Colonel Frank Seely School
Exampro A-level Physics
(7407/7408)
Name:
Class:
3.4.2.2 The Young modulus
Author:
Date:
Time:
337
Marks:
276
Comments:
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Colonel Frank Seely School
Q1.Figure 1 and Figure 2 both show the side view of a steel bolt.
Figure 1
(a)
Figure 2
Show on Figure 1 forces acting on the bolt which would produce a tensile strain.
(1)
(b)
The ultimate tensile stress of steel is 5.0 × 108 Pa, the elastic limit is 2.5 × 108 Pa and
the Young modulus of steel is 2.0 × 1011 Pa.
Defining any terms used, state what is meant by:
(i)
tensile stress;
...............................................................................................................
...............................................................................................................
(1)
(ii)
tensile strain;
...............................................................................................................
...............................................................................................................
(1)
(iii)
the elastic limit.
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...............................................................................................................
(1)
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Colonel Frank Seely School
(c)
When the main engines of a space shuttle are fired, they develop a total thrust of
4.5 × 106 N. In a test firing the shuttle is held to the launch pad by 8 steel bolts each of
diameter 9.0 × 10–2 m. Using data given in (b):
(i)
calculate the strain for each bolt during the test;
(4)
(ii)
determine the minimum number of bolts that could have been used when
carrying out the test.
(3)
(Total 11 marks)
Q2.A gymnast does a hand-stand on a horizontal bar. The gymnast then rotates in a vertical circle
with the bar as a pivot. The gymnast and bar remain rigid during the rotation and when
friction and air resistance are negligible the gymnast returns to the original stationary
position.
Figure 1 shows the gymnast’s position at the start and Figure 2 shows the position after
completing half the circle.
Figure 1
(a)
Figure 2
The gymnast has a mass of 70 kg and the centre of mass of the gymnast is 1.20 m
from the axis of rotation.
acceleration of free fall, g = 9.8 m s–2
(i)
Show clearly how the principle of conservation of energy predicts a
speed of 6.9 m s–1 for the centre of mass when in the position shown in Figure
2.
(3)
(ii)
The maximum force on the arms of the gymnast occurs when in the position
shown in Figure 2.
Calculate the centripetal force required to produce circular motion of the
gymnast when the centre of mass is moving at 6.9 m s–1.
(2)
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Colonel Frank Seely School
(iii)
Determine the maximum tension in the arms of the gymnast when in the position
shown in Figure 2.
(1)
(iv)
Sketch a graph to show how the vertical component of the force on the bar
varies with the angle rotated through by the gymnast during the manoeuvre.
Assume that a downward force is positive.
Include the values for the initial force and the maximum force on the bar.
Only show the general shape between these values.
(2)
(b)
The bones in each forearm have a length of 0.25 m. The total cross-sectional area of
the bones in both forearms is 1.2 × 10–3 m2 . The Young modulus of bone in
compression is 1.6 × 1010 Pa.
Assuming that the bones carry all the weight of the gymnast, calculate the reduction in
length of the forearm bones when the gymnast is in the start position shown in Figure
1.
(3)
(Total 11 marks)
Q3.The four bars A, B, C and D have diameters, lengths and loads as shown. They are all made
of the same material.
Which bar has the greatest extension?
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Colonel Frank Seely School
(Total 1 mark)
Q4.(a)
State what is meant by the yield stress of a material.
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(3)
(b)
A steel piano wire has a diameter of 1.8 × 10−3 m and a length of 1.55 m. When
tightened to emit a note of the required frequency it extends by 1.3 × 10−3 m. The
Young modulus of the steel is 2.1 × 1011 Pa.
(i)
Calculate the force exerted on the frame of the piano by this wire.
(3)
(ii)
Calculate the strain energy stored in this stretched wire.
(2)
(Total 8 marks)
Q5.The diagram below shows the rotor-blade arrangement used in a model helicopter. Each of
the blades is 0.55 m long with a uniform cross-sectional area of 3.5 × 10−4 m2 and negligible
mass. An end-cap of mass 1.5 kg is attached to the end of each blade.
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Colonel Frank Seely School
(a)
(i)
Show that there is a force of about 7 kN acting on each end-cap when the blades
rotate at 15 revolutions per second.
(3)
(ii)
State the direction in which the force acts on the end-cap.
...............................................................................................................
(1)
(iii)
Show that this force leads to a longitudinal stress in the blade of about 20 MPa.
(2)
(iv)
Calculate the change in length of the blade as a result of its rotation.
Young modulus of the blade material = 6.0 × 1010 Pa
(2)
(v)
Calculate the total strain energy stored in one of the blades due to its extension.
(2)
(b)
The model helicopter can be made to hover above a point on the ground by directing
the air from the rotors vertically downwards at speed v.
(i)
Show that the change in momentum of the air each second is Aρv2, where A is
the area swept out by the blades in one revolution and ρ is the density of air.
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Colonel Frank Seely School
(2)
(ii)
The model helicopter has a weight of 900 N. Calculate the speed of the air
downwards when the helicopter has no vertical motion.
Density of air = 1.3 kg m−3
(3)
(Total 15 marks)
Q6.(a)
Explain why an engineer needs to consider the yield stress of a metal such as steel
when deciding on its suitability for use in the construction of a building or a bridge.
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(2)
(b)
In order to prevent the collapse of walls of old buildings a metal rod is often used to tie
opposite walls together, as shown below.
In one case a steel tie rod of diameter 19 mm is used as shown above. When the nuts
are tightened, the rod extends by 1.5 mm. The Young modulus of steel is 2.1 × 1011
Pa.
Calculate:
(i)
the force exerted on the walls by the rod;
(3)
(ii)
the elastic strain energy in the rod when it is extended by 1.5 mm.
(2)
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Colonel Frank Seely School
(Total 7 marks)
Q7.(a)
The Young modulus is defined as the ratio of tensile stress to tensile strain. Explain
what is meant by each of the terms in italics.
tensile stress .................................................................................................
........................................................................................................................
........................................................................................................................
tensile strain ...................................................................................................
........................................................................................................................
(3)
(b)
A long wire is suspended vertically and a load of 10 N is attached to its lower end. The
extension of the wire is measured accurately. In order to obtain a value for the Young
modulus of the material of the wire, two more quantities must be measured. State
what these are and in each case indicate how an accurate measurement might be
made.
quantity 1 ........................................................................................................
method of measurement ...............................................................................
........................................................................................................................
quantity 2 ........................................................................................................
method of measurement ................................................................................
........................................................................................................................
(4)
(c)
Sketch below a graph showing how stress and strain are related for a ductile
substance and label important features.
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Colonel Frank Seely School
(2)
(Total 9 marks)
Q8.As part of a quality check, a manufacturer of fishing line subjects a sample to a tensile test.
The sample of line is 2.0 m long and is of constant circular cross-section of diameter 0.50
mm. Hooke’s law is obeyed up to the point when the line has been extended by 52mm at a
tensile stress of 1.8 × 108 Pa.
The maximum load the line can support before breaking is 45 N at an extension of 88 mm.
(a)
Calculate
(i)
the value of the Young modulus,
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
the breaking stress (assuming the cross-sectional area remains constant),
...............................................................................................................
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...............................................................................................................
(iii)
the breaking strain.
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(5)
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Colonel Frank Seely School
(b)
Sketch a graph on the axes below to show how you expect the tensile stress to vary
with strain. Mark the value of stress and corresponding strain at
(i)
the limit of Hooke’s law,
(ii)
the breaking point.
(4)
(Total 9 marks)
Q9.One end of a steel wire of length 1.2 m and 2.0 mm diameter is attached to a rigid beam. A 25
g mass is attached to the free end of the steel wire and placed against the underside of the
beam as shown.
The 25 g mass is released and falls freely until the wire becomes taut. The kinetic energy of
the falling mass is converted to elastic potential energy in the wire as the wire extends to a
maximum of 1.0 mm. Energy converted to other forms is negligible.
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Colonel Frank Seely School
For maximum extension of the wire, complete parts (a) to (e).
(a)
Show that the elastic potential energy stored by the extended wire is 0.29 J.
........................................................................................................................
........................................................................................................................
(b)
Calculate the tension in the wire.
........................................................................................................................
........................................................................................................................
(c)
Calculate the stress in the wire.
........................................................................................................................
........................................................................................................................
(d)
Calculate the strain of the wire.
........................................................................................................................
........................................................................................................................
(e)
Hence, calculate the Young modulus for the steel of the wire.
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(Total 9 marks)
Q10.(a)
(i)
Draw and label suitable apparatus required for measuring the Young modulus of
a material in the form of a long wire.
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Colonel Frank Seely School
(ii)
List the measurements you would make when using the apparatus described in
part (i).
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(iii)
Describe briefly how the measurements listed in part (ii) would be carried out.
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(iv)
Explain how you would calculate the Young modulus from your measurements.
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(13)
(b)
A uniform heavy metal bar of weight 250 N is suspended by two vertical wires,
supported at their upper ends from a horizontal surface, as shown.
One wire is made of brass and the other of steel. The cross-sectional area of each
wire is 2.5 ×10–7 m2 and the unstretched length of each wire is 2.0 m.
the Young modulus for brass = 1.0 × 1011 Pa
the Young modulus for steel = 2.0 × 1011 Pa
(i)
If the tension, T, in each wire is 125 N, calculate the extension of the steel wire.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Estimate how much lower the end A will be than the end B.
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(3)
(Total 16 marks)
Q11.(a)
(i)
Define the Young modulus for a material.
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...............................................................................................................
