Download 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
Module 2 – Understanding Motion
In our real world, motion is fundamental – everything moves. We walk, run, jump, breathe.
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
=
distance
-----------------------------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
=
40 kilometres
-------------------------------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)
Vector
(magnitude and direction)
distance
(e.g. 10 m)
displacement
(e.g. 10 m east)
speed
(e.g. 10 ms–1)
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:
Vy to be the vector representing your velocity.
Vw to the vector representing the velocity of the wind.
Vwy to be the vector representing your final velocity.
thus
Vy + Vw
=
Vwy
Consider if you were running into the wind. It is harder running when the wind is against you
because
Vwy
=
Vy – Vw
This can be represented using vectors.
Vy
Vw
In the olympics wind ‘assistance’ is considered when deciding whether new records are
allowed.
TPP7160 – Preparatory physics
2.4
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.
Vw
=
wind velocity
r
you
Vy =
ty
oci
l
t ve
n
a
ult
res
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
2.72 + 6.72
=
r2
r2
= 2.72 + 6.72
r2
= 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.
2.7
tan # = ------6.7
=
0.4030
# = 21.9$
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
=
change in velocity
----------------------------------------------------------------------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 two-second period then
Acceleration
=
80 km/hr – 60 km/hr ' change in velocity(
------------------------------------------------- -------------------------------------------% change in time &
2 second
=
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 – 80
-----------------2
=
–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 merry-go-round you can feel the changes that are taking place as you swing
around. The speed (or the magnitude of the velocity) of the merry-go-round 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 =
a
=
change in velocity
-------------------------------------------change in time
v–u
----------t
(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 =
u+v
-----------2
=
distance travelled
------------------------------------------ (average velocity is used when velocity
change in time
changes over the period of time)
s
t
Making s the subject of the formula above,
!"
s
=
+v
'u
------------( t
% 2 &
s
=
+ ) u + at *(
' u--------------------------- t
%
&
2
s
=
ut + ut + at 2
-----------------------------2
s
=
1
ut + --- at2
2
substituting for v
=
u + at (from equation 1)
=
v–u
----------t
(Equation 2)
This is referred to as the 2nd equation of motion.
Finally consider the equations
s =
) u + v *t
-----------------2
a
and
we rearrange both equations to make t the subject of the formula
t
=
2s
-----------u+v
and
t
(v + u) (v – u)
=
=
v–u
----------a
2s
v–u
equating for t we have, ------------ = ----------u+v
a
2as(cross-multiply)
v2 – u2 =
2as
!
u2 + 2as
This is referred to as the 3rd equation of motion.
v2 =
(Equation 3)
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
=
1
ut + --- at2
2
v2 =
u2 + 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
a
(acceleration due to gravity)
=
9.8 ms–2
t
(time)
=
4s
(dropped)
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
An image of what’s
going on is always
helpful.
t = 4s
As we visualise the
event, the actual
concept here is
acceleration.
v=?
2.10
TPP7160 – Preparatory physics
We know that acceleration
a
=
v–u
----------t
9.8
=
v–0
----------4
or
!
or
change in velocity
-------------------------------------------time
=
(a = g in this case)
9.8 × 4 = v
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
v
=
0 + 9.8 " 4
=
39.2 ms–1
(relating the variables we are
interested in.)
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)
v–u
As you can see I chose to use a = ----------- in the example above because it describes our
t
central concept of acceleration.
Remember the 3 equations of motion are simplified versions describing the concepts of linear
v–u
motion. The first equation of motion v = u + at came from simplifying a = ----------- .
t
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
displacement
average velocity = ----------------------------------time
s=?
t= 4s
v = 39.2 ms
displacement = average velocity " time
–1
g = 9.8 ms–2
u+v
s = % ------------& " time
# 2 $
0 + 39.2
= % -------------------& " 4
#
2 $
= 78.4 m in the direction of the
acceleration.
OR we could have used the equation of motion
s =
o
s
ut
1
+ --- gt2where u = 0,
2
1
= --- × 9.8 " 42
2
= 78.4 m
o
v2 =
or
!
u
2
+ 2as
(39.2)2 = 0 + (2 × 9.8 × s)
or
2as = v2
2
v
s = -----2a
2
s = 78.4m
39.2
= ---------------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
Time of flight = 2t
(time to reach maximum height = time to
fall down to starting point)
Height of ball
=
t
=
10
=
u=0
10 m
t
s
s
t
C
v=?
t2
=
1 2
ut + --- at ,
2
1
2
--- " 9.8 " t
2
20
t
s
t
0
9.8 ms–2
v
=
u + at,
v
=
9.8 "!1.42
v
=
13.9 ms–1
u = 0,
a = 9.8 ms–2
t = 1.42 s
2.04
=
1 2
ut + --- at
2
=
1
2
--- " 9.8 " t
2
u=0
20 m
=
=
Final velocity (v)
time it takes to drop
through a displacement of 10m
t
= 1.43 s
time of flight = 2t
= 2.86 s
s
u
a
2.13
!After falling back to its starting
point, its final velocity = 13.9 ms–1
v
=
u + at
v
=
9.8 "!'()'
=
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
distance
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 post-test to see how you are going so far.
Module 2 – Understanding motion
Post-test 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.
2.15
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/ that of earth?
6
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:
Upward force exerted by desk
Backward force exerted
by friction between book
and desk
Table’s surface
Forward force e
Forwar
by hand
Downward force exerted by gravity
Actual diagram
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 non-moving bodies?
2.2.4 Forces acting horizontally and vertically
Consider again your study book sitting on your desk.
Upward force exerted by desk
Backward force exerted
by friction between book
and desk
Table’s surface
Forward force exerted
Forwar
by hand
Downward force exerted by gravity
Actual diagram
A ‘free body diagram’
Module 2 – Understanding motion
2.21
The forces acting on it would be as shown – downward force (Fw) due to gravity (known as
weight = mg) and an upward force (Fn) 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
or
mg + Fn
( is the symbol for summation)
=
0
Remembering that forces are vectors with direction, Fn and weight have equal magnitude but
opposite directions
i.e.
Fn = –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 (Fp) and the frictional
frictional force (Ff) between the book and the surface. This means that there are four forces
acting on the book.
Fn
Fp
table’s surface
Ff
Fw
Since the book is being pushed forward, there must be a net force in the forward direction.
Summing horizontal forces,
F =
Fp + Ff
=
Fnet
Fnet (in this case, Fp > Ff 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.
TPP7160 – Preparatory physics
2.22
The result can be summarised by the equation:
Ff
=
" s Fn
Ff is the magnitude of the frictional force
where
Fn 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, Fk = " kFn where Fk is the force of kinetic friction "k is the coefficient
of kinetic friction.
The coefficient of friction being a ratio of the magnitude of two forces is a unit-less 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 are much
smaller.
Note that if Fp = Fk i.e. net force = 0 there is no acceleration, and the book moves forward with
constant velocity.
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
Ff
Fn
becomes
Fpush
Ffrictional
Actual
Fw
Fp
mg
Free body diagram
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
Frictional force, Ff
=
15.68 N.
=
1.6 # 9.8
=
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:
Ff
=
" s Fn
Fn
=
Fn
=
$ Fn =
Ff
=
15.68 N
Ff
----"s
"s
=
0.35
15.68
------------0.35
Fn is force exerted by the wall and is unknown.
where
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
2.24
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
F sin q
Vertical component of F=
q
F cos q
F sin!&!'!10 sin 35%
Horizontal component of F=
=
5.73 N
F cos &!'!10 cos 35%
=
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
C
30 N
ulta
nt
20 N
30º
D
B
30 N
20 N
69º
Given
Figure 2.1:
A
Res
140º
B
Module 2 – Understanding motion
2.25
The resultant force on B is 30 N and directed at 69° to
the horizontal.
