Download 2 Mechanics

2
2.1
Mechanics
Kinematics
Assessment statements
2.1.1 Define displacement, velocity, speed, and acceleration.
2.1.2 Explain the difference between instantaneous and average values of
speed, velocity and acceleration.
2.1.3 Outline the conditions under which the equations for uniformly
accelerated motion may be applied.
2.1.4 Identify the acceleration of a body falling in a vacuum near the Earth’s
surface with the acceleration g of free fall.
2.1.5 Solve problems involving the equations of uniformly accelerated motion.
2.1.6 Describe the effects of air resistance on falling objects.
2.1.7 Draw and analyse distance–time graphs, displacement–time graphs,
velocity–time graphs and acceleration–time graphs.
2.1.8 Calculate and interpret the slopes of displacement–time graphs and
velocity–time graphs, and the areas under velocity–time graphs and
acceleration–time graphs.
2.1.9 Determine relative velocity in one and two dimensions.
What is kinematics?
Kinematics is the study of motion. As we have already seen, distance is a scalar and
displacement is a vector quantity. Distance travelled is a simple measure of length
while displacement is defined as the distance travelled in a particular direction
from a specified origin.
When we want to describe how fast something is moving we say that:
distance
speed _______
time
displacement
velocity ___________
time
These simple equations can be used for two situations; firstly when the body is
moving at a constant speed, and secondly in order to find the average speed. In the
second case we need to know the total distance and the total time.
Worked example
Calculate your average speed if your journey to school takes half an hour and the
distance is 10 km.
Solution
10 km
speed ______
0.5 h
distance
Examiner’s hint: Speed ________
time
20 km hⴚ1
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Mechanics
This works even though your car or bus has to slow down, speed up and
sometimes stop according to the traffic conditions.
If your vehicle is changing speed, or direction, it must be changing velocity, which
means that it is accelerating. Acceleration is defined as the change in velocity per
unit time. Acceleration a can be written as:
v
a ___
t
where v is change in velocity and t is change in time.
Worked example
A car starts from rest and accelerates steadily to a speed of 40 m s1 in a time of
5.0 s. Find the magnitude of the acceleration of the car.
Solution
1
40 m s
a _______
5.0 s
8.0 m sⴚ2
The units of velocity and acceleration are different, and you must always be careful
to measure speed and velocity in m sⴚ1 but acceleration in m sⴚ2.
If a ball is rolling down a slope, its speed will be increasing. We can find the average
speed of the ball by dividing the total distance by the total time, as seen earlier.
Another way to describe the motion is by using the instantaneous speed of the
ball; the speed at any instant in time. You can find this by drawing a distance–time
graph of the motion and measuring the tangent at a particular instant – we will
see how this works later on in this chapter.
The equations for uniformly accelerated motion
It is essential that you remember
that ‘s ’ in these equations stands
for distance travelled, and not for
speed.
If speed or velocity is changing at a steady rate, then the acceleration is constant. In
these situations we can use the equations of uniformly accelerated motion, which
we will call the suvat equations. These are a set of equations with five variables as
follows:
s
distance
(m)
u
initial velocity (m s1)
v
final velocity
(m s1)
a
acceleration
(m s2)
t
time
(s)
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There are four suvat equations and their power is that if we know any three of
the five variables, by using the equations carefully, we can solve for the other two
unknown quantities.
These are the equations:
v u at
vt
_____
su
2
s ut _12at 2
v 2 u 2 2as
Worked examples
1 A ball is thrown vertically upwards with an initial velocity of 20 m s1. Taking
the acceleration due to gravity as 10 m s2 and neglecting air resistance, find:
(a) the time taken to return to the
thrower’s hand
v 0 m s1
(b) the maximum height reached.
Figure 2.1 The maximum speed will
be reached just after the ball leaves the
girl’s hand.
a g 10 m s2
u 20 m s1
Solution
(a) s ?
u 20 m s1
v0
a 10 m s2
t?
Examiner’s hint: First list the suvat
variables.
Then choose the best equation, the one
with the relevant variables.
v u at
0 20 10t
10t 20
t2s
Examiner’s hint: Always write down
the equation you are using, sometimes
you can get a mark for just writing down
the correct equation.
Examiner’s hint: Next plug in the
numbers and solve for the unknown.
Time to return to the thrower’s hand will be twice this 4 s.
(b) s ?
u 20 m s1
v0
a 10 m s2
t2s
In this case the final velocity, when the
ball reaches the maximum height, will
be zero.
The acceleration will be negative, if we
consider the ball travelling upwards,
because it will be slowing down.
This type of problem involving
gravity is always symmetrical; the
distance, time and speed going
up are the same as those coming
back down.
Examiner’s hint: To find the
maximum height we again list the
variables for the motion upwards.
vt
_____
su
2
20
______
s 02
2
s 20 m
Examiner’s hint: Choose an
equation and write it down.
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Mechanics
2 A stone is dropped into the sea from a cliff 45 m high. Taking the acceleration
due to gravity as 10 m s2 and neglecting air resistance find:
(a) the speed of the stone when it hits the sea
(b) the time taken to reach the sea.
Examiner’s hint: This time the initial
speed is zero and the acceleration is
positive. Again we list the variables:
(a) s 45 m
u0
v?
a 10 m s2
t?
u 0 m s1
a g 10 m s2
Figure 2.2 Acceleration
is positive in this case. The
stone accelerates until it hits
the water.
v 30 m s1
v2 u2 2as
v2 0 (2 10 45)
900
v 30 m s1
Examiner’s hint: To find the time in
this case we could use any of the other
three equations.
(b) s 45 m
u0
v 30 m s1
a 10 m s2
t?
s ut _12at 2
45 0 (0.5 10 t 2)
45 5t 2
t2 9
t 3s
Try this last problem for yourself using the other two equations to make sure you
get the same value for the time taken.
To view an animation of motion in
2D, visit heinemann.co.uk/hotlinks,
enter the express code 4266S and
click on the Weblink 2.1.
In the two examples above, the acceleration of the bodies was g, the acceleration
due to gravity. This varies from place to place on the Earth’s surface, but is often
taken as 9.81 m s2, or approximated to 10 m s2.
The suvat equations work for any situation where the acceleration of the body
is constant, and you must learn to spot the type of problem where the suvat
equations will provide the solution.
Exercises
1 In an experiment a small steel ball is filmed and timed as it falls, from rest, next to a metre ruler.
Calculate the time taken for the ball to fall
(a) from the zero mark to the 90 cm mark on the ruler
(b) from the 90 cm mark to the end of the ruler.
Neglect air resistance and take g 9.8 m s2.
2 A car is travelling at 20 m s1 when a dog runs out into the road. If the driver’s reaction time is
0.40 s and the car decelerates steadily at a rate of 20 m s2, calculate the minimum distance
travelled before the car can come to rest.
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In the real world, bodies do not fall in a vacuum they experience the effects
of air resistance. As long as the body is small and hard, and the height is not too
great, then we can ignore these effects. Otherwise air resistance or drag becomes
important; obviously the motion of a falling body such as a feather or a leaf will be
significantly changed by the air through which it falls. If a feather or a leaf were to
fall in a vacuum they would drop just like a stone.
The acceleration due to gravity is
slightly lower off the coast of India
and slightly higher in the South
Pacific, owning to the nature of the
rocks in the Earth’s crust.
