My weight on Mars is 597.6 N
Question 4:
In outer space – provided I am not within the gravitational field of any planet, the gravitational force is zero, so my weight is zero, even though my mass is still
72 kg.
!" In outer space I am weightless.
2.88
TPP7160 – Preparatory physics
Activity 2.5
Question 1:
(a) The mass of the car m = 800 kg since the coefficient of kinetic friction
0
k
= 0.5
Frictional force
F f
= 0
k
F n
where F
n
= normal force
= mg
F f
= 0.5 # 800 × 9.8
= 3920 N
This force acts along the road in a direction opposite to the direction of motion of the car.
(b) If a is the acceleration of the car, then using F =
ma
–3920 = 800 a (we assign the direction of motion to be positive)
a
= –
3920

800
= –4.9 ms
–2
!"The car decelerates at 4.9 ms
–2 in the direction of motion.
(c) Since initial velocity is 30 ms
–1
and final velocity is zero, using
v
2
=
u
2
+ 2as
0 = 30
2
+ 2 # (–4.9) # s using
s
=
2
2
#
4.9
= 91.8 m
v
= u + at
0 = 30 – 4.9t
t
=
4.9
= 6.12s
! It will take 6.1s and a stopping distance of approximately 92 m for the car to come to rest.
Module 2 – Understanding motion
2.89
Question 2:
F n
= 50 N
F f
F = 36 N
W = 50 N
Since the box moves at constant speed there is no acceleration.
To move the box at constant speed, there is no net force and the applied force is just equal to the frictional force.
So
F f
= 36 N
The normal force F
n
= 50 N
Hence the coefficient of kinetic friction 0
k
0
k
=
F f

F n
=
50
= 0.72
2.90
TPP7160 – Preparatory physics
Activity 2.6
Question 1:
(a) Drawing to scale:
C
lt an re su force t
65 N
N q
A
83 N
Scale 1 cm
10 N
By measurement the resultant force = 105 N angle ' = 52(
!"
The resultant force is 105N at N 52° E
(b) By calculation
AC
2
=
AB
2
+ BC
2
AC = 83
2
+ 65
2
= 105.4
tan
1
CAB
=
65

83
1
CAB
= 0.78
= 38(
! ' = 90° – 38° = 52°
!"The resultant force is 105 N at a bearing of 052°
B
Note: Remember that force is a vector and answers for vectors must always include both magnitude and direction unless otherwise stated.
Module 2 – Understanding motion
2.91
Question 2:
(a) Given:
A
Drawing to scale at 4mm : 1 N
We find the resultant by adding the 2 vectors.
θ
Drawing to scale. 4mm 1N
Resultant by measurement = 114mm
!
Resultant force = 28.6 N
By calculation: using the cosine rule (see page ‘2.25’) a
2
= b
2
+ c
2
– 2bc cos side a)
2.92
TPP7160 – Preparatory physics
Applying the cosine rule to our problem above b
2
= a
2
+ c
2
–2ac cos
= 15
2
+ 15
2
– 2(15)(15) cos 145°
= 450 – (450)(–0.819)
= 450 + 368.55
b = 818.5
= 28.6
!" The magnitude of the resultant is 28.6 N
Activity 2.7
Question 1:
F n
F f
Wsin35º
35º
W
Wcos35º
The forces acting on the car are:
(i) its weight W = 1.2 # 10
4
N acting vertically down
(ii) the frictional force F
f
acting up the slope.
(a) Along the slope, the component of the car’s weight is W sin 35° and this is the force that tends to roll the car downhill.
!" W sin 35° = 1.2 # 10
4
# sin 35(
= 1.2 # 10
4
# 0.574
= 6.88 # 10
3
N
!"The force tending to cause the car to roll down slope
= 6.88 # 10
3
N
(b) At right angles to the slope, the component of the car’s weight is W cos 35°
= 1.2 # 10
4
# cos 35(
= 1.2 # 10
4
# 0.819
= 9.83 # 10
3
N
So2"Force at right angles to the slope
= 9.83 # 10
3
N
Module 2 – Understanding motion
2.93
Question 2:
Since the wagon is well oiled and frictionless (not in real life) we can disregard the friction between the wagon and the slope.
Thus the only force acting down the slope is the component of the weight down the slope, W sin '&
F n
110 N
Wsin
θ
θ
W
Since the wagon is pulled steadily uphill we can say that there is no acceleration.
!"The force uphill = the component of the weight downhill.
F = mg sin '" (since W = mg)
110 = 75 # 9.8 sin '" sin ' =
110

75
#
9.8
sin ' = 0.15
' = 8.6(
! The slope of the hill is 8.6°.
Question 3:
F n
89 N
Wsin35º
35º
W
Let W be the weight of the trolley. Then resolving the force W along the plane and at right angles to it, we have:
2.94
TPP7160 – Preparatory physics
Along the plane, the component of W down the slope
= W sin 35(
But 89 = W sin 35("since friction is neglected
W = sin
89

35(
= 155 N
! The weight of the trolley is 155 N
Question 4:
normal reaction plank
W cos20°
20º
W
Force exerted normally on plank = W cos 20°
= mg cos 20°
= 20 #"g #"cos 20(
= 184.2 N
!"""The child exerts a force of 184.2 N perpendicular to the plank.
Module 2 – Understanding motion
2.95
Activity 2.8
Question 1:
F f
F n
W
sin
42°
W
W cos 42°
42°
Let W Newtons be the weight of the block.
Then resolving W along the plane and normal to it, force along the plane = W sin 42( force normal to the plane = W cos 42(
Since the block is on the point of sliding,
But frictional force
So i.e.
Force down the plane
F f
= frictional force
F f
=
0 F
n
(where F
n
= W sin 42°
= normal reaction and
0 = coeff of friction)
W sin 42( =
0 W cos 42(
0 = sin cos
42(

42(
= tan 42°
= 0.9
!" The coefficient of static friction 0" 3" 0.9
2.96
TPP7160 – Preparatory physics
Question 2:
(a)
F n
50
0
N
40 °
W
sin
40°
F f
W
=
mg
W cos 40°
(b) The net force or resultant force up the plane = 18 N (up plane is + ve)
500 –
F f
– 25 g sin 40 = 18!
"!
F f
= 500 – 25g sin 40° – 18 (where F
f
is the frictional force)
(c) F
n
= 25 g cos 40 since the object is balanced on the surface of the slope.
(d) from the equation in (b)
F f
= 500 – 25g sin 40° – 18 whence
F f
F f
= 500 – 157.5 – 18
= 324.5 N
The coeff of kinetic friction #
k
=
F

F f n
=
25 $ 9.8
$ cos 40
.
=
187.7
= 1.73
Module 2 – Understanding motion
2.97
Question 3:
F n
F
N
W sin 8°
8°
Wcos 8°
W
=
mg
Resolving W at right angles to the plane,
Wcos 8° = 300 g cos 8
= 2911 N
" F
n
= 2911N
The frictional force, F
f
= 350 N down the slope since it opposes the uphill motion
" since F
f
= #F
n
(where # is the coeff of sliding friction)
# =
F

