Download n X ab E - Firefly

1
In the equation X =
, X represents a physical variable in an electric or a gravitational field, a
is a constant, b is either mass or charge and n is a number.
Which line, A to D, in the table provides a consistent representation of X, a and b according to
the value of n?
The symbols E, g, V and r have their usual meanings.
n
X
a
b
A
1
E
charge
B
1
V
mass
C
2
g
G
mass
D
2
V
G
charge
(Total 1 mark)
2
(a)
State, in words, Coulomb’s law.
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(2)
Page 1 of 62
(b)
The graph shows how the electric potential, V, varies with , where r is the distance from a
point charge Q.
State what can be deduced from the graph about how V depends on r and explain why all
the values of V on the graph are negative.
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(2)
Page 2 of 62
(c)
(i)
Use data from the graph to show that the magnitude of Q is about 30 nC.
(2)
(ii)
A +60 nC charge is moved from a point where r = 0.20 m to a point where r = 0.50 m.
Calculate the work done.
(2)
work done ................................................... J
(iii)
Calculate the electric field strength at the point where r = 0.40 m.
electric field strength .......................................... V m−1
(2)
(Total 10 marks)
3
In stars, helium-3 and helium-4 are formed by the fusion of hydrogen nuclei. As the temperature
rises, a helium-3 nucleus and a helium-4 nucleus can fuse to produce beryllium-7 with the
release of energy in the form of gamma radiation.
The table below shows the masses of these nuclei.
Nucleus
Mass / u
Helium-3
3.01493
Helium-4
4.00151
Beryllium-7
7.01473
Page 3 of 62
(a)
(i)
Calculate the energy released, in J, when a helium-3 nucleus fuses with a helium-4
nucleus.
energy released ................................................... J
(4)
(ii)
Assume that in each interaction the energy is released as a single gamma-ray
photon.
Calculate the wavelength of the gamma radiation.
wavelength ................................................. m
(3)
Page 4 of 62
(b)
For a helium-3 nucleus and a helium-4 nucleus to fuse they need to be separated by no
more than 3.5 × 10–15 m.
(i)
Calculate the minimum total kinetic energy of the nuclei required for them to reach a
separation of 3.5 × 10–15 m.
total kinetic energy ................................................... J
(3)
(ii)
Calculate the temperature at which two nuclei with the average kinetic energy for that
temperature would be able to fuse.
Assume that the two nuclei have equal kinetic energy.
temperature .................................................. K
(3)
(c)
Scientists continue to try to produce a viable fusion reactor to generate energy on Earth
using reactors like the Joint European Torus (JET). The method requires a plasma that has
to be raised to a suitable temperature for fusion to take place.
(i)
State two nuclei that are most likely to be used to form the plasma of a fusion reactor.
1 .............................................................................................................
2 .............................................................................................................
(2)
Page 5 of 62
(ii)
State one method which can be used to raise the temperature of the plasma to a
suitable temperature.
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(1)
(Total 16 marks)
4
The electric potential at a distance r from a positive point charge is 45 V. The potential increases
to 50 V when the distance from the charge decreases by 1.5 m. What is the value of r?
A
1.3 m
B
1.5 m
C
7.9 m
D
15 m
(Total 1 mark)
5
(a)
Define the electric potential at a point in an electric field.
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(3)
Page 6 of 62
(b)
Figure 1 shows part of the region around a small positive charge.
Figure 1
(b)
(i)
The electric potential at point L due to this charge is + 3.0 V. Calculate the magnitude
Q of the charge. Express your answer to an appropriate number of significant figures.
answer = ................................. C
(3)
(ii)
Show that the electric potential at point N, due to the charge, is +1.0 V.
(1)
(iii)
Show that the electric field strength at point M, which is mid-way between L and N, is
2.5 Vm–1.
(1)
(c)
R and S are two charged parallel plates, 0.60 m apart, as shown in Figure 2.
They are at potentials of + 3.0 V and + 1.0 V respectively.
Figure 2
(i)
On Figure 2, sketch the electric field between R and S, showing its direction.
(2)
Page 7 of 62
(ii)
Point T is mid-way between R and S.
Calculate the electric field strength at T.
answer = .......................... Vm–1
(1)
(iii)
Parts (b)(iii) and (c)(ii) both involve the electric field strength at a point mid-way
between potentials of + 1.0 V and + 3.0 V. Explain why the magnitudes of these
electric field strengths are different.
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(1)
(Total 12 marks)
Page 8 of 62
6
In the research into nuclear fusion one of the most promising reactions is between deuterons
and tritium nuclei
in a gaseous plasma. Although deuterons can be relatively easily
extracted from sea water, tritium is difficult to produce. It can, however, be produced by
bombarding lithium-6
with neutrons. The two reactions are summarised as:
+ energy
+ energy
Masses of reactants:
= 1.008665u
= 2.013553u
= 3.016049u
= 4.002603u
= 6.015122u
1u is equivalent to 1.66 × 10–27 kg or 931 MeV
(a)
(i)
Explain why the atomic mass unit, u, may be quoted in kg or MeV.
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(2)
Page 9 of 62
(ii)
Calculate the maximum amount of energy, in MeV, released when 1.0 kg of lithium-6
is bombarded by neutrons.
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energy released ...................................... MeV
(5)
(iii)
Suggest why the lithium-6 reaction could be thought to be self-sustaining once the
deuteron-tritium reaction is underway.
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(1)
(b)
(i)
In order to fuse, a deuteron and a tritium nucleus must approach one another to
within approximately 1.5 × 10–15 m.
Calculate the minimum total initial kinetic energy that these nuclei must have.
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minimum total kinetic energy of nuclei ............................................. J
(3)
Page 10 of 62
(ii)
Show that a temperature of approximately 4 × 109 K would be sufficient to enable this
fusion to occur in a gaseous plasma.
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(3)
(iii)
Explain in terms of the forces acting on nuclei why the deuteron-tritium mixture must
be so hot in order to achieve the fusion reaction.
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(4)
(Total 18 marks)
7
Which one of the following statements about electric potential and electric field strength is
correct?
A
Electric potential is zero whenever the electric field strength is zero.
B
Electric field strength is a scalar quantity.
C
Electric potential is a vector quantity.
D
Electric potential due to a point charge varies as
charge.
where r is the distance from the point
(Total 1 mark)
Page 11 of 62
8
The diagram shows a uniform electric field of strength 10 V m–1
A charge of 4 µC is moved from P to Q and then from Q to R. If the distance PQ is 2 m and QR
is 3 m, what is the change in potential energy of the charge when it is moved from P to R?
A
40 µJ
B
50 µJ
C
120 µJ
D
200 µJ
(Total 1 mark)
9
X and Y are two points in an electric field a distance d apart. The potential difference between X
and Y is V. A particle carrying a charge Q is accelerated by that field from X to Y in a time t. The
gain in kinetic energy of the particle is
A
QV
B
C
D
QVd
(Total 1 mark)
Page 12 of 62
10
Which one of the following arrangements of charge will produce zero electric field strength and
zero electric potential at the point labelled P?