Res
u
ltan
t
20 N
Figure 2.2
30 N
69º
30º
B
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
The horizontal component for the resultant vector, Rx is the sum of all the horizontal
components, 25.98 + (–15.32) = 10.66
(Note: negative sign denotes that it is in
the opposite direction, i.e. to the left)
The vertical component for the resultant vector, Ry 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
Ry = 27.86
Rx = 10.66
Using Pythagoras’ rule, the magnitude of the resultant vector is:
R
=
2
2
Rx + Ry
10.66 ! 2 + 27.86 ! 2
=
=
29.82
=
30
30º
2.26
TPP7160 – Preparatory physics
Using trigonometry, the direction of the resultant vector is given by:
Ry
27.86
tan & = ----- = ------------- = 2.61
10.66
Rx
& = 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,
A B̂ D
#
B Â C
=
=
=
=
140° – 30°
110°
180° – 110°
70°
=
b2 + c2 –2bc Cos Â
=
202 + 302 – 2(20)(30) cos 70°
=
400 + 900 – 410.42
=
889.57
Applying the Cosine Rule, where
a2
C
b = 20 N
a
70º
A
c = 30 N
889.57
=
a
B
#
=
29.8 N
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.
Using
sin B̂
----------b
=
sin Â
----------a
sin B̂
=
b sin Â
---------------a
20 sin 70"
= ----------------------29.8
#
B̂
$ %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
Horizontal Component (x)
Vertical Component (y)
80 cos 0"
=
80
80 sin 0"
=
0
100
100 cos 45"
=
70.7
100 sin 45"
=
70.7
110
110 cos 150"
=
–95.3
110 sin 150"
=
55
160
160 cos 200"
=
–150.4
160 sin 200"
=
–54.7
Ry = 71N
Rx = –95 N
Resultant
Note that Rx 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
=
=
=
2
2
Rx + Ry
Ry = 71 N
–95 ! 2 + 71 ! 2
118.6 N
Rx = -95 N
Figure 2.3
2.28
TPP7160 – Preparatory physics
Direction of R,
tan & =
tan & =
Ry
----Rx
71
--------–95
tan & =
–0.747
&
143"
=
Resultant, R
Ry
θ
Rx
original reference
line
Figure 2.4
since x-component is negative and y-component 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 Ff represent the frictional force, and Fn represent the force exerted by the earth
normal to the plane
Fn
0N
10
30o
Ff
Fn
mg
Ff
mg
(a) In the horizontal direction
The frictional force must be equal to the horizontal component of the pulling force.
Therefore Ff
=
100 cos 30"
=
86.6 N
(b) In the vertical direction
The downward force (weight) must be equal to the sum of the upward forces.
mg
=
Fn + F sin 30"
40 ' 9.8
=
Fn + 100 sin 30"
Fn
=
342 N
Therefore
# 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 sin 51 °
F
F
51°
51°
Bookshop
229 N
F cos 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
=
229
-----------------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.
B
θ
mg sin θ
α
θ
θ
mg mg cos θ
α
( + A = 90"
and ( + B = 90"
# B̂ = Â = &
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 Fp is mg sin &.
Fp
θ
Fp
=
562 sin 20"
=
192 N
20°
20°
mg
Fn
Force perpendicular to the inclined plane Fn is mg cos &.
Fn
=
562 cos 20"
=
528 N
Activity 2.7
Question 1: A car weighing 1.2 ' 104 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 well-oiled, friction-less 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
Ff
=
) Fn
where s is the coefficient of static friction, Ff is the frictional force which is equal to W sin !
and Fn 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 snow-covered 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. cross-sectional 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 cross-sectional area of their parachute or the position of their limbs and body.
Drag
or
air resistance
height
wt
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 moon-earth
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 Nm2kg–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 1 m 2
----------------d2
where
=
6.67 # 10 –11 # 5.3 # 5.3
--------------------------------------------------------12
=
0.000 000 001
=
1 # 10–9 N (a very small force.)
m1
=
mass 1 in kg
=
5.3 kg
m2
=
mass 2 in kg
=
5.3 kg
d
=
distance in m =
G
=
universal gravitational constant
1m
2. Calculate the mass of the earth, assuming it is a sphere of radius 6370 km.
Solution:
Let m1 be the mass of the earth, and m2 the mass of an object on its surface. The weight of
force on that object will be m2g.
So given
F
=
Gm 1 m 2
----------------d2
=
Gm 1 m 2
----------------d2
m2
object
d
m2 g
Simplifying
m1
=
g # d2
-------------G
m1
Earth
2.38
TPP7160 – Preparatory physics
=
9.8 # & 6370 # 10 3 ' 2
----------------------------------------------6.67 # 10 –11
=
6.0 # 1024 kg.
(where d is the radius of the
earth, assuming the object
is relatively small).
$% Mass of the earth is 6.0 # 1024 kg.
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?
(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 # 1024 kg,
while Mars has a radius of 3 440 km and a mass 0.11 that of
earth:
Question 2:
(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 # 1027 kg and a radius of
7.14 # 107 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
=
(v
m -----(t
(v
a is acceleration and equals, ------ , a change in velocity over a
(t
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.
(p
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.
2.40
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.
padding on the dash-board of a vehicle.
Low force, long time
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,Pi
(b) Final momentum,Pf
=
=
mu
mv
=
=
0.04 # 40
=
0.04 # –40 =
1.6 kg ms–1
–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
(c) The change in momentum, (P =
2.41
Pf – Pi
=
–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 # 104 ms–1 and its mass is 6 # 1024 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 # 105 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
Question 4:
If a child (20 kg) is an unrestrained passenger, and also stops
in the same time, what average force is acting on the child?
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 ice-skaters.
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
m2
m1
m1
m2
u1
v2
v1
u2 = 0
2.43
Elastic
Collision
After collision
Before Collision
v
m2
m1
m1
u1
v
Inelastic
Collision
u2 = 0
After collision
Before 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)
bullet)
=
momentum (block and
(note: this is an inelastic collision)
In SI units –
m1 u1 + m2 u2
=
(m1 + m2)v
0
=
10.006
0.006 u1
=
4.0024
u1
=
667 ms–1
0.006 u1 + 10
0.40
! Initial velocity of the bullet is 667 ms–1.
(where u1 is the initial velocity of the bullet)
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
7
–1
kg and speed 1.5
10 ms , 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
Post-test 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 m1, m2 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
=
1
--- ma2
2
(c) F
=
m
a
(d) F
=
m
a/t
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.
2.48
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 Non-linear 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 y-direction)
Using the equation of motion s =
ut + 1/2 at 2
s = 1/2 gt2
t=
=
2s
----g
2 50
--------------9.8
where
s = vertical displacement (50 m)
u = initial vertical velocity (zero)
a = acceleration due to gravity, (9.8 ms–2)
= 3.18 s
OR substituting into the equation we get
Module 2 – Understanding motion
50 = 0 + 1/2
9.8
t2
2 50
t2 = --------------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.
2.51
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.
uy
–1
s
u=
20
m
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,
The initial vertical component of velocity will be,
ux =
u cos 40°
=
20 cos 40°
=
15.32 ms–1
uy =
–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) + 4.9t2
where s
9.8
t2
= 50m
uy = –u sin 40°
a = 9.8 ms–2 (using positive sign for
downward direction)
!
4.9t2 –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
2
t
=
– $ – 12.86 % & $ – 12.86 % – $ 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
= uxt
s
= u cos 40° × t
(horizontal component of velocity is a constant
since it is not affected by gravity)
= 15.32 × 4.8
= 73.5 m
(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 lead-weighted 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.