(ii)
Explain what is meant by the elastic limit for a wire.
...............................................................................................................
...............................................................................................................
(2)
(b)
A wire supported at its upper end, hangs vertically. The table shows readings
obtained when stretching the wire by suspending masses from its lower end.
(i)
load / N
0
2.0
4.0
6.0
7.0
8.0
9.0 10.0 10.5
extension /
mm
0
1.2
2.4
3.6
4.2
4.9
5.7
7.0
8.0
Plot a graph of load against extension.
(One sheet of graph paper should be provided)
(ii)
Indicate on your graph the region where Hooke’s law is obeyed.
(iii)
The unstretched length of the wire is 1.6 m and the area of cross-section
8.0 × 10–8 m2. Calculate the value of the Young modulus of the material.
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(8)
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Colonel Frank Seely School
(c)
(i)
By considering the work done in stretching a wire, show that the energy stored
is given by
Fe, where F is the force producing an extension e.
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(ii)
Calculate the energy stored in the wire in part (b) when the extension is 4.0 mm.
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(4)
(Total 14 marks)
Q12.
A uniform wooden beam of mass 35.0 kg and length 5.52 m is supported by two
identical vertical steel cables A and B attached at either end, as shown in Figure 1.
Figure 1
(a)
Calculate
(i)
the weight of the beam,
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(ii)
the tension in each cable.
.............................................................................................................
.............................................................................................................
(2)
(b)
Each unstretched cable has a diameter of 8.26 mm and a length 2.50 m. Calculate
the extension of each cable when supporting the beam.
The Young modulus for steel = 2.10 × 1011 Pa
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(4)
(c)
An object of mass 20.0 kg is hung from the beam 1.00 m from cable A, as shown in
Figure 2.
Figure 2
(i)
Show that the new tension in cable A is 332 N.
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.............................................................................................................
(ii)
Calculate the new tension in cable B.
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(6)
(Total 12 marks)
Q13.
(a)
The graph shows the variation of tensile stress with tensile strain for two wires X
and Y, having the same dimensions, but made of different materials. The materials
fracture at the points FX and FY respectively.
You may be awarded marks for the quality of written communication provided in your
answer to the following questions.
State, with a reason for each, which material, X or Y,
(i)
obeys Hooke’s law up to the point of fracture,
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.............................................................................................................
(ii)
is the weaker material,
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(iii)
is ductile,
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(iv)
has the greater elastic strain energy for a given tensile stress.
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(8)
(b)
An elastic cord of unstretched length 160 mm has a cross-sectional area of 0.64 mm2.
The cord is stretched to a length of 190 mm. Assume that Hooke’s law is obeyed for
this range and that the cross-sectional area remains constant.
the Young modulus for the material of the cord = 2.0 × 107 Pa
(i)
Calculate the tension in the cord at this extension.
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(ii)
Calculate the energy stored in the cord at this extension.
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(5)
(Total 13 marks)
Q14.
(a)
State Hooke’s law for a material in the form of a wire.
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......................................................................................................................
(2)
(b)
A rigid bar AB of negligible mass, is suspended horizontally from two long, vertical
wires as shown in the diagram. One wire is made of steel and the other of brass.
The wires are fixed at their upper end to a rigid horizontal surface. Each wire
is 2.5 m long but they have different cross-sectional areas.
When a mass of 16 kg is suspended from the centre of AB, the bar remains horizontal.
the Young modulus for steel = 2.0 × 1011 Pa
the Young modulus for brass = 1.0 × 1011 Pa
(i)
What is the tension in each wire?
.............................................................................................................
(ii)
If the cross-sectional area of the steel wire is 2.8 × 10–7 m2, calculate the
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Colonel Frank Seely School
extension of the steel wire.
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(iii)
Calculate the cross-sectional area of the brass wire.
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(iv)
Calculate the energy stored in the steel wire.
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(7)
(c)
The brass wire is replaced by a steel wire of the same dimensions as the brass wire.
The same mass is suspended from the midpoint of AB.
(i)
Which end of the bar is lower?
.............................................................................................................
(ii)
Calculate the vertical distance between the ends of the bar.
.............................................................................................................
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(2)
(Total 11 marks)
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Colonel Frank Seely School
Q15.
(a)
When determining the Young modulus for the material of a wire, a tensile stress
is applied to the wire and the tensile strain is measured.
(i)
State the meaning of
tensile stress .......................................................................................
.............................................................................................................
tensile strain ........................................................................................
.............................................................................................................
(ii)
Define the Young modulus ..................................................................
.............................................................................................................
(3)
(b)
The diagram below shows two wires, one made of steel and the other of brass, firmly
clamped together at their ends. The wires have the same unstretched length and the
same cross-sectional area.
One of the clamped ends is fixed to a horizontal support and a mass M is suspended
from the other end, so that the wires hang vertically.
(i)
Since the wires are clamped together the extension of each wire will be the
same.
If ES is the Young modulus for steel and EB the Young modulus for brass, show
that
where FS and FB are the respective forces in the steel and brass wire.
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(ii)
The mass M produces a total force of 15 N. Show that the magnitude of the
force
FS = 10 N.
the Young modulus for steel = 2.0 × 1011 Pa
the Young modulus for brass = 1.0 × 1011 Pa
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(iii)
The cross-sectional area of each wire is 1.4 × 10–6 m2 and the unstretched
length is 1.5 m. Determine the extension produced in either wire.
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(6)
(Total 9 marks)
(a)
State Hooke’s law for a material in the form of a wire and state the conditions
under which this law applies.
Q16.
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......................................................................................................................
(2)
(b)
A length of steel wire and a length of brass wire are joined together. This combination
is suspended from a fixed support and a force of 80 N is applied at the bottom end, as
shown in the figure below.
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Each wire has a cross-sectional area of 2.4 × 10–6 m2.
length of the steel wire = 0.80 m
length of the brass wire = 1.40 m
the Young modulus for steel = 2.0 × 1011 Pa
the Young modulus for brass = 1.0 × 1011 Pa
(i)
Calculate the total extension produced when the force of 80 N is applied.
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(ii)
Show that the mass of the combination wire = 4.4 × 10–2 kg.
density of steel = 7.9 × 103 kg m–3
density of brass = 8.5 × 103 kg m–3
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(7)
(c)
A single brass wire has the same mass and the same cross-sectional area as the
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Colonel Frank Seely School
combination wire described in part (b). Calculate its length.
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(2)
(Total 11 marks)
Q17.
(a)
(i)
Describe the behaviour of a wire that obeys Hooke’s law.
.............................................................................................................
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(ii)
Explain what is meant by the elastic limit of the wire.
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(iii)
Define the Young modulus of a material and state the unit in which it is
measured.
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(5)
(b)
A student is required to carry out an experiment and draw a suitable graph in order to
obtain a value for the Young modulus of a material in the form of a wire.
A long, uniform wire is suspended vertically and a weight, sufficient to make the wire
taut, is fixed to the free end. The student increases the load gradually by adding
known weights. As each weight is added, the extension of the wire is measured
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Colonel Frank Seely School
accurately.
(i)
What other quantities must be measured before the value of the Young
modulus can be obtained?
.............................................................................................................
.............................................................................................................
(ii)
Explain how the student may obtain a value of the Young modulus.
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(iii)
How would a value for the elastic energy stored in the wire be found from the
results?
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(6)
(Total 11 marks)
Q18.
The diagram below shows a dockside crane that is used to lift a container of mass
22000 kg from a cargo ship onto the quayside. The container is lifted by four identical ‘lifting’
cables attached to the top corners of the container.
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Colonel Frank Seely School
(a)
When the container is being raised, its centre of mass is at a horizontal distance 32 m
from the nearest vertical pillar PQ of the crane’s supporting frame.
(i)
Assume the tension in each of the four lifting cables is the same. Calculate the
tension in each cable when the container is lifted at constant velocity.
answer ........................... N
(2)
(ii)
Calculate the moment of the container’s weight about the point Q on the
quayside, stating an appropriate unit.
answer ...........................
(3)
(iii)
Describe and explain one feature of the crane that prevents it from toppling
over when it is lifting a container.
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(2)
(b)
Each cable has an area of cross–section of 3.8 × 10–4 m2.
(i)
Calculate the tensile stress in each cable, stating an appropriate unit.
answer ..................................
(3)
(ii)
Just before the container shown in the diagram above was raised from the ship,
the length of each lifting cable was 25 m. Show that each cable extended by 17
mm when the container was raised from the ship.
Young modulus of steel = 2.1 × 1011 Pa
(2)
(Total 12 marks)
Q19.
A rubber cord is used to provide mechanical resistance when performing fitness
exercises. A scientist decided to test the properties of the cord to find out how effective it
was for this purpose. The graph of load against extension is shown in the figure below for a
0.50 m length of the cord.
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Curve A shows loading and curve B shows unloading of the cord.
(a)
State which feature of this graph confirms that the rubber cord is elastic.
......................................................................................................................
(1)
(b)
Explaining your method, use the graph (curve A) to estimate the work done in
producing an extension of 0.30 m.
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answer = ................................... J
(3)
(c)
Assuming that line A is linear up to an extension of 0.040 m, calculate the Young
modulus of the rubber for small strains.
The cross-sectional area of the cord = 5.0 × 10–6 m2
The unstretched length of the cord = 0.50 m
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Colonel Frank Seely School
answer = ................................ Pa
(3)
(d)
The scientist compared this cord with a steel spring that reached the same extension
for the same maximum load without exceeding its limit of proportionality.