(Total 1 mark)
Page 13 of 62
11
(a)
Figure 1 shows the electron gun that accelerates electrons in an electron microscope.
Figure 1
(i)
Draw, on Figure 1, electric field lines and lines of equipotential in the region between
the anode and cathode. Assume that there are no edge effects and that the holes in
the plates do not affect the field.
Clearly label your diagram.
(3)
(ii)
Calculate the kinetic energy, speed and momentum of an electron as it passes
through the hole in the anode.
mass of an electron
=
9.1 × 10–31 kg
charge of an electron
=
–1.6 × 10–19 C
(4)
Page 14 of 62
(b)
By calculating the de Broglie wavelength of electrons coming through the anode of this
device, state and explain whether or not they will be suitable for the investigation of the
crystal structure of a metal.
Planck constant
=
6.6 × 10–34 J s
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(4)
(Total 11 marks)
Page 15 of 62
12
(a)
The diagram below shows part of a precipitation system used to collect dust particles in a
chimney. It consists of two large parallel vertical plates maintained at potentials of +25 kV
and –25 kV.
The diagram below also shows the electric field lines between the plates.
(i)
Add arrows to the diagram to show the direction of the electric field.
(1)
(ii)
Explain what is meant by an equipotential surface.
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(1)
(iii)
Draw and label on the diagram equipotentials that correspond to potentials
of –12.5 kV, 0 V, and +12.5 kV.
(2)
Page 16 of 62
(b)
A small dust particle moves vertically up the centre of the chimney, midway between the
plates.
(i)
The charge on the dust particle is +5.5 nC. Show that there is an electrostatic force
on the particle of about 0.07 mN.
(2)
(ii)
The mass of the dust particle is 1.2 × 10–4 kg and it moves up the centre of the
chimney at a constant vertical speed of 0.80 m s–1.
Calculate the minimum length of the plates necessary for this particle to strike one of
them. Ignore air resistance.
(4)
(Total 10 marks)
13
The Earth has an electric charge. The electric field strength outside the Earth varies in the same
way as if this charge were concentrated at the centre of the Earth. The axes in the diagram below
represent the electric field strength E and the distance from the centre of the Earth r. The electric
field strength at A has been plotted.
(a)
(i)
Determine the electric field strength at B and then complete the graph to show how
the electric field strength varies with distance from the centre of the Earth for
distances greater than 6400 km.
(3)
Page 17 of 62
(ii)
State how you would use the graph to find the electric potential difference between
the points A and B.
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(1)
(b)
The permittivity of free space ε0 is 8.9 × 10–12 F m–1 .
(i)
Calculate the total charge on the Earth.
(2)
(ii)
The charge is distributed uniformly over the Earth's surface. Calculate the charge per
square metre on the Earth's surface.
(2)
(Total 8 marks)
14
A physicist wants to design an experiment in which two free protons collide to produce two
delta-plus (∆+) particles. This is an allowed reaction and is fully represented by the equation
p+ + p+ → ∆+ + ∆+
Two options are available that are shown as A and B in the diagram below.
A
accelerated proton
stationary proton
accelerated proton
accelerated proton
B
In A an accelerated proton collides with a stationary proton and in B two accelerated protons,
each with the same energy, collide head on.
the charge on a proton
the rest mass of a proton
the permittivity of free space ε0
(a)
= +1.6 × 10−19 C
= 1.7 × 10−27 kg
= 8.9 × 10−12 F m−1
State the baryon number of a ∆+ particle.
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(1)
(b)
The radius of a proton is 1.5 × 10−15 m.
(i)
Calculate the minimum total kinetic energy that the accelerated protons need so that
they will touch each other.
(3)
Page 18 of 62
(ii)
State what happens in situation A when the energy is less than your answer to part
(i).
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(1)
(iii)
State and explain what happens in situation B when the energy is less than your
answer to part (i).
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(3)
(iv)
Explain why the protons can undergo fusion if this energy is exceeded.
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(2)
(c)
Calculate the minimum total kinetic energy, in J, of the protons that will allow the two
protons to collide and produce the two ∆+ particles.
speed of electromagnetic radiation in free space = 3.0 × 108 m s−1
rest mass of a ∆+ particle
= 2.2 × 10−27 kg
(3)
(Total 13 marks)
Page 19 of 62
15
The diagram shows four point charges at the corners of a square of side 2a. What is the electric
potential at P, the centre of the square?
A
B
C
D
(Total 1 mark)
16
(a)
This part of the question is about protons.
(i)
Calculate the electrostatic potential energy, in J, of two protons at a distance apart of
1.0 × 10–15 m.
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Page 20 of 62
(ii)
Two protons moving in opposite directions at the same initial speed collide head-on
with each other. The least distance apart of the two protons is 1.0 × 1015 m. By
considering conservation of energy, estimate the initial kinetic energy, in MeV, of each
proton.
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(5)
(b)
State the quark composition of
(i)
a proton,
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(ii)
a positive pion, π+.
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(2)
(c)
A proton collides with another proton moving in the opposite direction at the same speed,
creating a positive pion and a further particle X in the process. This process is represented
by the equation
p + p → p + p + π+ + X.
(i)
State the charge, Q, and baryon number, B, of X.
Q ...........................................................................................................
B ...........................................................................................................
(ii)
State the identity and quark composition of X.
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Page 21 of 62
(iii)
Explain why two protons with initial kinetic energies as in part (a)(ii) could not produce
the reaction in part (c).
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(5)
(Total 12 marks)
17
Which one of the following statements about electric field strength and electric potential is
incorrect?
A
Electric potential is a scalar quantity.
B
Electric field strength is a vector quantity.
C
Electric potential is zero whenever the electric field strength is zero.
D
The potential gradient is proportional to the electric field strength.
(Total 1 mark)
18
(a)
Show that the kinetic energy of an α particle travelling at 2.00 × 107 m s–1 is 1.33 × 10–12 J
when relativistic effects are ignored.
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(2)
(b)
Calculate the closest distance of approach for a head-on collision between the α particle
referred to in part (a) and a gold nucleus for which the proton number is 79. Assume that
the gold nucleus remains stationary during the collision.
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(4)
Page 22 of 62
(c)
State one reason why methods other than α particle scattering are used to determine
nuclear radii.
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(1)
(Total 7 marks)
19
(a)
(i)
Define electric field strength, and state whether it is a scalar quantity or a vector
quantity.
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(ii)
Complete the diagram below to show the electric field lines in the region around two
equal positive point charges. Mark with a letter N the position of any point where the
field strength is zero.
(6)
(b)
Point charges A, of +2.0 nC, and B, of –3.0 nC, are 200 mm apart in a vacuum, as shown
by the figure. The point P is 120 mm from A and 160 mm from B.
Page 23 of 62
(i)
Calculate the component of the electric field at P in the direction AP.