Linear velocity at A (v)
A
s
q
angular velocity (ω)
B
r
In moving from A to B along the arc, the linear displacement along the arc is s while the
angular displacement is '($clockwise). If the radius is r, then the angular displacement is
defined as
" or
s
'( = r
arc length#
--------------------------radius !
the unit for angular
displacement '(is radians
In moving through 360º, the arc length = circumference
2$r
'( = --------r
Hence
=
2 $ radians
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,
!
2$
& = -----T
rad s–1
2$r
v = --------- ms–1
T
........................................equation 2
.........................................equation 3
2$
2$r
------ = --------&
v
v = r&
.........................................equation 4
This equation relates the linear velocity to the angular velocity.
Module 2 – Understanding motion
2.55
Since T is the time taken for 1 revolution, and frequency f is the number of revolutions per sec
1
' f = --T
1
Substituting for --- in ()we have
T
or
&
=
2$f
2$
T
& = ------ = 2$f
........................................ equation 5
Remember the merry-go-rounds that were common in playgrounds once? If you rode on one
of these the faster the equipment rotated the faster your tangential or linear velocity was when
you tried to jump off.
Consider two merry-go-rounds – one for young children and one for older children. The
linear velocity on exit than the smaller apparatus is usually slower than that on the larger
apparatus. This is always true even if you are rotating at the same speed as rotational speed is
not dependent on distance from the centre. The linear velocity, however, is dependent on r, as
shown by v = r *
V = Rω
θ
R
If the angular velocity changes, then an object must be accelerating along the circle.
2.56
TPP7160 – Preparatory physics
Angular acceleration
change in &
----------------------------- (measured in radians per second per second)
t
=
&2 – &1
, = ------------------- ........................................ equation 6
t
The corresponding linear acceleration which is always directed towards the centre is called the
centripetal acceleration, and is given by
v2
ac = ----- = r&2
r
ac = r&2
or
v
ac
r
(using equation 4)
Think about these questions:
•
•
Why do clothes in the rapidly spinning tub of a spin dryer ‘stick’ to the sides of the tub but
falls to the bottom when the tub stops spinning?
Why does the water not fall out of a can when it is being spun?
Example:
A spin drier (of a washing machine), with a diameter 30 cm, revolves at 900 rpm.
(a) What is the angular velocity of the spin drier in radians per second?
(b) Calculate the linear velocity of the spin drier.
(c) What is the centripetal acceleration of the spin drier?
Solution:
(a) The angular velocity is 900 revolutions per minute.
900 rpm =
900
--------60
=
15 rev per second
=
15 + 2$ rad s–1
=
94 rad s–1
(1 rev = 2$ radians)
' Angular velocity is 94 rad s–1
(b) Linear velocity,
v
=
r
=
0.15 + 94
= 14.1 ms–1
(c) Centripetal acceleration, ac
=
v2
----r
=
14.1 2
-----------0.15
=
1330 ms–2
Module 2 – Understanding motion
2.57
Example:
A car negotiates a round-about 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
ac =
v2
----r
v2 = acr
v
=
ac r
=
6 + 30
ac = 6ms–2 and r = 30m
= 13.4 ms–1
For cars, speed is usually expressed in km/hr so 13.4 ms–1 = 48 km/hr.
You could multiply by 3.6 to convert from ms–1 to kmh–1 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 + 1011 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 + 108 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.
Fc
=
mac
=
mv 2
--------r
mr&2
=
v
Example:
ac
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:
The rock is undergoing a centripetal acceleration given by the equation,
ac = &2r
& =
where
=
therefore ac =
3 revs–1
= 3 +)2$)rads–1)
18.8 rads–1
(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.
Fc
' The tension in the string T
=
mac
-
7N
= Fc
=
0.02 + 353.4 =
= 7N
7.07
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 + 106 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 + 1011m. Estimate
the mass of the Sun if the Earth has a period of 365 days and
G = 6.67 + 10–11 Nm2 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
axis
r
r
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).
θ
The units of torque are in Newton-metres (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’.
F/ /
r
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?
Module 2 – Understanding motion
2.61
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.
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 see-saw 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 see-saw?
Solution:
Let x be the distance of the father from the pivot as shown in the diagram.
2.62
TPP7160 – Preparatory physics
4m
x
0.4 m
F2
F1
45 g
25 g
F3
85 g
Pivot
clockwise torque
anticlockwise
torques
No forces other than F1, F2 and F3 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.
!"#"(F1 $"2) + (F2 $ 1.6) + F3 (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
support
F1
F2
L
--2
Beam
L
--4
Shoes
C
A
200 N
L
--4
450 N
B
Module 2 – Understanding motion
2.63
Solution:
L
If L is the length of the beam, its centre of gravity is at --- from each end. As the ‘Shoes’ sign is
2
L
L
shown to be --- from an end of the beam, it must be --- from the centre of the beam.
4
4
Since the object is in equilibrium "F = 0 and "!"= 0
In the vertical plane the forces in equilibrium are represented by the equation
F1 + F2 – 200 N – 450 N
F1 + F2
=
=
0
( (–) sign for downward forces)
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,
3L
–L
------ 200 – ------ 450 + LF2 =
2
4
0
( (–) sign for clockwise torques about an axis
through A)
Solving for F2 we have,
%
F2
= 438
F2
= 438 N
To find F1 we substitute F2 = 438N into the equation (1)
F1 + F2 = 650
F1 = 650 – F2
%"
%"
F1 = 212 N
The two supporting forces are 438 N and 212 N.
2.64
!"#
TPP7160 – Preparatory physics
Activity 2.16
F1
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?
1
(b) If a child weighing 140N sat on the beam --- way from its
4 use to carry the
lower end, what force must each person then
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 F1 vertically as shown. Find:
(a) the torque being applied to the crowbar by you if F1 = 250 N.
(b) the force F2 subsequently exerted on the rock.
(c) What is the weight of the rock?
(d) Comment on the magnitudes of F1 and F2 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 long-handled spanner than with a short-handled
spanner.
Module 2 – Understanding motion
2.65
Section review
In this section we have examined Newton’s three laws and how they apply to non-linear
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.
2.66
TPP7160 – Preparatory physics
Post-test 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?
Module 2 – Understanding motion
60
°
A
q
=
B
30 °
2.67
2.68
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
Average Power
=
2.69
work done
-----------------------------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 $ 104 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:
Fs cos !
(a) Work done
=
In this instance
is zero
Work done
F
=
1.5 " 104 " 10
=
1.5 " 105 Nm
1.50 × 104 N
# The work done is 1.5 " 105 Joules
(b) Power
=
work
-----------time
=
1.5 " 10 5
---------------------15
=
1 " 104 Js–1 or W
# 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 " 104 = 23330 N
Power
=
work
-----------time
=
23330 " 10
--------------------------15
=
15553 W
=
15.6 kW
=
force " distance
-------------------------------------time
#!$he maximum power should be greater than 15.6 kW
2.70
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.
2.72
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 " 1013J?
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
=
1
--- mv2
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
35km
35000 m
-------------- = ----------------------1hr
( 60 " 60 ) sec s
=
9.72 ms–1
60 " 1000
60 km * hr = -----------------------3600
= 16.67 ms–1
Module 2 – Understanding motion
2.73
m
Let m be the mass of the first car, the mass of the second car will be ---2
1
= 612.5 " m
KE (1st car) = --- m (9.72)2
2
= 47.2m Joules
KE (2nd car) =
1 m
--- ( ---- ) (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.
2.74
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
Energy
constant value
PE
A
KE
C
B
A
v maximum
C
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
Loss in kinetic energy
=
= constant
gain in potential energy
Kinetic energy is given by the formula, KE
Loss in Kinetic energy
=
Gain in potential energy =
1
--- mv2
2
=
1
--- " m " (202 – 82)
2
(mg) (h2–h1)
h2
V2 = 8 ms−1
h1
V1 = 20 ms−1
where h2 – h1 is the change in height.
