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Physics Unit 3 2015
3. Energy
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3. Energy
Study Design
• apply laws of energy and momentum conservation in isolated systems;
• apply the concept of work done by a constant force
– work done = constant force × distance moved in direction of net force
– work done = area under force-distance graph;
• analyse transformations of energy between: kinetic energy; strain potential energy; gravitational
potential energy; and energy dissipated to the environment considered as a combination of heat,
sound and deformation of material
– kinetic energy, i.e. ½mv2; elastic and inelastic collisions in terms of conservation of kinetic energy
– strain potential energy, i.e. area under force-distance graph including ideal springs obeying
Hooke’s Law, ½kx2
– gravitational potential energy, i.e mgΔh or from area under force-distance graph and area under
field-distance graph multiplied by mass;
Introduction
……………………………….
Energy exists in many different forms, for example Kinetic Energy, Gravitational Potential Energy,
Electrical Energy, Elastic Potential Energy. A fundamental principle of nature is that energy cannot
be created or destroyed, only transformed or transferred to another body. A body that has energy
may transfer some, or all, of its energy to another body. The total amount of energy remains
constant (conserved), even if it has been transformed to another type of energy. For example,
during a car crash, the car originally has kinetic energy. After the collision it will have less kinetic
energy than it began with. The lost kinetic energy will have been transferred to sound energy, heat
energy or energy of deformation.
Energy, Work and Power**
The amount of energy transferred is called
work. The body losing energy does work,
the body gaining energy has work done on
it.
Work = Force  displacement
Work is a scalar quantity,
The unit of Work is the Joule.
Work can also be expressed in units of
power  time, this is the origin of the unit
kilowatt hours. Often the energy being
used is converted into heat energy.
**Power is the rate of doing work, and
therefore the rate of increase of
energy. The average power, P, is the
total work done divided by the total
W
Fd
time interval. P =
=
= Fv
Δt
Δt
if the change in velocity is not from
zero, then Psupplied = F v
Power is a scalar quantity with the
units of Joule sec-1 or Watt.
** Not strictly on the course, but can be
a useful idea.
A joule is the amount of work done
when a force of 1 Newton acts through
a distance of 1 metre.
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Types of Energy
Kinetic Energy (KE) is the energy a body
possesses due to its motion. This will be
equal to the work done by a force, F, on a
mass, m, to give it from rest a velocity, v, over
a displacement, x.
v2 = u2 + 2ax, but u = 0.
 v2 = 2 a x = 2
x
 mv2 = F x
F x is the work done and
Potential Energy (PE), (U), is the
energy possessed by a body due to:
Its state (elastic potential energy)

e.g. a compressed spring, or
Its position (gravitational potential

energy) e.g. a raised mass.
mv2 is the kinetic
energy gained by the body. So KE =
mv2.
The work done on a body is equal to the
change in KE of a body, so
WD =
= mv2 - mu2
Gravitational Potential Energy.
When changes in height 'h' are small
compared to the radius of the earth, the
potential energy Ug of a body near the
earth's surface is given by Ug = mgh.
Efficiency of Energy Conversions.
In most real life situations when energy is transferred from one object to another, not all of the
energy is transferred in a useful form. Some of the energy is turned into types of energy that are
not desired, such as heat, sound or light. The efficiency of the energy transfer is a measurement
of how much of the energy is transferred to the desired form of energy.
Final energy
100
%
% Efficiency =

1
Initial energy
Collisions
Elastic collisions
If the collision is elastic, then both
momentum and energy is
conserved in the collision.
 Efinal = Einitial and pfinal = pinitial
Inelastic collisions
If the collision is inelastic, momentum
is conserved, but some energy is ‘lost’
to the environment. The energy that
is ‘lost’ to the environment is usually
transformed into heat energy, sound
energy and energy of deformation.
 pfinal = pinitial but Einitial > Efinal
Energy and Graphs
The area under a force-displacement graph shows the work done. If the force is constant then the
area under the graph is given by W = F  d where F is the force, d is the distance over which the
force acts. This is just the familiar work formula. If the force is not constant then the area under
the graph must be determined.
This assumes that the force and the displacement are in the same direction. If they aren't then the
work is the product of the resolved part of the force (in the direction of motion)  the displacement.
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A delivery van of mass 1200 kg, travelling south at 20 m s–1, collides head-on with a power pole.
The impact crushes the crumple zone of the van by 0.60 m bringing the van to rest against the
pole.
Example 1 2004 Question 4 (3 marks)
Calculate the average force that the pole exerts on the van.