(i)
On the figure above, draw the load-extension line for this spring up to a load of
50 N and label it C.
(1)
(ii)
With reference to the spring, explain what is meant by limit of proportionality.
.............................................................................................................
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(1)
(Total 9 marks)
Q20.
A cable car system is used to transport people up a hill. The figure below shows a
stationary cable car suspended from a steel cable of cross-sectional area 2.5 × 10–3 m2.
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Colonel Frank Seely School
(a)
The graph below is for a 10 m length of this steel cable.
(i)
Draw a line of best fit on the graph.
(2)
(ii)
Use the graph to calculate the initial gradient, k, for this sample of the cable.
answer = ............................ N m–1
(2)
(b)
The cable breaks when the extension of the sample reaches 7.0 mm. Calculate the
breaking stress, stating an appropriate unit.
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answer = .....................................
(3)
(c)
In a cable car system a 1000 m length of this cable is used. Calculate the extension of
this cable when the tension is 150 kN.
answer = ..................................m
(2)
(Total 9 marks)
Q21.
The table below shows the results of an experiment where a force was applied to a
sample of metal.
(a)
On the axes below, plot a graph of stress against strain using the data in the table.
Strain
/ 10–3
0
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
Stress
/108 Pa
0
0.90
2.15
3.15
3.35
3.20
3.30
3.50
3.60
3.60
3.50
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(3)
(b)
Use your graph to find the Young modulus of the metal.
answer = ...................................... Pa
(2)
(c)
A 3.0 m length of steel rod is going to be used in the construction of a bridge. The
tension in the rod will be 10 kN and the rod must extend by no more than 1.0mm.
Calculate the minimum cross-sectional area required for the rod.
Young modulus of steel = 1.90 × 1011 Pa
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Colonel Frank Seely School
answer = ...................................... m2
(3)
(Total 8 marks)
Q22.
The figure below shows a stress-strain graph for a copper wire.
(a)
Define tensile strain.
........................................................................................................................
........................................................................................................................
(1)
(b)
State the breaking stress of this copper wire.
answer = ................................ Pa
(1)
(c)
Mark on the figure above a point on the line where you consider plastic deformation
may start.
Label this point A.
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Colonel Frank Seely School
(1)
(d)
Use the graph to calculate the Young modulus of copper. State an appropriate unit for
your answer.
answer = .....................................
(3)
(e)
The area under the line in a stress-strain graph represents the work done per unit
volume to stretch the wire.
(i)
Use the graph to find the work done per unit volume in stretching the wire to a
strain of 3.0 × 10–3.
answer = .....................................J m–3
(2)
(ii)
Calculate the work done to stretch a 0.015 kg sample of this wire to a strain
of3.0 × 10–3.
The density of copper = 8960 kg m–3.
answer = .....................................J
(2)
(f)
A certain material has a Young modulus greater than copper and undergoes brittle
fracture at a stress of 176 MPa.
On the figure above draw a line showing the possible variation of stress with strain for
this material.
(2)
(Total 12 marks)
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Colonel Frank Seely School
Q23.The diagram below shows a tower crane that has two identical steel cables. The length of
each steel cable is 35 m from the jib to the hook.
(a)
Each cable has a mass of 4.8 kg per metre. Calculate the weight of a 35 m length of
one cable.
weight = ................................. N
(2)
(b)
The cables would break if the crane attempted to lift a load of 1.5 × 106 N or more.
Calculate the breaking stress of one cable.
cross-sectional area of each cable = 6.2 × 10−4 m2
breaking stress = ................................ Pa
(2)
Page 35
Colonel Frank Seely School
(c)
When the crane supports a load each cable experiences a stress of 400 MPa. Each
cable obeys Hooke’s law. Ignore the weight of the cables.
Young modulus of steel = 2.1 × 1011 Pa
(i)
Calculate the weight of the load.
weight = ................................. N
(2)
(ii)
The unstretched length of each cable is 35 m.
Calculate the extension of each cable when supporting the load.
extension = ................................. m
(3)
(iii)
Calculate the combined stiffness constant, k, for the two cables.
stiffness constant = ........................... Nm−1
(2)
(iv)
Calculate the total energy stored in both stretched cables.
energy stored = ................................... J
(2)
(Total 13 marks)
Q24.A student adds a series of masses to a vertical metal wire of circular cross–section and
measures the extension of the wire produced. The figure below is a force-extension graph
of the data.
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Colonel Frank Seely School
(a)
Mark on the figure the point P, the limit beyond which Hooke’s law is no longer
obeyed.
(1)
(b)
Outline how the student can use these results and other measurements to determine
the Young modulus of the wire.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
When the wire has been extended to A, the masses are removed one by one and the
extension re-measured.
Draw, on the figure above, the shape of the graph that the student will obtain.
(1)
(d)
Explain why the graph has the shape you have drawn.
........................................................................................................................
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Colonel Frank Seely School
........................................................................................................................
........................................................................................................................
(2)
(e)
The metal wire is used to make a cable of diameter 6.0 mm. The Young modulus of
metal of the cable is 2.0 × 1011 Pa.
Calculate the force necessary to produce a strain of 0.20% in the cable.
force = ........................ kN
(3)
(f)
The cable is used in a crane to lift a mass of 600 kg.
Determine the maximum acceleration with which the mass can be lifted if the strain in
the cable is not to exceed 0.20%.
acceleration = ........................ m s−2
(3)
(g)
An engineer redesigns the crane to lift a 1200 kg load at the same maximum
acceleration.
Discuss the changes that could be made to the cable of the crane to achieve this,
without exceeding 0.20% strain.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
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Colonel Frank Seely School
........................................................................................................................
........................................................................................................................
(3)
(Total 16 marks)
Q25.The diagram shows how the stress varies with strain for metal specimens X and Y which are
different. Both specimens were stretched until they broke.
Which of the following is incorrect?
A
X is stiffer than Y
B
X has a higher value of the Young modulus
C
X is more brittle than Y
D
Y has a lower maximum tensile stress than X
(Total 1 mark)
Q26.The term ultrasound refers to vibrations in a material that occur at frequencies too high to be
detected by a human ear. When ultrasound waves move through a solid, both longitudinal
and transverse vibrations may be involved. For the longitudinal vibrations in a solid, the
speed c of the ultrasound wave is given by
where E is the Young modulus of the material and ρ is the density. Values for c and ρ are
given in the table below.
Page 39
Colonel Frank Seely School
Substance
c/ms
ρ / kg m
glass
5100
2500
sea water
1400
1000
−1
−3
Ultrasound waves, like electromagnetic radiation, can travel through the surface between
two materials. When all the energy is transmitted from one material to the other, the
materials are said to be acoustically matched. This happens when ρc is the same for both
materials.
(a)
Calculate the magnitude of the Young modulus for glass.
Young modulus = ...............................
(1)
(b)
State your answer to (a) in terms of SI fundamental units.
(1)
(c)
The passage states that ’when ultrasound waves move through a solid both
longitudinal and transverse vibrations may be involved’.
State the difference between longitudinal and transverse waves.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(d)
Show that when two materials are acoustically matched, the ratio of their Young
moduli is equal to the ratio of their speeds of the ultrasound waves.
Page 40
Colonel Frank Seely School
(2)
(e)
The wave speed in a material X is twice that in material Y. X and Y are acoustically
matched.
Determine the ratio of the densities of X and Y.
X = ............................... Y = ...............................
(1)
(f)
Ultrasound waves obey the same laws of reflection and refraction as electromagnetic
waves.
Using data from Table 1, discuss the conditions for which total internal reflection can
occur when ultrasound waves travel between glass and sea water.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 10 marks)
Q27.If lengths of rail track are laid down in cold weather, they may deform as they expand when
Page 41
Colonel Frank Seely School
the weather becomes warmer. Therefore, when rails are laid in cold weather they are
stretched and fixed into place while still stretched. This is called pre-straining.
The following data is typical for a length of steel rail:
Young modulus of steel =
cross sectional area of a length of rail =
amount of pre-strain =
2.0 × 1011 Pa
7.5 × 10–3 m2
2.5 × 10-5 for each kelvin rise in
temperature the rail is expected to
experience.
A steel rail is laid when the temperature is 8 °C and the engineer decides to use a pre-strain
of 3.0 × 10-4.
(a)
Calculate the tensile force required to produce the pre-strain in the rail required by the
engineer.
tensile force = ............................... N
(3)
(b)
Calculate the elastic strain energy stored in a rail of unstressed length 45 m when
pre-strained as in part (a)
elastic strain energy = ............................... J
(2)
(c)
Calculate the temperature at which the steel rail becomes unstressed.
temperature = ............................... °C
(2)
(d)
Explain why the engineer does not use the highest observed temperature at the
Page 42
Colonel Frank Seely School
location of the railway track to determine the amount of pre-strain to use.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 9 marks)
Page 43
Colonel Frank Seely School
M1.(a)
two forces shown producing tension (see examples below)
B1
(2)
(b)
(i)
tension / area
or force / area (perpendicular to the surface on which the force acts)
or F / A with terms defined (condone loose definition of area)
B1
(2)
(ii)
extension / original length
or extension per unit length
B1
(2)
(iii)
the point / stress / strain / force or extension up to which
either all energy supplied in stretching is returned when (deforming)
forces are removed
or the object returns to its original shape when (deforming forces) are
removed
or beyond which further stress causes plastic deformation or permanent
deformation
B1
(2)
Page 44
Colonel Frank Seely School
(c)
(i)
force on each bolt = 0.56 MN (ie total force divided by 8)
or
cross-sectional area each bolt = πr2 plus correct substitution 6.4 × 10–3 m2
C1
stress = 8.8 × 107 or
stress / strain = Young modulus
C1
strain = 4.4 × 10–4
A1
(4)
(ii)
area of steel needed = maximum force / UTS
or maximum force / elastic limit
C1
4.5 × 106 / 5.0 × 108 = 9.0 × 10–3 m2
or 4.5 × 106 / 2.5 × 108 = 18.0 × 10–3 m2
C1
number of bolts = 9.0 × 10–3 / area of a bolt (allow e.c.f.)