#
mg (h2–h1)
1
2
2
= --- m ( 20 – 8 )
2
2
h2–h1
2
( 20 – 8 ) " m
= ----------------------------------2 " 9.8 " m
=
336
---------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.
2.76
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
Post-test 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.
2.78
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
=
1
ut + --- at2
2
x + 9
=
1
(2 " 9) + ( --- " 3 " 4)
2
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
#!
v2
=
u2 + 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).
2.80
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
85°
resultant
start
8.2 m
3.5 m
30 °
Scale 1 cm
≡
1m
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:
(i) (a) Graphical solution
C
resu
N
t
ltan
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)
6 ms–1
67
A
θ
B
14 ms–1
(i) (b)
Mathematical solution AC2 =
=
142 + 62
=
196 + 36
AC =
#
AB2 + BC2
resultant velocity =
15.23
15 ms–1
=
232
2.82
TPP7160 – Preparatory physics
6
tan ' = ------ = 0.4286
14
since
!
'
= 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
=
displacement
------------------------------velocity
=
100
--------14
=
7.14 secs
15.2 ms-1
(iii) During this time the river has carried the boat
downstream at 6 ms–1
!
Displacement
14 ms-1
=
velocity # time
=
6 # 7.14
=
42.9 m
x
23.2º
Alternatively you can use the angle to find the
x
displacement
tan 23.2° = --------100
!
x
6 ms-1
100 m
= 42.9 m
Therefore the boat is 42.9 m downstream.
Question 3:
90 − θ
45 km h -1
θ
340 km h -1
scale: 1 cm
40 km h–1
340
Using this scale the plane’s velocity of 340 kmh–1 $ " --------40
45
and the wind velocity of 45 kmh–1 $ " -----40
$" 8.5 cm
$" %&%cm
By measurement, the resultant = 8.60 cm and ' =
7.5°
!
8.6 × 40
The resultant velocity of the plane
=
= 344 kmh–1 at a bearing of 082.5°
Module 2 – Understanding motion
2.83
By calculation,
resultant =
2
340 + 45
=
117625
=
343 kmh–1
2
45
tan ' = --------340
!
–1 45
' = tan --------- = 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
u
=
3.0 ms–1
final velocity
v
=
3.5 ms–1
acceleration
=
change in velocity over time
a
=
v–u
----------t
=
3.5 – 3.0
--------------------4
= 0.13 ms–2
where t = 4 secs
in the direction the car is
moving
Question 2:
initial velocity of car
u
=
0
final velocity of car
v
=
10 ms–1
)v
=
v–u
=
10 – 0
=
10 ms–1
change in velocity
time taken
t
=
3 secs
acceleration
a
=
) v/t
10
-----3
3.3 ms–2
=
=
down the hill
2.84
TPP7160 – Preparatory physics
Question 3:
initial velocity of train
u =
50 ms–1
final velocity of train
v =
10 ms–1
)v =
change in velocity
Time taken
t
acceleration
a =
=
=
10 – 50
=
–40 ms–1
15 secs
–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 =
final velocity
v
acceleration
a =
=
15 ms–1
0 ms–1
–2 ms–2
(negative because bus is slowing)
To find distance travelled s, use the equation
i.e.
v2 =
u2 + 2as
0
=
(15)2 + 2 × (–2) # s
s
=
15
-------4
=
56.25 m
(since t is not involved)
2
!"
!"The bus will travel for 56.3 m before it stops.
Question 2:
60 km/hr
=
60 # 1000
------------------------ ms–1
3600
=
600
--------36
=
100
--------6
(convert to SI units)
=
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
i.e.
u
=
16.7ms–1
s
=
100 m
v
=
0
To find acceleration use
2.85
66
To convert ms–2 to kmh–1s–1
(i.e. km per hour per sec)
v2 =
u2 + 2as
0
=
(16.7)2 + (2a # 100)
a
=
16.76
– --------------200
=
–1.39 ms–2 or –5 kmh–1s–1
1.39 m
1.39 km
= ----------------- = -----------------------------------------------------1s # 1s
1h
1000 # - ------------------. # 1s
+ 60 # 60,
2
1.39 km
= - --------------------. × 3600 h–1s–1
+ 1000 ,
= 5 kmh–1s–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 free-fall
g = 9.8 ms–2
After two seconds, velocity
at)
v = 0 + 2 # 9.8
u =
(using v
=
0
u +
= 19.6 ms–1
1
1
s = 0 # 2 + --- # 9.8 # 22 (using s = ut + --2
2
Distance travelled
at2)
= 19.6 m
The jumper would have fallen 19.6 m in 2 secs.
Question 4:
Time of flight of cat
= 0.5 secs
g = 9.8 ms–2
Acceleration
using
1
s = ut + --- at2
2
1
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
= 18 ms–1
Velocity v at end of this distance
v2 =
using
182 =
u2 + 2as
0 + 2 # a # 240
2
a =
60 kmh–1
Speed limit,
18
--------480
324
= --------480
= 0.68ms–2
60 # 1000metres
= ------------------------------------------------1 # 60 # 60 sec onds
3
=
60 # 10
-------------------- ms–1
3600
=
100
--------- = 16.6 ms–1
6
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
3600
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
F
where
=
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
1
The force acting on me is --- of that on Earth.
6
Fm =
So
!
My weight
=
705.6
------------- N
6
117.6 N on the moon
Question 3:
On Mars the gravitational force
Fmars =
!
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
Frictional force
0k =
0.5
Ff = 0 k Fn
Ff
where Fn = normal force
= mg
=
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
–3920
a
= ma
= 800 a (we assign the direction of motion to be positive)
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
v2
=
u2 + 2as
0
=
302 + 2 # (–4.9) # s
s
=
30
---------------2 # 4.9
=
91.8 m
2
using
v
= u + at
0
= 30 – 4.9t
t
=
30
------- = 6.12s
4.9
! It will take 6.1s and a stopping distance of approximately 92 m for the car
to come to rest.
Module 2 – Understanding motion
Question 2:
Fn = 50 N
F = 36 N
Ff
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
Ff
=
36 N
The normal force Fn
= 50 N
Hence the coefficient of kinetic friction 0k
0k
=
Ff
-----Fn
=
36
-----50
=
0.72
2.89
2.90
TPP7160 – Preparatory physics
Activity 2.6
Question 1:
(a) Drawing to scale:
C
t
an
t
l
u
res orce
f
65 N
N
q
A
B
83 N
Scale 1 cm
10 N
By measurement the resultant force
angle '
!"
=
=
105 N
52(
The resultant force is 105N at N 52° E
(b) By calculation
AC2 =
AB2 + BC2
AC =
83 + 65
=
tan 1CAB
1CAB
2
2
105.4
65
= -----83
=
0.78
=
38(
! ' = 90° – 38° = 52°
!"The resultant force is 105 N at a bearing of 052°
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
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.
Resultant by measurement =
!
Resultant force
4mm
114mm
1N
= 28.6 N
By calculation: using the cosine rule (see page ‘2.25’)
a2 = b2 + c2 – 2bc cos Â
(where a, b, c are the sides of a triangle ABC and  is the angle opposite
side a)
2.91
2.92
TPP7160 – Preparatory physics
Applying the cosine rule to our problem above
b2 = a2 + c2 –2ac cos B̂
= 152 + 152 – 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:
Fn
Ff
Wsin35º
35º
W
Wcos35º
The forces acting on the car are:
(i) its weight W = 1.2 # 104 N acting vertically down
(ii) the frictional force Ff 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 # 104 # sin 35(
=
1.2 # 104 # 0.574
=
6.88 # 103 N
!"The force tending to cause the car to roll down slope
=
6.88 # 103 N
(b) At right angles to the slope, the component of the car’s weight is W cos 35°
=
1.2 # 104 # cos 35(
=
1.2 # 104 # 0.819
=
9.83 # 103 N
So2"Force at right angles to the slope
=
9.83 # 103 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 '&
Fn
110 N
Wsin θ
W
θ
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 '"
110 =
(since W = mg)
75 # 9.8 sin '"
sin ' =
110
------------------75 # 9.8
sin ' =
0.15
' =
8.6(
! The slope of the hill is 8.6°.