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(ii)
Calculate the component of the electric field at P in the direction PB.
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(iii)
Hence calculate the magnitude and direction of the resultant field at P.
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(6)
(c)
(i)
Explain why there is a point X on the line AB in part (b) at which the electric
potential is zero.
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(ii)
Calculate the distance of the point X from A.
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(4)
(Total 16 marks)
Page 24 of 62
20
The electrical field strength, E, and the electrical potential, V, at the surface of a sphere of radius
r carrying a charge Q are given by the equations
A school van de Graaff generator has a dome of radius 100 mm. Charge begins to leak into the
air from the dome when the electric field strength at its surface is approximately 3 × 106 V m–1.
What, approximately, is the maximum potential to which the dome can be raised without
leakage?
A
3 × 104 V
B
3 × 105 V
C
3 × 106 V
D
3 × 107 V
(Total 1 mark)
21
Two horizontal parallel plate conductors are separated by a distance of 5.0 mm in air. The lower
plate is earthed and the potential of the upper plate is + 50 V.
Which line, A to D, gives correctly the electric field strength, E, and the potential, V, at a point
midway between the plates?
electric field strength E/V m–1
potential V/V
A
1 × 104 upwards
25
B
1 × 104 downwards
25
C
1 × 104 upwards
50
D
1 × 104 downwards
50
(Total 1 mark)
Page 25 of 62
22
Two charges, P and Q, are 100 mm apart.
X is a point on the line between P and Q. If the potential at X is 0 V, what is the distance from P
to X?
A
40 mm
B
45 mm
C
50 mm
D
60 mm
(Total 1 mark)
23
Which one of the following statements about a charged particle in an electric field is correct?
A
No work is done when a charged particle moves along a field line.
B
No force acts on a charged particle when it moves along a field line.
C
No work is done when a charged particle moves along a line of constant potential.
D
No force acts on a charged particle when it moves along a line of constant potential.
(Total 1 mark)
Page 26 of 62
24
An α particle travels towards a gold nucleus and at P reverses its direction.
Which one of the following statements is incorrect?
A
The electric potential energy of the α particle is a maximum at P.
B
The kinetic energy of the α particle is a minimum at P.
C
The total energy of the α particle is zero.
D
The total energy of the α particle has a constant positive value.
(Total 1 mark)
25
Which one of the following statements about electric field strength and electric potential is
incorrect?
A
Electric potential is a scalar quantity.
B
Electric field strength is a vector quantity.
C
Electric potential is zero whenever the electric field strength is zero.
D
The potential gradient is proportional to the electric field strength.
(Total 1 mark)
Page 27 of 62
26
The diagram shows two charges, +4 µC and –16 µC, 120 mm apart. What is the distance from
the +4 µC charge to the point between the two charges, where the resultant electric potential is
zero?
A
24 mm
B
40 mm
C
80 mm
D
96 mm
(Total 1 mark)
27
The diagram shows two charges, +4 µC and –16 µC, 120 mm apart. What is the distance from
the +4 µC charge to the point between the two charges where the resultant electric potential is
zero?
A
24 mm
B
40 mm
C
80 mm
D
96 mm
(Total 1 mark)
Page 28 of 62
28
The first artificially produced isotope, phosphorus
aluminium isotope,
(a)
, was formed by bombarding an
, with an α particle.
Complete the following nuclear equation by identifying the missing particle.
(1)
(b)
For the reaction to take place the α particle must come within a distance, d, from the centre
of the aluminium nucleus. Calculate d if the nuclear reaction occurs when the α particle is
given an initial kinetic energy of at least 2.18 × 10–12 J.
The electrostatic potential energy between two point charges Q1 and Q2 is equal
to
where r is the separation of the charges and ε0 is the permittivity of free space.
answer = .......................................m
(3)
(Total 4 marks)
29
Two identical positive point charges, P and Q, separated by a distance r, repel each other with a
force F. If r is decreased so that the electrical potential energy of Q is doubled, what is the force
of repulsion?
A
0.5 F
B
F
C
2F
D
4F
(Total 1 mark)
Page 29 of 62
30
An electron and a proton are 1.0 × 10–10 m apart. In the absence of any other charges, what is
the electric potential energy of the electron?
A
+2.3 × 10–18J
B
–2.3 × 10–18J
C
+2.3 × 10–8J
D
–2.3 × 10–8J
(Total 1 mark)
31
The table shows the binding energy per nucleon for two nuclei.
(a)
(i)
nucleus
binding energy per
nucleon/10–12J
helium-4
1.1332417
beryllium-8
1.1314027
Explain what is meant by the total binding energy of a nucleus.
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(1)
(ii)
It is more usual to quote binding energies of nucleons in MeV rather than J.
Calculate the total binding energy, in MeV, of a beryllium-8 nucleus.
binding energy .........................................................MeV
(3)
Page 30 of 62
(b)
(i)
Calculate the change in mass that occurs when two helium-4 nuclei fuse to form a
beryllium-8 nucleus.
mass change .........................................................kg
(2)
(ii)
Two helium-4 nuclei are initially separated by a large distance and are travelling
toward one another. The helium nuclei become influenced by the strong force when
their centres are separated by a distance of 3.82 × 10–15 m.
Calculate the total initial kinetic energy of the nuclei needed for them to reach this
separation.
kinetic energy ..........................................................J
(3)
(iii)
Explain why the kinetic energy calculated in part (b)(ii) will not enable the helium
nuclei to fuse and produce a beryllium-8 nucleus.
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(3)
(Total 12 marks)
Page 31 of 62
32
Two identical positive point charges, P and Q, are separated by a distance of 4.0 m.
The resultant electric potential at point M, which is mid-way between the charges, is 25.0 V.
What would be the resultant electrical potential at a point 1.0 m closer to P?
A
8.3 V
B
12.5 V
C
33.3 V
D
37.5 V
(Total 1 mark)
33
Two charges, each of + 0.8 nC, are 40 mm apart. Point P is 40 mm from each of the charges.
What is the electric potential at P?
A
zero
B
180 V
C
360 V
D
4500 V
(Total 1 mark)
Page 32 of 62
34
When a charge moves between two points in an electric field, or a mass moves between two
points in a gravitational field, energy may be transferred.
Which one of the following statements is correct?
A
No energy is transferred when the movement is parallel to the direction of the field.
B
The energy transferred is independent of the path followed.
C
The energy transferred is independent of the start and finish points.
D
Energy is transferred when the movement is perpendicular to the field lines.
(Total 1 mark)
35
The diagram shows a negative ion at a point in an electric field, which is represented by the
arrowed field lines.
Which one of the following statements correctly describes what happens when the ion is
displaced?
When the negative ion is displaced
A
to the left the magnitude of the electric force on it decreases.
B
to the right its potential energy increases.
C
along the line PQ towards Q its potential energy decreases.
D
along the line PQ towards P the magnitude of the electric force on it is unchanged.
(Total 1 mark)
Page 33 of 62
36
The diagram below shows the field lines and equipotential lines around an isolated positive point
charge.