= 1.4 bolts
or = 2.8 bolts
A1
answer = 2 bolts (c.a.o.)
or answer = 3 bolts (c.a.o.)
B1
(3)
[11]
M2.(a)
(i)
loss of PE = gain of KE or mgh = ½mv2
allow for statement of conservation of energy
(energy can not be destroyed but can be converted from one form to
another)
B1
correct height used (2.4 m or 2 × 1.2 seen in an equation)
Page 45
Colonel Frank Seely School
B1
correct substitution including values for h and g (no u.p.)
B1
(3)
(ii)
F = mv / r
2
(allow mrω2
C1
2800 N (2780 N) or
2700 N (2740 N) if using v = 6.86 m s–1
A1
(2)
(iii)
(ii) + 690 (3500 N or 3460 N)
(3400 N or 3430 N if using v = 6.86 m s–1)
B1
(1)
(iv)
graph shape down up down up (condone linear); minima at 90° and 270°
M1
graph starts at 690 (N); this point labelled;
maximum labelled consistent with answer to (iii),
zero at 90 and 270 (allow any shape between these points)
A1
(2)
(b)
stress = F / A and strain = extension /original length and E = stress / strain
or
E = Fl / Ae
C1
correct substitution using 690 N (condone 700 N)
or substitution with e.c.f. from graph
C1
allow e.c.f. for use of g without substitution if penalised in (i)
8.9 × 10–6 – 9.1 × 10–6 m
Page 46
Colonel Frank Seely School
A1
allow only 1 mark if candidate divides by 2 at any stage
(3)
[11]
M3.A
[1]
M4.(a)
The force per unit area
B1
at which the material extends considerably / a lot / plastically /
or strain increases considerably etc
NOT doesn’t return to its original shape / permanently deformed
B1
for no (or a small) increase in) force / stress
B1
(3)
(b)
(i)
strain = 8.4 × 10–4 (1.3 × 10–3 / 1.55 seen) (allow if in E = FL / AΔL)
B1
or area of cross section = 2.54 × 10–6
or π (0.9 × 10–3)2
stress = E × strain (explicit or numerically) and
stress = F / A or E = FL / AL
C1
force = 440 – 450 N(cao)
A1
(3)
(ii)
Energy = ½ F Δ l or ½ stress × strain × volume
C1
0.29 J ecf for F from (b)(i)
A1
(2)
[8]
M5.(a)
(i)
15 rev / s = 30π rad / s or v = 51 / 52 m s–1 [could appear in subst]
Bl
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Colonel Frank Seely School
F = mw2r [or mv2 / r & v = ωr]
Bl
appropriate sub leading to 7.33 kN
[2+sf evaluation mandatory]
Bl
(ii)
to centre of rotor OWTTE
Bl
(iii)
stress = F / A
Bl
correct substitution from ai
Bl
(iv)
0.55 × 2.09 × 107 / 6 ×1010
[or ε = 3.3 × 10–4 ]
Cl
= 0.192 mm
Al
(v)
½ × 7.32 × 103 × 1.92 × 10–4
[ecf]
Cl
= 0.702 J
Al
(b)
(i)
volume pushed down [per second] = Av [mass = ρ × volume]
Bl
Change of momentum [per second] = mass pushed down per second × v
Bl
(ii)
Upward force = 900 N OWTTE [penalise use of 900g]
OR area swept out by blades = π × 0.552
Cl
900 = (0.55)2 π1.3v2
Cl
= 27 m s–1
Al
[15]
Page 48
Colonel Frank Seely School
M6.(a)
below yield stress material behaves elastically
or returns to original length when forces are removed
B1
above the yield stress: (condone ‘at the yield stress’)
material behaves plastically/is permanently deformed / is ductile
B1
extends considerably / has large strain / extension
B1
for very small increases in stress / force
B1
Max 2
(b)
(i)
Strain = 3.33 × 10–4 or
seen
C1
E = stress / strain and stress = F / A;
or E = Fl / AΔl
C1
or π(9.5 × 10–3)2 seen
A = 2.8 × 10 m or
–4
2
C1
Stress = 7.0 × 107 Pa
C1
2 max for C marks
Force = 19.6 to 19.8 (20) kN
A1
(3)
(ii)
Strain energy = ½FΔl or ½ their (b)(i) × (1.5 × 10–3)
C1
condone incorrect power or no 10–3 for C mark
or ½ σε × volume
14.6 to 14.9 (15) J (e.c.f.)
A1
(2)
[7]
M7.(a)
tensile stress =
tensile strain =
(1)
(1)
mention of tensile and original (1)
Page 49
Colonel Frank Seely School
(3)
(b)
diameter of wire (1)
in several places [or repeated] (1)
using a micrometer (1)
(original) length of wire (1)
using a metre rule (or tape measure) (1)
(max 4)
(c)
(2)
[9]
M8.(a)
(i)
(ii)
strain = 0.026 (1)
E = 6.92 × 109 Pa (1)
A = 1.96 × 10 (m ) (1)
–7
2
stress = 230 × 10 Pa (1)
8
(iii)
breaking strain = 0.044 (1)
5
(b)
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Colonel Frank Seely School
shape overall (1)
(i)
straight line (1)
0 to (0.026, 1.8) (1)
(ii)
curve (1)
to (0.044, 2.3) (1)
max 4
[9]
M9.(a)
appropriate discussion of energy conservation (1)
Δ p.e. = 2.5 × 10–2 × 9.8 × 1.2 (1) (= 0.29 J)
(b)
F=
(c)
A = 3.1 × 10 (m ) (1)
(1) = 590 N (1)
–6
2
stress = 1.9 × 108 Pa (1)
(d)
strain =
= 8.3 × 10–4 (1)
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Colonel Frank Seely School
(e)
(1) = 2.3 × 1011 Pa (1)
E=
[9]
M10.(a)
(i)
diagram to show:
(long) wire fixed at one end (1)
mass / weight at other end (1)
measuring scale (1)
mark on wire, or means to measure extension (1)
max 3
[alternative for two vertical wires:
two wires fixed to rigid support (1)
mass / weight at end of one wire (1)
other wire kept taut (1)
spirit level and micrometer or sliding vernier scale (1)]
(ii)
measurements:
length of the wire between clamp and mark (1)
diameter of the wire (1)
extension of the wire (1)
for a known mass (1)
max 3
(iii)
length measured by metre rule (1)
diameter measured by micrometer (1)
at several positions and mean taken (1)
(known) mass added and extension measured
by noting movement of fixed mark against vernier scale
(or any suitable alternative) (1)
repeat readings for increasing (or decreasing) load (1)
max 5
(iv)
graph of mass added / force against extension (1)
gradient gives
correct use of data in
(1)
where A is cross-sectional area (1)
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Colonel Frank Seely School
[if no graph drawn, then mean of readings
and correct use of data to give 2max) (1)
max 2
(13)
The Quality of Written Communication marks are awarded for the quality of
answers to this question.
(b)
(i)
for steel (use of
e=
gives) e =
(1)
(1)
= 5.0 × 10–3 m (1)
(ii)
extension for brass would be 10 × 10-3(m) (or twice that of steel) (1)
end A is lower by 5 mm √ (allow C.E. from (i))
max 3
[16]
M11.(a)
(i)
the Young modulus: tensile stress / tensile strain (1)
(ii)
maximum force or load which can be applied without wire being
permanently deformed
[or point beyond which (when stress removed,) material does not
regain original length] (1)
2
(b)
(i)
graph:
suitable scale (1)
correct points (1) (1)
best straight line followed by curve (1)
(ii)
indication of region or range of Hooke’s law (1)
(iii)
(use of E =
)
values of F and e within range or correct gradient (1)
to give E =
(1)
= 3.3(5) × 1010 Pa (1)
8
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Colonel Frank Seely School
(c)
(i)
work done = force × distance (1)
= average force × extension (= ½Fe) (1)
[or use work done = area under graph
area = ½ base × height]
(ii)
energy stored =
(1)
= 13.(4) × 10-3 J (1)
4
[14]
M12.
(35 × 9.81) = 343 N
(a)
(i)
(ii)
tension in each cable
= 172 N (1)
2
(b)
area of cross-section (=
)=
= 5.36 × 10–5 (m2)
=
4
(c)
(i)
moments about T2, (cable B) gives
5.52 T1 (1) = 343 × 2.76 (1) + 196 × 4.52 (1)
T1 =
(ii)
(1) (= 332 N)
T1 + T2 = 343 + 196 = 539 (N) (1)
T2 = 539 – 332 = 207 N (1)
(allow C.E. for. value of T1, from (i))
Page 54
Colonel Frank Seely School
[or moments about T1 gives 5.52 T2 = (343 × 2.76) + (196 × 1.) (1)
T2 =1143/5.52 = 207 N (1)
6
[12]
M13.