Question 3:
Fn
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
=
But
W sin 35(
89 =
W sin 35("since friction is neglected
W =
89
---------------sin 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:
Ff
Fn
42
°
W cos 42°
W
sin
W
42°
Let W Newtons be the weight of the block.
Then resolving W along the plane and normal to it,
=
W sin 42(
force normal to the plane =
W cos 42(
force along the plane
Since the block is on the point of sliding,
Force down the plane
But frictional force
=
Ff =
So
W sin 42( =
i.e.
0 =
=
frictional force
0 Fn
Ff = W sin 42°
(where Fn = normal reaction and
0 = coeff of friction)
0 W cos 42(
sin 42(
----------------- = tan 42°
cos 42(
0.9
!" The coefficient of static friction 0" 3" 0.9
2.96
TPP7160 – Preparatory physics
Question 2:
(a)
50
Fn
0N
°
W
sin
40
Ff
W cos 40°
W = mg
40 °
(b) The net force or resultant force up the plane = 18 N (up plane is + ve)
500
–
Ff
–
25 g sin 40 = 18!
"! Ff = 500 – 25g sin 40° – 18
(c) Fn = 25 g cos 40
(where Ff is the frictional force)
since the object is balanced on the surface of the slope.
(d) from the equation in (b)
Ff = 500 – 25g sin 40° – 18
whence
Ff =
500 – 157.5 – 18
Ff =
324.5 N
Ff
The coeff of kinetic friction #k = -----Fn
=
324.5
----------------------------------------- .
25 $ 9.8 $ cos 40
=
324.5
------------187.7
=
1.73
Module 2 – Understanding motion
2.97
Question 3:
Fn
F
W sin 8°
Ff = 350 N
8°
Wcos 8°
W = mg
Resolving W at right angles to the plane,
Wcos 8° =
" Fn
300 g cos 8
=
2911 N
=
2911N
The frictional force, Ff
=
since Ff =
"
"
350 N down the slope since it opposes the uphill
motion
#Fn(where # is the coeff of sliding friction)
# =
Ff
-----Fn
# =
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 =
Ball
W – Air resistance
= 4.9 N – 0.8 N =
Air resistance
4.1 N
If a is the resulting acceleration,
F = ma (Newton’s 2nd Law)
4.1 =
"
a =
=
0.5 a
4.1
------0.5
8.2 ms–2
W = 0.5g
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 –
GMm
F = ------------2
d
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 $ 106 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
GMm
F = ---------------------------------% 6.519 $ 10 6 & 2
6.67 $ 10 – 11 $ 6 $ 10 24 $ 15
= ------------------------------------------------------------------% 6.519 $ 10 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 $ 1024 kg
Mass of Mars
=
0.11 $ 6 $ 1024 kg
=
6.6 $ 1023 kg
200
A 200 N object on the Earth has a mass --------- = 20.4 kg
9.8
(a)On Mars, let the force of gravity on the object be mg/ where g/ is the
acceleration due to the gravity on Mars.
mg/ =
G $ Mm
------------------2
d
mg/ =
6.67 $ 10
$ 6.6 $ 10 $ 20.4
----------------------------------------------------------------------------6 2
% 3.440 $ 10 &
–11
=
=
"!
23
6.67 $ 6.6 $ 20.4
---------------------------------------2
3.440
object
m
d 3440km
6.67 $ 6.6 $ 20.4
---------------------------------------2
3.44
Mars M
It’s weight on Mars = 75.89 N
(b)Since the mass on Earth is 20.4 kg
75.89
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 =
GMm
------------2
d
where d = 7370 km
–11
24
=
6.67 $ 10
$ 6 $ 10 $ 70
-------------------------------------------------------------------6 2
% 7.370 $ 10 &
=
6.67 $ 6 $ 7 $ 10
--------------------------------------------2
12
7.370 $ 10
=
5.157 $ 10
=
5.157 × 102 N or 516 N
14
14 – 12
(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/ =
GM
-------2
d
where g/ is acceleration due to gravity on the surface of Jupiter
G =
6.67 $ 10–11 Nm2 kg–2
M =
1.9 $ 1027 kg
d =
7.14 $ 107 m
Substituting into the formula
–11
g/ =
27
$ 1.9 $ 10
6.67 $ 10
-----------------------------------------------------------7 2
% 7.14 $ 10 &
16
=
6.67 $ 1.9 $ 10
---------------------------------------2
14
7.14 $ 10
=
0.248 $ 102
=
24.8 ms–2
Module 2 – Understanding motion
(b)Since the weight of an object on earth
=
2.101
9.8 $ mass
and on Jupiter
= 24.8 $ mass
24.8
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:
m
=
6 $ 1024 kg
Linear velocity
v
=
3 $ 104 ms–1
Momentum
P
= mv
(a) Mass of earth
=
18 $ 1028 = 1.8 $ 1029 kg ms–1
(b) Mass of snail
m
=
0.1 kg
Velocity
v
=
1 $ 10–6 ms–1
Momentum
P
= mv
(c) Mass of rifle bullet
=
1 $ 10–1 $ 1 $ 10–6
=
1 $ 10–7 kg ms–1
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
=
9.1 $ 10–31 kg
Velocity
v
=
2 $ 105 ms–1
Momentum
P
=
mv
=
9.1 $ 10–31 $ 2 $ 105
=
18.2 $ 10–26
=
1.82 $ 10–25 kg ms–1
Question 2:
(a)The momentum of the cricket ball before impact Pi = mu =
=
After impact, the final momentum is
Pf =
0.14 $ 38
5.32 kg ms–1
mv
= –38 $ 0.14
=
–5.32 kg ms–1
(negative signs indicate movement in the
opposite direction)
Impulse
Ft
or
=
change in momentum
= mv – mu
=
–5.32 – 5.32
=
–10.64 kg ms–1 (in the opposite direction)
– 10.64 Ns
(b)The impulsive force exerted on the ball is given by F,
where Ft =
" F
F
=
change in momentum
mv – mu – 10.64
--------------------- = ---------------t
0.6
= –17.73 N
" The average force the ball exerted on the bat is 17.73 N
v–u
(c)To find the acceleration, a = ----------- =
t
– 38 – 38
---------------------0.6
– 76
= --------0.6
= – 126.6 ms–2 (in the opposite direction)
Module 2 – Understanding motion
2.103
Question 3:
u = 40 km hr–1
(a) Velocity
=
40000
--------------- ms–1
3600
=
11.1 ms–1
It comes to rest in 0.25 m, "
using v2 =
to find a
0
=
(11.1)2 + 2(a)(0.25)
2a $ 0.25
=
– (11.1)2
a
"
u2 ' 2as
– 123.46
= ------------------0.5
a =
The car’s deceleration
–246.9 ms–2
v–u
a = ----------t
using
"! t =
=
" The time required to stop the car
to find t
v–u
----------a
0 – 11.1
------------------– 246.9
= 0.045 s
(b) Mass of child
m =
20 kg
Initial momentum of child
Pi =
mu
=
20 $ 11.1
=
222 Ns
change in momentum
Since child comes to rest in 0.045 sec, Force = ------------------------------------------------------time
F =
mv – mu
-------------------t
0 – 222 Ns
= ---------------------------0.045s
=
(since Ff = 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
v2 = u2 + 2as
v2 = 2as (u = 0 at the height of 0.32m
above the ground)
v2 =
=
"
v =
2 $ 9.8 $ 0.32
9.8 $ 0.64
2.5 ms–1
Therefore momentum on reaching the ground is Pf = mv
Pf = 2.5 $ 55
= 137.7 kg ms–1
Ft = mv – mu
(b) The impulse
= 137.7 – 0 (since the dancer
comes to rest)
= 137.7 kg ms–1
" The impulse needed to stop
= 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
Whenever time is lenthened the
force of impact decreases.