F f n
# =
350

2911
= 0.12
" The coefficient of sliding friction between the snow and sled is 0.12.
Activity 2.9
Question 1:
(a) The force of gravity acting on the ball is 0.5 $ 9.8 N = 4.9 N
The air resistance is 0.8 N.
Hence the net force acting downward on the ball is
F = W – Air resistance
Ball
Air resistance
= 4.9 N – 0.8 N = 4.1 N
If a is the resulting acceleration,
F = ma (Newton’s 2nd Law)
4.1 = 0.5 a
W = 0.5g
" a =
0.5
= 8.2 ms
–2
2.98
TPP7160 – Preparatory physics
(b) If the ball is falling with constant velocity (terminal velocity), there is no net force acting.
Hence air resistance = force of gravity on ball
= 0.5 $ 9.8
= 4.9 N
(c) If there was no air resistance, the acceleration of the ball would be 9.8 ms
–2
.
Question 2:
(a) The balls will each reach terminal velocity when the upward force of the air resistance equals the force due to gravity on the ball.
If, as seems likely, the mass of the hollow ball is less than that of the solid one, the force of gravity on the hollow ball will be more quickly equalled by the air resistance, and so the hollow ball will reach terminal velocity first.
(b) The heavier ball will reach the ground first, because it reaches a higher terminal velocity than the hollow ball, having had a net accelerating force acting on it for a longer time.
Question 3:
If the tennis ball reaches terminal velocity before it finishes falling 20 storeys its velocity will remain constant as it falls a further 30 storeys, hence the final velocity after falling 50 storeys is the same as that after falling 20 storeys.
Activity 2.10
Question 1:
(a) The mass is unchanged that is, 15 kg.
(b) The weight of an object is the force of gravitational attraction between the object and the earth, and is given by the formula –
F
=
d
2
G = universal gravitational constant
M = mass of the earth (kg)
m = mass of the object (kg)
At a distance of 149 km above the earth’s surface,
d = 6370 + 149 km
= 6519 km
= 6.519 $ 10
6
m
Module 2 – Understanding motion
2.99
So, the weight of the object at a distance of 149 km above the earth’s surface is
F =
=
% 6.519
$ 10
6
&
2
6.67
$
%
10
– 11
6.519
$
$
6 $
10
10
24
$ 15

6
&
2
= 141 N
" weight of the object at this height
(compare this with its weight on earth)
Question 2:
The mass of the Earth = 6 $ 10
24 kg
Mass of Mars = 0.11 $ 6 $ 10
24
kg
= 6.6 $ 10
23
kg
A 200 N object on the Earth has a mass
200
20.4 kg
9.8
=
(a)On Mars, let the force of gravity on the object be mg
/ acceleration due to the gravity on Mars.
where g
/
is the mg
/
= mg
/
=
G
$
d
Mm
2
6.67
$ 10
– 11
$ $ 10
% 3.440
$ 10
6
&
2
23
$ 20.4
object
m
=
6.67
6.6
$ 20.4
3.440
2
=
6.67
$ 6.6
$ 20.4

3.44
2
"!
It’s weight on Mars = 75.89 N
(b)Since the mass on Earth is 20.4 kg
d 3440km
Mars M
The acceleration due to gravity is g
/
=
20.4
= 3.72 ms
–2
" Acceleration due to gravity on Mars is 3.72 ms
–2
2.100
TPP7160 – Preparatory physics
Question 3:
(a) The gravitational force is given by
F
=
= where d = 7370 km
d
2
6.67
$ 10
– 11
$ 6 10
7.370
$ 10
24
$ $ 70

6
2
% &
=
6.67
$ 6 7 $ 10
14
7.370
2
$ 10
12
= 5.157
$ 10
14 – 12
= 5.157 × 10
2
N or 516 N
(b) At the surface of the earth his weight in Newtons is
70 $ 9.8 = 686 N
So his weight at 1000 km above ground is about 0.75 of his weight on the Earth’s surface.
Question 4:
(a)The acceleration due to gravity is given by
g
/
=
d
2 where g
/
is acceleration due to gravity on the surface of Jupiter
G =
6.67 $ 10
–11
Nm
2 kg
–2
M = 1.9 $ 10
27
kg
d = 7.14 $ 10
7
m
Substituting into the formula
g
/
=
6.67
$ 10
– 11
$
% 7.14
$ 10
7
&
2
$ 10
27
=
6.67
1.9
$ 10
16
7.14
2
$ 10
14
= 0.248 $ 10
2
= 24.8 ms
–2
Module 2 – Understanding motion
2.101
(b)Since the weight of an object on earth = 9.8 $ mass and on Jupiter = 24.8 $ mass
Factor difference = = 2.53
9.8
" Your weight on Jupiter will be approximately 2.5 times larger.
Question 5:
It all depends on how you’re measuring the mass of the metal. If you’re using a beam balance – it makes no difference what altitude you use – the mass of the metal remains constant.
If however you are weighing on a device which utilises the force of gravity e.g. a spring balance, buying at a higher altitude, where the weight is smaller, and selling at the lower altitude, would appear to be the way to go.
Do you think the weight will vary very much at sea level and on Mt Everest?
Activity 2.11
Question 1:
(a) Mass of earth
Linear velocity
Momentum
(b) Mass of snail
Velocity
Momentum
m
= 6 $ 10
24
kg
v = 3 $ 10
4
ms
–1
P = mv
= 18 $ 10
28
= 1.8 $ 10
29 kg ms
–1
m
= 0.1 kg
v
= 1 $ 10
–6
ms
–1
P = mv
= 1 $ 10
–1
$ 1 $ 10
–6
= 1 $ 10
–7
kg ms
–1
(c) Mass of rifle bullet 15g = 0.015 kg
Velocity
v
= 600 ms
–1
Momentum P = mv
= 0.015 $ 600
= 9.0 kg ms
–1
2.102
TPP7160 – Preparatory physics
(d) Mass of electron
Velocity
Momentum
= 9.1 $ 10
–31
kg
v
= 2 $ 10
5
ms
–1
P = mv
= 9.1 $ 10
–31
$ 2 $ 10
5
= 18.2 $ 10
–26
= 1.82 $ 10
–25
kg ms
–1
Question 2:
(a)The momentum of the cricket ball before impact P
i
= mu = 0.14 $ 38
= 5.32 kg ms
–1
After impact, the final momentum is P
f
= mv
= –38 $ 0.14
= –5.32 kg ms
–1
(negative signs indicate movement in the opposite direction)
Impulse = change in momentum
Ft = mv – mu
= –5.32 – 5.32
= –10.64 kg ms
–1
(in the opposite direction)
or – 10.64 Ns
(b)The impulsive force exerted on the ball is given by F, where Ft = change in momentum
" F =
mv
–
mu

t
=
–
0.6
F = –17.73 N
" The average force the ball exerted on the bat is 17.73 N
(c)To find the acceleration, a =
v
–
t u
=
– –
0.6
38
=
– 76