(a)
(i)
X (1)
stress (force)
strain (extension) for the whole length (1)
(ii)
Y (1)
has lower breaking stress (or force/unit area is less) (1)
(iii)
Y (1)
exhibits plastic behaviour (1)
(iv)
Y (1)
for given stress, Y has greater extension
[or greater area under graph] (1)
8
QWC 2
(b)
(i)
(use of E =
gives)
=
(1)
(1) for data into correct equation, (1) for correct area
= 2.4 N (1)
(allow C.E. for incorrect area conversion)
(ii)
(use of energy stored = ½Fe gives) energy =
(1)
= 36 × 10–3 J (1)
(allow C.E. for value of F from (i))
5
[13]
Page 55
Colonel Frank Seely School
M14.
(a)
extension proportional to the applied force (1)
up to the limit of proportionality
[or provided the extension is small] (1)
2
(b)
(i)
8 × 9.81 = 78 (5) N (1)
(allow C.E. in (ii), (iii) and (iv) for incorrect value)
(ii)
(use of E =
gives) 2.0 x 1011 =
(1)
ΔL = 3.5 × 10–3 m (1)
(iii)
similar calculation (1)
to give AS = 5.6 × 10–7 m2 (1)
[or AB = 2AS (1) and correct answer (1)]
(iv)
(use of energy stored = ½ Fe gives) energy stored
= ½ × 78.5 × 3.5 × 10–3 (1)
= 0.14 J (1)
7
(c)
(i)
end A is lower (1)
(ii)
= ½ 3.5 × 10–3 = 1.8 × 10–3 m (1)
(1.75 × 10–3 m)
2
[11]
M15.
(a)
tensile stress: force/tension per unit cross-sectional area or
Page 56
Colonel Frank Seely School
with F and A defined (1)
tensile strain: extension per unit length or
the Young modulus:
with e and l defined (1)
(1)
3
(b)
(i)
(ii)
= 2 (1)
F = 2FB (1)
FS + FB = 15 N (1) gives FS = 10 N
[or any alternative method]
(iii)
e=
=
(1)
= 5.36 ×10–5m (1)
6
[9]
(a)
Hooke’s law: the extension is proportional to the force applied (1)
up to the limit of proportionality or elastic limit
[or for small extensions] (1)
M16.
2
(b)
(i)
(use of E =
gives) ΔLs =
(1)
= 1.3 × 10–4 (m) (1) (1.33 × 10–4 (m))
ΔLb =
= 4.7 × 10–4 (m) (1) (4.66 × 10–4 (m))
total extension = 6.0 × 10–4 m (1)
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Colonel Frank Seely School
(ii)
m = ρ × V (1)
ms = 7.9 × 103 × 2.4 × 10–6 × 0.8 = 15.2 × 10–3 (kg) (1)
mb = 8.5 × 103 × 2.4 × 10–6 × 1.4 = 28.6 × 10–3 (kg) (1)
(to give total mass of 44 or 43.8 × 10–3 kg)
7
(c)
(use of m = ρAl gives) l =
(1)
= 2.2 m (1) (2.16 m)
(use of mass = 43.8 × 10–3 kg gives 2.14 m)
2
[11]
M17.
(a)
(i)
the extension produced (by a force) in a wire is directly
proportional to the force applied (1)
applies up to the limit of proportionality (1)
(ii)
elastic limit:
(iii)
the Young modulus: ratio of tensile stress to tensile strain (1)
unit: Pa or Nm–2 (1)
the maximum amount that a material can be
stretched (by a force) and still return to its original
length (when the force is removed) (1)
[or correct use of permanent deformation]
5
(b)
(i)
length of wire (1)
diameter (of wire) (1)
Page 58
Colonel Frank Seely School
(ii)
graph of force vs extension (1)
reference to gradient (1)
gradient =
(1)
[or graph of stress vs strain, with both defined
reference to gradient
gradient = E]
area under the line of F vs ΔL (1)
[or energy per unit volume = area under graph of stress vs strain]
6
[11]
M18.
(a)
(i)
weight of container (= mg = 22000 × 9.8(1)) = 2.16 × 105 (N) (1)
tension (= ¼ mg) = (5.39) 5.4 × 104 (N) or divide a weight by 4 (1)
(ii)
moment (= force × distance) = 22000 g × 32 (1) ecf weight in (a) (i)
= 6.9 or 7.0 × 106 (1) N m or correct base units (1) not J, nm, NM
(iii)
the counterweight (1)
provides a (sufficiently large) anticlockwise moment (about Q)
or moment in opposite direction ( to that of the container to
prevent the crane toppling clockwise) (1)
or
left hand pillar pulls (down) (1)
and provides anticlockwise moment
or
the centre of mass of the crane(‘s frame and the counterweight)
is between the two pillars (1)
which prevents the crane toppling clockwise/to right (1)
7
(b)
(i)
(tensile) stress
ecf (a) (i) (1)
Page 59
Colonel Frank Seely School
= 1.4(2) × 108 (1) Pa (or N m–2) (1)
(ii)
extension =
(1)
and (= 1.7 × 10–2 m) = 17 (mm) (1)
=
5
[12]
M19.
(a)
returns to original length/shape/position/state/zero
extension/no permanent extension (1)
1
(b)
(12 to 14 big squares/318 small squares ± 8 area of 1 big square
= 10N × 0.05m = 0.50/small = 2 × 0.01 = 0.02)
statement of method that refers to area (1)
accept triangle if area is mentioned
5.0 to 8.0 (J) or clear attempt to calculate correct area (1)
triangle OK here 5.1 (J) for single triangle is max 2
6.0 to 7.0 (J) (1)
3
(c)
(E =)
(1)
(=)
(1) also gets first mark
incorrect values of F and ΔL get first mark only
2.5 × 107 (Pa) (1)
or (stress = F/A =) 10/5 × 10–6 (= 2.0 × 106 Pa) (1)
Page 60
Colonel Frank Seely School
(strain = ΔL/L =) 0.04/0.5 (= 0.08) (1)
2 × 106/0.08 gets both marks
E correctly evaluated from incorrect value of stress and
incorrect value of strain gets 1 mark only
use of 50 (N) and 0.04 (m) giving 1.25 × 108 (Pa) gains 2 marks
use of 5(N) and 0.4 (m) is max 2
2.5 × 107 (Pa) (1)
3
(d)
(i)
straight line through origin finishing at the same point as the
rubber ± 1 small division (1)
1
(ii)
point beyond which
graph is no longer linear
or force no longer proportional to extension
or Hooke’s law limit (1)
1
[9]
M20.
(a)
(i)
straight best fit line from 0 → (at least) extension of 4.0 × 10–3 m (1)
smooth curve near points after 5.0 × 10–3 m (1)
2
(ii)
their
(ignore powers of ten) (1)
= 5.1 × 107
and x axis interval ≥ 3.0 (1) (5.06 to 5.14 × 107 N m–1) ecf from graph
allow error in calculation ± 2%
2
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Colonel Frank Seely School
(b)
load = 2.8 × 105 or
=
(1) 2.8 only
= 1.1 × 108 (Pa) 110 (MPa) (1) (1.12 × 108)
(M)Pa, pascals, N m–2 (1)
3
(c)
(Δl =
)=
(1) (= 2.94 × 10–3 m for 10 m)
gives 0.29(4) (m) (1) ecf
or reads a reasonable extension for 150 kN from the graph (1)
and multiples by 100 (= 0.29) (ecf) (1)
2
[9]
M21.
(a)
Suitable scale on both axes (eg not going up in 3s) and > ½ space used
≥ points correct (within half a small square)
line is straight up to at least stress = 2.5 × 108 and curve
is smooth beyond straight section
3
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Colonel Frank Seely School
(b)
understanding that E = gradient (= Δy/Δx)
allow y/x if line passes through origin
= 1.05 × 1011 (Pa) (allow 0.90 to 1.1) ecf from their line in (a)
if answer outside this range and uses a y value ≥ 2
when values used from table;
•
two marks can be scored only if candidates line passes
through them
•
one mark only can be scored if these points are not on their line
2
(c)
correct rearrangement of symbols or numbers ignoring incorrect
powers of ten, eg A =
correct substitution in any correct form of the equation,
eg =
allow incorrect powers of ten for this mark
= 1.6 × 10–4
(1.5789) (m2)
3
[8]
M22.
(a)
extension divided by its original length
do not allow symbols unless defined
1
(b)
1.9 × 108 (Pa)
1
(c)
point on line marked ‘A’ between a strain of 1.0 × 10−3 and 3.5 × 10−3
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Colonel Frank Seely School
1
(d)
clear evidence of gradient calculation for straight section
eg 1.18 (1.2) × 108/1.0 × 10−3
= 120 GPa and stress used > 0.6 × 108 Pa
allow range 116 − 120 GPa
Pa or Nm−2 or N/m2
3
(e)
(i)
clear attempt to calculate correct area (evidence on graph is sufficient)
(32 whole squares + 12 part/2 = 38 squares)
(38 × 10000 = ) 380000 (J m−3)
allow range 375000 to 400000
2
(ii)
V = m/ρ or 0.015/8960 or 1.674 × 10−6 (m3)
380 000 × 1.674 × 10−6 = 0.64 (0.6362 J)
ecf from ei
2
(f)
straight line passing through origin (small curvature to the right only above 160
MPa is acceptable) end at 176 MPa
(allow 174 to 178)
straight section to the left of the line for copper (steeper gradient)
2
[12]
M23.(a)
(W = mg)
= 4.8 × 35 × 9.81
=1600 (1648 N)
Allow g=10 : 1680 (1700 N)
g = 9.8
1646 N
max 1 for doubling or halving.
Max 1 for use of grammes
2
(b)
(stress = tension / area)
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Colonel Frank Seely School
For first mark, forgive absence of or incorrect doubling /
halving.