t = 0.02 s
Long collision time means small
collision force and vice versa.
t = 0.04 s
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
m 1u 1
Pi (wagon 1) =
=
20000 $ 5 kg ms–1
Initial momentum of stationary wagon Pi (wagon 2) =
m1 + m 2
Combined mass after impact
m 2u 2
=
0
=
40000 kg
Let velocity after impact be v
Total momentum before collision = Total momentum after collision
m1u1 + m2u2
"
=
(m1+m2)v
20000 × 5 + 0 =
40000 × v
"v
=
20000 $ 5
-----------------------40000
v
=
2.5 ms–1
Velocity after impace 2.5 ms–1 in the same direction as wagon 1
Question 2:
u1 = 28 ms −1
u2 = 0 ms −1
Initial momentum of moving car
Pi(1) = m1u1
=
1245 $ 28 kg ms–1
Initial momentum of other car
Pi(2) = m2u2
v
=
( m1 + m2 )
0
Let v be the velocity after impact
"! Total inital momentum =
m1u1 + m2u2
=
1245 $ 28 + 0 =
"
v
Total final momentum
(m1 + m2) v
(1245 + 2163) v
=
1245 $ 28
-----------------------3408
=
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 v1 and v2
0 = m2 × u2
Since both are initially at rest the initial momentum
= 0
m2 × u1 = 0
After pushing, final momentum = 80 v1 + 60 v2
Total momentum before impact = Total momentum
after impact
0 = 80 v1 + 60 v2
m1v1
80 v1 = 60 v2
v1
! ----v2
=
– 60
--------80
v1
! ----v2
=
–3
-----4
m2 v2
the ratio of their velocities v1:v2 # –3:4
(the negative sign indicates that the velocities are in opposite directions)
Since
4
v2 = --- v1
3
the 60 kg student has greater speed.
(smaller mass greater speed and vice versa)
Question 4:
Let the velocity of recoil of the nucleus be v1
and the velocity of the ejected particle be v2
Since
Pf = Pi
momentum of ejected particle + momentum of remaining nucleus
!
momentum of remaining nucleus =
(m1 – m2) v1
=
=
0
– momentum of ejected particle
– m2 v 2
i.e. (3.8 " 10–25 – 6.6 " 10–27) v1 =
–6.6 " 10–27 " 1.5 " 107
(380 " 10–27 – 6.6 " 10–27) v1 =
–6.6 " 10–27 " 1.5 " 107
Module 2 – Understanding motion
(373.4 " 10–27) v1
=
2.107
–6.6 " 1.5 " 10–20
–20
v1
– 6.6 " 1.5 " 10
= -----------------------------------------–27
373.4 " 10
7
– 9.9 " 10
= ------------------------373.4
=
–0.0265 " 107
=
–2.65 " 105 ms–1
The velocity of recoil is 2.65 × 105 ms–1 in the opposite direction to the
ejected particle.
Question 5:
m2 = 128 kg
m1 = 95 kg
u2 = ?
u1 = 2.8 ms-1
Let the speed of the opposition player be u2 ms–1
Then momentum of tackler,
m 1u 1
=
95 " 2.8 kg ms–1
Momentum of other player,
m2 u 2
=
128 " u2
Since they both come to rest, their final momentum must be zero.
Total initial momentum
=
Total final momentum
!m1u1 + m2u2
=
0
95 "!2.7 + 128 "!u2
=
0
128 " u2
=
–95 " 2.8
u2
– 95 " 2.8
= ----------------------128
u2
=
2.08 ms–1
!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,
uy = 0 since the initial velocity was in the horizontal direction
1
s = ut + --- gt2
(since a = g = 9.8ms–2)
2
1
2
ux
s = --- " 9.8 " t
15 ms-1
2
100
=
4.9 t2
t2
=
100
--------4.9
!t
=
4.5 seconds
s = 100m
x
Target
!time of flight = 4.5 s
Now let us consider the horizontal motion where
ux = 15 ms–1
In this time the plane travelling horizontally at 15 ms–1 will cover a distance x,
given by
x = ux × t
= 15 × 4.5
= 67.5 m
Hence it must release the load 67.5 m before the target position.
Question 2:
vy = 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
uy =
3.45 sin 75$
Horizontal component
ux =
3.45 cos 75$
If t is time of flight of the ball (i.e. time taken to reach the ground),
1
consider the vertical motion using s = uyt + --- gt2
2
1
0 = 3.45 sin 75$!t + --- " –9.8 " t2
2
(assign (–) for downward motion)
0 = 3.33t – 4.9t2
Module 2 – Understanding motion
4.9t2 – 3.33t
= 0
t(4.9t – 3.33) =
t = 0 or
0
4.9t – 3.33
=
0
4.9t
=
3.33
t =
Therefore time of flight
(b) If height reached =
2.109
=
(t = 0 is not an option here)
3.33
---------- = 0.68
4.9
0.68 secs
h, then h occurs when final upward velocity vy = 0
Again considering the vertical motion,
using
v2
=
u2 + 2as
or
vy2
=
uy2 + 2gh
0
=
(3.45 sin 75°)2 + 2 (–9.8)h
2 × 9.8 × h
=
(3.45 sin 75°)2
h
=
& 3.45 sin 75 %
-------------------------------19.6
h
=
11.105
---------------- = 0.566
19.6
(taking down as negative)
2
!
The highest point reached = 0.566 m
Question 3:
Initial velocity of the line
u = 20 ms–1
Vertical component
uy = 20 sin 45$
Horizontal component
ux = 20 cos 45$
u = 20 ms-1
uy
Time of flight t is found by considering the
vertical motion
1
using
s = uyt + --- gt2
2
45º
ux
x
1
s = (20 sin 45$%t + --- gt2
2
i.e.
0 =
14.142 t – 4.9 t2
t(4.9t – 14.142) = 0
or
!
t
4.9t – 14.142
i.e.
=
0 or
= 0
14.14
t = ------------4.9
t = 2.89 secs
(taking down as –ve)
2.110
TPP7160 – Preparatory physics
During this time it travels horizontally over a distance x.
Considering the horizontal motion, since velocity is constant
x
=
ux × 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:
ux = 11.3 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 uy =
using
s =
s =
0
1
uyt + --- gt2,
2
1 2
--- gt where t is time of flight
2
4.9 t2
0.25
=
---------- =
4.9
= 11.3 ms–1
0.25 =
t
The horizontal velocity
ux
Distance travelled horizontally x
0.226 secs
= ux × 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 take-off 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 take-off the velocity of the jumper can be resolved in two
components
v cos ' horizontally and
v
v sin '!vertically
vsinθ
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 take-off and landing in the pit.