0.6
= – 126.6 ms
–2
(in the opposite direction)
Module 2 – Understanding motion
2.103
Question 3:
(a) Velocity u = 40 km hr
–1
=
40000

3600
ms
–1
= 11.1 ms
–1
It comes to rest in 0.25 m, " using v
2
=
u
2
' 2as to find a
0 = (11.1)
2
+ 2(a)(0.25)
2a $ 0.25
= – (11.1)
2
a
=
–
0.5
" The car’s deceleration a = –246.9 ms
–2 using a =
v
–
u

t
to find t
"! t =
v
–
a u
=
0
–
– 11.1
246.9
" The time required to stop the car = 0.045 s
(b) Mass of child
Initial momentum of child
m = 20 kg
P i
=
mu
= 20 $ 11.1
= 222 Ns
Since child comes to rest in 0.045 sec, Force = time
F =
t
=
0 – 222 Ns
0.045s
(since F
f
= 0)
= –4940 N
(it is a decelerating force)
" The average force acting on the child is 4940N.
2.104
TPP7160 – Preparatory physics
Question 4:
(a) Dancer falls through 0.32 m, using
v
2
= u
2
+ 2as
v
2
= 2as
(u = 0 at the height of 0.32m above the ground)
v
2
= 2 $ 9.8 $ 0.32
= 9.8 $ 0.64
v = 2.5 ms
–1
"
Therefore momentum on reaching the ground is P
f
= mv
P f
= 2.5 $ 55
= 137.7 kg ms
–1
(b) The impulse Ft = mv – mu
" The impulse needed to stop
= 137.7 – 0
(since the dancer comes to rest)
= 137.7 kg ms
–1
= 137.7 Ns
(c) If the stopping time is 0.055 secs and since
Ft = impulse
F $ 0.055 = 137.7
F
= 2503.6 N
" The average force exerted on the body is approximately 2500 N
Question 5:
Padded dashboards lengthen the stopping time by cushioning the impact as it resists a force
t = 0
t = 0.02 s
t = 0.04 s
Whenever time is lenthened the force of impact decreases.
Long collision time means small collision force and vice versa.
However if a padded dashboard is only one of the safety features in a car. Driving at a safe speed and wearing seat belts would give drivers and passengers a better chance of survival in any collision.
Module 2 – Understanding motion
2.105
Activity 2.12
Question 1:
Initial momentum of moving wagon
P i
(wagon 1) = m
1 u
1
= 20000 $ 5 kg ms
–1
Initial momentum of stationary wagon
P i
(wagon 2) = m
2 u
2
= 0
Combined mass after impact
Let velocity after impact be v
m
1
+ m
2
= 40000 kg
Total momentum before collision = Total momentum after collision
m
1
u
1
+ m
2
u
2
= (m
1
+m
2
)v
20000 × 5 + 0 = 40000 × v
" v =
20000 $
40000
5
v = 2.5 ms
–1
" Velocity after impace 2.5 ms
–1
in the same direction as wagon 1
Question 2:
u
1
= 28
ms
− 1
u
2
= 0
ms
− 1
v
Initial momentum of moving car
P i
(1) =
m
1 u
1
= 1245 $ 28 kg ms
–1
Initial momentum of other car
P i
(2) =
m
2 u
2
= 0
(
m
1
+
m
2
)
Let v be the velocity after impact
"! Total inital momentum = Total final momentum
m
1 u
1
+ m
2 u
2
= (m
1
+ m
2
) v
1245 $ 28 + 0 = (1245 + 2163) v
"
v
=
1245 $
3408
28
= 10.23 ms
–1
"!The final velocity of the combined masses is 10.23 ms
–1
2.106
TPP7160 – Preparatory physics
Question 3:
Let their final velocities be v
1
and v
2
Since both are initially at rest the initial momentum
= 0
After pushing, final momentum = 80 v
1
+ 60 v
2
Total momentum before impact = Total momentum after impact
0 = 80 v
1
+ 60 v
2
80 v
1
= 60 v
2
0 =
m
2
×
u
2
!
v

v
1
2
=
– 60

80
!
v

v
2
=
–
4
the ratio of their velocities v
1
:v
2
#
–3:4
(the negative sign indicates that the velocities are in opposite directions)
Since
v
2

3
1
the 60 kg student has greater speed.
(smaller mass greater speed and vice versa)
Question 4:
m
2
×
u
1
= 0
Let the velocity of recoil of the nucleus be v
1 and the velocity of the ejected particle be v
2
Since
P f
= P i
momentum of ejected particle + momentum of remaining nucleus = 0
!
momentum of remaining nucleus = – momentum of ejected particle i.e. (3.8 " 10
–25
(380 " 10
–27
(m
1
– m
2
) v
1
= – m
2
v
2
– 6.6 " 10
–27
) v
1
= –6.6 " 10
–27
" 1.5 " 10
7
– 6.6 " 10
–27
) v
1
= –6.6 " 10
–27
" 1.5 " 10
7
Module 2 – Understanding motion
2.107
Question 5:
(373.4 " 10
–27
) v
1
= –6.6 " 1.5 " 10
–20
v
1
=
– 6.6
"
1.5
"
10
–20

–27
373.4
"
10
=
–
"
10
7
373.4
= –0.0265 " 10
7
= –2.65 " 10
5 ms
–1
The velocity of recoil is 2.65 × 10
5 ms
–1
in the opposite direction to the ejected particle.
m
2
= 128 kg
u
2
= ?
m
1
= 95 kg
u
1
=
.
ms
1
Let the speed of the opposition player be u
2 ms
–1
Then momentum of tackler,
m
1
u
1
= 95 " 2.8 kg ms
–1
Momentum of other player,
m
2
u
2
= 128 " u
2
Since they both come to rest, their final momentum must be zero.
Total initial momentum = Total final momentum
!m
1
u
1
+ m
2
u
2
= 0
95 "!2.7 + 128 "!u
2
= 0
128 " u
u
2
2
= –95 " 2.8
=
– 95
"
2.8

128
= 2.08 ms
–1
u
2
!The initial velocity of the other player = 2.08 ms
–1
in the direction opposite to the tackler.
2.108
TPP7160 – Preparatory physics
Activity 2.13
Question 1:
The time taken for the load to strike the ground is its time of flight = time taken to fall from 100 m.
Consider the vertical motion,
u y
= 0 since the initial velocity was in the horizontal direction
1