Which one of the following statements concerning the work done when a small charge is moved
in the field is incorrect?
A
when it is moved from either P to Q or S to R, the work done is the same in each case
B
when it is moved from Q to R no work is done
C
when it is moved around the path PQRS, the overall work done is zero
D
when it is moved around the path PQRS, the overall work done is equal to twice the work
done in moving from P to Q
(Total 1 mark)
Page 34 of 62
37
A uniform electric field of electric field strength E is aligned so it is vertical. An ion moves
vertically through a small distance Δd from point X to point Y in the field.
There is a uniform gravitational field of field strength g throughout the region.
Which line, A to D, in the table correctly gives the gravitational potential difference, and the
electric potential difference, between X and Y?
Gravitational potential
difference
Electric potential
difference
A
gΔd
EΔd
B
gΔd
C
EΔd
D
(Total 1 mark)
38
Two horizontal parallel plate conductors are separated by a distance of 5.0 mm in air. The lower
plate is earthed and the potential of the upper plate is +50 V.
Which line, A to D, in the table gives correctly the electric field strength, E, and the potential, V, at
a point midway between the plates?
electric field strength E / Vm−1
potential V / V
A
1.0 × 104 upwards
25
B
1.0 × 104 downwards
25
C
1.0 × 104 upwards
50
D
1.0 × 104 downwards
50
(Total 1 mark)
Page 35 of 62
Mark schemes
1
2
C
[1]
(a)
force between two (point) charges is
proportional to product of charges ✓
inversely proportional to square of distance between the charges ✓
Mention of force is essential, otherwise no marks.
Condone “proportional to charges”.
Do not allow “square of radius” when radius is undefined.
Award full credit for equation with all terms defined.
2
(b)
V is inversely proportional to r [or V ∝ (−)1 / r ] ✓
(V has negative values) because charge is negative
[or because force is attractive on + charge placed near it
or because electric potential is + for + charge and − for − charge] ✓
potential is defined to be zero at infinity ✓
Allow V × r = constant for 1st mark.
max 2
(c)
(i)
Q(= 4πɛ0 rV ) = 4πɛ0 × 0.125 × 2000
OR gradient = Q / 4πɛ0 = 2000 / 8 ✓
(for example, using any pair of values from graph) ✓
= 28 (27.8) (± 1) (nC) ✓
(gives Q = 28 (27.8) ±1 (nC) ✓
2
(ii)
at r = 0.20m V = −1250V and at r = 0.50m V = −500V
so pd ΔV = −500 − (−1250) = 750 (V) ✓
work done ΔW (= QΔV) = 60 × 10−9 × 750
= 4.5(0) × 10−5 (J) (45 μJ) ✓
(final answer could be between 3.9 and 5.1 × 10−5)
Allow tolerance of ± 50V on graph readings.
[Alternative for 1st mark:
ΔV =
(or similar substitution using 60 nC
instead of 27.8 nC:
use of 60 nC gives ΔV = 1620V) ]
2
Page 36 of 62
(iii)
✓ = 1600 (1560) (V m−1) ✓
=
[or deduce E =
by combining E =
from graph E =
with V =
✓
= 1600 (1560 ± 130) (V m−1) ✓ ]
Use of Q = 30 nC gives 1690 (V m−1).
Allow ecf from Q value in (i).
If Q = 60 nC is used here, no marks to be awarded.
2
[10]
3
(a)
(i)
(Mass change in u=)
1.71× 10−3 (u)
or (mass Be−7) ‒ (mass He−3) ‒ (mass He−4) seen with numbers
C1
2.84 × 10−30 (kg)
or Converts their mass to kg
Alternative 2nd mark:
Allow conversion of 1.71 × 10−3 (u) to MeV by
multiplying by 931 (=1.59 (MeV)) seen
C1
Substitution in E = mc2
condone their mass difference in
this sub but must have correct value for c2 (3×108)2 or 9×1016
Alternative 3rd mark:
Allow their MeV converted to joules (× 1.6 × 10−13) seen
C1
2.55 × 10−13 (J) to 2.6 × 10−13 (J)
Alternative 4th mark:
Allow 2.5 × 10−13 (J) for this method
A1
4
Page 37 of 62
(ii)
Use of E=hc / λ
ecf
C1
Correct substitution in rearranged equation with λ subject ecf
C1
7.65 × 10−13 (m) to 7.8 × 10−13 (m)
ecf
A1
3
(b)
(i)
Use of Ep formula:
C1
Correct charges for the nuclei and correct powers of 10
C1
2.6(3) × 10−13 J
A1
3
(ii)
Uses KE = 3 / 2 kT: or halves KET, KE= 1.3 × 10−13 (J) seen
ecf
C1
Correct substitution of data and makes T subject ecf
Or uses KET value and divides T by 2
C1
6.35 × 109 (K) or 6.4 × 109 (K) or 6.28 × 109(K) or 6.3 × 109
(K) ecf
A1
3
(c)
(i)
Deuteron / deuterium / hydrogen−2
B1
Triton / tritium / hydrogen−3
B1
2
Page 38 of 62
(ii)
Electrical heating / electrical discharge / inducing a current in
plasma / use of e−m radiation / using radio waves (causing
charged particles to resonate)
B1
1
[16]
4
5
D
[1]
(a)
work done [or energy needed] per unit charge
[or (change in) electric pe per unit charge]
on [or of] a (small) positive (test) charge
in moving the charge from infinity (to the point)
[not from the point to infinity]
3
(b)
gives Q (= 4πε0rV) = 4π × 8.85 × 10–12 × 0.30 × 3.0
(i)
= 1.0 × 10–10 (C)
to 2 sf only
3
(ii)
use of V ∞
gives VM =
(= (+) 1.0 V)
1
(iii)
=
(= 2.50 V m–1)
1
(c)
(i)
uniformly spaced vertical parallel lines which start and end on plates
relevant lines with arrow(s) pointing only downwards
2
(ii)
= 3.3(3) (V m–1)
1
Page 39 of 62
(iii)
part (b) is a radial field whilst part (c) is a uniform field
[or field lines become further apart between L and M but are equally spaced between
R and S]
1
[12]
6
(a)
(i)
mass and energy have equivalent values
B1
E = mc2 mentioned
B1
MeV is energy unit (and kg that of mass)
B1
max2
(ii)
clear attempt to substitute amu values into equation
C1
5.135 × 10–3 (u) or 4.78 (MeV) seen
C1
mass of 1 lithium nucleus = 9.98 × 10–27 (kg)
C1
total number of nuclei in 1 kg = 1.00 × 1026
C1
total energy given out = 4.78 × 1026 MeV
A1
5
(iii)
neutrons needed (for the lithium reaction) can
come from the other (deuterium-tritium) reaction
B1
1
Page 40 of 62
(b)
(i)
potential energy equation (E =
or used
) quoted
C1
correct substitutions
C1
1.5(3) × 10–13 (J)
A1
3
(ii)
ke = 3/2 kT
C1
0.75/0.765 × 10–13 (J) or half of (b) (i)
or 4 × 109 (K) used
C1
3.7 × 109 (K) or total energy 1.6 × 10–13 (J)
A1
3
(iii)
each nucleus carries a positive charge
B1
(electrostatically) repel each other
B1
strong nuclear force
B1
this has a range of nucleus diameters
B1
high temperature needed for high kinetic energy
B1
max4
[18]
Page 41 of 62
7
8
9
10
11
D
[1]
C
[1]
A
[1]
C
[1]
(a)
(i)
Lines of equipotential parallel to the plates
B1
Field lines perpendicular to plates, evenly spaced
and with arrows upwards
B1
Lack of clear labelling of at least one of the types
of line loses 1 mark
Either field shown to be uniform
B1
3
(ii)
KE = 8.8 × 10–17 J
B1
Use of ½ mv2
C1
Speed = 1.4 × 107 m s–1
ecf
A1
Momentum =1.27 × 10–23 kg m s–1
ecf
B1
4
Page 42 of 62
(b)
Use of de Broglie wavelength = h/mv
C1
5.2 × 10–11 m
ecf
A1
diffraction of electrons necessary
M1
will work because wavelength is of same order as atomic
separation (not just wavelength is too small)/argument
consistent with their (a) (ii).