t is dependent upon the vertical component v sin '!as shown by the following:
Using the equation of motion,
1
s = ut + --- gt2
2
s = 0,
since
1 2
--- gt =
2
2
since
ut
t
---t
=
2u
-----g
!t
=
2u
-----g
=
2v sin '
----------------g
x =
v cos ' t
x
2v sin '
v cos ' × ----------------g
=
2
2v sin ' cos '
= -------------------------------g
The value of '!which optimizes the distance covered horizontally under these
(
conditions is 45$!or --- since the product sin '!cos ' is maximum then. So the ideal
4
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:
2
ac =
Centripetal acceleration
=
v
----r
2r
Since the Earth orbits the Sun once every 365 days (or 365 × 24 × 60 × 60 secs)
=
centripetal acceleration
ac =
2(
------------------------------ rad s–1 (1 revolution = 2 (!rad)
365 " 86400
2
2(
11
+ -----------------------------, " 1.5 " 10
) 365 " 86400*
–7 2
=
& 1.99 " 10 % " 1.5 " 10
=
5.95 " 10 ms
–3
11
–2
Question 2:
(a) No of rev/s
s =
!Angular velocity
=
(b) Centripetal acceleration ac =
1500 rev s–1
1500 " 2( rad s–1 (1 rev = 2( rad)
2r
=
(1500 " 2()2 " 0.15
=
13323966 ms–2
=
1.3 × 107 ms–2
(Note: It is common practice to designate centripetal accelerations by comparing
them to g)
7
1.3 " 10
!At 1500 rev s–1 the contents in the test tube is under ---------------------- = 1.3 × 106
9.8
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 " 2(
! = ------------------------ rads–1
60
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
2
(b) Centripetal acceleration
ac =
v
----r
=
& 52.36 % 2
------------------ = 27415.6 ms–2
0.1
=
2.7 × 104 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
-
radius of orbit =
3.00 " 108 ms–1
0.751
------------- = 7.51 " 10–3 m
100
2
Centripetal acceleration ac =
v
----r
8 2
=
& 3.00 " 10 %
------------------------------–3
7.51 " 10
=
9 " 10
--------------------------–3
7.51 " 10
16
19
= 1.2 " 10 ms–2
Activity 2.15
Question 1:
=
1.657 " 10–27 kg
=
2m
r
=
1m
Velocity of protons v
=
2.0 " 106 ms–1
Mass of proton,
m
diameter of magnet
radius,
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)
2(r
dis tan ce
Period,
T = ---------------------- " = --------v
velocity
=
\
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
Fc =
mv
--------r
6 2
–27
=
& 1.657 " 10 % & 2.0 " 10 %
------------------------------------------------------------------1
=
1.657 " 4 " 10–27 " 1012 N
=
6.628 " 10–15 N
The force exerted on each proton is 6.6 × 10–15 N
Question 2:
0.9 m
Fc
0.8 kg
The maximum allowable centripetal force
=2N
(the maximum tension
allowed on the string)
2
Centripetal force
Fc =
mv
--------r
2
0.8 " v
-----------------0.9
m =
0.8 kg and r = 0.9 m
2
!
Hence
1.8
------0.8
3
= --= 1.5 ms–1
2
= time for one revolution
v2 =
v
The period,
=
T
The circumference of the path
Period
!
=
2( r
=
2( " 0.9 m
circumference
T = -------------------------------------velocity
=
2( " 0.9
-------------------1.5
Minimum period =
3.77 secs
2(r
= --------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.
(a) If MS =
Fc =
mass of Sun and ME
=
mass of Earth
GM S M E
------------------2
d
Also, if the angular velocity of the earth is ,
Fc
=
mr
2
2
! ! F c = ME d
If we equate these two equations for Fc
2d
ME
2
=
d3 =
GM E M S
----------------2
d
G MS ............................................ equation (A)
2(
= -----T
Now
2d3
From equation (A) above
=
2(
-----------------------------365 " 86400
=
2
3
2(
11
+ -----------------------------, " & 1.5 " 10 %
) 365 " 86400*
=
1.339 × 1020
GMS =
!
MS
(since period, T = 365 days)
2 3
d
= 1.339 × 1020
2 3
20
1.339 " 10
d
= ------------ = ----------------------------– 11
G
6.67 " 10
MS = 2.0 × 1030 kg
Mass of the Sun = 2.0 × 1030 kg
2.116
TPP7160 – Preparatory physics
Activity 2.16
Question 1:
F2
B
l
F1
2l
l cos25°
25°
A
400 N
If the plank is in equilibrium and is uniform then!/F = 0 .................. (1)
or the sum of forces
i.e.
F1 + F2 – 400
=
0
=
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)
F1 " 2l cos 25° =
F1
Since
=
400
--------2
=
F1 + F2 = 400 N
400 l cos 25°
(taking clockwise direction as tve)
200 N
!
F2 = 200 N
So each person exerts an upward force of 200 N.
l
If a child sat at --- from A and its weight was 140 N,
2
/F
=
0
F1 + F2 – 400 – 140
=
0
F1 + F2
=
540
then as before
Taking moments as before about the end B, /0! = 0
,!
Module 2 – Understanding motion
F1
2.117
3
2l cos 25° – 140 --- l cos 25° – 400 l cos 25° = 0 (taking anticlockwise as (–))
2
!
2F1 – 210 – 400
!
F1 =
610
--------2
F2 =
235 N
Question 2:
=
=
0
305 N
F1
F2 sin 55°
F1 sin 55°
F2
(a) The torque you applied to
where
"1 = F r
#
F = F1 sin 55° (by resolving F perpendicular to the crow bar)
#
!$" 1 = F1 sin 55° × 1.6 (Note: the #$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.
!$The rock will also be acted upon by a clockwise torque given by
But
"2 =
F2 sin 55° × 0.4
"2 =
"1
F2 sin 55° × 0.4 = 328
!$ F2 = 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
F2
(d) Since -----F1
=
1001
------------ * 4
250
1
F 1 * --- F 2
4
! $ F 2 * 4F 1 or
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 " $% $F#r and F1 #r1 = F2 #r2
If F2 >> F1, then r2 << r1
As can be seen r2 = 0.4 m
(note >> means much larger)
r1 = 1.6 m
r1 = 4r2
i.e.
or
r1
F2
4
------ = ---- = --1
r2
F1
These ratios
F2 : F1 + r1 : r2 + 4 : 1
will be useful in determining how efficiently we can use our crowbar.
Question 3:
F2
F1
A
B
1.4
1
550 N
Let the tensions in the ropes be F1 and F2
Neglecting the weight of the scaffold, at equilibrium,
&F$ %$ 0$
!$F1 + F2 –550 = 0
'()
and
&"$ %$ $0
F2 × 2.4 – 550 × 1 =
0 (taking
(2)
as (–))
Module 2 – Understanding motion
F1 + F2
=
!$
550
F2
2.4
=
F2 =
=
Since F1 + F2
2.119
550
1
550
--------2.4
229 N
= 550
F1
= 550 – 229
F1
=
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
is twice as long as another then using the same force, Torque = F × 2r you can
#
double the torque.
axis
r
2r
τ = F⊥ × r
F
τ = F⊥ × 2 r
So a longer handled spanner would give a larger torque
F
TPP7160 – Preparatory physics
Activity 2.17
Question 1:
(a)
sta
irs
2.120
4m
4.5 m
Work done against gravity
=
weight
=
210
=
840 J
vertical distance raised
4
(b) If you rushed up the stairs carrying the books, the total weight raised
=
your weight + weight of books
=
75
=
735 + 210
=
945 N
9.8 + 210
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
total work done
If the climb takes 30 secs, average power = ------------------------------------time taken
6603
Power = ------------ Js–1
30
$
%
220 W
Activity 2.18
m = 1 kg
s = 400 m
2.122
TPP7160 – Preparatory physics
m = 1 kg
Mass of lunch
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:
mgh
The gravitational P.E. of the pumped water =
h =
where
9.8 ms–2
300m and g =
To store 2
1013 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 × 1013
13
!
m =
=
2 10
---------------------9.8 300
6.8
109 kg
v =
220 km h–1
Activity 2.19
Question 1:
(a) The skiers’ velocity
K.E. of skier
=
220 1000
--------------------------- ms–1
3600
=
61 ms–1
1
= --- mv2
2
=
1
--- (72)
2
=
134000 J (or 134 kJ or 1.34 × 105 J)
(61)2
Module 2 – Understanding motion
(b) The runner’s velocity
v =
44 km h–1
=
44 1000
-----------------------3600
=
12.2 ms–1
1
--- mv2
2
1
= --72
2
K.E. of runner
2.123
=
=
(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,
134000
is twenty-five times that of the runner i.e. ------------------ = 25.0
5358
Question 2:
(a) Let the original velocity of the car be v ms–1 and its mass be m
1
Then its original K.E. is --- mv2
2
If v is increased by a factor of 4
new velocity
v1 =
4v
1
= --- mv12
2
New K.E.