A car is on top of a hill at X, h metres above the top of a cliff. The top of the cliff is H metres above
the water level. The brakes are released and the car begins to roll back down the hill. When the car
reaches the cliff at Y it is projected horizontally and travels a horizontal distance, d metres, from
the cliff edge. It enters the water at Z.
Take the acceleration due to gravity as g downwards. (For the following questions, ignore air
resistance.)
Example 2 2004 Question 12
(2 marks)
Which of the expressions (A – D) gives the speed of the car at point Y?
A.
2gh
B.
2gH
C.
2g(H+ h)
D.
2g(H - h)
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Jo is riding on a roller-coaster at a fun fair. Part of the structure is shown below
When Jo is at X her velocity is 10 ms-1 in a horizontal direction, and at Y it is 24 ms-1 in a horizontal
direction.
Example 3 2000 Question 14
(3 marks)
What is the height difference (h) between points X and Y? Assume that friction and air resistance
are negligible.
A car – truck crash can be modelled as a ‘head-on’ collision between a truck of mass 4000 kg
travelling at 15 ms-1 and a stationary car of mass 1000 kg.
After the collision the truck continues moving forward at 10 ms-1.
Example 4 1998 Question 4 (3 marks)
Calculate the speed of the car immediately after the collision.
Example 5 1998 Question 5 (2 marks)
Calculate the combined total kinetic energy of the truck and car immediately before the collision.
Example 6 1998 Question 6 (2 marks)
Calculate the combined total kinetic energy of the truck and car immediately after the collision.
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Example 7 1998 Question 7 (3 marks)
Compare the magnitudes of the total kinetic energies before and after the collision as calculated in
Questions 5 and 6. Explain any differences that you observe.
A particle of mass m, travelling south-east at constant speed v, hits a wall and then travels northeast at the same speed v.
Example 8 1980 Question 16
(1 mark)
How much work has been done on the particle by the wall?
Example 9 1980 Question 17
(1 mark)
The acceleration of the particle while it is in contact with the wall is
A.
zero
B.
towards the east
C.
towards the west
D.
towards the north
E.
changing from south-east to north-east
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Hooke’s Law
Extending a spring is an example of a situation where the force on an object is not constant. As the
spring gets compressed, the force required to further extend it increases. The mathematical
equation that represents this type of situation is called Hooke’s Law
F = kx
Where F is the magnitude of the force required, x is the extension (or compression) of the spring,
a k is called the spring constant. The spring constant has a specific value for each individual
spring. It depends on the size, thickness and material from which the spring is made. It is very
important to use x, as this demonstrates that you are only interested in the change in the length of
the spring.
The equation is illustrated in the graph below
Force on the
spring
(N) potential energy (strain
Elastic
F=kx
energy) is the energy stored in
F
any
material
that
has been
The
energy
stored
in the spring
1
stretched =
or compressed
from its
base

height
2
normal shape.
= 21 . x. kx
2bands are
Springs and
k( x)x
= 21elastic
Extension of the
good examples.
(m) work done,
The energy stored in the spring isspring
also the
which is the area under the graph.
The energy stored in the
extended spring can be
determined by calculating the
area under the force-distance
graph. This can be done directly
from the graph, or using the
formula derived in the box to the
left.
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This question is straight off the 2011 mid-year exam. It is difficult and requires some serious
thinking about the solution.
Physics students are conducting an experiment on a spring which is suspended from the ceiling.
Ignore the mass of the spring.
Without the mass attached, the spring has an unstretched length of 40 cm. A mass of 1.0 kg is
then attached. When the 1.0 kg mass is attached, with the spring and mass stationary, the spring
has a length of 70 cm.
Example 10 2011 Question 16
What is the spring constant, k, of the spring?
To find the spring constant use the formula mg = kx.
(always use x, in this case it is 70 – 40 = 30 cm, therefore x = 0.3 m.)
This give 1.0 x 10 = k x 0.30
k=
10
= 33.3 Nm-1
0.3
You must remember to use metres as your default measurement.
(Average score 1.1/2)
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The spring is now pulled down a further 10 cm from 70 cm to 80 cm and released so that it
oscillates. Gravitational potential energy is measured from the point at which the spring is released
(80 cm on previous figure).
Example 11 2011 Question 17
Which of the graphs (A – H) best shows the variation of the kinetic energy of the system plotted
against the length of the stretched spring?
The kinetic energy of the system, is considered as the KE of the mass. At both extremes of the
masses motion, it momentarily comes to rest, so the KE at these points is zero. The mass has its
maximum speed in the middle of the motion, therefore the max KE occurs in the middle (at 70 cm)
Therefore graph D is best.