= (0.5 ×) 1.5 × 106 / 6.2 × 10−4 OR = 1.5 × 106 / (2 ×) 6.2 × 10−4
= 1.2 × 109 (1.21 GPa)
Forgive incorrect prefix if correct answer seen.
2
(c)
(i)
(weight = stress × area)
max 1 mark for incorrect power of ten in first marking point
= 400 ×(106 ) × 6.2 × 10−4 (= 248 000 N)
max 1 mark for doubling or halving both stress and area
( × 2 = ) 5.0 × 105 (496 000 N)
Forgive incorrect prefix if correct answer seen.Look out for YM
÷ 400k Pa which gives correct answer but scores zero.
2
(ii)
OR correct substitution into a correct equation (forgive incorrect
doubling or halving for this mark only
OR alternative method:
strain = stress / E
then ΔL = L × strain
If answer to 4ci is used, it must be halved, unless area is
doubled, for this mark
Any incorrect doubling or halving is max 1 mark.
Allow 0.07
3
(iii)
OR correct substitution into F=kΔL
ecf ci and cii (answer 4c(i)
÷ answer 4c(ii) )
Allow halving extension for force on one cable
= 7.4(4)× 106
(Nm−1)
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Colonel Frank Seely School
Correct answer gains both marks
2
(iv)
Correct answer gains both marks
= ½ × 496 000 × 6.667 × 10−2
ci, cii, ciii
OR
½ × 7.4(4) × 106 × (6.667 × 10−2 )2
ecf
= 1.6(5) × 104 (J)
Forgive incorrect prefix if correct answer seen.
Doubling the force gets zero.
2
[13]
M24.(a)
P at the end of linear section ✓
1
(b)
Measure original length and diameter ✓
1
Determine gradient of linear section to obtain F / extension ✓
1
1
Alternative:
Convert to stress–strain graph and determine gradient.
(c)
Line from A
Parallel to straight section of original
Ending at horizontal axis ✓
1
(d)
Plastic deformation has produced permanent extension / re-alignment of bonds
in material hence intercept non-zero ✓
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Colonel Frank Seely School
1
Gradient is same because after extension identical forces between bonds ✓
1
(e)
0.2% is a strain of 0.002
Stress = 2.0 x 1011 x 0.002 =
1
4 x 108 ✓
1
= 11.3 kN ✓
1
(f)
Maximum force = 11300 N
Weight of mass = 600 x 9.81 = 5886 N ✓
1
Accelerating force must be less than
11300 – 5886 = 5423 N ✓
1
a (= F / m = 5423 / 600)
= 9.0 m s–2 ✓
1
(g)
To lift double the load at the same acceleration, would require double the force,
✓
The first mark is for discussing the effect on the force
1
To produce the same strain either use:
•
double the diameter of wire – so the stress stays the same and therefore
the strain is the same for the same wire, ✓
•
a wire with double the Young modulus – so that double the stress
produces the same strain for the same diameter. ✓
1
1
The other two are for discussing the two alternative methods
of keeping the strain the same
[16]
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Colonel Frank Seely School
M25.D
[1]
M26.(a)
6.5 × 1010 Pa ✓
1
(b)
kg m s ✓
-1
-2
1
(c)
Direction of movement of particles in transverse wave perpendicular to energy
propagation direction✓
1
Parallel for longitudinal✓
1
(d)
ρ1c1=ρ2c2✓
E=ρc or ρc =
2
seen
1
1
(e)
[
and cx = 2cy ]
0.5✓
1
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Colonel Frank Seely School
(f)
speed of the wave in seawater is less than speed of the wave in glass✓
1
argument to show that watern glass
1
so tir could be observed when wave moves from water to glass ✓
1
[10]
M27.(a)
Use of Young Modulus =
✓
The first mark is for calculating the tensile stress
1
To give tensile stress = 2 × 1011 × 3.0 × 10-4 = 6.0 × 107✓
The second mark is substituting into the tensile force equation
1
Use of tensile stress =
To give tensile force = 6.0 × 107 × 7.5 × 10-3 = 4.5 × 105 N ✓
The third mark is for the correct answer
1
(b)
Use of strain = extension / original length
To give extension = 3.0 × 10-4 × 45 = 1.4 × 10-2 m
(1.35 × 10-2) ✓
The first mark is for calculating the extension
1
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Colonel Frank Seely School
Use of energy stored = ½ F e
To give
Energy stored = ½ × 4.5 × 105 × 1.4 × 10-2
= 3.2 × 103 J ✓
(3.04 × 103)
The second mark is for the final answer
1
(c)
Temperature change = pre-strain / pre-strain per K
= 3.0 × 10-4 / 2.5 × 10-5 = 12 K✓
The first mark is for the temperature change
1
Temperature = 8°C + 12 = 20 °C ✓
The second mark is for the final answer
1
(d)
So that the rail is not always under stress✓
1
as the rail spends little time at the highest temperature✓
Or
To reduce the average stress the rail is under ✓
as zero stress will occur closer to average temperature / the rail will be under
compressive / tensile stress at different times✓
1
[9]
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Colonel Frank Seely School
E1.(a)
The question asked for the forces required. Many showed only one force. Some showed
only arrows below the surface. Many candidates appeared to have had little
experience of drawing force diagrams.
(b)
There was frequently a lack of precision in stating the meaning of the terms. Although
definition of any terms used was asked in the question, many simply quoted, for
example, F / A. In (ii) the ‘length’ involved was often not defined clearly. In part (iii)
many candidates wrote that the elastic limit is ‘the point at which the material would
not return to its original length when the force is removed’. This should have been
expressed as the maximum force for which the material would return to its original
length when the force is removed. The phrase ‘when the force is removed’ was often
omitted from otherwise good answers.
(c)
(i)
Working was often hard to follow in many responses. Candidates should be
encouraged to state briefly what they are trying to determine when they use a
formula as often there seemed no logic to what was being attempted. With
some explanation it might have been possible to gain some credit. The ‘divide
by 8’ aspect of the problem often appeared in odd places in the process and
lack of analysis of the problem often led to this being done twice. Many failed
because they did not know how to calculate area of cross section and others did
not read data accurately and used the diameter given as the radius. Obtaining a
strain as if one bolt had been used and then dividing by 8 was an unconvincing
approach used by many.
(ii)
The best answers showed that candidates had good problem solving ability and
that they could set out a sensible argument. The obvious approach is to
determine the cross sectional area of steel required and divide by the area of
one bolt to determine the number of bolts (rounding up). The use of elastic limit
or breaking stress was acceptable.
E2.(a)
(i)
The majority of the candidates appreciated and tried to use the fact that the
change in PE would be equal to the gain in KE. The most common error was to
use 1.2 m as the change in height of the centre of mass. Even though this gave
an incorrect answer few seemed to go back to check why their answer was
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Colonel Frank Seely School
incorrect.
(b)
E4.(a)
(b)
(ii)
Most candidates knew the correct equation and used correct data. Some
however tried to use F = mrω2, substituting 6.9 m s−1 for ω. A significant
proportion of the candidates incurred a significant figure penalty in this part.
(iii)
This part was not well done. Many candidates gave the answer as the weight of
the gymnast mg (690 N) not appreciating that this had to be added to the force
determined in (a)(ii). Some subtracted the weight.
(iv)
The graph was partly a test in graphing data that had previously been
determined and appreciating that in the initial position the force downwards is
the weight of the gymnast and realising that the vertical force on the bar would
be zero when the gymnast is in the horizontal position. Those who determined
the weight in (a)(iii) could gain full marks here as error carried forward.
Candidates were not expected to realise that the force acts upwards before this
position is reached.
Most candidates were able to quote a correct equation or series of equations.
Candidates could gain full credit for using the weight or the value they had plotted on
the graph for 0°. Many, however, used the value from part (a)(ii) or an incorrect value
from (a)(iii). The ‘record’ for arm-stretching during this manoeuvre was 1013 m, arrived
at without comment!
It was expected that within the answer candidates would show that they understood the
meaning of stress. Only a minority of the candidates did this. The idea that small
increases in stress would produce large increases in strain was not commonly given.
Many simply stated that the material would become plastic.
(i)
Most candidates were able to gain 2 or 3 marks for this question. Some made
mistakes in arithmetic following use of the correct formula and substitution.
Others used the wrong formula for area of cross section, 2πr2 or πr2L being
seen frequently.
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Colonel Frank Seely School
(ii)
E5.(a)
(i)
This was usually well done but a significant proportion of the candidates
misread the question and gave the strain rather than the strain energy as the
answer. Some made the going hard by using ½ stress × strain × volume. Others
simply determined ½ stress × strain thinking that this was the strain energy.
Candidates here (and elsewhere in the paper) must recognise that questions
beginning with the words ‘Show that...’ demand a clear and well-explained
exposition of the calculation or proof. Examiners are not satisfied with muddled
and incomplete answers or answers that simply re-state the question. Explicitly,
in this part, candidates were expected to quote their answer to more than the
one significant figure in the question. They were also expected to show how
they converted 15 revolutions per second into a value that could be used in their
equation. Failures to convince often lead to loss of marks.
(ii)
Although examiners were generous here, the common answer ‘to the centre’ is
not adequate for a candidate at A2 level in a physics examination; the centre of
what? Strong candidates often illustrated the answer on a labelled diagram of
the helicopter itself and drew the examinerߢ s attention to this amendment.
Candidates should take every opportunity to make their answers clear.