2
2
(since a = g = 9.8ms
–2
)
s =
1

2
"
9.8
"
t
2
u x
15 ms
1
100 = 4.9 t
2
t
2
=
100

4.9
!
t =
4.5 seconds
!time of flight = 4.5 s
s = 100m
x
Target
Now let us consider the horizontal motion where
u x
= 15 ms
–1
In this time the plane travelling horizontally at 15 ms
–1
will cover a distance x, given by x = u
x
× t
= 15 × 4.5
= 67.5 m
Hence it must release the load 67.5 m before the target position.
Question 2:
v y
= 0 at the highest point
u
height
(a) Velocity of ball u = 3.45 ms
–1 at 75° to the horizontal
Vertical component of velocity
u y
Horizontal component
u x
=
=
3.45 sin 75$
3.45 cos 75$
If t is time of flight of the ball (i.e. time taken to reach the ground), consider the vertical motion using s =
u y
t +
gt
2
2
0 = 3.45 sin 75$!t +
1

2
" –9.8 " t
2
(assign (–) for downward motion)
0 = 3.33t – 4.9t
2
Module 2 – Understanding motion
2.109
4.9t
2
– 3.33t = 0
t(4.9t – 3.33) = 0
t = 0 or 4.9t – 3.33 = 0
4.9t = 3.33
(t = 0 is not an option here)
t =
4.9
= 0.68
Therefore time of flight = 0.68 secs
(b) If height reached = h, then h occurs when final upward velocity v
y
= 0
Again considering the vertical motion, using
v
2
=
u
2
+ 2as or
v y
2
=
u y
2
+ 2gh
0 = (3.45 sin 75°)
2
+ 2 (–9.8)h
(taking down as negative)
2 × 9.8 × h =
(3.45 sin 75°)
2
h =
&
3.45
sin 75
%
2
19.6
!
h =
19.6
= 0.566
The highest point reached = 0.566 m i.e.
or
!
Question 3:
Initial velocity of the line u = 20 ms
–1
u
= 20 ms
1
Vertical component
u y
= 20 sin 45$
u y
Horizontal component
u x
= 20 cos 45$
Time of flight t is found by considering the vertical motion using s = u
y
1

2
2
s
= (20 sin 45$%t +
1

2
gt
2
45º
u x x
0 = 14.142 t – 4.9 t
2
t(4.9t – 14.142) = 0
(taking down as –ve) i.e.
t
= 0 or
4.9t – 14.142 = 0
t
=
14.14

4.9
t = 2.89 secs
2.110
TPP7160 – Preparatory physics
During this time it travels horizontally over a distance x.
Considering the horizontal motion, since velocity is constant
x =
u x
× t
= 20 cos 45° × 2.89
= 14.142 " 2.89
= 40.9 m
!The cast will be dropped at a horizontal distance of 40.9m from the point of casting.
Question 4:
u x
1 ms
1
Target
0.25 m
x
During its flight the dart falls 0.25 m vertically below the target as shown above
Consider the vertical motion,
Since its initial vertical velocity u
y
using
= 0
s = u
y
1

2
2
,
s =
1

2
gt
2
where t is time of flight
0.25 = 4.9 t
2
The horizontal velocity
u t = x
4.9
= 11.3 ms
–1
= 0.226 secs
Distance travelled horizontally x = u
x
× t
= 11.3 " 0.226
= 2.55 m
The player is 2.55 m away from the board directly infront of it.
Question 5:
There are three factors involved in attempting to increase the length of the competitor’s jump.
(i) The speed of the competitor at the takeoff point.
(ii) The angle at which the competitor projects his/her body relative to the horizontal.
Module 2 – Understanding motion
2.111
(iii)Air resistance
At the point of takeoff the velocity of the jumper can be resolved in two components
v cos
'
horizontally and
v sin
'
!vertically
v
vsin θ where
'!
is the angle of projection.
θ
v cosθ
The horizontal distance covered is given by
x
= v cos
'
!t where t is the time of flight i.e. the time between takeoff and landing in the pit.
t is dependent upon the vertical component v sin
'
!as shown by the following:
Using the equation of motion, since
s = ut + gt
2
2
s = 0,
2
gt
2
= ut
t
2

t
=
g
!t
=
g
=
2v sin
g
' since x = v cos
'
t x
= v cos
'
×
2v sin
g
'
=
2 sin
' cos
'
g
The value of
'
!which optimizes the distance covered horizontally under these conditions is 45$!or since the product sin
4
'!
cos
'
is maximum then. So the ideal take off should be at some angle close to this.
Practically, however, the problem is more complicated, because such a change in direction at the point of take off implies a large impulsive force to effect the change in momentum, and the final optimum take off angle and velocity may have to be a compromise depending on the physique of the jumper.
2.112
TPP7160 – Preparatory physics
Activity 2.14
Question 1:
Centripetal acceleration
a c
=
r
2
=
2
r
Since the Earth orbits the Sun once every 365 days (or 365 × 24 × 60 × 60 secs)
=
2(

365
"
86400
rad s
–1
(1 revolution = 2 (!rad)
centripetal acceleration a
c
=
+
)

365 86400
,
*
2
"
1.5
"
10
11
=
&
1.99
"
10
– 7
%
2
"
1.5
"
10
11
= 5.95
"
10
– 3 ms
– 2
Question 2:
(a) No of rev/s
!Angular velocity
s
= 1500 rev s
–1
= 1500 " 2( rad s
–1
(1 rev = 2( rad)
(b) Centripetal acceleration
a
c
=
2
r
= (1500 " 2()
2
" 0.15
= 13323966 ms
–2
= 1.3 × 10
7 ms
–2
(Note: It is common practice to designate centripetal accelerations by comparing them to g)
!At 1500 rev s
–1 the contents in the test tube is under
1.3
"
10
7

9.8
= 1.3 × 10
6 times the acceleration due to gravity g.
Question 3:
(a) At this speed of 5000 revolutions per minute, a point on the edge of the disc has an angular velocity
! =
5000
60
"
2(
rads
–1
It’s linear velocity v = r
=
0.1
"
5000
"
2(

60
= 52.36 ms
–1
So tangential speed of a point on the edge of the disc = 52.4 ms
–1
Module 2 – Understanding motion
2.113
(b) Centripetal acceleration
a c
=
=
&
2
v

r
0.1
%
2
= 27415.6 ms
–2
= 2.7 × 10
4
ms
–2
Such a sander would not be permitted to be used. The sanding disc would almost certainly disintegrate, because of the centripetal force overcoming the intermolecular strength holding the molecules together. In any case, the practical use of such a sander would be almost nil.
Question 4:
The velocity of the protons
 3.00 " 10
8
ms
–1 radius of orbit =
0.751

100
= 7.51 " 10
–3 m
2
Centripetal acceleration a
c
=
r
=
&
3.00
"
10
8
%
7.51
"
10
–
3
2
=
9
"
10
16