(Average score 0.4/1)
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Example 12 2011 Question 18
Which of the graphs (A – H) best shows the variation of the total energy of the system of spring
and mass plotted against the length of the stretched spring?
The total energy of the system must remain unchanged because there are not any forces adding
energy to the system by doing work.
Therefore the graph needs to be a straight horizontal line
Therefore graph C is best.
(Average score 0.7/1)
Example 13 2011 Question 19
Which of the graphs (A – H) best shows the variation of the gravitational potential energy of the
system of spring and mass (measured from the lowest point as zero energy) plotted against the
length of the stretched spring?
Gravitational potential energy is given by mgh. This is a linear function, so the graph will be a
straight line. It must have zero GPE at 80 cm.
Therefore graph B is best.
(Average score 0.5/1)
Example 14 2011 Question 20
Which of the graphs (A – H) best shows the variation of the spring (strain) potential energy of the
system of spring and mass plotted against the length of the stretched spring? Give reasons for
choosing this answer for the spring (strain) potential energy.
Spring (strain) energy is given by ½ k(x)2. The spring energy increases as the length of the spring
changes from 60 cm to 80 cm, as x increases from 20 cm to 40 cm. The graph is not linear, (from
E = ½ k(x)2), hence the parabolic curve. The curve should tend to zero when x  0, i.e 40 cm
length.
Therefore graph F is best.
(Average score 0.8/3)
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A novelty toy consists of a metal ball of mass 0.20 kg hanging from a spring of spring constant
k = 10 N m-1. The spring is attached to the ceiling of a room as shown below. Ignore the mass of
the spring.
0
cm
40
55
60
65
10 cm
Without the ball attached, the spring has an unstretched length of 40 cm. When the ball is
attached, but not oscillating, the spring stretches to 60 cm.
Example 15 2008 Question 12
How much energy is stored in the spring when the ball is hanging stationary on it? You must show
your working.
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The ball is now pulled down a further 5 cm and released so that it oscillates vertically over a
range of approximately 10 cm.
Gravitational potential energy is measured from the level at which the ball is released. Ignore air
resistance.
Use Graphs A–E in answering Questions 13 and 14.
Example 16 2008 Question 13
Which of the graphs best represents the shape of the graph of kinetic energy of the system as a
function of height?
Example 17 2008 Question 14
Which of the graphs best represents the gravitational potential energy of the system as a
function of height?
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In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second floor, and slides
from rest to the ground floor below, as shown below. The slide has a linear length of 6.0 m, and is
designed to provide a constant friction force of 50 N on the box. The box reaches the end of the
slide with a speed of 8.0 m s–1.
Example 18 2004 Pilot Question 14
What is the height, h, between the floors?
(4 marks)
(4 marks)
The box then slides along the frictionless floor, and is momentarily stopped by a spring of stiffness
30 000 N m–1.
Example 19 2004 Pilot Question 15
(3 marks)
How far has the spring compressed when the box has come to rest?
(3 marks)
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A car, equipped with a driver’s air bag, hits a large tree while travelling horizontally at 54 km h-1
(15 m s-1). The air bag is designed to protect the driver’s head in a collision.
For Questions 7 – 9, model this as a collision involving the driver’s head (mass 8.0 kg).
Tests show that the graph of retarding force on the driver’s head versus compression distance of
the air bag is as shown below.
Example 20 2000 Question 7
(4 marks)
Calculate the maximum compression distance of the air bag in this collision.
Example 21 2000 Question 8
(2 marks)
Which one of the graphs (A–D) best represents the retarding force versus compression distance if
the collision was with the hard surface of the steering wheel rather than the air bag?
Example 22 2000 Question 9
(3 marks)
Explain your answer to Question 8, giving specific reasons for choosing the graph that you
selected as the best answer.
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In a horizontal pinball machine, a 0.10 kg ball rests against a plunger of mass 0.10 kg which can
be pulled back against a light spring as shown.
The spring constant is 250 Nm-1.
Example 23 1983 Question 27
(1 mark)
What force must the player exert on the plunger to compress the spring by 0.040 m?
Example 24 1983 Question 28
(1 mark)
In compressing the spring, how much work has been done?
Example 25 1983 Question 29
(1 mark)
When the player releases the handle, with what velocity will the ball leave the plunger?
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The spring constant (k) of a spring can be defined as the force per unit distance required to extend
the spring. Thus a strong spring will have a large value of k, while a weak spring will have a small
value. The force-extension curve of a spring is shown. Take g = 10 ms-2
Example 26 1981 Question 21
(1 mark)
What is the value of k, the spring constant? (Give both magnitude and unit)
Example 27 1981 Question 22
(1 mark)
A 10 kg mass hangs at rest from the spring. How much energy is stored in the spring?