(iii)
Again, a ‘show that’ which often did not. Those who quote the equation p = F / A
for ‘stress’ without defining their symbols must not be surprised if the examiner
assumes that p stands for ‘pressure’. The safest route is to supply a word
equation in cases like this.
(iv)
There were many good answers here, but solutions were frequently marred by
arithmetic errors and by negligent candidates who calculated the strain and then
omitted to go on to evaluate the required change in length.
(v)
Again, this was well done by many, but there was a significant number of
candidates who attempted to go via the route
strain energy per unit volume = ½ stress × strain.
This is indeed a possible method, but these candidates too often failed to
multiply in the volume of the helicopter blade.
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Colonel Frank Seely School
(b)
E6.(a)
(b)
(i)
Only a few candidates were able to prove this simple relationship. Those who
did stated clearly the volume of air being forced downwards every second, could
use this to evaluate the mass of air being forced down, and could go on to show
the change in momentum of this air. Otherwise candidates simply floundered
around trying unsuccessfully to ‘spot’ a relationship that would yield the result
they needed.
(ii)
To the examiners’ surprise very many found the evaluation of the air speed
beyond them. Although many recognised that the change in momentum per unit
time has to equal the weight of the helicopter, candidates still turned the weight
of the aircraft back to mass, and failed to spot the squared velocity term. This
was a disappointing failure by many candidates.
Although there were many good answers this was not well understood by the majority of
the candidates. Many candidates seemed unaware of the term and wrote only of
breaking or bending. Weak answers implied that for stresses below the yield stress
the material (bridge or building) would not ‘deform’ at all.
(i)
Most candidates were able to gain one or two marks here by making some
progress toward the answer. Common errors were use of the diameter value in
πr2 or failure to convert the mm dimensions to m. Weak candidates often related
the ‘original length’ to 19 mm.
(ii)
Allowing the error carried forward there were many correct answers. A minority
simply calculated the strain and some thought that ½ stress × strain gave the
energy stored. Using ½ stress × strain × volume made it hard going but those
using it usually succeeded. Some following this route used the increase in
volume rather than the total volume of the rod.
E8.In part (a)(i) there were many good answers with the calculation clearly set out and the answer
quoted correctly to 2 significant figures and with the correct unit. Some candidates
incorrectly attempted to calculate the stress for a force of 45N while others had difficulty in
converting from mm to m in determining the strain. In part (a)(ii) about 25% of the
candidates could not calculate correctly the cross-sectional area of the wire since they used
the diameter and not the radius, or they failed to convert the radius from mm to m. Most
candidates were able to calculate correctly the value for the breaking strain.
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Colonel Frank Seely School
The sketch graph in part (b) was often carelessly drawn. Sketch graphs gaining full marks
showed a straight line up to the limit of Hooke's law followed by a curved region up until the
breaking point. The nature of the curved region of the graph was treated generously in the
marking. Most candidates marked correctly their values of stress and corresponding strain
on the axes.
E9.Answers to this question were not good overall. In part (a) candidates appreciated that an
energy transformation was taking place but did not state the exact process involved. In part
(b) very few candidates were able to calculate the tension in the wire and the vast majority
simply calculated the tension when it is in equilibrium. Parts (c), (d) and (e) produced better
answers and the only common error was to confuse stress and strain. Disappointingly, a
unit for strain was also incorrectly given by a significant number of candidates.
E10.Almost all candidates gained reasonable marks on part (a)(i) even though some of the
descriptions were lacking in detail. Most of the diagrams were reasonably drawn with the aid
of a ruler. Candidates who drew freehand usually produced inferior diagrams which failed to
gain all the available marks. A variety of methods were shown, usually two wires hanging
vertically, linked together by means of some vernier arrangement and a spirit level. Also
shown was a horizontal wire on a bench. Although this is not such an accurate method, it
was accepted but many candidates showed the mark on the wire as being about half way
along. This of course is only acceptable if the length of the wire is measured to that point,
but this was usually overlooked in the description. The least satisfactory method was
suspending a single wire with a ruler alongside although this did gain some marks. There
were an alarming number of diagrams which showed a completely unrelated length of wire,
a ruler, an isolated hook with a mass attached and a micrometer. Needless to say, such
efforts gained no marks.
In part (ii) candidates could have saved themselves considerable time and effort by reading
the question carefully and just listing the measurement they would make. Many candidates
listed the area of cross section as a measurement. This was not acceptable since area is a
derived quantity and it is the diameter which is measured. Many candidates also listed the
‘width’ of the wire, which again was not accepted.
The descriptions in part (iii) were, on the whole, quite reasonable, although most effort
seemed to go into describing how the length of the wire and its diameter were measured
and not giving sufficient attention to the experiment, i.e. measuring the extension for each
mass added and increasing the total mass to a certain value. There were very few
references to repeating the readings while unloading. This particular section of the question
was also used to award the quality of written communication marks and most candidates
scored well on this.
The descriptions in part (iv) of how to use the measurements to give the Young modulus
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Colonel Frank Seely School
was reasonably done with about 50% of the candidates drawing a graph of force vs
extension or stress vs strain and using the gradient accordingly. Candidates who only used
one set of values to give one value of the Young modulus were not awarded all the available
marks.
The calculation in part (b)(i) was performed satisfactorily, with the majority of candidates
calculating the correct extension for the steel wire. Marks were lost in part (ii) when the
answer was given without any reasoning.
E11.High marks were gained for this question on the Young modulus. The definitions in part (a)
were usually correct, although it is worth reminding candidates that when defining the
Young modulus it is essential to use tensile stress and tensile strain and not just stress and
strain. The description of elastic limit was sometimes vague, but the examiners sensed that
the candidates knew what it was, even if their wording was not perfectly clear. It should be
emphasised however that the wire can only regain its original length when the load or force
is removed. A large number of candidates referred to the wire regaining its original shape
rather than length; the shape of a wire does conjure up a different picture to the length of a
wire.
The graph in part (b) was usually well drawn, although a significant number of candidates
did omit the zero point, which was an important point to plot. Surprisingly in part (iii),
although candidates had indicated correctly on the graph the extent of Hooke’s law, they
used a load of 10N in the calculation. This load extended the wire well beyond the region of
Hooke’s law. Many candidates not only omitted the units in the calculation, but also used
incorrect units, N m2 being a popular alternative.
In part (c), because the expression for the energy stored was given in the question, showing
that the work done in stretching the wire = ½Fe proved to be more difficult than expected. In
the calculation in part (ii) the usual error encountered was not converting the 4 mm to
metres.
E12.
In part (a) some candidates did not gain both marks as a result of a significant figure
penalty. The calculation in part (b) was generally done well but some candidates calculated
the area of cross-section incorrectly or used the weight of the beam instead of the tension in
each cable.
Only a minority of candidates made progress in part (c)(i). The majority were unable to give
the correct equation using the Principle of Moments. Some candidates worked out the
correct answer without providing a satisfactory explanation, using little more than
knowledge of the answer to guide them. However, a large number of candidates knew how
to proceed in part (ii) and were able to gain full credit.
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Colonel Frank Seely School
E13.
Examiners were pleased to find that part (a) was answered satisfactorily and that
candidates not only chose the correct wire but were very often able to provide the correct
reason for doing so. Many candidates gained full marks, while a large number only lost one
or two marks.
Part (i) was usually correct, although reasons such as ‘the graph is a straight line’ were not
accepted. A ‘constant gradient’ was accepted but few candidates gave this as a reason,
most giving the proportionality of the quantities involved. In part (ii) answers such as ‘Y
broke before X’ was not accepted. Examiners were looking for a reason in terms of lower
breaking stress.
Answers to part (iii) were not so good and candidates who did not know the correct answer
attempted an answer in terms of the gradients of the curves or the bending of curve Y as the
tensile strain increased. Part (iv) gave the most trouble. Many candidates again tried an
explanation in terms of the gradient, but a significant number followed the correct track and
gave a reason in terms of the area under such a graph. Unfortunately the majority of these
candidates referred to the area under the whole curve, whereas it should have been the
area under the curve at a given tensile stress. Surprisingly, many candidates, even when
using a given stress, gave the area under X as being greater than that under Y.
The final calculation in part (b) did not cause too much difficulty and, provided the initial
equation for the Young modulus was correct, candidates produced a correct answer with
correct units. One common error which again arose from not reading the question
thoroughly, was using the extended length of the elastic cord as the extension. Converting
the cross-sectional area of the cord from mm2 to m2 caused some problems, but this error
was carried forward after the initial penalty had been imposed. The calculation in part (ii)
was also done well by those who knew the expression for the energy stored, or were aware
that it was given in the data sheet. Some answers, resulting from a carry forward of an
incorrect force in part (i) gave energies amounting to several million joules. This attracted no
comment.
E14.
Responses to part (a) were extremely disappointing. The impression gained by
examiners was that many candidates had not heard of Hooke’s Law because they
attempted to state it in terms of current and voltage. Others, realising that it had something
to do with solids, attempted an answer in terms of a wire returning to its original shape and
length when the force was removed. It should be pointed out that merely stating F
e
gained no credit unless the symbols were defined. Of those candidates who gave the
correct version of Hooke’s Law, most failed to gain the second mark by not giving the
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Colonel Frank Seely School
condition under which it was valid, i.e. up to the limit of proportionality. Validity up to the
elastic limit was not accepted.