Imagine what happens to the speed of
a skydiver who jumps out of a plane,
falls free for a while and finally opens a
parachute.
Initially the skydiver’s speed is low, so the air resistance is much smaller than
the weight, and the skydiver accelerates. As the speed increases, the air resistance
increases until a point is reached when the drag force is equal and opposite to
the weight. Then the skydiver will fall at a constant, maximum speed called
the terminal velocity. When the parachute opens, the air resistance increases
significantly and a new, slower terminal velocity is reached, enabling a safe landing.
Figure 2.3 A velocitytime graph of a
parachute jump.
B
velocity
(ms1)
C
A
D
O
E
Examiner’s hint: When you are
asked to sketch a graph, you should draw
a pair of labelled axes with units, and a
line showing the trend. You do not have
to show numbers on a sketch graph but
you should draw it clearly with a ruler
and pencil.
time (s)
From O to A, the velocity is low and acceleration fairly constant.
From A to B, air resistance becomes greater as the velocity increases so the rate of
acceleration decreases.
From B to C, acceleration is zero, the skydiver is falling at constant terminal
velocity.
At C, the parachute opens and velocity rapidly decreases.
From D to E, a new, lower, terminal velocity is reached.
At E, the skydiver lands on the ground.
For a person in free fall, terminal
velocity is about 180 km h1 or
50 m s1, similar to a speeding car.
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Mechanics
In 1944, the plane in which Nicholas Alkemade was flying, was set on fire and his
parachute destroyed. Nicholas chose to jump without a parachute rather than stay in
the burning plane. Although he fell nearly 6000 m, he was extremely lucky because he
blacked out, landing relaxed, through pine trees and into snow. He suffered only minor
sprains, cuts and bruises and survived until 1987!
To view an animation of the lunar
lander, visit Heinemann.co.uk/hotlinks,
enter the express code 4266S and click
on Weblink 2.2.
Graphing motion
We have already seen some examples of motion graphs, but now we will look at
them in more detail. Note that time always goes on the x-axis.
Distance or displacement vs time
Figure 2.4 Four distancetime
graphs.
The car shown in these graphs is:
1. at rest, not moving
2. moving at a constant speed
3. accelerating
4. decelerating
1
d
2
O
t
distance
(m)
O
time (s)
d
d
3
O
4
O
t
t
The next graph shows displacement against time; remember that displacement is
a vector quantity. The way we show the difference between scalars and vectors on
a graph is to draw the scalar above the x-axis only, and the vector both above and
below.
Figure 2.5 Displacementtime graph.
O–A constant velocity forwards
A–B decelerating
B–C stopped
C–D accelerating and changing
direction
D–E constant velocity back to starting
point
E–F constant velocity backwards
F–G decelerating
G–H stopped
H–I accelerating and again changing
direction
I–J
constant velocity back to starting
point
B
A
displacement
O
(m)
C
D
J
E
time (s)
I
F
G
H
The slope or gradient of a distance–time graph gives the speed, while for a
displacement–time graph the gradient gives the velocity. In Figure 2.5 above, the
car is travelling at constant velocity from O to A because the gradient is constant.
From A to B, the velocity is decreasing because the gradient is decreasing; in other
words the car is decelerating.
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40 s
2s
20 m
40 m
B
A
instantaneous velocity
402 sm 20 m s1
instantaneous velocity
204 sm 5 m s1
Figure 2.6 The instantaneous velocity
is found by by measuring the slope of
the tangent to the curve.
A The tangent line is steep at this point
indicating a high instantaneous
velocity.
B Because the curve has flattened out
at this point, the tangent is less steep
and the instantaneous velocity is
lower.
Velocity vs time
It is vitally important that you do not mix up the various different types of motion
graph. The next ones we will consider are speed and velocity against time. Try and
describe what is happening to a car moving as follows:
Figure 2.7 Velocity–time graph.
A
B
velocity
(ms1)
O
C
time (s)
The car starts from rest when the speed is zero, and then from O to A its speed
increases at a steady rate. This is known as constant acceleration. Then from A
to B the car is not stopped, but is moving at constant speed. From B to C we have
constant deceleration and finally the car again comes to rest. As long as the car is
travelling in a straight line in the same direction, this graph would be exactly the
same with either speed or velocity on the y-axis.
The gradient of a velocity–time graph gives the acceleration. So in Figure 2.7
above from O to A the acceleration is constant because the gradient is constant;
the graph is a straight line. From A to B the gradient is zero because the rise, or
y term is zero; so the acceleration is zero. From B to C the gradient is steeper than
from O to A and this tells us that the rate of deceleration was more rapid than the
rate of acceleration.
The slope or gradient of a graph
y2 y1 ___
y
rise ______
___
run x2 x1 x
Now we will analyse the same graph using numbers.
10
A
Figure 2.8 Finding distance travelled
from a velocitytime graph.
O–A constant acceleration
B
1
10 m s 2 m s2
_______
5s
velocity
(m s1)
A–B
B–C
C
O
5
15 17 time (s)
constant maximum, velocity of
10 m s1 for 10 s
constant deceleration
1
10 m s 5 m s2
_______
2s
The area under a velocity–time graph gives the displacement. In this case the
area is the sum of a triangle, a rectangle and a smaller triangle. Remember that the
area of a triangle is half the base times the height.
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Mechanics
Displacement (_12 5 s 10 m s1) (10 s 10 m s1) (_12 2 s 10 m s1)
25 m 100 m 10 m
135 m
You need to be able to sketch and label these types of graph for various situations.
Exercises
3 A girl throws a ball up in the air and then catches it. Sketch a speed–time graph and then a
velocity–time graph for the motion of the ball.
4 A ball is dropped onto the floor and bounces twice. Sketch a velocity–time graph and then a
speed–time graph for the motion of the ball.
Acceleration vs time
The last type of motion graph we are going to look at is acceleration against time.
If the acceleration is uniform or constant, then these graphs will be simply vertical
and horizontal straight lines. An acceleration–time graph for the car in Figure 2.8
would look like this:
Figure 2.9 Acceleration–time graph
O–A constant acceleration of 2 m s2
for 5 s
A–B zero acceleration for 10 s
B–C constant negative acceleration or
deceleration of 5 m s2 for 2 s
acceleration
(ms2)
2
O
A
5
B C
15
17 time (s)
5
The area under an acceleration–time graph gives the change in velocity.
So from O to A, the change in velocity the area of the rectangle
(5 s 2 m s2)
10 m s1
From A to B, the area under the graph is zero, so the change in velocity is zero.
You can check for yourself that the area from B to C is 10 m s1 in the opposite
direction, and that this all fits with the data in Figure 2.8.
If the acceleration is not constant, we cannot use the suvat equations to solve
the problems. The area under the graph will still tell us the change in velocity,
however, as shown in the next example.
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Worked example
Find the change in velocity of a body that accelerates as shown in the graph below.
10
B
Figure 2.10 Accelerationtime graph.
The change in velocity is the area of
triangle ABC plus area of rectangle
OACD.
acceleration
(m s2)
2
C
A
D
6 time (s)
O
Solution
Change in velocity (_12 6 s 8 m s2) (6 s 2 m s2)
24 m s
1
12 m s
1
36 m s
1
To view The moving man simulation,
visit Heinemann.co.uk/hotlinks, enter
the express code 4266S and click on the
Weblink 2.3.