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The diagram below shows a section of a roller coaster track. The roller coaster car is travelling at
constant speed towards the right.
Example 28 2001 Question 7
(2 marks)
Which of the graphs (A–E) best shows the gravitational potential energy of the roller coaster car
against the horizontal distance from the start of the track?
Example 29 2001 Question 8
(2 marks)
Which of the graphs (A–E) best shows the kinetic energy of the roller coaster car against the
horizontal distance from the start of the track? (Neglect frictional effects.)
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Example 30 2001 Question 9
(1 mark)
Which of the graphs (A – E) best shows the total energy of the roller coaster car against the
horizontal distance from the start of the track? (Neglect frictional effects.)
Jane, an adventurous tourist from New Zealand (and mate of Tarzan), decides to try the new sport
of falling off a bridge with a ‘bungee’ attached to her ankles. The ‘bungee’ is a stretchy rope which
can be considered as an ideal spring with a spring constant of 17 Nm-1.
Jane has a mass of 50 kg, and in jumping from the bridge, her centre of mass falls through a
distance of 13 m before the ‘bungee’ starts to exert a force on her. Take g= 10ms-2.
Example 31 1990 Question 25
(1 mark)
Assuming air resistance is negligible, what is Jane’s speed when the ‘bungee’ first exerts an
upward force on her?
Example 32 1990 Question 26
(1 mark)
How much energy is stored in the ‘bungee’ when she has fallen a further 4 m?
In a braking test a car of mass 1000kg was travelling down a hill on a straight road which has a
constant slope of 1 in 10 as shown in the diagram.
The car was travelling at 20m/s at A where a constant braking force was applied so that the car
came to a stop at B, 100m from A.
Example 33 1988 Question 6
(1 mark)
What is the kinetic energy of the car at A?
Example 34 1988 Question 7
(1 mark)
What is the change in gravitational potential energy of the car in going from A to B?
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Example 35 1988 Question 8
(1 mark)
What is the magnitude of the constant braking force acting on the car as it moved from A to B?
The graph above gives the velocity-time relationship for a block of mass 4.0 kg which slides across
a rough, horizontal floor, coming to rest after 1.0 sec.
Example 36 1986 Question 1
(1 mark)
What is the magnitude of the frictional force of the floor on the block?
Example 37 1986 Question 2
(1 mark)
What is the magnitude of the frictional force of the block on the floor?
Example 38 1986 Question 3
(1 mark)
What distance did the block travel in the 1.0 s time interval?
Example 39 1986 Question 4
(1 mark)
What was the work done on the block by the floor?
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A bow used in archery consists of a curved section, often made of wood, with a string fastened
between its ends as shown in figure 1. To load the bow, an arrow is held against the string and
pulled back to its full extent as shown in figure 2. The arrow is then released to fly towards its
target.
As the distance PQ in Figure 2 is increased, an increasing force is exerted on the arrow by the
string. Figure 3 shows the force exerted on the arrow versus PQ as PQ is increased from 0.15m to
0.60m.
Example 40 1986 Question 19
(1 mark)
Find the amount of work required to increase PQ from 0.15m to 0.60m.
Assume that all the energy stored in the bow when PQ is 0.60 m is transferred to the arrow when it
is released. The mass of the arrow is 0.60 kg.
Example 41 1986 Question 20
(1 mark)
Calculate the speed of the arrow when it leaves the string.
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A light spring of natural length. 0.40 m, hangs from a fixed support. When a mass of 0.40 kg is
hung from the spring the spring, the spring is stretched by 0.10 m, as shown in the figure.
Example 42 1986 Question 27
(1 mark)
What is the spring constant of the spring?
Example 43 1986 Question 28
(1 mark)
What work is done on the spring by the 0.40 kg mass in extending it from its natural length by
0.10m?
A box of mass 20 kg is pulled 3.0 m along a rough floor with a force of 100N. The friction force
acting on the box is a constant 30 N.
Example 44 1981 Question 11
(1 mark)
What is the change in kinetic energy of the box?
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A spring behaves so that the restoring force it exerts is related to the compression by the
relationship
F = 200 x
Where F is the magnitude of the restoring force (in N) and x is the compression (in m).
A body of mass 0.50 kg travelling at 2.0 ms-1 approaches the spring which is fixed to a wall as
shown below. Friction can be neglected.
The body comes to rest instantaneously at a time t1, when the spring is compressed by a total
amount x1, It then rebounds.
Example 45 1977 Question 21
Calculate the value of x1.