A1
4
[11]
12
(a)
(i)
shows arrows from + to –
Bl
(ii)
surface of constant potential / no work done in moving charge
on surface OWTTE
Bl
(iii)
3 correct lines between plates, straight, labelled, +12.5 kV on left
Bl
outwards curvature at edge of plates
Bl
(b)
(i)
F = Vq / d or 50000 × 5.5 × 10–9 / 4
Bl
= 0.0690 [mN]
[0.0688]
Bl
Page 43 of 62
(ii)
a = F / m = 0.069 × 10–3 / 0.12 × 10–3
= 0.575 / 0.573 m s–2
Cl
use of appropriate kinematic equation
Cl
t = √2 × 2 / 0.575 = (2.63) s
Cl
so length must be 0.8 × 2.63 = 2.11 m [gets mark ecf from
third mark if number quoted]
allow alternative energy approach
Bl
[10]
13
(a)
(i)
E at 2R = 20 to 21 (NC–1) i.e. no up
B1
(i.e. have used inverse square law possibly misreading the E axis)
correct curvature with line through given point
must not increase near tail
(ignore below 6400 km)
B1
no intercept on distance axis and through correctly
calculated point
B1
(ii)
determine the area under the graph
B1
between A and B or between the points
ignore any reference to V = Ed)
B1
(b)
(i)
E = q/4πε0r2 (Q = 84 × 4π 8.9 × 10–12 (6400 000)2
C1
(3.8–3.9) × 105 C
A1
Page 44 of 62
(ii)
surface area of the Earth = 5.15 × 1014 (m2)
C1
or:
charge per square metre = total charge/ surface area of
Earth
(may be seen as a numerical substitution with wrong area)
738 – 760 pC (m–2) ecf for Q from (b)(i)
A1
NB
(i) answer is the same when unit is left in km since r2
cancels so condone
(ii) Use of E = q / 4πε0r followed by area = 4πr gives
correct value but no marks
[8]
14
(a)
1
Bl
(1)
(b)
(i)
Ek = Ep when the protons touch
or Ek = q2 / 4πε0r
or separation when they touch = 3.0 × 10–15 m
or V = q / 4πε0r
Cl
Ek = (1.6 × 10–19)2 / 4π (8.9 × 10–12) (3.0 × 10–15)
or
Ek = (1.6 × 10–19)2 / 4π (8.9 × 10–12) (1.5 × 10–15)
Cl
7.6(1) × 10–14 J (cao)
Al
(3)
(ii)
incident proton will stop and the stationary proton will move off at
velocity / speed of the incident proton
or
All KE / momentum is transferred to the stationary particle
NB not they will not touch
Bl
(1)
Page 45 of 62
(iii)
protons travel in the opposite directions or velocity is reversed
Ml
with initial speeds
Al
total momentum before = 0 so momentum after must be 0
or
provided they have said that speeds are the same
total KE is same before and after the collision
or
the collision is elastic
Bl
(3)
(iv)
mention of strong nuclear force
or
the repulsive force is overcome
Cl
the strong nuclear force is greater than the electrostatic repulsion
or
the strong nuclear force is effective when the protons touch
Al
(2)
(c)
E = mc2
Cl
mass increase = {(2 × 2.2) – (2 × 1.7)} × 10–27 kg = 1.0 × 10–27 kg
or
calculates initial energy equivalence of 2 protons
or
final energy equivalence of 2 delta + particles
Cl
8.6 or 9 × 10–11 J (i.e. allow 1sf) c.a.o.
(NB Adding on the answer to (b) (i) is correct but it has no influence
on the answer to 2 sf so its absence is condoned)
Al
(3)
[13]
15
A
[1]
Page 46 of 62
16
(a)
(i)
(use of EP =
gives) EP =
(1)
= 2.3 × 10–13 (J) (1)
(ii)
EK at least distance apart = 0
EK of (each) proton = 0.5 × 2.3 ×10–13 (J) (1)
= (1.15 ×10–13(J)) = 0.72 MeV (1) (0.719 MeV)
5
(b)
(i)
(ii)
uud (1)
(1)
2
(c)
(i)
(ii)
(iii)
Q = –1(e) (1)
B = 0 (1)
(1)
(1)
mass of extra particles produced from total initial kinetic energy (1)
extra mass possible in (a) = 1.4 MeV / c2 (1)
pions rest mass in (b) >> extra mass in (a) (1)
max 5
[12]
17
18
C
[1]
(a)
m = 4.0026 × 1.66 × 10–27 (kg) (1) (= 6.6 × 10–27 kg – electron masses are not significant)
kinetic energy
= 0.5 × 6.65 × 10–27 × (2.00 × 107)2 (1)
(= 1.33 × 10–12 J)
2
(b)
loss in k.e. = gain in p.e. (1)
loss of ke. [or 1.33 × 10–12] =
R=
(1)
(1)
=2.73 × 10–14 m (1)
4
Page 47 of 62
(c)
any valid point including:
strong force complicates the process (*)
scattering caused by distribution of protons not whole nucleon distribution (*)
α particles are massive causing recoil of nucleus which complicates results (*)
(*) any one (1)
1
[7]
19
(a)
(i)
force per unit positive charge (1)(1)
[force on a unit charge (1) only]
vector (1)
(ii)
overall correct symmetrical shape (1)
outward directions of lines (1)
spacing of lines on appropriate diagram (1)
neutral point, N, shown midway between charges (1)
6
(b)
(i)
(1)
= 1250 V m–1 (1)
(ii)
EPB =
= 1050Vm–1 (1)
(iii)
allow e.c.f. from wrong numbers in (i) and (ii)
E=
θ = tan-1
(1) 1630Vm–1 (1)
= 50.0° to line PB and in correct direction (1)
max 6
Page 48 of 62
(c)
(i)
potential due to A is positive, potential due to B is negative (1)
at X sum of potentials is zero (1)
(ii)
= 0 (1)
gives AX (= x) = 0.080m (1) (only from satisfactory use of potentials)
4
[16]
20
21
22
23
24
25
B
[1]
B
[1]
A
[1]
C
[1]
C
[1]
C
[1]
Page 49 of 62
26
27
A
[1]
A
[1]
28
Al + α →
(a)
P+
n
1
(b)
kinetic energy lost by the α particle approaching the
nucleus is equal to the potential energy gain
2.18 × 10–12 =
r = 2.75 × 10–15 (m)
3
[4]
29
30
31
D
[1]
B
[1]
(a)
(i)
energy released when the separate nucleons combine to form
the nucleus
or energy needed to separate the nucleus into individual nucleons
owtte
B1
1
Page 50 of 62
(ii)
BE in J = 8 × 1.1314027 × 10–12 (9.05122 × 10–12)
C1
BE in eV = 5.6570135 × 107 eV or BE/nucleon
= 7.07 × 106 MeV
C1
56.570135 (MeV) (condone 3 sf consistent with electron
charge)
A1
3
(b)
(i)
change in BE = 0.0147120 (× 10–12) J
C1
use of E = mc2 with their energy 1.635 × 10–31 kg
A1
2
(ii)
use of charge on alpha particles = 2 e
C1
attempt to substitute in PE =
C1
2.4(2.39) × 10–13 J
A1
3
Page 51 of 62
(ii)
the mass of Be > mass of 2 He nuclei
B1
explains that when they touch there is zero KE
only mass available is that of the two alpha particles
B1
extra KE provides the increase in mass of the beryllium-8
compared with the 2 He nuclei
B1
3
[12]
32
33
34
35
36
37
38
C
[1]
C
[1]
B
[1]
D
[1]
D
[1]
A
[1]
B
[1]
Page 52 of 62
Examiner reports
1
2
This question proved to be somewhat easier, despite the rather abstract phrasing of the stem.