Relative velocity
Imagine you are riding in a train travelling at 140 km h1. Another train overtakes
you at a speed of 150 km h1. It looks like the other train is moving quite slowly.
This is because relative to you, the speed of the other train is only 10 km h1.
Think about the following scenario; Diego is standing on the platform and
Maria, on a train, passes through. We can talk about the relative motion in two
ways; the motion of Maria relative to Diego, or the motion of Diego relative to
Maria. The magnitude, or speed, will be the same but the direction will clearly
be different.
Maria is on the train travelling to the right with
a constant velocity 7 m s1
Figure 2.11 Relative velocity.
The velocity of Maria, relative to Diego is
7 m s1 to the right.
The velocity of Diego relative to Maria is
7 m s1 to the left.
7 m s1
Diego is on the platform
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Mechanics
Head-on car collisions are likely to have
serious consequences. One reason is
that the relative speeds will add up,
and so the resulting damage will also
increase.
So far we have been looking at relative motion in one dimension. To understand
how it works in two dimensions, imagine you are in a car driving down a long
straight road. A bird is flying at right angles to the road and you can see it out of
the window.
motion of car
relative to bird
motion of bird
relative to car
at 90° to the road
Figure 2.12 The motion of the bird
relative to the car is different from the
motion of the car relative to the bird.
You will learn a lot more about
relative motion if you study
Option D which includes Einstein’s
theory of Special Relativity.
car driving along
a straight road
at constant velocity
Practical hints on drawing graphs
When you are asked to sketch or plot a motion graph, think about the following
points:
1 Is this a graph of distance, displacement, speed, velocity or acceleration? They
are all usually sketched against time but they are all different graphs!
2 Is this a scalar or a vector graph? In other words do you need to draw it just
above the line, or both above and below?
3 Remember to label the axes with both the name and the unit; sometimes you
will get a mark for just doing that alone.
4 Does the graph start from the origin (the zero-zero point)?
5 Is the graph going to be a straight line or a smooth curve?
6 Make sure you plot the graph with a sensible scale; use numbers like 2, 4 and 5,
not numbers like 3 and 7.
7 Aim for your graph to fill at least half the available space.
8 Measure the gradient or tangent over at least half the space.
9 Link what you need to find to the gradient, intercept or area under the graph.
10 Always remember the equation of a straight line is of the form y mx c and
link it whenever possible to any graph you come across.
Much of this applies to other types of physics graphs, not just motion graphs.
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2.2
Forces and dynamics
Assessment statements
2.2.1
Calculate the weight of a body using the expression W mg.
2.2.2
Identify the forces acting on an object and draw free-body diagrams
representing the forces acting.
2.2.3
Determine the resultant force in different situations.
2.2.4
State Newton’s first law of motion.
2.2.5
Describe examples of Newton’s first law.
2.2.6
State the condition for translational equilibrium.
2.2.7
Solve problems involving translational equilibrium.
2.2.8
State Newton’s second law of motion.
2.2.9
Solve problems involving Newton’s second law.
2.2.10 Define linear momentum and impulse.
2.2.11 Determine the impulse due to a time-varying force by interpreting a
force–time graph.
2.2.12 State the law of conservation of linear momentum.
2.2.13 Solve problems involving momentum and impulse.
2.2.14 State Newton’s third law of motion.
2.2.15 Discuss examples of Newton’s third law.
Force, weight and mass
In simple terms a force is a pull or a push. Force is a vector quantity and is
measured in newtons. There are many different types of force, including
• friction
• normal or supporting or contact force
• tension and compression
• air resistance or drag
• upthrust or buoyancy
• lift
• thrust
• weight.
Figure 2.13 Force can be a pull or a
push.
Weight is the pull of gravity on a body. Assuming that the body is on the surface
of the earth, it is directed towards the centre of the earth. Usually we call this
direction ‘down’.
Strictly speaking, weight and mass are not the same thing. Weight
is a vector measured in newtons, and mass is a scalar measured
in kilograms. Mass can be described as the amount of matter in a
body.
In everyday speech, weight and mass are taken to mean the same,
but in physics, we need to remember that there is a difference.
If you went to the moon, your mass would not change, but your weight
would decrease because the pull of gravity is less on the moon.
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Mechanics
Astronauts on the moon were able to
take giant steps because their weight
was less (although their mass hadn’t
changed).
Weight is related to mass in a
simple way:
W mg
W is the weight measured in N
m is the mass measured in kg
g is the acceleration due to
gravity measured in m s2
Worked example
What is the weight of a 1.0 kg mass on the surface of the earth where g 9.8 m s2?
Solution
W mg
W 1.0 kg 9.8 m s2
9.8 N
If a body is in translational
equilibrium, it means that the forces
acting on the body, in all three
dimensions, are balanced.
Balanced forces
Here, translating means moving from one place to another in a straight line. If
a body is in translational equilibrium then it will not be accelerating. This is the
basic idea described in Newton’s first law.
We have already seen some free-body diagrams in Chapter 1, but here are some
more examples.
The helium balloon is at rest, so must be in translational
equilibrium.
upthrust weight tension
upthrust
The weight of the balloon will be small but
cannot be zero. If the string breaks, then the tension
becomes zero and the balloon will accelerate upwards.
Figure 2.14 Notice that the direction of the
tension in the string must be downwards to
balance the forces.
tension
weight
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Figure 2.15 If the person pulls harder
and just succeeds in lifting the block,
then the contact force becomes zero
and the pulling force is equal to the
weight.
pulling force
contact
force
FN
Ffr
x
Figure 2.16 Inclined plane the weight is resolved into x and y
components, parallel and perpendicular
to the incline. The three forces acting on
the block are mg (the weight), FN anf Ffr.
θ
weight
θ
y
mg
The free-body diagram must show only the three real forces acting on the block.
Since the block is accelerating, the forces are not balanced down the incline.
x > Ffr
y FN
where Ffr is the frictional force and FN is the normal (or perpendicular) force.
Using geometry, you can see that the angle of the incline is equal to the angle
between the weight vector and the y-component. This is because there are 180° in
a triangle and the normal force is at 90° to the incline.
Using trigonometry, we can then deduce that:
x
sin ___
mg
x mg sin y
cos ___
mg
y mg cos Newton’s laws of motion
The last recorded outbreak of the Black Death in England was in 1665. In those days people thought
that the plague would remain local, so they returned to their villages, often spreading the infection.
In 1665 Isaac Newton was a young Cambridge professor, working on some problems regarding
the physics of motion. The plague forced him into isolation at his family home in Woolsthorpe and
he worked alone there for eighteen months. During this time he made several completely original
contributions to mathematics and science, effectively inventing calculus and revolutionizing our ideas
about gravitation and light.
Isaac Newton
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Mechanics
Constant velocity means two
things, constant speed and
constant direction; in other words
constant speed in a straight line.
Aristotle (384322 BC) was one of
the most important of the ancient
Greek philosophers and he tutored
Alexander the Great. Aristotle
thought that a heavy object would
fall more quickly than a light one,
and that a force was necessary for
an object to move.
Galileo (15641642) was born in
Pisa, Italy and in many ways was
the founder of modern physics.
Galileo thought that a heavy object
and a light object should fall at the
same rate and that a body could
move forever if no force were
acting.
Who was right, Aristotle or Galileo?