(1 mark)
Example 46 1977 Question 22
(1 mark)
Which of the graphs below best represents the kinetic energy of the body as a function of the
compression, x1 of the spring?
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A piece of elastic 0.200 metre long is extended by 0.040 metre when a mass of 0.050 kg is hung
on it. Take g= 10 N kg-1.
Example 47 1975 Question 33
(1 mark)
What force does the mass exert on the elastic?
(State magnitude and unit)
Example 48 1975 Question 34
(1 mark)
How much work is done extending the elastic?
(State magnitude and unit)
Example 49 1975 Question 35
(1 mark)
A student pulls the mass downwards through a further 0.040 metre. How much work does the
student do?
(State magnitude and unit)
Example 50 1975 Question 36
(1 mark)
A piece of the same elastic 0.400 metre long has the same mass hung on it. By how much will it be
extended?
(State magnitude and unit)
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In a pinball machine the plunger is pulled to compress the spring.
When it is released, the spring projects the steel ball.
Experiments performed on the mechanism yield the graph of spring displacement d against the
compressing force F.
Example 51 1973 Question 35
(1 mark)
What is the spring constant of the spring before it is fully compressed?
Example 52 1973 Question 36
(1 mark)
How much work is done in compressing the spring to a point where the force is 5.0 Newton?
Example 53 1973 Question 37
(1 mark)
How much additional work is done in increasing the force to 8.0 Newton?
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Solutions
Example 1 2004 Question 4
As the crumple distance is given in the
question, you know,
m = 1200 kg, u = 20 m/s, v = 0 m/s, x = 0.6 m
and F = ?.
Use WD = F x d = ∆KE.
 F x 0.6 = ½mv2 – ½mu2
 F x 0.6 = -½ x 1200 x 202
1
×1200× 202
F= 2
0.6
 F = 4 x 105 N
(ANS)
Examiner’s comment (1.7/3)
This question could be addressed by either
impulse-momentum or work-energy. The fact
that the crumple distance was stated in the
question meant that the work-energy
approach was simpler.
Example 2 2004 Question 12
The total energy at X will be the same as the
total energy at Y, since there is no friction
acting.
Let’s take PE to be zero at Y
At Y
TE = ½ mv2
At X
TE = mgh
 ½mv2 = mgh
 v2 = 2gh
 v = 2gh
A
(ANS)
Examiner’s comment (1.5/2)
The average score for Question 12 indicates
a satisfactory understanding of the
conversion of gravitational potential energy to
kinetic energy.
Example 3 2000 Solution Q14
This question is based on the conservation
of energy.
Etop = 21 mv 2 +mgh
Subbing in the values we know
* Etop = 21 m×102 +m×10×h
And we do the same for the energy at the
bottom
Ebottom = 21 mv 2
Subbing in the values we know
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Ebottom = 21 m× 242
The energy is conserved; therefore the
energy Jo has at the top will equal the
energy Jo has at the bottom. So we can
write;
Ebottom = E top
* 21 m× 242 = 21 m×102 + m×10 ×h
because m is present in all part of the
equation we can take it out as a common
factor.
** m 21 ×242 = m 21 ×102 +10×h the mass




will now cancel on both sides.
1
2
× 242 = 21 ×102 +10×h
* h =
1
2
×242 - 21 ×102
10
*h = 23.8 m
(ANS)
Examiner’s comment (1.35/3)
With friction and air resistance forces being
ignored, the gain in kinetic energy equals the
loss in gravitational potential energy. Thus,
when the energy equation was set-up and
values for the initial and final speeds
substituted, the height was calculated.
The average score indicated that most
students experienced some difficulty with this
concept. The most common error was in
neglecting Jo’s initial kinetic energy.
Example 4 1998 Solution Q4
Momentum is conserved in all collisions.
 pf = pi. Where pi - sum of the initial
momentums.
 pi = (4000  15) + (1000  0) = 60 000 Ns.
 pf = 60 000 = (4000  10) + (1000  v)
 60 000 = 40 000 + 1 000v
 20 000 = 1 000v
 v = 20 m/s
(ANS)
Examiner’s comment (2.1/3)
The average mark for this question indicated
that the majority of students understood
conservation of momentum.
Example 5 1998 Solution Q5
KEtotal = KEtruck + KEcar
= (½mv2 )truck + (½mv2 )car
= ½  4000  152 – 0
= 450 000 J
= 4.5  105 J (ANS)
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Examiner’s comment
Seventy-eight per cent of students obtained
full marks for this question.