3
7.51
" 10
= 1.2 " 10
19 ms
–2
Activity 2.15
Question 1:
Mass of proton,
m
= 1.657 " 10
–27 kg diameter of magnet = 2 m radius,
r
= 1 m
Velocity of protons v = 2.0 " 10
6 ms
–1
The circumference of the path of the protons = 2( r = 2( × 1
= 2( metres
(a) The time to cover 1 revolution (a distance equal to 2( r)
Period,
T =
dis tan ce
"
velocity
=
2(r

v
=
2(

6
2
"
10
=
( " 10
–6 secs
=
(!micro seconds or 3.14 .s
2.114
TPP7160 – Preparatory physics
(b) The force exerted by the magnet on each proton = centripetal force
2
F c
=
r
=
&
1.657
"
10
– 27
% 2.0 10
6
%
2
1
= 1.657 " 4 " 10
–27
" 10
12
N
= 6.628 " 10
–15
N
The force exerted on each proton is 6.6 × 10
–15
N
Question 2:
0.9 m
F c
0.8 kg
The maximum allowable centripetal force = 2 N (the maximum tension allowed on the string)
Centripetal force
F c
=
2
mv

r m
= 0.8 kg and r = 0.9 m
! 2
=
0.8
"
v
2

0.9
Hence
The period,
v
2
=
1.8

0.8
v
=
3

2
= 1.5 ms
–1
T = time for one revolution
The circumference of the path = 2( r
= 2( " 0.9 m
Period T =
velocity
=
2(
"
1.5
0.9
!
Minimum period = 3.77 secs
=
v
Module 2 – Understanding motion
2.115
Question 3:
The centripetal force exerted by the Sun on the Earth can be expressed in two ways.
= mass of Earth (a) If M
S
= mass of Sun and M
E
F c
=
GM

2
M
E d
Also, if the angular velocity of the earth is ,
F c
= mr
2
! ! F c
= M
E
d
2
If we equate these two equations for F
c
M
E
2
d =
GM

d
2
M
S
2
d
3
=
G M
S ............................................
equation (A)
Now =
T
=
2(

365
"
86400
(since period, T = 365 days)
2
d
3
=
+
)

365 86400
,
*
2
" &
1.5
"
10
11
%
3
= 1.339 × 10
20
From equation (A) above GM
S
=
2
d
3
= 1.339 × 10
20
!
M
S
=
2
G
3
=
1.339
"
10
20
– 11
6.67
"
10
M
S
= 2.0 × 10
30
kg
Mass of the Sun = 2.0 × 10
30
kg
2.116
TPP7160 – Preparatory physics
Activity 2.16
Question 1:
F
2
B
F
1
2l
l
l cos25°
A
25°
400 N
If the plank is in equilibrium and is uniform then!/F = 0 .................. (1) or the sum of forces = 0 i.e.
F
1
+ F
2
– 400 = 0 (taking downward as negative)
Also if we take moments about the end B, (let the length of the plank be 2l)
/0! 1! 0!2222222222222222222222!(2) or!sum of torques = 0
(Torque = Force × perpendicular distance to the pivot)
F
1
" 2l cos 25° = 400 l cos 25°
(taking clockwise direction as tve
)
Since
F
1
= = 200 N
2
F
1
+ F
2
= 400 N
!
F
2
= 200 N
So each person exerts an upward force of 200 N.

2 then as before
/F
= 0
F
1
+ F
2
– 400 – 140 = 0
F
1
+ F
2
= 540
Taking moments as before about the end B, /0! = 0
Module 2 – Understanding motion
2.117
F
1
2l cos 25° – 140 l cos 25° – 400 l cos 25° = 0
2
(taking anticlockwise as (–))
!
2F
1
– 210 – 400 = 0
!
Question 2:
F
1
=
610

2
F
2
= 235 N
= 305 N
F
1
F
1 sin 55 °
F
2 sin 55 °
F
2
(a) The torque you applied to where
"
1
= F#r
F# = F
1
sin 55° (by resolving F perpendicular to the crow bar)
!$"
1
= F
1
sin 55° × 1.6 (Note: the #$distance = 1.6 sin 55°)
= 250 sin 55° × 1.6
= 327.66
= 328 Nm
(b) This torque turns the whole crowbar clockwise about the pivot.
!$The rock will also be acted upon by a clockwise torque given by
"
2
=
F
But
"
2
= "
1
2
sin 55° × 0.4
F
2
sin 55° × 0.4 = 328
!$ F
2
= 1001.03 N
! The force subsequently exerted on the rock = 1001 N upwards.
(c) The weight of the rock = 1001 N
2.118
TPP7160 – Preparatory physics
(d) Since
F
F
2

1
=
1001

250
*
4
! $
F
2
*
4F
1
or
F
1
*
F
4
2
By using a simple machine such as a crowbar (in this case as a 1st class lever) we have used a force equal to about one quarter the weight of the rock to lift it.
Since " $% $F#r and F
1
#r
1
= F
2
#r
2
If F
2
>> F
1
, then r
2
<< r
1
As can be seen r
2
= 0.4 m
r
1
= 1.6 m i.e.
r
1
= 4r
2
(note >> means much larger) or
F

F
1
=
r

r
2
=
1
These ratios
F
2
: F
1
+
r
1
: r
2
+
4 : 1 will be useful in determining how efficiently we can use our crowbar.
Question 3:
F
2
F
1
A B
1.4
1
550 N
Let the tensions in the ropes be F
1
and F
2
Neglecting the weight of the scaffold, at equilibrium,
&F$ %$ 0$ '() and
&"$ %$ $0
!$F
1
+ F
2
–550 = 0
F
2
× 2.4 – 550 × 1 = 0 (taking
(2)
as (–))
Module 2 – Understanding motion
2.119
F
1
+ F
2
= 550
!$ F
2
2.4 = 550 1
F
2
=
2.4
= 229 N
Since F
1
+ F
2
= 550
F
1
= 550 – 229
F
1
= 321 N
! The rope at B has a tension of 321 N and the rope at A has a tension of 229 N.
Question 4:
It is easier to remove a subborn tyre nut with a long handled spanner than a short handled one. Since Torque = F#$r and as can be seen in the diagram, if a spanner double the torque.
axis
r
τ =
F
⊥
×
r
2r
τ =
F
⊥
× 2
r
F
So a longer handled spanner would give a larger torque
F
2.120
TPP7160 – Preparatory physics
Activity 2.17
Question 1:
(a) st air s
4 m
4.5 m
Work done against gravity
= weight vertical distance raised
= 210 4
= 840 J
(b) If you rushed up the stairs carrying the books, the total weight raised
= your weight + weight of books
= 75 9.8 + 210
= 735 + 210
= 945 N
The weight is raised through 4 m. Hence work done
= 945 4
= 3780 J
If this work is done in 15 s
Power =
3780

15
= 252 Js
–1
or W
Module 2 – Understanding motion
2.121
Question 2:
F cos60°
60º
F
dock
Resolving F along the horizontal (in the direction of the motion),
F cos 60° = 250 cos 60,
The boat has to move 25 m to its mooring in the horizontal direction along the dock.
Thus work done on the boat = F cos 60° × s
= 250 cos 60° × 25
= 3125 J
Question 3:
Weight of bushwalker = 710 N
Work done in climbing 9.3 m = mgh
= 710 9.3
= 6603 J
If the climb takes 30 secs, average power = total work done
time taken
Power =
6603