If two variables are directly
proportional to each other it
means that if one increases, the
other will increase in proportion,
or at the same rate. A graph of the
two variables will be a straight line
through the origin.
Newton’s first law states that a body will either remain at rest, or move with
constant velocity, unless acted on by an unbalanced force.
We have already seen how this law works, and it is simply common sense that
something does not start to move unless there is resultant force acting on it. More
difficult to understand is the idea that a body can be in equilibrium, even though it
is moving. It can be difficult to accept that a body can move without a force being
exerted, as this goes against both common sense and the teaching of Aristotle.
Newton’s second law states that the rate of change of momentum of a body is
proportional to the applied force and takes place in the direction in which the
force acts. This law tells us what happens when an unbalanced force acts and is
expressed in the most important equation of basic mechanics:
F
m
a
F ma
is the resultant or unbalanced force measured in N
is the mass measured in kg
is the acceleration measured in m s2
Note that the earlier definition of weight is a special case of this general law.
For a constant mass, this means that force is directly proportional to acceleration.
If we take a constant out of a simple equation, then we get a relationship that is
directly proportional. For example if:
y mx
and m is a constant, then
yx
Conversely if two things are directly proportional to each other and we put in a
constant then we get an equality or equation. For example if:
Fx
F kx
and k is a constant, then
Worked example
Find the acceleration of a block of mass 500 g pulled across a table with a force of
12 N, if there is a frictional force of 6 N.
Figure 2.17 The resultant force 6 N
to the right.
force of
friction
pulling force
12 N
500 g
6N
Solution
The force in Newton’s second law is
always the resultant or unbalanced
force.
F ma
F
a __
m
6N
a _____
0.5 kg
a 12 m sⴚ2
to the right.
Practical hints on calculations
The basic method for numerical physics problems is not complicated.
1 Choose the correct equation
2 Usually you can find the equations in the Data Booklet but sometimes you
simply have to memorise them.
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3 Make sure you know what the symbols (letters) stand for.
4 When solving the problem, always write down the equation you are using sometimes you will get a mark for just writing the equation.
5 Plug in the numbers.
6 Make sure the units are consistent; if not, change them all to SI units.
7 Give your final answer to the correct number of significant figures and make
sure you have included the correct unit.
8 Before you move on, ask yourself whether the answer you have given is sensible
and realistic.
Exercises
5 A train of mass 1.5 105 kg is travelling at 40 m s1 when the brakes are applied and it
decelerates steadily. The train travels a distance of 250 m before coming to a halt.
(a) Calculate the deceleration of the train.
(b) Find the average braking force.
To view a simulation of forces in one
dimension – The ramp, visit
Heinemann.co.uk/hotlinks, enter the
express code 4266S and click on the
Weblink 2.4.
6 A large helium balloon is attached to the ground by two fixing ropes. Each rope makes an angle
of 50° with the ground. There is a force F vertically upwards of 2.15 103 N. The total mass of the
balloon and its basket is 1.95 102 kg.
(a) State the magnitude of the resultant force when it is attached to the ground.
(b) Calculate the tension in either of the fixing ropes.
(c) The fixing ropes are released and the balloon accelerates upwards. Calculate the magnitude
of this initial acceleration.
(d) The balloon reaches a terminal velocity 10 s after take-off. The upward force F remains
constant. Describe how the magnitude of the air resistance on the balloon varies during the
first 10 s of its flight.
Momentum and impulse
The momentum of a body is a property that depends on its mass and its velocity.
Linear momentum is defined as the product of mass and velocity. Linear simply
means in a straight line.
Note that the definition refers to velocity, not speed.
momentum mass velocity
p mv
p
m
v
Momentum is a vector quantity.
is the momentum measured in kg m s1
is the mass measured in kg
is the velocity measured in m s1
Given that mass usually remains constant, this is often written:
p mv
Worked example
Find the momentum of a car of mass 600 kg travelling at a constant velocity of
72 km h1.
Solution
72 1000 m s1
72 km h1 _________
(60 60)
20 m s1
p mv
p 600 kg 20 m s1
p 12 000 kg m s1
Examiner’s hint: The velocity is not
given in SI units so we must convert.
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2
Mechanics
Conservation of momentum
Momentum is important for solving problems involving collisions and explosions.
We say that momentum is conserved in these interactions. This means that the
total momentum before an event is equal to the total momentum after the event.
Worked example
A ball of mass 1.0 kg is dropped and hits the floor with a velocity of 8 m s1.
It bounces back with a velocity of 6 m s1. Find the change of momentum.
Figure 2.18
8 ms1
6 ms1
For a vector problem like this,
we must use a convention for
direction. Usually we take the
directions up or to the right as
positive and down or to the left as
negative.
Figure 2.19 Vector quadrant.
Solution
The change in momentum final momentum initial momentum
p mv mu
p (1.0 kg 6 m s1) (1.0 kg 8 m s1)
p 6 kg m s1 8 kg m s1
p 14 kg m s1
This idea can be expressed as a problem-solving equation in different ways:
m1v1 m2v2
The law of conservation of linear
momentum states that the total
momentum remains constant in
any interaction, providing there is
no external force.
m1v1 m2v2 m3v3
m1v1 m2v2 m3v3
The version we choose depends on the problem, whether it is a collision or
explosion and whether the two bodies stick together after the collision.
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Worked example
A car of mass 600 kg travelling at 20 m s1 collides head on with a truck of mass
2400 kg. If both vehicles come to rest immediately after the crash what was the
velocity of the truck?
Solution
m1v1 m2v2 m3v3
(600 kg 20 m s1) (2400 kg v2) (3000 kg 0 m s1)
12 000 kg m s1 2400v2 0
12 000
v2 ______
2400
v2 5 m s1
Examiner’s hint: Since the bodies
involved stick together after the collision,
use this equation.
Examiner’s hint: The final velocity,
v3, is zero, and we will take the direction
of motion of the car as positive.
6000
Impulse
5000
The change in momentum of a body is also known as the impulse.
Impulse is defined as the product of a force and the time during which it acts.
4000
The unit for impulse is Ns and this can also be used as a unit of momentum.
Imagine what happens if a girl kicks a football and a boy kicks a brick of the same
mass, say 500 g. Both the ball and the brick are accelerated from rest to a velocity
of 10 m s1.
The difference is that the brick is harder than the ball so the girl’s foot is in contact
with the ball for a longer time interval. If the girl’s foot is in contact with the ball
for 0.2 s then we can calculate the force exerted like this:
Ft mv
F 0.2 s 0.5 kg 10 m s1
5
F ___
0.2
F 25 N
If the boy’s foot is in contact with the brick for 0.01 s then the force would be
much greater:
Ft mv
F 0.01 s 0.5 kg 10 m s1
5
F ____
0.01
F 500 N
Area under line A
force (N)
Impulse Ft mv
B
1
2
1.0s 500 N
250 N s
3000
Area under line B
2000
1
2
0.1s 5000 N
250 N s
1000
500
0 0.1
A
0.5
1.0
time (s)
1.5
This graph shows a simplified version of
the forces exerted on a person involved
in a car crash. Line A shows the force
exerted by the steering wheel on a
driver wearing a seat belt. Line B shows
the force if the driver were not wearing
a seat belt. In both cases, the impulse,
or change in momentum, seen as the
area under the graph, is the same.