=
1
--- m(4 v)2
2
1
1
= --- m × 16v2 or 16 ×( --- mv2)
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
d1 =
F Newtons
1
where d1 is the stopping
--- mv2
2
distance
In the second case, if d2 is stopping distance (braking force remaining
constant)
1
F
d2 = 16 ( --- mv2)
2
!
d2 : d1
=
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 = 8m
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
1
(b) If v is the velocity on impact, K.E. = --- mv2
2
1
117.6 = --- × 1.5
v2
2
117.6 2
---------------------1.5
v2 =
=
!$
156.8
12.5 ms–1
v =
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 =
2176
---------------5 9.8
=
44.4 m
h
!$ 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.
1
Then its K.E. at the lowest point of its swing = --m
(4)2 J
2
At the end of the swing this is all converted to P.E.
If h is the maximum height in metres to which the ball rises,
K.E. = P.E.
1
--- × m × 16 =
2
! h =
=
m × 9.8 × h
8
------9.8
0.8 m
v = 400 cms−1
Question 4:
m =
Mass of nitrogen molecule
Its K.E. at sea level is 6.2
10–26 kg
4.7
10–21 J
If v is the average velocity, then
6.2
=
1
--- mv2
2
10–21 =
1
--- × 4.7
2
K.E
10–26 × (v2)
–21
v2
!
!
=
2 6.2 10
----------------------------------–26
4.7 10
=
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. =
=
It’s
K.E. =
since final P.E. =
mgh
10–26
4.7
1
--2
4.7
initial K.E.
1
m gh = --- m v2
2
9.8
10–26
h
(514)2
2.126
TPP7160 – Preparatory physics
2
9.8
h =
514
----------2
h
=
514
---------------2 9.8
=
13479 m
2
Question 5:
u=0
v2
35m
25m
v1
(a) Let the mass of carriage =
At top of first hill P.E. =
m kg
m
9.8
35
At the bottom of the hill all of this is converted into K.E.
If v1 is the resulting velocity at the bottom of the hill,
K.E. =
1
--- m v12
2
=
1 2
--- v1 =
2
P.E.
m gh
9.8
35
v 12
=
19.6
v1
=
26.2 ms–1
35
(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).
1
At the top of the 25 m hill P.E. = m
9.8
25 and K.E. = --- mv22
2
(If v2 is the velocity at the top of the second hill)
1
At the bottom, initial
K.E. = --- m
(26.2)2 and initial P.E. = 0
2
Module 2 – Understanding motion
2.127
Since energy is conserved,
Total K.E. + P.E. =
constant
1
--- mv22 + m gh =
2
1
--- m v12
2
2
v2
------2
or
2
=
/ 26.2
----------0 – 9.8
- 2 .
=
(343 – 245)
=
98
v22
=
196
v2
=
14 ms–1
25
2
v2
------2
!$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
--- mv22 + mgh2 = mgh1
2
whereby
1 2
--- v2 =
2
v22
g(h1 – h2)
= 2g(35–25)
= 2 × 9.8 × 10
v2
=
196
= 14ms–1
Question 6:
m = 2000 kg
100 m
100 m
u = 12 ms- 1
h
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
=
K.E. =
=
(where h = 100 sin 25°)
2000
9.8 100 sin 25, = 828,300 J
1
--- mv2
2
1
(12)2 = 144,000 J
--- 2000
2
2.128
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.
Work
F
Then
!
= Force × distance.
100 =
F
=
972300 J
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. =
If the braking force
1
--- mv2
2
=
1
--- m
2
=
F Newtons
13.92 J
Then since it skids 15 m, work done by the brakes in bringing it to a stop
Force × distance = Change in K.E.
!
F
1
15 = 0 – --- mv2
2
=
1
– --- m
2
13.92
=
1
–( --2
13.9
-----------15
=
–(6.4 m) N
2
m)
(negative sign for opposite direction)
Assuming this force to be constant, we can use this formula in the second case.
Here
v =
=
K.E. =
150 km h–1
41.7 ms–1
1
(41.7)2
--- m
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
=
1
--- × m(41.7)2
2
d
=
' 41.7 )
----------------2 6.4
1
136 metres
2
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.
1
F × d = --- mv2
2
v1 = 50
if
d1 = 15
1
F × d1 = --- m (50)2
2
v2 = 150
if
(1)
d2 = ?
1
F × d2 = --- m (150)2
2
(2)
dividing equation (2) by equation (1)
2
1
--- m ' 150 )
2
F d2
150
2
- = 9
--------------- = ---------------------- = ---------2
F d1
1
2
50
--- m ' 50 )
2
d2
----- = 9
d1
or
d2 = 9 × d1
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
1
(K.E. = --- mv2)
2
2.130
TPP7160 – Preparatory physics
Solutions to post-tests
Post-test 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=
1
2
ut + --- at
2
3. v2=
u2 + 2as
If the body starts from rest these equations become
1. v=
at since u = 0
2. s=
1
2
--- at
2
3. v2=
2as
Module 2 – Understanding motion
2.131
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.
PE = max
KE = 0, v = 0
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.
KE
+ PE
vmax
PE = 0
KE = max
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.
Post-test 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.
2.132
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 m1, m2 are the masses of the bodies and d is their distance apart,
Gm 1 m 2
- where G is a constant called the Universal Gravitational
F = ----------------2
d
Constant.
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:
(a) F
=
a
Question 3:
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.)
m
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
F sin θ
One Example of resolving
the vector F into its #$
components
θ
F cosθ
F sin φ
This is another example of
resolving the same vector F
into its #$components
φ
F cosφ
Components are useful when we want to find the resultant of a number of forces acting on a
body, or the resultant force acting in a certain direction.
2.134
TPP7160 – Preparatory physics
F
Fn
θ
F cosθ
W sin φ
W cosφ
φ
W
Eg. (i) The weight W can be resolved in to W sin 2 along the slope and W cos2 # to the slope.
Then the normal reacton Fn = W cos 2 .
(ii) The effective part of the force F pulling the object up the slope = F cos 3$
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:
Fn
applied force F
q
n
Wsi
Ff
tion
c
i
r
F
s
Wco
weight of body W
Fcosq
Module 2 – Understanding motion
Question 7:
2.135
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
22
-----------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
Post-test 2.3
Question 1:
Uniform circular motion occurs when an object moves in a circular path with
uniform speed.
angular displacement " #
Angular velocity is defined as ------------------------------------------------------- = --------- .
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?
Fc
Fc
P1
P2
v
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 Fc. This force is needed
to keep the person ‘P’ revolving at the same speed v as the wheel. Remember
without Fc 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 P2 and P1.
In a rotating frame of reference, such as for a person inside the wheel, apart from
the force Fc 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
2r
where m is the mass of the body concerned, or
g/
=
2r
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.
60
°
A
q
=
B
30 °
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
3
vA = v sin 60$ or ------- v = 0.866 v
2
In case B the vertical component is
v
vB = 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
vA = v cos 60$ or --2
3
vB = v cos 30$ or ------- v
2
B therefore has the larger horizontal velocity
Module 2 – Understanding motion
2.139
Post-test 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.