30
Js
–1
$ % 220 W
Activity 2.18
m = 1 kg s = 400 m
2.122
TPP7160 – Preparatory physics
Mass of lunch m = 1 kg
Weight of lunch W = mg = 1 × 9.8 N
(a) Work done by gravity = F × s
= 9.8 400 J
= 3920 J
(b) This is the amount of gravitational potential energy lost.
Question 2:
(a) Since P.E. = mgh, the car that is twice as high would have twice the P.E. of the other car.
(b) Similarly the car with twice the mass would have twice the potential energy of the other, since potential energy is proportional to mass and height above ground.
Question 3:
The gravitational P.E. of the pumped water =
mgh
where h = 300m and g = 9.8 ms
–2
To store 2 10
13
J of energy, an amount of water of mass m must be pumped
300m above the level of the river.
!$
m 9.8 300 = 2 × 10
13
!
m
=
2 10
13
9.8
300
= 6.8 10
9
kg
Activity 2.19
Question 1:
(a) The skiers’ velocity v = 220 km h
–1
=
220 1000
ms
–1
3600
= 61 ms
–1
K.E. of skier = mv
2
2
=
1
2
(72) (61)
2
= 134000 J (or 134 kJ or 1.34 × 10
5
J)
Module 2 – Understanding motion
2.123
(b) The runner’s velocity v = 44 km h
–1
=
44 1000

3600
= 12.2 ms
–1
K.E. of runner
1
= mv
2
2
=
2
72 (12.2)
2
= 5358 J (or 5.4 kJ)
(c) Comparing the K.E. of the two athletes, it is found that the K.E. of the skier, is twentyfive times that of the runner i.e.
134000

5358
= 25.0
Question 2:
(a) Let the original velocity of the car be v ms
–1
and its mass be m

2
2
If v is increased by a factor of 4 new velocity
v
1
= 4 v
New K.E.

2
1
2
=
2
m(4 v)
2
=
1

2
m × 16v
2
1

2
2
)
The new K.E. is now 16 times what it was at the original velocity.
(b) To bring the car to rest the brakes must transform all the car’s K.E. to other forms of energy mainly heat.
If the braking force
Then in the first case F d
1
= F Newtons
=
1

2
mv
2 where d
1 distance
is the stopping
In the second case, if d
2 constant)
is stopping distance (braking force remaining
F d
2
=
1
16 ( mv
2
2)
!
d
2
: d
1
= 16:1
Provided the braking force is constant in both cases, the stopping distance increases by a factor of 16. Therefore the work that the car’s brakes need to perform to stop the car at the new speed is 16 times more than the work done previously.
2.124
TPP7160 – Preparatory physics
Activity 2.20
Question 1:
(a) Mass of hail m = 1.5 kg
Height above ground h = 8 m
Potential energy P.E. = mgh
= 1.5 9.8 8
= 117.6 J
On reaching the ground this is converted to kinetic energy
So K.E. on striking ground = 117.6 J
(b) If v is the velocity on impact, K.E. = mv
2
2
117.6 = × 1.5 v
2
2
v
2
=
117.6
2
1.5
= 156.8
!$
v = 12.5 ms
–1
Velocity of impact is 12.5 ms
–1
.
Question 2:
Mass of rocket m = 5 kg
Let h be the height it is able to reach
Then at this height P.E. = mgh
= 5 9.8 h
At this height all initial K.E. is converted to P.E.
so 2176 = 5 9.8 h
h =
2176

5 9.8
= 44.4 m
!$ Maximum height the rocket can reach is 44.4 m.
Module 2 – Understanding motion
2.125
Question 3:
Suppose the mass of the ball is m kg. At its lowest point, its velocity is maximum while at its highest point its velocity is momentarily zero.
Then its K.E. at the lowest point of its swing =
At the end of the swing this is all converted to P.E.
1

2
m (4)
2
J
If h is the maximum height in metres to which the ball rises,
K.E. = P.E.
1
 ×
m
× 16 =
2
m
× 9.8 × h
! h =
8

9.8
= 0.8 m
v
=
400 cms
− 1
Question 4:
Mass of nitrogen molecule
Its K.E. at sea level is 6.2 10
–21
J
m = 4.7 10
–26
kg
If v is the average velocity, then K.E
=
1
2
mv
2
6.2 10
–21
= × 4.7 10
2
–26
× (v
2
)
v
2
=
2 6.2

4.7
10
10
– 21
– 26
= 263829
v
= 263829
!
v
= 514 ms
–1
! Average velocity of a nitrogen molecule is 514 ms
–1
If the molecule could rise and reach a maximum height h metres
P.E. = mgh
= 4.7 10
–26
9.8 h
It’s K.E. =
2
4.7 10
–26
(514)
2 since final P.E. = initial K.E.
m
gh =
1
 m
2
v
2
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TPP7160 – Preparatory physics
9.8 h =
514
2

2
h
=
514
2

2 9.8
= 13479 m
Question 5:
u
= 0
v
2
35m
25m
v
1
(a) Let the mass of carriage = m kg
At top of first hill P.E. = m 9.8 35
At the bottom of the hill all of this is converted into K.E.
If v
1
is the resulting velocity at the bottom of the hill,
K.E. = P.E.
2
v
1
2
=
m gh
1

2
v
1
2
= 9.8 35
v
1
2
= 19.6 35
At the bottom, initial
v
1
= 26.2 ms
–1
(b) As the carriage ascends the next hill, all this energy is transformed into P.E.
and K.E. (since energy is conserved and the second hill is lower than the first).
At the top of the 25 m hill P.E. = m 9.8 25 and K.E. =
1
2
mv
2
2
(If v
2
is the velocity at the top of the second hill)
K.E. = m (26.2)
2
2 and initial P.E. = 0
Module 2 – Understanding motion
2.127
Since energy is conserved,
Total K.E. + P.E. = constant or
2
v
2
2
+
m
gh =
2
2
v

2
=
/

26.2
2
v
0
.
1
2
2
– 9.8
25
2
v
2

2
= (343 – 245)
= 98
v
2
2
= 196
v
2
= 14 ms
–1
!$The carriage’s velocity at the top of the second hill = 14ms
–1
.
Note: We could also have used the P.E. on the top of the 1st hill
1

2
mv
2
2
+ mgh
2
= mgh
1 whereby
1

2
v
2
2
=
g(h
1
– h
2
)
v
2
2
= 2g(35–25)
= 2 × 9.8 × 10
v
2
= 196
= 14ms
–1
Question 6:
m = 2000 kg
100 m
u
= 12 ms
1
h
100 m
25º
The truck has both potential and kinetic energy. If it is brought to rest in 100 m the loss in P.E. and K.E. is given by
P.E.
= mgh (where h = 100 sin 25°)
= 2000 9.8 100 sin 25, = 828,300 J
K.E. =
1