Clearly, the greater force, shown in Line
B, will be more damaging to the driver.
Figure 2.20 It is not a good idea to
kick a brick. For the same change in
momentum, if the time of contact is
smaller, the force will be greater. We say
that the time is inversely proportional
to the force.
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2
Mechanics
Remember that Newton’s second law is:
F ma
As we saw earlier in the chapter:
v
a ___
t
Substituting for a in the second law gives:
____
F mv
t
Knowing that:
mv p
We can re-write Newton’s second law as:
p
F ___
t
Notice this can be rearranged to give the impulse equation:
Ft p mv
In words we say that impulse is the change in momentum while force is the rate
of change of momentum.
Newton’s third law of motion
Newton’s third law states that if body A exerts a force on body B, then body B
exerts an equal and opposite force on body A.
Clearly there must always be two different bodies involved in the third law.
An easy way to demonstrate this is with two spring balances.
N
0
5
10
15
20
25
30
A
Education is not the same in
every country and depends on
differences in language, politics,
culture and religion, among many
other things. In all educational
systems there is however,
agreement in the validity of the
ideas behind Newton’s three laws
of motion.
15 N
N
0
5
10
15
20
25
30
15 N
Figure 2.21 Two spring balances.
The reading on A will always be the
same as the reading on B.
B
There are examples of third law pairs of forces all
around us; here are just a few examples.
• When you kick a football (or a brick), your
foot exerts a force on the ball and the ball
exerts an equal and opposite force on your
foot. The ball moves, hopefully towards the
goal, and you can feel the force on your foot.
• When a cannon is fired, the cannon ball
moves forward but the ball exerts an equal
and opposite force on the cannon, so it recoils.
• If you play tennis, your racket exerts a force
on the ball and the ball exerts an equal and
opposite force on your racket.
• The Earth exerts a force on the Moon and the
Moon exerts an equal and opposite force on
the Earth. We know the Earth pulls the Moon
because it remains in orbit around us and we
know the Moon pulls on the Earth because it
influences the tides in the oceans.
The photo shows the racket strings
being stretched and deformed.
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2.3
Work, energy and power
Assessment statements
2.3.1 Outline what is meant by work.
2.3.2 Determine the work done by a non-constant force by interpreting a
force-displacement graph.
2.3.3 Solve problems involving the work done by a force.
2.3.4 Outline what is meant by kinetic energy.
2.3.5 Outline what is meant by change in gravitational potential energy.
2.3.6 State the principle of conservation of energy.
2.3.7 List different forms of energy and describe examples of the
transformation of energy from one form to another.
2.3.8 Distinguish between elastic and inelastic collisions.
2.3.9 Define power.
2.3.10 Define and apply the concept of efficiency.
2.3.11 Solve problems involving work, energy and power.
Physical work
Sometimes in science we use words in a different sense to the way they are
normally used in the English language. Although a word often looks, sounds and is
spelt the same, it sometimes has a different meaning in physics. One example is the
definition of work in the study of physics.
Work is defined as the force times the distance moved in the direction of the force.
This means that if there is no movement in the direction of the force, then no
work is done. If your car is stuck in the mud or the sand, you can push it as hard
as you like, and get completely exhausted, but if you do not move it then you have
done no work (on the car)!
Work is a scalar quantity and is measured in joules. It is independent of the mass
of the body being moved and the path taken. It only depends on the magnitude of
the force and the distance moved in the direction of the force.
The equation for work is:
W Fscos W
F
s
is the work measured in J
is the force measured in N
is the distance moved measured in m
is the angle between the applied force and the direction of motion
Worked example
applied force
40 N
A force of 40 N acting at 60° to the
horizontal pulls a block of mass 10 kg
a distance of 2.0 m across a smooth
surface. How much work is done?
1 kg
60°
Figure 2.22 Work done pulling a
block.
x
2m
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2
Mechanics
Solution
The size of the mass makes no difference to the work done.
If the surface is smooth, there is no friction, so the resultant force in the xdirection is the x-component of the 40 N force.
Since the x-component is adjacent to 60° angle:
x
cos 60° ___
40
x 40 cos 60° N
W Fscos
W 40 cos 60° N 2 m
W 40 J
When a series of weights are hung on a spring causing it to stretch, work is done.
The force is the weight and the distance moved is the extension of the spring. As
long as the spring is not stretched beyond a certain point, called the elastic limit,
then the force is directly proportional to the extension. A graph of force against
extension is a straight line through the origin.
F
force
(N)
O
extension (cm)
x
Figure 2.23 A graph of force against
extension for a stretched spring.
The work done by this nonconstant force is found from the area under the
graph.
work done _12 x F
W _12 x kx
W _12kx 2
The gradient k _Fx
F kx
k is the spring constant.
In physics, energy and work are very closely linked; in some senses they are the
same thing. Both are scalar quantities and both are measured in joules. If you have
energy, you can do work.
In general, we say that:
work done energy converted
Energy is defined as the ability to
do work.
W E
When work is done, energy is converted or changed to different types. In the case
of the spring being stretched, the work done increases the elastic potential energy
of the spring.
elastic potential energy _12 kx 2
Kinetic energy
Examiner’s hint: Kinetic energy is
often abbreviated to KE or Ek.
Kinetic energy (Ek) is the energy a body has because it is moving. The word kinetic
refers to motion.
Imagine a block of mass m accelerated from rest by a resultant force, F. After
travelling a distance s the mass has an acceleration a.
Examiner’s hint: Sometimes in an
exam you will be asked to derive an
equation, such as the kinetic energy
derivation here. Many students find this
difficult and there is really only one way
to improve your skills. You simply have to
practise and repeat them with the book
closed and then practise some more!
The work done, W Fs
but from Newton’s second law F ma
so that W mas (substituting for F)
From the suvat equations, v u 2 2as
since u 0 then v 2 2as
as _12v 2
Since W mas E
and the energy converted is kinetic energy
kinetic energy mas m _12 v 2
Ek _12 mv 2
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Worked example
Find the kinetic energy of a truck of mass 4000 kg travelling at a speed of 30 m s1
Solution
Ek _12mv2
Ek _12 4000 kg (30 m s1)2
Ek 1.8 106 J
Potential energy
Gravitational potential energy (Ep) is the energy a body has because of its position,
in particular, its vertical height relative to a given point.
Examiner’s hint: Potential energy is
often abbreviated to PE or Ep.
Now imagine a mass m lifted a vertical height h. Here work is being done against
the pull of gravity.
the work done, W Fs
here the force is the weight mg
and the distance is the height h
so the work W mgh
The work is increasing the gravitational potential energy of the mass.
W Ep
Ep mg h
Worked example
The Leaning Tower of Pisa is 56 m high.
(a) Find the gravitational potential energy of a 10 kg cannon ball at the top of the
tower.
(b) If the cannon ball is dropped from the top of the tower, what will be its speed
when it hits the ground?
(Neglect air resistance and take g 10 m s2)
Solution
(a)
Ep mg h
Ep 10 kg 10 m s2 56 m
Ep 5600 J
(b)
Ep Ek
5600 J _12 mv 2
5600 J _12 10 kg v 2
5600
v 2 ____
5
v 33 m s1
Examiner’s hint: We could solve
this using the suvat equations, but
if air resistance is neglected, all the
gravitational potential energy will be
changed to kinetic energy.