Example 6 1998 Solution Q6
KEtotal = KEtruck + KEcar
= (½mv2 )truck + (½mv2 )car
= ½  4000  102 + ½  1000  202
= 400 000J
= 4.0  105J (ANS)
Examiner’s comment
The majority of students were able to
correctly calculate the final kinetic energy.
Any errors were usually due to incorrect
calculations, such as omitting to square the
velocity or combining the masses in some
way. The expected answer was
consequential upon the answer to Question
5. Sixty-seven per cent of students obtained
full marks for this question.
Example 7 1998 Solution Q7
The final KE is less than the initial KE. Initial
KE was 4.5  105 J whilst the final KE was
4.0  105J. Therefore, this collision was
inelastic, because kinetic energy has been
lost.
This kinetic energy has been transformed into
other forms, - sound, heat, and energy of
deformation.
To get full marks on this question you needed
to include your data from questions 5 and 6.
Don’t assume that the examiner will go back
and read this as part of your answer to this
question.
Examiner’s comment
This question was generally well done, with
students receiving an average mark of 2.0/3,
and 51 per cent of students scoring the full 3
marks. Students were expected to refer to
their answers for Questions 5 and 6, stating
that the final kinetic energy was less than the
initial kinetic energy. They then had to explain
that this was due to kinetic energy being
transformed into other forms, such as heat,
sound and deformation.
The most common errors were due to:
• Students who felt that simply identifying it as
an inelastic collision would be sufficient to
gain the 3 marks.
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• Students who treated the terms ‘energy’ and
‘kinetic energy’ as identical. Hence, they had
energy ‘disappearing’, rather than energy
being transformed. Students need to
understand that energy, like momentum, is
always conserved. However, energy can be
transformed between kinetic energy, potential
energy etc.
Example 8 1980 Solution Q16
WD = KE.
In this case the final velocity is the same as
the initial velocity.
 V = 0
 KE = 0
 WD = 0 J (ANS)
Example 9 1980Solution Q17
The acceleration is in the direction of the
change in velocity.
The change in velocity is given by
FINAL - INITIAL
In this case this will be perpendicularly up.
 D (ANS)
The other way of considering this question is
to think about is: where is this force coming
from. The only force acting (to change the
direction of the ball) is the force of the wall on
the ball. The wall can only provide a normal
reaction force. Hence this force is in a
northerly direction.
Example 15 2008 Question 12, 55%
The energy stored in the spring is 21 kx2
= 21 × 10 × (0.2)2
(use metres)
= 0.02 J
(ANS)
Example 16 2008 Question 13, 40%
The ball will be stationary (momentarily) at
the top and the bottom of the oscillation.
Therefore the KE will be zero at these points.
The KE will be a maximum at the midpoint.
 D (ANS)
Example 17 2008 Question 14, 50%
The gravitational potential energy is
measured from the point of release.
 at the bottom, PE = 0.
 A (ANS)
Physics Unit 3 2015
3. Energy
26 of 30
Example 18 2004 Solution Pilot Q 14
Initial energy = mgh. We need to find ‘h’.
Work done to overcome friction
 F  d = 50  6
= 300J
KE gained = ½mv2 = ½  30  82 = 960J
Now using the kinetic energy of the
head as being the work done on the
air bag you get.
½mv2 = ½kx2
 ½  8  152 = ½  8104  x2
So total work done by gravity
300 + 960 = 1260J
 mgh = 1260
1260
h=
= 4.2 m
30  10
x =
Examiner’s comment (0.9/4)
The most straightforward approach was to
equate the gravitational potential energy at
the top of the ramp to the sum of the kinetic
energy at the bottom plus the energy lost to
work done against friction on the way down
the ramp. Very few students managed to deal
with the work done against friction.
Example 19 2004 Solution Pilot Q 15
Elastic potential of the spring is given by
E = ½kx2 .
 ½  30 000  x2 = 960
(since no energy is lost sliding across the
floor)
 x = 0.25m
Examiner’s comment (01.6/3)
By equating the kinetic energy of the box to
the energy stored in the spring students were
able to determine the compression of the
spring.
Example 20 2000 Solution Q7
This question requires you to
understand that the kinetic energy of
the head has gone into compressing
the bag. Therefore: Ek = Ec. 
1
mv 2  21 kx2 at about this point you
2
realise that you need to know the
value of k, which is the gradient of
the Force-Distance graph.
F
k =
x
16000
k=
0.2
= 8  104 Nm-1
8×152
8×10 4
= 0.15m
(ANS)
Examiner’s comment (1.16/4)
This question required students to
understand that the area under a forcedistance graph represented the work done, or
the change in kinetic energy of the driver’s
head.