85% of the students knew that only alternative C gave a consistent expression i.e. g = GM / r2.
Statements of Coulomb’s law were generally satisfactory in part (a) and marks were high.
Occasionally there was confusion with the law of gravitation (distance between masses rather
than charges) and reference to indirect proportion (which was not acceptable) rather than inverse
proportion.
In part (b), inverse proportion was generally recognised as the relationship between V and r
shown on the graph. Many students knew that the negative values of potential were caused by
the charge Q being negative. Alternative arguments that approached an answer via the definition
of potential usually failed because the positive nature of the charge being moved was not stated.
No doubt it was confusion with gravitational potential, which is always attractive, that caused a
significant number of students to conclude that electric potential is always negative.
There was a good spread of marks across the three calculations in part (c). Part (c)(i) was
usually answered correctly, either by substitution of a data point from the graph or by use of the
gradient. Part (c)(ii) was most easily approached by reading the potentials corresponding to r =
0.20 m and r = 0.50 m from the graph, leading to ΔV = 750V, and then applying ΔW = QΔV. The
principal failing in the answers that started from first principles, by calculating the two potentials
from the Q value in part (i), was to use 60 nC as both the source of the potentials and as the
charge being moved. Direct substitution of the charge from part (i) into E = Q / 4πε0r2 gave the
most straightforward answer in part (iii). Although the use of E = V / d gave the correct numerical
answer here, it should be recognised that this equation is only valid in a uniform field whereas
the field in this question is radial. Both marks were therefore awarded for this approach only
when the use of E = V / r had been justified.
3
(a)
(b)
(c)
(i)
These calculations were well known and competently completed by the vast majority
of candidates. Where mistakes occurred these were more common in part (i) with
significant number of candidates failing to convert their mass to kg or forgetting to
square the speed of light in E = m c2.
(ii)
As above.
(i)
These calculations proved to be good discriminators with only the better candidates
able to achieve all 6 marks. There were a significant number of non-attempts, 10%
for this part.
(ii)
There were a significant number of non-attempts, 23% for this part. There were lots of
mistakes in the formula for potential energy with r2 instead of r. Candidates were
unsure about how to proceed in this part with many neglecting to divide the total
kinetic energy by 2.
(i)
Just over 30% of candidates could recall that H-2 and H-3 were most likely. The most
common answer seen was hydrogen and helium.
(ii)
Candidates enjoyed more success in this part with over 50% able to state a method
used to heat the plasma in the JET reactor.
Page 53 of 62
4
5
1/r for a point charge, but made appreciable
This question tested the relationship V
mathematical demands because it required candidates to deal with a change in V. Rather fewer
than half of the responses were correct, with distractor C as the most popular incorrect answer.
The definition of electric potential in part (a) was generally well known. Where students did not
score all three marks this was down to oversight; typically either omitting to mention that the
charge involved in the definition is positive or that the definition involves the work done per unit
charge.
In part (b) (i), most students successfully applied V = Q /4πε0r in order to determine the
magnitude of the charge (1.0 × 10–10 C) from the value of V when r = 0.30 m. As the data in the
question is given to two significant figures, an answer was expected to two significant figures.
Some students need to appreciate that the number of significant figures they should quote in an
answer needs to be limited to the least precise data they are working with, not the most (in the
Data Sheet (see Reference Material) ε0 is given to three significant figures). At the same time, in
these circumstances the answer should never be abbreviated to one significant figure
(1 × 10–10 C), as was the case in many answers.
The mark in part (b) (ii) was gained easily, usually by applying V = Q/4πε0r, although more
perceptive students saw that V ∝ 1/r could lead to a more concise answer. Part (b) (iii) caused a
little more difficulty for some students. Application of E = Q/4πε0r2 with r = 0.60 m was the
obvious route. The pitfall for many was that, by first finding V at M (by the same method as
before), they then had to apply E = V/d to find the field strength. This last equation specifically
applies to a uniform field and it therefore cannot be used here. Surprisingly, there were many
students who, having obtained an incorrect charge in part (b) (i) as a result of an arithmetical slip,
did not revisit part (i) when they could not show either of the required values in parts (ii) and (iii).
Many good attempts to represent the electric field between two plates were seen in part (c) (i),
but careless sketching, such as field lines stopping short of the plates, often meant that it was not
possible to award both marks. Because this was the field between two plates at different positive
potentials, some students were thrown off course both when sketching the field and when the
uniform field strength had to be found in part (c) (ii). In part (c) (iii) the respective radial and
uniform fields were usually recognised but a precise statement that identified which was which
was required to gain the mark.
Page 54 of 62
6
Credit was not given for a statement that mass and energy are the same thing in answer to part
(a) (i).
The most common mark for this part was one out of the two available.