There is a famous story about
Galileo dropping cannon balls of
different weights from the top
of the Leaning Tower of Pisa to
demonstrate that heavy objects and
lighter objects fall at the same rate.
This story may or may not be true.
To view a more recent version
of Galileo’s experiment, visit
Heinemann.co.uk/hotlinks, enter
the express code 4266S and click on
Weblink 2.5.
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2
Mechanics
Conservation of energy
The idea of one form of energy changing into another is summarised in a very
important principle or law of physics, the principle conservation of energy.
We usually classify energy as the following forms:
The principle of conservation of
energy states that energy cannot
be created or destroyed; only
changed from one form to another.
1 Kinetic energy
2 Gravitational potential energy
3 Elastic potential energy
All of the energy on Earth has
come originally from the Sun. Is this
statement true? How do you know?
4 Thermal or heat energy
5 Light energy
6 Sound energy
7 Chemical energy
8 Electrical energy
9 Magnetic energy
10 Nuclear energy
Examiner’s hint: In an exam do not
say that energy is ‘lost’ but instead say it
has been changed or converted to other
forms.
In any given transformation, some of the energy is almost inevitably changed to
heat.
There are energy changes happening all around us all of the time. Here are a few
examples to think about.
• A bouncing ball
The gravitational potential energy is changed to kinetic energy and back again
with each bounce. While the ball is in contact with the ground, some energy is
stored as elastic potential energy. Eventually the ball comes to rest and all the
energy has become low grade heat.
• An aeroplane taking off
As fuel is burned in the engines, chemical energy is converted to heat, light and
sound. The plane accelerates down the runway and its kinetic energy increases.
There will be heat energy due to friction between the tyres and the runway. As the
plane takes off and climbs into the sky, its gravitational potential energy increases.
• A nuclear power station
Nuclear energy from the uranium fuel changes to thermal energy that is used to
boil water. The kinetic energy of the steam molecules drives turbines, and as the
kinetic energy of the turbines increases, it interacts with magnetic energy to give
electrical energy.
• A laptop computer
The electrical energy form the mains or chemical energy from the battery is
changed to light on the screen and sound through the speakers. There is kinetic
energy in the fan, magnetic energy in the motors, and plenty of heat is generated.
• A human body
Most of the chemical energy from our food is changed to heat to keep us alive.
Some is changed to kinetic energy as we move, or gravitational potential energy if
we climb stairs. There is also elastic potential energy in our muscles, sound energy
when we talk and electrical energy in our nerves and brain.
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There is a useful relationship between kinetic energy and momentum which will
help you to solve numerical problems.
mv 2
Ek _12 mv 2 ____
2
2 2
m
v
Multiplying top and bottom by m gives Ek _____
2m
Momentum
p mv
Squaring gives
Finally substituting in the above equation
p2 m 2v 2
p2
Ek ___
2m
Collisions
When we looked at momentum earlier, we mentioned that it was important for
collisions. Now we must distinguish between two different types of collisions.
When elastic collisions occur, momentum is conserved and kinetic energy is also
conserved. This means that no energy is changed to heat in an elastic collision. In
fact there is really no such thing as a perfectly elastic collision, although, of course,
some collisions are more elastic than others. After an elastic collision the bodies
always move separately.
Billiard ball collisions are highly elastic
because the balls are so hard. In a
perfectly elastic collision, one ball could
stop, and the other move off with the
same speed as the first. In practice, this
cannot happen because some of the
kinetic energy changes to heat and
sound.
Particle physicists believe some
collisions between sub-atomic
particles are perfectly elastic.
During inelastic collisions, momentum is conserved, but kinetic energy is not
conserved. In other words, some of the kinetic energy is converted to heat and
other forms of energy. After an inelastic collision the bodies may stick together.
Examiner’s hint: In exam questions
about collisions and explosions,
especially multiple-choice, assume that
momentum is always conserved.
Kinetic energy can sometimes increase, for example, if there is an input of
chemical energy from a fuel or explosive material.
Power
Power is a measure of how fast work is done or how quickly energy is converted.
Power is a scalar quantity.
energy or work
power _____________
time
If the energy or work is measured in joules and the time in seconds, then the
power will be measured in watts (W).
Power is defined as the rate of
doing work.
1 W 1 J s1
A 100 W light bulb converts electrical energy to heat and light at the rate of 100 J
every second.
Sometimes you will see power given in kW or even MW.
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2
Mechanics
Worked example
Calculate the power of a worker in a supermarket who stacks shelves 1.5 m high
with cartons of orange juice, each of mass 6.0 kg, at the rate of 30 cartons per
minute.
Solution
work
power _____
time
Fs
power __
t
(30 60 N) 1.5 m
power _________________
60 s
power 45 W
This engraving shows a steam engine
designed by James Watt.
James Watt (17361819) was
a Scotsman whose work on
the steam engine was a major
influence on the Industrial
Revolution.
From the days of Aristotle, through
the time of Galileo, Newton,
Watt and Joule right up until
Einstein in the early 20th century,
scientists worked mostly alone.
Nowadays, scientists often work in
multinational teams. These teams
may make important discoveries
in fields such as particle physics, or
work on solving the great global
problems like HIV/Aids and global
warming.
There is another way to calculate power.
Fs F _s Fv
power __
t
t
In other words, power is equal to force times velocity, providing the velocity is
constant.
Efficiency
One of the most efficient devices
is the electric transformer. They
can be over 99% efficient because,
since there are no moving parts,
there are no energy losses due to
friction or air resistance.
Efficiency is a ratio of how much work, energy or power we get out of a system
compared to how much is put in.
useful output
efficiency ___________
total input
Since it is a ratio, there is no unit, and we can express efficiency as a percentage by
multiplying by 100%.
No real machine or system can ever be 100% efficient, because there will always be
some energy changed to heat due to friction, air resistance or other causes.
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Worked example
A car engine has an efficiency of 20% and produces an average of 25 kJ of useful
work per second. How much energy is converted to heat per second?
Solution
useful output
efficiency ___________
total input
25 000 J
0.2 _________
total input
Examiner’s hint: The efficiency is
20% which equals 0.2.
Examiner’s hint: 25 kJ 25 000 J.
25 000 125 000 J
total input ______
0.2
If the total input is 125 kJ, and 25 kJ of useful work are obtained then the heat
produced must be:
125 kJ 25 kJ 100 kJ
Exercises
7 This question is about driving a metal bar into the ground.
Large metal bars can be driven into the ground using a heavy falling object.
object
mass 2.0 103 kg
bar
mass 400 kg
In the situation shown, the object has a mass 2.0 × 103 kg and the metal bar has a mass of 400 kg.
The object strikes the bar at a speed of 6.0 m s1. It comes to rest on the bar without bouncing.
As a result of the collision, the bar is driven into the ground to a depth of 0.75 m.
(a) Determine the speed of the bar immediately after the object strikes it.
(b) Determine the average frictional force exerted by the ground on the bar.
8. This question is about estimating energy changes for an escalator (moving staircase).
The diagram below represents an escalator. People
B
step on to it at point A and step off at point B.
(a) The escalator is 30 m long and makes an angle
of 40° with the horizontal. At full capacity,
30 m
48 people step on at point A and step off at
point B every minute.
(i) Calculate the potential energy gained
by a person of weight 7.0 × 102 N in
40°
moving from A to B.