The average mark for this question was
disappointing with the most common error
being to interpret this question as relating to
impulse-momentum rather than work-energy.
It was clear that students were so used to
analysing collisions in terms of impulsemomentum, that this change in emphasis
caught them quite unawares.
It was disappointing to note that
approximately 15% of students showed no
working at all, giving only the final answer,
which was sometimes correct but often not.
Example 21 2000 Solution Q8
A
(ANS)
Steel would have a large k value, A
has the largest value for its gradient.
Examiner’s comment (1.44/2)
Collision with the hard surface of the steering
wheel would result in a shorter compression
distance. In order for the area under the
graph to remain as 900 J, the retarding force
would need to be much larger.
Example 22 2000 Solution Q9
The specific reasons for choosing
graph A needed to cover:
• A collision with a harder surface
would result in a smaller
compression distance.
• The material must have 900J of
work done on it, and therefore the
area under the graph remains
constant.
• Hence, the required graph must
have a shorter compression
distance and a larger force.
Physics Unit 3 2015
3. Energy
Examiner’s comment (1.04/3)
The specific reasons for choosing graph A
needed to cover:
• a collision with a harder surface would result
in a smaller compression distance
• the area under the graph remains constant
(900 J) no matter whether the collision is with
the air bag or steering wheel
• hence, the required graph must have a
shorter compression distance and a larger
force.
Alternatively, students could have addressed
this answer via an understanding that harder
surfaces are ‘stiffer’ and have a steeper
force-distance graph gradient. The aspect of
the area under the graph remaining at 900 J
was essential no matter which method
students used.
The average mark here was disappointing,
students found this concept to be quite
difficult. Very few students noted that the area
under the force-compression graph had to be
the same (900 J) as in the original problem.
Most students attempted to address this via
an impulse-momentum approach, which
made it difficult for them to gain full marks.
Example 23 1983 Solution Q27
F = kx
 F = 250 x 0.04
 F = 10N
(ANS)
Example 24 1983 Solution Q28
WD = ½kx2
= ½ x 250 x 0.042
= 0.20 J (ANS)
Example 25 1983 Solution Q29
When the ball leaves the plunger,
KEball + KEplunger = PEstored in sprng
 0.20 = ½(mball + mplunger)v2
v=
0.20 × 2
(0.1+ 0.1)
 v = 1.41 ms-1
(ANS)
Example 26 1981 Solution Q21
Using F = kx, the spring constant is the
gradient of the graph as the graph is F vs x.
50
k=
0.1
 k = 500 Nm-1
(ANS)
27 of 30
You must remember to include the units as
they were specifically asked for in the
question.
Example 27 1981 Solution Q22
Energystored = ½kx2, where k is the spring
constant and x is the extension of the spring
(not the length of the spring).
Use F = kx to find the extension. The force
being applied is the weight of the mass.
 mg = kx
mg
x=
500
10 ×10
x=
500
 x = 0.2
Now use
E = ½ x 500 x 0.22
 E = 10 J
(ANS)
Example 28 2001 Solution Q7
Gravitational energy is given by mgh. The
graph will have the same shape as the track,
because ‘mg’ is constant, with ‘h’ being the
only variable.
 A (ANS)
Examiner’s comment (1.55/2)
This question was well answered.
Example 29 2001 Solution Q 8
There isn’t any energy being supplied (or lost)
by the roller coaster.
 the total energy should remain constant.
This means that when the roller coaster loses
PE it must gain KE.
the KE graph must be such that when you
add it to the PE graph you get a constant
value.
 B (ANS)
Examiner’s comment (1.41/2)
The requirements of this question were well
understood by most students.
Example 30 2001 Solution Q 9
See the explanation for Question 8.
 E (ANS)
Examiner’s comment (0.77/1)
This question was well understood.
Physics Unit 3 2015
3. Energy
Example 31 1990 Solution Q25
The loss in gravitational potential energy =
gain in kinetic energy.
 ∆mgh = ∆ ½mv2
 gh = ½ v2
 v = 2gh
=
2×10×13
= 260
= 16.1 ms-1
(ANS)
Example 32 1990 Solution Q26
The energy stored in the ‘bungee’ is given by
E = ½kx2
 E = ½ x 17 x 42
= 136 J (ANS)
Example 33 1988 Solution Q6
The KE of the car = ½mv2
 KE = ½ x 1000 x 202
= 2.0 x 105 J (ANS)
Example 34 1988 Solution Q7
Change in potential energy is given by mgh.
PE = mgh
= 1000 x 10 x 10
= 1.0 x 105 J
(ANS)
Example 35 1988 Solution Q8
The work done by the braking force is given
by WD = F x d, where ‘F’ is the braking force,
and ‘d’ is the distance it acts.