Only a limited number of candidates gained all five marks in part (a) (ii), but several gained two
for finding a value of 4.78 MeV by converting the mass defect from values in terms of u to MeV.
Although the mark to part (a) (iii) was frequently gained, there was considerable benefit of doubt
applied with candidates not saying that it was the deuterium-tritium reactions which supplied the
neutrons needed for the lithium reactions.
Part (b) (i) was answered poorly. Most candidates did not use the electric potential energy
equation – force or potential were more common.
A limited number of candidates used the 3/2 kT equation in answer to part (b) (ii).
A few candidates started with 4 × 109 K and worked backwards but rarely saw that this gave
approximately half the ke in (b) (i) as would be expected. Part (b) (iii) was generally answered
well.
8
This question proved to be more demanding in the examination than in the pre-test; the facility
fell from 56% to 48%. Energy changes associated with the movement of a charge in an electric
field is not well understood by weaker candidates, because 24% selected distractor A
(4μC × 10 Vm–1) which suggests guesswork. The question discriminated well.
11
(a)
(b)
(i)
Most of the candidates could draw the field using both lines of equipotential and
electric lines. A few omitted to label the lines. A more common mistake was to draft
the diagram carelessly so that it was not clear that the field was apparently uniform.
(ii)
Weaker candidates got very tangled in this calculation, attempting to use
½ mv2
to calculate kinetic energy rather than using it to calculate speed once they had found
the kinetic energy by using the potential difference in the field.
The calculation in this part was done quite well. Few candidates could go on to explain
whether or not the de Broglie wavelength made the electrons suitable for the investigation
of metallic crystal structures. Some had no idea what the typical values for atomic
separations are in metallic crystals. More surprisingly, those who did know the separations
tended to be unclear about whether the wavelength was too big or too small or broadly
applicable.
Page 55 of 62
12
(a)
(b)
13
(a)
(b)
(i)
The vast majority were able to assign the correct direction to the electric field.
(ii)
Descriptions of what is meant by an equipotential surface could have been better.
There was lack of clarity in the answers and poorer responses were made in terms of
potential difference or field strength.
(iii)
Parallel lines for the equipotentials were common and correct for the first mark.
However, only about half the candidates were able to go on to show the correct
shape of the equipotential at the edge of the plates.
(i)
This was a ‘Show that’ question and candidates must endeavour to show complete
solutions if they expect to obtain full credit. Although the vast majority of candidates
obtained the correct answer, a significant number simply wrote down numbers that
occurred in the question, arriving at a number similar to the suggested answer.
Examiners wished to see a clear link between the potential gradient between the
plates and ratio of force per unit charge. If this link was missing, candidates obtained
little credit.
(ii)
This question required a sustained effort by candidates to carry out a number of steps
in a calculation to arrive at the correct answer. It was done well by many and showed
that the traditional skills of the physicist are still accessible by significant numbers of
those taking the examination. A common fault was to try to go down an erroneous
route involving a spurious centripetal force acting on the dust particle, presumably the
candidates thought that magnetic field theory was the appropriate physics here.
Additionally, energy arguments, although appropriate, did not score well as
candidates often failed to appreciate that they were dealing with a final and not an
average velocity. However, despite the common success with the question, the
working was usually shown in a poor and scrambled way. Examiners had to work
hard to follow a chain of logic, only rarely expressed in clear steps and seldom
presented clearly on the page.
This question was generally well answered.
(i)
Most candidates knew the correct curvature but many failed to produce the correct
graph because they either misread the scale or applied a 1 / r law instead of a 1 / r2
law.
(ii)
This was frequently answered correctly but many simply stated ‘ find the area under
the graph’ without stating any limits.
(i)
The process was generally well known and there were many correct answers
although too many significant figures cost a number of candidates a mark here. The
most common error was however failure to convert the radius of the Earth to m.
(ii)
Even though the formulae are now on the formula sheet a large number used an
incorrect formula for the surface area of the Earth which, at this level, is matter of
some concern. Many gained one mark for dividing the charge by what they thought
was the area. Those who used 6400 km again without conversion were able to get
full credit. A significant number of candidates had a unit penalty applied here by
giving the answer in C m−1.
Page 56 of 62
14
(a)
The majority of the candidates were able to state this correctly. 1 / 3 was a common
incorrect answer.
(b)
(i)
Weak candidates simply quoted ½ mv2 so clearly did not identify this as a fields
problem. Many who were thinking along the right lines did not realise that the
separation would be twice the radius when the protons touched.
(ii)
There were very few correct answers to this part. The majority restated in some form
that since the energy is less than the answer to part(i) they would not touch. Others
with the right idea did mention the fact that they repelled but were unable to look
beyond this and describe subsequent events.
(iii)
Responses were usually along similar lines to those in part(ii). Appreciation that they
would travel in opposite directions was a mark gained by many but very few went
beyond this.
(iv)
Many candidates simply stated that the protons would fuse or react so were simply
repeating the stem of the question. It was not well known that the particles needed to
overcome coulomb repulsion sufficiently for them to touch or get close enough for the
strong nuclear force to act.
(c)
15
Weak candidates were again drawn, by the mention of kinetic energy in the question, to ½
mv2. Those who appreciated that they needed to use mc2 often failed to calculate the mass
increase and simply inserted the mass of one or possibly two delta+particles in the
equation. Those candidates who arrived at the correct energy by considering the mass
difference gained full credit although, strictly, the relatively small KE to overcome coulomb
repulsion should be added.
There is no doubt that this question made appreciable mathematical demands, but it is surprising
that the facility declined from 38% when pre-tested to 28% in this examination. This made it the
most demanding question on the paper. Candidates should have seen that the contributions to V
from the charges at top left and bottom right would effectively cancel. This then meant that the
total potential would be double that due to one of the remaining charges. Application of
Pythagoras leads to the distance r being √2 a. Guesswork is probably the explanation for as
many as 32% of the responses being for distractor D, which was no more than the simple
potential equation V = Q / 4πε0r with r = a.
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Few good answers were seen for part (a). Some candidates were unable to calculate the
potential energy correctly in part (i), and in part (ii), although the general principle was usually
known, candidates often did not realise the question asked for the initial kinetic energy of each
proton.
Part (b) was more successful and many candidates scored both marks. However, some sought
to include a strange quark in the positive pion.
The majority of candidates scored both marks in part (i) and many gave the correct quark
composition in part (ii) although they failed to identify the particle X. In part (iii) only one or two
candidates were able to give an adequate answer. Although some candidates stated that the
extra mass needed to be created from the kinetic energy of the initial protons, they usually failed
to use data to support this statement.
18
Only the weaker candidates made errors in part (a). These errors included finding the mass of
four nucleons or incorrect units.
Part (b) discriminated well. Errors occurred in the algebra, in the use of a calculator and in giving
the formula. Most candidates omitted the opening statement that kinetic energy is converted to
potential energy.
In answering part (c), some candidates stated why they would choose another technique rather
than explaining the limitations of alpha particle scattering.