A
(ii) Estimate the energy supplied by
the escalator motor to the people every minute when the escalator is working at full
capacity.
(iii) State one assumption that you have made to obtain your answer to (ii).
The escalator is driven by an electric motor that has an efficiency of 70 %.
(b) Using your answer to (a) (ii), calculate the minimum input power required by the motor to
drive the escalator.
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2
Mechanics
2.4
Uniform circular motion
Assessment statements
2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle
moving with constant speed in a circle is directed towards the centre of
the circle.
2.4.2 Apply the expression for centripetal acceleration.
2.4.3 Identify the force producing circular motion in various situations.
2.4.4 Solve problems involving circular motion.
Motion in a circle
If a body is travelling in a circular path, then it is constantly changing direction;
at no point is it travelling in a straight line. Even if it is travelling at constant or
uniform speed, a body that is moving round a circle must be changing velocity.
This is because velocity is a vector and a change in velocity can be a change of
speed or direction.
If the velocity of the body is changing, then, by definition the body must be
accelerating. This type of acceleration is called centripetal acceleration. The word
centripetal means centre-seeking and the direction of the acceleration is always
towards the centre of the circle.
v
Figure 2.25 A particle moving in a
circle. The instantaneous velocity, v, is
the tangent to the circle. The direction
of the centripetal force will be in the
same direction as the centripetal
acceleration.
a
The equation for centripetal acceleration is:
v2
a __
r
a centripetal acceleration m s2
v speed
m s1
r radius of the circle
m
Since the body moves with constant speed we can use the simple equation:
distance
speed _______
time
If the body goes round the entire circle once, then this can be written:
r
v 2____
T
2 r is the circumference of a circle.
T is the period (or time period) and is the time taken for one complete
revolution.
If we square this equation it becomes:
4 r
v 2 _____
T2
2 2
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Substituting for v 2 in the centripetal acceleration and cancelling one of the r terms
gives:
2r
a 4____
T2
If a body is accelerating, we know from Newton’s second law that it must be being
acted on by an unbalanced force. The force causing the centripetal acceleration is
called the centripetal force. The direction of the centripetal force is also towards the
centre of the circle; this is logical because bodies accelerate in the direction of the
resultant force.
To understand this, think about a car driving round a circular track. If the friction
is too low, for example, if the car skids on a patch of ice, then the car will fly off at
a tangent.
The centripetal force is simply an
extra name we give to a force that
is already there; it is not a new or
different force.
Figure 2.26 There must be friction
between the car tyres and the road to
keep the car going round.
v
Fc
The friction is the force that turns the car into the bends and it is directed towards
the center of the circle. The friction here is the centripetal force.
In the same way, when the Moon orbits the Earth there is a centripetal force from
the Moon towards the centre of the Earth. In this case the force is gravitational. If
you whirl a rubber bung or a stone tied to a piece of string around your head, then
the centripetal force that keeps the stone turning is the tension in the string. If the
string breaks and the tension becomes zero, the bung or stone flies off in a straight
line.
Examiner’s hint: If you are asked to
draw the force on a body moving in a
circle, always draw it pointing towards
the centre of the circle. Do not be
tempted to add another arrow pointing
outwards to balance the forces; they
are not balanced because the body is
accelerating.
If the equation for centripetal acceleration is:
v2
a __
r
and we know that
F ma
we can substitute for a to give:
mv 2
F ____
r
This is the equation for centripetal force.
Worked example
A ball of mass 50 g tied to the end of a piece of string is whirled by a student in a
horizontal circle of radius 1.2 m at constant speed. The ball makes 1.5 revolutions
per second.
(a) Find the acceleration of the ball.
(b) Find the force that the student must exert on the string.
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2
Mechanics
Solution
distance
speed _______
time
1.5
2 1.2 m
v ______________
1.0 s
v 11.3 m s1
(a)
Examiner’s hint: Since the ball is
moving in a circle, the acceleration will
be centripetal.
2
v
a __
r
(11.3 m s1)2
a ___________
1.2 m
107 m sⴚ2
2
mv
F ____
r
(b)
0.050 kg (11.3 m s1)2
____________________
1.2 m
5.3 N
Exercises
9
An aircraft accelerates from rest along a horizontal straight runway and then takes off. The mass
of the aircraft is 8.0 × 103 kg.
(a) The average resultant force on the aircraft while travelling along the runway is 70 kN. The
speed of the aircraft just as it lifts off is 75 m s1. Estimate the distance travelled along the
runway.
(b) The aircraft climbs to a height of 1250 m. Calculate the potential energy gained during the
climb.
(c) When approaching its destination, the pilot puts the aircraft into a holding pattern. This
means the aircraft flies at a constant speed of 90 m s1 in a horizontal circle, and the radius
of the circle is 500 m.
For the aircraft in the holding pattern,
(i) calculate the magnitude of the resultant force on the aircraft
(ii) state the direction of the resultant force.
10 A geostationary satellite, for example a communications satellite, orbits the Earth with a time
period of 24 hours. If the distance from the satellite to the centre of the Earth is 4.2 107 m,
calculate the acceleration of the satellite.
Practice questions
1 A ball, initially at rest, takes time t to fall through a vertical distance h. If air resistance is
ignored, the time taken for the ball to fall from rest through a vertical distance 9h is
A 3t
B 5t
C 9t
D 10t
2 A raindrop falling through air reaches a terminal velocity before hitting the ground. At
terminal velocity, the frictional force on the raindrop is
A zero.
B less than the weight of the raindrop.
C greater than the weight of the raindrop.
D equal to the weight of the raindrop.
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3 An athlete runs round a circular track at constant speed. Which one of the
following graphs best represents the variation with time t of the magnitude d of the
displacement of the athlete from the starting position during one lap of the track?
A d
0
0
C
B d
0
0
t
D d
d
0
0
t
0
t
4 A ball is dropped from rest at
time t 0 on to a horizontal
surface from which it rebounds.
The graph shows the variation
of time t with speed v of
the ball.
0
t
v
A
C
0
B
D
t
Which one of the following best represents the point at which the ball just loses contact
with the surface after the first bounce?
A
B
C
D
5 Juan is standing on the platform at a railway station. A train passes through the station
with speed 20 m s1 in the direction shown measured relative to the platform. Carmen
is walking along one of the carriages of the train with a speed of 2.0 m s1 measured
relative to the carriage in the direction shown. Velocity is measured as positive in the
direction shown on the diagram.
The velocity of Carmen relative
to Juan is
A 22 m s1
20 ms1
B 18 m s1
C 18 m s1
D 22 m s1
Carmen
2.0 ms1
Juan
platform
positive direction
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2
Mechanics
6 The graph below shows the variation with time of the distance moved by a car along a
straight road. During which time interval does the car have its greatest acceleration?
distance
moved
A
B
C
time
D
7 The variation with time t of the speed v of a car moving along a straight road is shown
below.
v
S1
0
0
S2
S3
t
Which area, S1, S2 or S3, or combination of areas, represents the total distance moved
by the car during the time that its speed is reducing?
A S1
B S3
C S1 S3
D S1 S2 S3
8 When a body is accelerating, the resultant force acting on it is equal to its
A change of momentum.
B rate of change of momentum.
C acceleration per unit of mass.
D rate of change of kinetic energy.