2
mv
2
= 2000 (12)
2
2
= 144,000 J
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TPP7160 – Preparatory physics
Total energy to be dissipated in brakes = work done to stop the truck
= (828300 + 144000) Joules
= 972300 J
This energy has to be dissipated by the brakes over a distance of 100 m.
Let F Newtons be the force applied by the brakes.
Then
!
Work = Force × distance.
F 100 = 972300 J
F
= 9723 N
Question 7:
50 kmh
–1
=
50 1000
ms
–1
3600
= 13.9 ms
–1
Let the mass of the car be m
Then K.E. =
1

2
mv
2
If the braking force
=
2
m 13.9
2
J
= F Newtons
Then since it skids 15 m, work done by the brakes in bringing it to a stop
Force × distance = Change in K.E.
!
F 15
1
= 0 – mv
2
2
= – m 13.9
2
2
= –(
1
2
13.9
2

15
m)
= –(6.4 m) N
(negative sign for opposite direction)
Assuming this force to be constant, we can use this formula in the second case.
Here
v
= 150 km h
–1
= 41.7 ms
–1
K.E. =
1
2
m (41.7)
2
Module 2 – Understanding motion
2.129
If d is the distance skidded under the braking force, then
Work done by F = change in K.E.
6.4m d = × m(41.7)
2
2
d
=
' )
2 6.4
2
1 136 metres
An easier way would be to say: The car travels 3 times as fast, therefore it will have 9 times the energy and take 9 times as long to stop.
This is shown mathematically below
Work done = change in K.E.
if if
F × d =
mv
2
2
v
1
= 50
F × d
1
=
1

2
m (50)
2
v
2
= 150
d
1
= 15
d
2
= ?
F × d
2
= m (150)
2
2 dividing equation (2) by equation (1)
F d

F d
2
1
=
1
m 150
2
' )

2
m 50
)
2
2
=
50
2
2
= 9
d
9
d
1
= or
d
2
= 9 × d
1
(1)
(2)
Question 8:
An object can possess potential energy without having momentum, if it is at rest, e.g. a piledriver poised above a pile.
However, since any object which has momentum must be moving,
(momentum = mv) such an object must also possess kinetic energy
(K.E.
=
1
2
mv
2
)
2.130
TPP7160 – Preparatory physics
Solutions to posttests
Posttest 2.1
Question 1:
Complete this summary of the previous section.
The rate of change of displacement with time is called velocity.
Velocity is a vector quantity i.e. both its magnitude and direction must be specified.
The magnitude of the velocity of a body is called its speed.
Speed is a scalar quantity.
If we divide total displacement by time taken the result is the average velocity in the direction of displacement. (displacement is also a vector as it is a change in position).
The velocity of a moving object at any particular time is called its instantaneous velocity.
The SI unit of velocity is metres sec
–1
.
If velocity is changing with time, the rate at which it changes is called the
acceleration of the body.
The SI unit of acceleration is metres sec
–2
.
If the velocity changes by a constant amount every second, the acceleration is said to be constant.
Gravitational acceleration at any particular point on the earth’s surface is an example of constant acceleration.
If an object undergoes a displacement s in time t, starting with an initial velocity,
u, and acceleration a in the direction of the displacement and after time t the velocity is v, the three equations of motion that relate these variables s, t, u, v and
a are:
1. v=
u + at
2. s=
ut +
1
 at
2
2
3. v
2
=
u
2
+ 2as
If the body starts from rest these equations become
1. v=
at since u = 0
2. s=
2
3. v
2
= 2as
2
Module 2 – Understanding motion
2.131
KE
+
PE = max
KE
= 0 ,
PE v
= 0
v
PE = 0 max
KE = max
Question 2: Can a body have a constant speed and a changing velocity? Can a body have a constant velocity and a changing speed? Explain your answers.
A body can have a constant speed and a changing velocity, because speed is the magnitude of a velocity, which may change direction while still not altering magnitude. An example is the motion of an object around a circular path at constant speed, where the direction is constantly changing along the circumference of the circle.
However, a body cannot have a constant velocity and a changing speed, because for the velocity to be constant, both its direction and its magnitude (i.e. its speed) must be constant.
Question 3: If a ball is thrown straight up into the air, describe the changes in its velocity and acceleration.
If a ball is thrown vertically up in the air, it has an initial velocity imparted by the thrower. As soon as it is thrown it experiences an acceleration downwards of 9.8 ms
–2
and so is eventually brought to rest at a certain height when KE = 0. It undergoes acceleration when it falls vertically downwards at 9.8 ms
–2
, its velocity increasing until it reaches the ground with a maximum final velocity.
Posttest 2.2
Question 1: Complete the following summary.
Newton’s first law of motion states:
Any object continues in its state of rest or uniform motion in a straight line, unless acted upon by an unbalanced force.
Force is a vector quantity, i.e. to specify a force we must give both its magnitude and direction. For acceleration of a body to occur there must be an unbalanced force acting.
The tendency of a body which causes it to resist any change in its state of rest or motion is called its inertia.
Newton’s second law of motion states:
The acceleration of a body is directly proportional to the resultant force, and
inversely proportional to the mass of the body.
In symbols, with suitable units Newton’s second law may be written as:
F
=
ma.
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TPP7160 – Preparatory physics
The unit of mass in the SI system of units is the kilogram.
The unit of force in the SI system of units is the Newton.
One Newton is the force which when applied to a mass of one kilogram results in an acceleration of one metre per second per sec.
The weight of a body is the force of gravity exerted on that body.
In symbols
w = mg
Momentum is the product of mass and velocity.
It is a vector quantity and is measured in Ns or kgms
–1
.
Impulse is a product of force and time.
The change in momentum of an object is equal to the impulse applied to it.
In any interaction between two or more isolated objects the total momentum does not change. This is referred to as the law of conservation of momentum.
Newton’s Law of Universal Gravitation states every body in the universe attracts every other body with a force which is directly proportional to the product of the masses of the two bodies, and inversely proportional to the square of the distance between the two bodies.
If m
1
, m
2
are the masses of the bodies and d is their distance apart,
F
=
Gm
1
m