19
Few of the definitions in part (a) included any reference to a positive charge, and “force on a unit
charge” seemed more common than the more correct “force per unit charge”. The only field
patterns tested in previous PH03 papers have been magnetic ones, so the need here to draw an
electric field caught out the vast majority of candidates. Examiners found that almost all of the
drawings more closely resembled the field between two current-carrying wires than that between
two point charges.
Attempts to answer part (b) were variable and it was not uncommon for candidates having a
good understanding of vector addition to score full marks. Misinterpretation of nanocoulomb was
a problem for some candidates. The direction of the resultant field caused some difficulty,
principally because candidates did not understand the fact that an electric field is directed
inwards towards a negative charge.
Only a minority of able candidates showed a clear understanding of the concept of electric
potential in part (c). Most answers tried to present arguments which were concerned with field
rather than potential. It needs to be more clearly accepted that potential is a scalar quantity and
therefore does not have an associated direction; potential values may add to zero but they can
never “balance”. In part (c)(ii) there were many intuitive answers of 80 mm, which quoted 3≤ of
200 mm as the reasoning. Credit was given only for answers which were properly reasoned in
terms of potential.
20
This question was also concerned with electric fields, set in the context of a van de Graaff
generator. Although the facility of this question moved from 57% in the pre-test to 69% in the
examination, the discrimination index deteriorated somewhat to a value of 0.38.
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23
24
25
26
Teachers preparing students for these examinations might like to ponder over why almost as
many candidates chose distractor D (implying a constant potential of 50V between the parallel
plates) as the correct answer. The facility of this question was only 38% and it was a weak
discriminator.
In the main, the candidates showed a good understanding of electric potential, causing the
question to have a facility of 75%. With a 40:60 division of the 100 mm separation, it is not
surprising that the most common wrong response was distractor D (60 mm) when the correct one
was A (40 mm). This question was the weakest discriminator on this paper.
This question, about a charged particle moving in an electric field, had a facility of 66% and was
a good discriminator. Incorrect responses were almost equally distributed between the incorrect
distractors.
The statement that is incorrect was to be chosen in this question, about the energy of an α
particle during a head-on encounter with a gold nucleus. The facility of this question was 62%,
the most common incorrect choice being distractor D (19%).
This question on electrostatics, was concerned with field strength and potential. This was the
worst discriminator in the test, and only 43% of candidates selected the correct response. A
principal reason for as many as 24% of them choosing distractor A (potential is a scalar) must be
that they had failed to notice that the question asked for the incorrect statement.
The subject of this question was the electric potential at a point in the field of two point charges.
Almost two-thirds of the candidates gave the correct answer but the question did not discriminate
very well.
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This question required candidates to appreciate that, for the total potential to be zero at the
chosen point, the magnitude of V due to the +4 µC charge should be the same as the magnitude
of V due to the –16 µC charge. This required (Q/r) to be the same and should give a distance
ratio of 1:4. 58% of the candidates were able to work this out correctly, which is 5% lower than
when this question was last used in an examination. Almost one in four of the candidates chose
distractor B, suggesting that the distance ratio would be 1:2. This question was the worst
discriminator in this examination.
Part (a) was very straightforward for most candidates but less than half could tackle part (b)
effectively.
Problems were seen at every stage. Some had no idea what was happening at all; some used
the wrong charge on the aluminium nucleus and used 27 × 1.6 × 10.19 C; and some even
changed the equation given in the question to the Coulomb law of force equation by introducing a
squared term for the separation.
29
30
The question on Coulomb’s law also needed candidates to know that V ∝ (1 / r) in an electric
field; if V is doubled r must have been halved. Therefore the force must have been increased by
a factor of 4. The facility of the question was 61%, and the most common incorrect choice
(distractor C) had the force increasing by a factor of 2.
Failure to realise that a negatively charged electron has an associated negative value of electric
potential caused many candidates to go wrong in this question. As in gravitation, the force
between an electron and a proton is attractive and so the potential is negative. 40% of the
candidates made the correct selection, B. 26% selected distractor A; the only difference between
A and B is the – sign in B.
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Many showed misconceptions in part (a) (i) with answers that suggested there is an energy input
needed to form a nucleus from individual nucleons.
Part (a) (ii) was a straightforward question for those who knew how to convert units and
understood the concepts of total binding energy and binding energy per nucleon and there was a
good proportion of correct answers.
Most realised the need to apply E = mc2 in part (b) (i) but many had difficulty identifying the
correct energy. For part (b) (ii), many students appreciated the need to use to potential energy
formula and/or realised the need to use a charge of 2e. However, completely correct answers
were rare.
There were very few correct answers to part (b) (iii). Most referred to energy needing to
overcome the strong force or electrostatic repulsion. Very few realised the need to provide
energy to produce the higher mass of the beryllium nucleus compared with two alpha particles
32
33
34
35
36
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This question required an understanding of electric potential as a scalar, so that the contributions
of the two charges to the total potential at any point around them can be found by simple
addition. Combined with an appreciation that V ∝ (1 / r), it follows that the potential at the
mid-point due to one charge is 12.5V − and that this becomes 25V when the distance is
decreased to 1.0m, and 8.3V when increased to 3.0m. Therefore the resultant potential at the
required point is 33.3V. Distractors B and D each attracted about one-fifth of the responses.
This question required students to understand that electric potential is a scalar, and that the
potential at a point close to two charges is therefore the sum of the potentials due to each of
them. This was understood by fewer than half of the candidates, and so the facility of the
question was only 45%. Almost a third of responses were for distractor A (zero) and almost a fifth
for distractor B (the potential due to one charge alone).
This question turned out to be the hardest in the test, with a facility less than 40%, possibly
because it required rather abstract thinking about energy transfer in fields. More than one quarter
of the candidates did not spot that the displacement described in distractor D amounts to
movement along an equipotential line, and so selected this as the correct answer.
This question was answered correctly by two-thirds of the candidates. No doubt it was
misunderstanding of the direction of the force that acts on a negative ion that caused 19% of the
candidates to select distractor B.
This question was the first of the two questions in this test where the students had to select a
“wrong” statement. This always highlights the need for careful reading of the question. When a
charge is moved completely around a closed path in an electric field the net work done is zero;
this correct statement was given in distractor C, which was chosen by 26%. Distractor D – the
correct response – directly contradicts distractor C. 67% made the right choice.
Familiarity with E = V / d should have enabled students to realise in this question
that potential difference can be calculated from (field strength) × ∆d, for both gravitational end
electric fields. 62% of students correctly gave answer A; the three other distractors were chosen
fairly randomly by between 10% and 17% of the entry.
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This question was another re-used question, and its facility of 43% this time was a slight
improvement over the previous occasion. Candidates’ responses showed that 75% of them
recognised that the electric field acts downwards, but those who chose distractor D were
evidently under the impression that there is a constant value of potential at all points between
two parallel plates. Similar weaknesses in understanding the properties of the field between
parallel plates were also evident in Section B of this Unit 4 test.
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