The calculation in part (b) was carried out quite successfully and many completely correct
answers were obtained. The usual error occurred in part (i) in not realising that the tension
due to the 16 kg was shared equally between the two wires. Others did not multiply by g, or
used g = 10 m s–2, which was not acceptable. The data sheet gives g = 9.81 m s–2 and this is
the value which was required. Incorrect answers to part (i) were allowed to be carried
forward into the remaining parts of the section. However, if the force in part (ii) was given as
16 kg or 8 kg, without conversion into a force (i.e. Newtons), then it was considered as a
Physics error and both marks were lost.
The majority of candidates deduced (or guessed) that end A of the bar would be lower than
end B, although some of the convoluted answers required significant interpretation to know
which end of the bar candidates were talking about. The calculation in part (ii) was usually
carried out correctly.
E15.
Normally, the question on elasticity realizes high marks. Not so this time, although
there were many completely correct answers. The three definitions were usually correct
although, as in past papers, the Young modulus was defined in terms of stress and strain,
rather than tensile stress and tensile strain. There also appeared this year, references to the
stiffness constant of the wire. It should be pointed out that the stiffness constant is not the
same as the Young modulus.
The algebra involved in parts (i) and (ii) of (b) caused problems and a significant number of
candidates failed on this section. However, part (iii) proved to be a straightforward
calculation for the majority of candidates.
E16.
Hooke’s law, in part (a), was generally known to candidates although many did not
state the condition under which it applied. Many introduced temperature into the argument.
The calculation in part (b) was usually correct with comparatively few candidates adding the
two lengths or adding the values of the Young moduli to perform just one calculation.
Questions on density, similar to those in part (b) (ii), are usually done well, and this question
was no exception. Full marks were quite common in part (b).
Part (c) also proved to be relatively easy with the large majority of candidates obtaining the
correct answer. Those who failed were usually those who tried to tackle it from a Young
modulus point of view.
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Colonel Frank Seely School
E17.
Few candidates gained the maximum mark of five for part (a). In part (i), the
necessary condition that the wire had not been stretched beyond the elastic limit or beyond
the limit of proportionality was usually omitted. In part (iii) the Young modulus was often
defined as stress/strain, which was not acceptable, and finally a large number of candidates
failed to give the unit of the Young modulus, many going for the easy option of stating that it
had no units.
Candidates who took the trouble to read the stem of the question carefully usually gave
good answers, but those who thought that part (b) required a description of the experiment
failed miserably. In part (i) the usual answer of measuring the cross-sectional area was not
accepted. The only answers that gained both allocated marks were the diameter and the
length of the wire. In part (ii) a graph of force vs extension, or stress vs strain (provided both
were defined) were acceptable, but many who opted for the stress vs strain graph then went
on in part (iii) to give the answer as the area under the graph, which, of course, is wrong,
unless they referred to the energy per unit volume.
E18.
A surprisingly large number of candidates divided the mass by four to get a .weight. of
5500 kg in part (a) (i). Many also forgot to divide by four in what should have been a fairly
uncomplicated question.
In part (a) (ii), many candidates simply multiplied the mass of 22000 kg by 32, indicating a
surprising confusion between weight and mass. For the unit mark there were many common
errors such as N, NM, Nm–1, Nm–2, J, nm, kg and Nkg–1.
A very easy mark for mentioning the .counterweight. was picked up by most candidates in
part (a) (iii). However, not many went on to discuss the .anticlockwise moment. that this
provides.
Most picked up the first two marks to part (b) (i), some as a result of the ecf for the tension.
Many candidates used wrong units; pa, PA, Nm–1, being common rather than Pa.
Those with an ecf in (b) (i) generally failed to get both marks to part (b) (ii) because they did
not arrive at 17 mm. This may have given some candidates a clue that one of their previous
answers was incorrect. The candidates who were successful on the first parts of the
question invariably scored both marks here.
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Colonel Frank Seely School
E19.
It was a little surprising that more candidates did not gain the mark to part (a). A
common incorrect answer was ‘curve B’.
In part (b), the majority of candidates did not know that the work done is the area under the
line or they did not know a suitable method to estimate area. The most common approach
was to treat the curve as a straight line and use W = ½ FΔL.
The most reliable method in this case is to count the number of large squares, treating all
part squares as half a square, and then multiply the total by the value of one square (0.5 J).
If this method is taught, the candidates will always get the answer that appears on the mark
scheme.
The typical Young modulus of rubber can be found in most data books. This is obtained for
small values of strain. The majority of candidates knew what to in part (c) and many gained
all three marks. Unfortunately, a significant number of candidates were careless in reading
the question and used 0.40 instead of 0.040 and therefore 50 N instead of 10 N.
The majority of candidates were successful in part (d) (i). A common error was to draw
another curve or a straight line at a tangent to initial slope of A.
Part (d) (ii) was another definition that caused problems for many candidates. Some failed
to gain credit by confusing limit of proportionality with elastic limit. Some neglected to say
‘point beyond which the load and extension will no longer be proportional’.
E20.
In part (a) (i), the line of best fit had to start very near to the origin and go between the
fifth and sixth points on the graph. Most candidates did this very well. Very few candidates
who attempted a freehand line made their line smooth or straight enough to gain the first
mark. A smooth curve was expected for the last few points on the graph. Most candidates
knew that this is how a spring is likely to behave and assumed a curve would be more
appropriate than another straight section.
Many candidates did not get the powers of ten correct or simply ignored them when
calculating the gradient in part (a) (ii). There was a general lack of care with the precision of
the gradient measurement. Often a line would not go exactly through the origin and this
would not be taken into account by the candidate. Gradients were often calculated from less
than half of the available length of the line.
Part (b) was done well. However, a surprising number of candidates misread the force as
2.6 or 2.7 or 2.85 × 105 N. Marks were often lost on the unit. The unit (Pa or N m–2) needed to
have a capital ‘N’ or ‘P’ and a lowercase ‘m’ or ‘a’.
For part (c), many candidates successfully used Hooke’s law to find the extension of a 10 m
length of the cable with a force of 150 kN. Many did not realise that they then needed to
multiply by 100 to get the extension of a 1000 m section. Some multiplied by 1000 instead of
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100. A considerable number thought they needed to divide by 1000. Surprisingly, very few
realised they simply could read off the extension from the graph for a 10 m length at 150 kN
and then multiply by 100. It was also common to see the Young modulus equation used, but
this was unnecessarily complicated and rarely yielded the correct answer. The suspicion is
that this question caught many candidates out, because a certain amount of manipulation of
numbers was necessary, in addition to substitution into an equation. This is a skill that will
be essential for those continuing to A2.
E21.
In part (a), suitable scales were chosen by nearly all candidates and points were
plotted very accurately by all but a few. From knowledge of material properties, it is sensible
to assume that the first section of the graph should be a straight line. However, many
candidates drew a curve for the first part. After this, any suitable best fit line was accepted if
it was smooth.
Part (b) was generally done very well, with most choosing points on their line correctly and
using suitably large values. A few candidates wrongly used the ultimate tensile stress
divided by the corresponding strain.
Most candidates were very successful on the calculation in part (c). The main problem was
arithmetic errors and mistakes on powers on ten. It is a good idea to make even high ability
students practice plenty of these questions when revising.
E22.
In part (a), many students confused strain with stress and there were many vague
descriptions rather than definitions, for example ‘amount of extension due to a force
applied’. The definition has to describe how a correct strain would be calculated. Therefore,
it is essential that the ‘original’ length is specified and the phrase ‘divided by’ rather than
‘compared to’ needs to be used. Some students used the word ‘from’ to convey ‘divided by’
for example, ‘the extension from the original length’. This was not accepted. ‘The ratio of the
extension from the original length’ was acceptable.
Only 1.9 × 108 was accepted as an answer to part (b). Few students strayed from 1.9.
However, many lost the mark by missing out the power of ten.
Most students did very well on part (c). Nearly all put the point in a sensible place just
beyond the linear section but a few did not include a suitable label and therefore they did not
get the mark.
Most students were very successful in part (d). Most chose the straight section for the
gradient calculation and most chose a large enough section of it. Some went to a stress of
1.3 or more where the line was clearly curving. The unit was usually correct but a few had
capital M or m–1. Surprisingly, a significant number missed the unit out altogether.
Part (e)(i) should have been a straightforward question. Most students identified the correct
area to evaluate but either did not use an accurate method or did not recognise the value of
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each square they counted. A very common answer was 38 rather than 38 × 105. Many
divided the area into two triangles and a rectangle, leaving out a significant area from their
calculation. With a number of similar examples of area approximation on past papers, it was
surprising that very few were able to score both marks here. Many students do not know
how to find the area under a curve to a sufficient accuracy.
Many students obtained the volume in answer to part (e) (i) but did not realise that they then
needed the value they calculated from part (e) (i).
In part (f), many students did not recognise that a higher Young Modulus would give a
steeper line and many who did realise this did not stop their line at 176 MPa. Many showed
excessive curvature not characteristic of a brittle material.
E23.(a)
Most were successful but a significant number did not multiply by g, perhaps not
understanding the difference between weight and mass.
(b)
A common error was to not half the force (or double the area). However, a high
percentage did realise that you had to do more than simply substitute the numbers
given.
(c)
(i)
Most correctly calculated the load on one of the cables but many did not realise
they needed to double their result to get the complete load.
(ii)
A lot of rounding errors were evident. 0.06 recurring was often rounded to 0.06
rather than 0.067. Many used their value for weight but did not halve it. Some
candidates therefore lost a mark because even though they got the correct
answer, they had not halved that weight and this was a physics error.
(iii)
Many did not understand that they should use the weight and the extension
previously calculated. Many thought that the total load divided by the extension
of one cable would give only half of the total stiffness constant. This is not the
case because the extension of each cable is the same.
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