d
2
Constant.
where G is a constant called the Universal Gravitational
Newton’s third law states that if two bodies A and B interact the force exerted by the body B on body A is equal and opposite to that exerted by body A on body
B.
Question 2:
If an unbalanced force F, acts on an object of mass, m, producing an acceleration, a, for a time, t, these quantities are related by:
(Choose one of the following.)
(a) F
=
a m
Question 3:
Which of the following statements are incorrect?
(b) The weight of an object has the SI units, the kilogram.
Question 4:
Which of the following is a true statement concerning friction?
(a) Frictional forces are independent of the size of the surface areas of contact.
Question 5:
If an unbalanced force acts on an object for a short time then:
(a) The impulse of the force equals the change in momentum of the system.
Question 6:
What do we mean by the components of a force and why is it useful to resolve a force into its components? Give an example to explain your answer.
Module 2 – Understanding motion
2.133
Because force is a vector, it is possible to resolve a force into two component forces at right angles to each other, so that their vector sum represents the original force in magnitude and direction. These two forces are called the components of that force. They are not unique. Indeed their magnitude and direction are determined by the angle one of them makes with the direction of the original force.
Given Vector F
One Example of resolving the vector F into its #$ components
F sin θ
θ
F cos θ
F sin φ
This is another example of resolving the same vector F into its #$components
φ
F cosφ
Components are useful when we want to find the resultant of a number of forces acting on a body, or the resultant force acting in a certain direction.
2.134
TPP7160 – Preparatory physics
F n
F
θ
F cosθ
W sin φ
φ
W
W cosφ
Eg. (i) The weight W can be resolved in to W sin 2 along the slope and W cos2 # to the slope.
Then the normal reacton F
n
= W cos 2 .
(ii) The effective part of the force F pulling the object up the slope = F cos
3$
For example suppose a body of mass M kg is resting on a rough plane inclined at 2, to the horizontal and that a force of F Newtons is applied up the plane at an angle 3 to the plane.
A force diagram may look like this:
F n
a pplied force
F
Fcos q q
Wsin
Friction
F f
Wcos weight of body
W
Module 2 – Understanding motion
2.135
Question 7:
Can a truck have the same momentum as a bullet? Explain your answer.
The momentum of a body is the product of its mass and its velocity.
Suppose that the bullet has a mass of 20g and a velocity of
1100 ms
–1
.
Then its momentum is
0.02 1100 = 22 kgm s
–1
If the truck has a mass of 2 tonnes i.e. 2000 kg then for its momentum to be the same as that of the bullet, its velocity would have to be
2000
= 0.011 ms
– 1
In short, in general, unless the bullet is very large, or the truck is moving very slowly, the momentum of the truck will exceed that of the bullet.
2.136
TPP7160 – Preparatory physics
Posttest 2.3
Question 1:
Uniform circular motion occurs when an object moves in a circular path with uniform speed.
Angular velocity is defined as
angular displacement
time taken
=
" #

"
t
.
The unit for angular velocity is radians/sec.
The uniform linear velocity v, of a body moving in a circle of radius r, is related to the angular velocity 4, by the equation v
=
r
Since the speed on the circle is uniform, there is no linear acceleration, but the velocity is changing in direction.
The acceleration of the moving object is directed towards the centre of the circle and is called the centripetal acceleration.
A body projected with velocity, v, at an angle to the horizontal has two component velocities; v cos horizontally and v sin vertically. Each of these components can be considered independent of the other.
Question 2:
If a torque, !, acts on an object it can produce:
(b) a change in the angular velocity.
Question 3:Many science fiction stories are based on space stations which appear to look like rotating wheels around a central axis. Why would this design simulate gravity and which way would be down?
F c
F c v
P
1
P
2
P
Module 2 – Understanding motion
2.137
If the space station is rotating about the central axis, it and everything it contains must be acted upon by a centripetal force to keep them on the same circular motion. If the rim of the space station consists of a large hollow tube, a person in that tube will be acted upon by the same centripetal force F c
. This force is needed to keep the person ‘P’ revolving at the same speed v as the wheel. Remember without F c
there is no centripetal acceleration and therefore no change in velocity
(change in direction) and P would just move straight forward instead of in a circular path as shown by P
2
and P
1
.
In a rotating frame of reference, such as for a person inside the wheel, apart from the force F c
exerted on the person by the outer rim of the wheel, the person experiences a ‘force’ acting outwards towards the rim away from the centre. The floor of the wheel presses on the person and the person reacts by pressing back on the floor. Without a support we would feel weightless anywhere.
Thus the person would feel an artificial gravitational force which would act radially away from the centre of station.
The relevant equation is
mg
/
= m
2
r
where m is the mass of the body concerned, or
g
/
=
2
r
The magnitude of the artificial acceleration due to gravity is therefore a function of both the angular velocity of rotation of the space station, and the radius of the circle. For g
/
to be 9.8 ms
–2
, there are many possible combinations for the product of
2 and r.
Question 4:
A golfer hits two shots as shown. Which hit
(a) has the shorter flight time;
(b) has the larger flight time;
(c) has the larger horizontal velocity.
A
q
=
60°
30
°
B
Time of flight is dependent solely on the vertical component of the velocity. To answer this question we must assume that the velocity imparted to the ball in each case is the same.
2.138
TPP7160 – Preparatory physics
Thus if v is the velocity, in case A, the vertical component is
v
A
= v sin 60$ or 
2
3
v = 0.866 v
In case B the vertical component is
v
B v
= v sin 30$ or = 0.5 v
2
Since in case A the initial vertical velocity is the greater, the time of flight will be larger for case A.
So the hit B will have the shorter flight time and hit A will have the larger flight time.
The horizontal velocities are
v
A
= v cos 60$ or
2
v
B
= v cos 30$ or
2
3
v
B therefore has the larger horizontal velocity
Module 2 – Understanding motion
2.139
Posttest 2.4
Question 1:
When work is done on a system the energy of the system is increased.
The energy of an object determined by its position is termed potential energy, while that which is determined by its motion is termed kinetic energy.
Energy is a scalar quantity and is measured in Joules (in SI units). When an object falls freely to the earth it gains kinetic energy and loses potential energy.
In a closed isolated system energy can change form but the total amount of energy is constant.
This principle is called the principle of conservation of energy.
If an object is moved a distance, s, against a force, F then work is done and is the product of force F and distance s. If the force is at an angle% , then the work is determined by Fs cos . Work is measured in Joules (SI units) and is a scalar quantity.
Work done over a set time interval is called power and is measured in Watts (or
Js
–1
) (SI units).
Question 2:
When work is done then: (Choose one of the following.)
(a) energy is transformed from one form to another
Question 3:
People commonly speak of ‘energy consumption’. Is this term appropriate or inappropriate? Explain.
Although we talk of energy consumption what we really mean is energy transformation, because energy cannot be destroyed or consumed. For example, if we wish to boil water to make coffee we may put a litre of water at 20° C into the electric jug and bring it to the boil. The electricity flowing through the heating coil is converted to heat. The molecules of water gain kinetic energy as a result of the heating, and this is seen by a rise in temperature of the water. If our jug is a 2 kW jug and it is switched on for 5 minutes we have converted
2 & 1000 & 5 × 60 Joules of electrical energy to heat energy. Our meter will record this as ‘consumption’ of electricity, because the electricity commission is only concerned with how much we used or consumed.
Question 4:
If you drop a ‘superball’ from a height could it bounce higher than this height? Explain.
No matter how superbly elastic the superball material is, some of its kinetic energy is converted to heat in the ball as the material is compressed and stretches again. Hence the superball loses K.E. on impact, and will not rise as high when the remaining K.E. is converted again to P.E.
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