In this case the work needs to stop the car at
B, so it needs to overcome the initial KE as
well as the gain in PE (from rolling down the
slope).
The total energy the work done needs to
overcome is
KE(2.0 x 105) + PE(1.0 x 105) = 3.0 x 105J
 F x 100 = 3.0 x 105
 F = 3.0 x 103 N
(ANS)
Example 36 1986 Solution Q1
Use F = ma
The gradient of the v-t graph is the
acceleration.
∴F = ma
Δv
=4 
Δt
5
= 4×
1
= 20 N (ANS)
28 of 30
Example 37 1986 solution Q2
From Newton’s third Law, FA on B = FB on A.
This is an example of the third law.
 20 N
(ANS)
Example 38 1986 Solution Q3
The distance travelled is the area under the
velocity-time graph.
d=½x5x1
= 2.5 m (ANS)
Example 39 1986 Solution Q4
WD = F x d
= 20 x 2.5
= 50 J
(ANS)
WD = KE
= ½ x 4 x 52 – ½ x 4 x 0 2
= 50 J
(ANS)
Example 40 1986 Solution Q19
This is a force distance graph, so the work
done is the area under the graph.
The best way to do this is to count squares
(even though I recommend that you use this
method as a last resort).
Each square is 10 x 0.05 = 0.5J.
Count the squares and put a small mark in
each one as you count it. Only count a part
square if it is greater than 50%.
I count 58 squares
 58 x 0.5 = 29.0 J
(ANS)
Example 41 1986 Solution Q20
The energy stored in the ‘spring’ is converted
into the KE of the arrow.
 ½mv2 = 29.0
v2 =
29.0× 2
0.6
 v = 9.8 m/s
(ANS)
Example 42 1986 Solution Q27
The spring constant is ‘k’ in the equation
F = kx
Here the force extending the spring is the
weight.
 mg = kx
 0.4 x 10 = k x 0.1
 k = 40 N/m
(ANS)
Physics Unit 3 2015
3. Energy
Example 43 1986 Solution Q28
The work done on the spring is the energy
stored in the spring.
The stored energy = ½kx2
= ½ x 40 x 0.12
= 20 x 0.01
= 0.2 J
(ANS)
Example 44 1981 Solution Q11
The change in KE is the same as the work
done on the box.
You must use the ‘net force’ to calculate the
work done.
 WD = F x d
= (100 – 30) x 3
= 210 J
(ANS)
Example 45 1977 Solution Q21
At x1, the initial KE of the block is stored as
PE of the spring.
 ½mv2 = ½kx2
 ½ x 0.5 x 22 = ½ x 200 x (x1)2
 x1 =
1
100
 x1 = 0.1 m (ANS)
Example 46 1977 Solution Q22
See the explanation to 1982, Question 32.
 B (ANS)
Example 47 1975 Solution Q33
The weight is causing the elastic to extend.
 mg = kx
 mg = 0.05 x 10
= 0.5 N
(ANS)
Make sure you include the correct units in
your answer.
Example 48 1975 Solution Q34
WD = ½kx2
To find use mg = kx
mg
k=
x
0.5
k=
0.04
= 12.5 Nm-1
 WD = ½kx2
29 of 30
= ½ x 12.5 x 0.042
= 0.01 J
(ANS)
Make sure you include the correct units in
your answer.
Example 49 1975 Solution Q35
The work done is given by WD = ½kx2, where
x is the extension (not the length of the
elastic).
 WD = ½kx2
= ½ x 12.5 x 0.082
= 0.04 J
Originally it took 0.01 J to extend the elastic
the initial 0.04 m, so the difference between
0.04 J and 0.01 J is the work done by the
student.
 0.04 -0.01
= 0.03 J
(ANS)
Make sure you include the correct units in
your answer.
Example 50 1975 Solution Q36
If the elastic is twice as long, then the mass
will extend the original length by twice as
much.
 0.04 x 2
 0.08 m
(ANS)
Make sure you include the correct units in
your answer.
Example 51 1973 Solution Q35
The spring constant is given by the gradient
of the graph.
5
k=
0.1
= 50 Nm-1
(ANS)
Example 52 1973 Solution Q36
WD = ½ kx2
Use the graph to find x when the force = 5.0
N.
x = 0.1 m
 ½ kx2 = ½ x 50 x 0.12
= 0.25 J (ANS)
Example 53 1973 Solution Q37
Increasing the force to 8 N will not alter the
compression.
0J
(ANS)
Unit 3 Physics 2012
Bialik College
Page 30 of 30