9 A ball of mass m, travelling in a direction at right angles to a vertical wall, strikes the
wall with a speed v1. It rebounds at right angles to the wall with a speed v2. The ball is
in contact with the wall for a time t. The magnitude of the force that the ball exerts on
the wall is
m (v1 v2)
A _________
t
D m (v1 v2)t
m (v1 v2)
B _________
t
C m (v1 v2)t
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10 A sphere of mass m strikes a vertical wall and bounces off it, as shown below.
momentum pB
momentum pA
The magnitude of the momentum of the sphere just before impact is pB and just after
impact is pA. The sphere is in contact with the wall for time t. The magnitude of the
average force exerted by the wall on the sphere is
(pB pA)
A ________
t
pB pA)
(________
B
t
(pB pA)
________
C
mt
pB pA)
(________
D
mt
11 A truck collides head on with a less massive car moving in the opposite direction to
the truck. During the collision, the average force exerted by the truck on the car is F T
and the average force exerted by the car on the truck is FC. Which one of the following
statements is correct?
A F T will always be greater in magnitude than FC.
B F T will always be equal in magnitude to FC.
C F T will be greater in magnitude than FC only when the speed of the car is less than
the speed of the truck.
D F T will be equal in magnitude to FC only when the speed of the truck is equal to the
speed of the car.
12 A small boat in still water is given an initial horizontal push to get it moving. The boat
gradually slows down. Which of the following statements is true for the forces acting on
the boat as it slows down?
A There is a forward force that diminishes with time.
B There is a backward force that diminishes with time.
C There is a forward force and a backward force both of which diminish with time.
D There is a forward force and a backward force that are always equal and opposite.
13 The graph below shows the variation with displacement 5
d of the force F applied by a spring on a cart.
4
The work done by the force in moving the cart through
a distance of 2 cm is
A 10 102 J
3
B 7 102 J
F/N
C 5 102 J
2
D 2.5 102 J
1
0
0
1
2
d/102 m
3
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2
Mechanics
14 Which one of the following is a true statement about energy?
A Energy is destroyed due to frictional forces.
B Energy is a measure of the ability to do work.
C More energy is available when there is a larger power.
D Energy and power both measure the same quantity.
15 An electric train develops a power of 1.0 MW when travelling at a constant speed of
50 m s1. The net resistive force acting on the train is
A 50 MN
B 200 kN
C 20 kN
D 200 N
16 Which of the following quantities is/are conserved in an inelastic collision in an isolated
system of two objects?
Linear momentum of system Kinetic energy of system
A
Yes
Yes
B
Yes
No
C
No
Yes
D
No
No
17 The diagram below represents energy transfers in an engine.
input energy
EIN
useful output energy
EOUT
wasted energy
EW
The efficiency of the engine is given by the expression
EW
A ___
EIN
EW
B ____
EOUT
EOUT
C ____
EIN
EOUT
D ____
EW
18 A machine lifts an object of weight 1.5 103 N to a height of 10 m. The machine has
an overall efficiency of 20%. The work done by the machine in raising the object is
A 3.0 × 103 J
B 1.2 × 104 J
C 1.8 × 104 J
D 7.5 × 104 J.
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19 An electric motor, with an input power of 250 W, produces 200 W of mechanical power.
The efficiency of the motor is
A 20%.
B 25%.
C 55%.
D 80%.
20 A stone of mass m is attached to a string and moves round in a horizontal circle of
radius R at constant speed V. The work done by the pull of the string on the stone in
one complete revolution is
A zero
B 2mV 2
2mV 2
C ______
R
2
mV
2
______
D
R
21 Two objects X and Y are moving away from the point P. The diagram below shows the
velocity vectors of the two objects.
velocity vector for object Y
P
velocity vector for object X
Which of the following velocity vectors best represents the velocity of object X relative
to object Y?
A
B
22 This question is about throwing
a stone from a cliff.
Antonia stands at the edge of a
vertical cliff and throws a stone
vertically upwards.
C
D
v 8.0 m s1
The stone leaves Antonia’s hand
with speed v 8.0 m s1.
The acceleration of free fall g is
10 m s2 and all distance
measurements are taken from the
sea
point where the stone leaves
Antonia’s hand.
(a) Ignoring air resistance, calculate
(i) the maximum height reached by the stone.
(ii) the time taken by the stone to reach its maximum height.
The time between the stone leaving Antonia’s hand and hitting the sea is 3.0 s.
(b) Determine the height of the cliff.
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2
Mechanics
23 This question is about the collision between two railway trucks (carts).
(a) Define linear momentum.
In the diagram on the right,
truck A is moving along a
horizontal track. It collides
with a stationary truck B, and
on collision, the two join
together. Immediately before
the collision, truck A is moving
with speed 5.0 m s1.
Immediately after collision, the
speed of the trucks is v.
5.0 m s1
A
B
immediately before
f
collision
v
A
B
immediately after
f collision
The mass of truck A is 800 kg
and the mass of truck B is 1200 kg.
(b) (i) Calculate the speed v immediately after the collision.
(ii) Calculate the total kinetic energy lost during the collision.
24 An elevator (lift) starts from rest on the ground floor and comes to rest at a higher floor.
Its motion is controlled by an electric motor. A simplified graph of the variation of the
elevator’s velocity with time is shown below.
0.80
0.70
0.60
0.50
velocity
0.40
( ms1)
0.30
0.20
0.10
0.00
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0 11.0 12.0
time (s)
The mass of the elevator is 250 kg. Use this information to calculate
(a) the acceleration of the elevator during the first 0.50 s
(b) the total distance travelled by the elevator
(c) the minimum work required to raise the elevator to the higher floor
(d) the minimum average power required to raise the elevator to the higher floor
(e) the efficiency of the electric motor that lifts the elevator, given that the input power
to the motor is 5.0 kW.
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25 This question is about momentum and the kinematics of a proposed journey to Jupiter.
(a) State the law of conservation of momentum.
A solar propulsion engine uses solar power to ionize atoms of xenon and to accelerate
them. As a result of the acceleration process, the ions are ejected from the spaceship
with a speed of 3.0 104 m s1.
spaceship
mass 5.4 102 kg
xenon ions
speed 3.0 104 m s1
(b) The mass (nucleon) number of the xenon used is 131. Deduce that the mass of one
ion of xenon is 2.2 1025 kg.
(c) The original mass of the fuel is 81 kg. Deduce that, if the engine ejects 7.7 1018
xenon ions every second, the fuel will last for 1.5 years. (1 year 3.2 107 s)
(d) The mass of the spaceship is 5.4 × 102 kg. Deduce that the initial acceleration of
the spaceship is 8.2 × 105 m s2.
The graph below shows the variation with time t of the acceleration a of the spaceship.
The solar propulsion engine is switched on at time t 0 when the speed of the
spaceship is 1.2 103 m s1.
10.0
9.5
a/105 m s1
9.0
8.5
8.0
0.0
1.0
2.0
3.0
t/107 s
4.0
5.0
6.0
(e) Explain why the acceleration of the spaceship is increasing with time.
(f) Using data from the graph, calculate the speed of the spaceship at the time when
the xenon fuel has all been used.
(g) The distance of the spaceship from Earth when the solar propulsion engine is
switched on is very small compared to the distance from Earth to Jupiter. The fuel
runs out when the spaceship is a distance of 4.7 1011 m from Jupiter. Estimate
the total time that it would take the spaceship to travel from Earth to Jupiter.
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