1 Electrons in a beam are accelerated from rest by a potential difference V between two vertical plates

before entering a uniform electric field of electric field strength E between two horizontal parallel
plates, a distance 2d apart.

The path of the electrons is shown in Fig. 2.1. The electron beam travels a horizontal distance x
parallel to the plates before hitting the top plate. The beam has been deflected through a vertical
distance d.

For different values of the accelerating p.d. V, the horizontal distance x is recorded. A table of results
is shown with a third column giving values of x2 including the absolute uncertainties.

V/V x / cm x2 / cm2
500 3.3 ± 0.1 10.9 ± 0.7
600 3.6 ± 0.1 13.0 ± 0.7
700 3.9 ± 0.1 15.2 ± 0.8
800 4.2 ± 0.1 17.6 ± 0.8
900 4.5 ± 0.1 20.3 ± 0.9
1000 4.7 ± 0.1

(i) Complete the missing value in the table, including the absolute uncertainty.

[1]

(ii) Fig. 2.2 shows the axes for a graph of x2 on the y-axis against V on the x-axis. The first four points
have been plotted including error bars for x2. Use data from the table to complete the graph.

[2]

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 (iii) The separation of the horizontal plates is 4.0 ± 0.1 cm.
Use the graph to determine a value for E. Include the absolute uncertainty and an appropriate unit
in your answer

E = _ _ _ _ _ _ _ _ _ _ _ _ _ _ ±_ _ _ _ _ _ _ _ _ _ _ unit_ _ _ _ _ _ _ _ _ _ _ [4]

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 2 Fig. 24 shows two horizontal metal plates in a vacuum.

Fig. 24

The diagram is not drawn to scale.

Electrons travelling horizontally enter the space between the charged plates and are deflected
vertically.

The potential difference between the plates is 4000 V.

The distance between the plates is 0.08 m.
The initial speed of the electrons is 6.0 × 107 m s–1.
The vertical deflection of the electrons at the far end of the plates is x.

(i) Show that the vertical acceleration a of an electron between the plates is 8.8 × 1015 m s–2.

[3]

(ii) The length of each plate is 0.12 m.

Show that the time t taken by the electron to travel this length is 2.0 × 10–9 s.

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 [1]

(iii) Calculate the vertical deflection x of the electron.

x = ..................................................... m [2]

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 3 Fig. 21.2 shows two parallel vertical metal plates connected to a battery.

The plates are placed in a vacuum and have a separation of 1.2 cm. The uniform electric field strength
between the plates is 1500 V m–1. An electron travels through holes X and Y in the plates. The
electron has a horizontal velocity of 5.0 × 106 m s–1 when it enters hole X.

(i) Draw five lines on Fig. 21.2 to represent the electric field between the parallel plates.

[2]

(ii) Calculate the final speed of the electron as it leaves hole Y.

speed = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ m s–1 [3]

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 4 Two different minerals acquire opposite charges when they are crushed into tiny particles. These
oppositely charged mineral particles fall from a conveyor belt through the uniform electric field between
two vertical parallel plates, as shown in Fig. 1.2.

The potential difference across the plates is 60 kV. The separation between the plates is 25 cm and
each plate has length 1.8 m. The mineral particles fall through the air between the plates with a
terminal velocity of 1.2 ms–1. Each mineral particle has a charge of magnitude 1.5 × 10–13 C and a
mass of 8.0 × 10–7 kg.

(i) Calculate the horizontal electric force experienced by a positively charged mineral particle as it falls
between the plates.

force = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ N [2]

(ii) Calculate the horizontal displacement of a positively charged mineral particle after a 1.8 m fall

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 through the electric field of the plates. Ignore any horizontal drag forces due to air.

displacement = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ m [3]

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 5(a) The figure below shows the path of a proton moving in a region occupied by both an electric field and
a magnetic field.

The direction of the electric field lines is perpendicular to the direction of the magnetic field lines.

The uniform electric field is directed upwards, with electric field strength E = 0.90 N C–1.

The uniform magnetic field is directed out of the plane of the paper, with magnetic flux density B = 5.0
× 10–5 T.

At point X the proton is moving horizontally to the right. The magnitude of the magnetic force at X is
5.6 × 10–19 N.

At point Y the proton is moving vertically downwards. The magnitude of the magnetic force at Y is 3.9
× 10–19 N.

The electric forces acting on the proton at X and Y are not shown in the figure.

Show that the magnitude of the constant electric force acting on the proton is about 10–19 N.

[1]

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 (b)

(i) Suggest why the magnetic force acting on the proton has a different magnitude at X than at Y.

[1]

(ii) At X, the motion of the proton is instantaneously equivalent to motion in a circle at a constant
speed.

Calculate the radius of this circular motion.

radius = ..................................................... m [4]

(iii) 1 Calculate the magnitude of the resultant force on the proton at Y.

resultant force = ...................................................... N [2]

2 Explain why the motion of the proton at Y is not instantaneously equivalent to motion in a circle
at a constant speed.

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 [2]

6(a) The diagram below shows a fairground ride. Each rider is secured in a seat suspended by a rod.

The distance from the top of the rod to the base of the seat is 11.1 m.

The rod is attached to the edge of a disc of diameter 7.8 m.

To test the equipment a sandbag is attached to the seat and the ride is started.

The combined mass of the seat and the sandbag is 12 kg.

The rod makes an angle of 35° with the vertical.

(i) Draw an arrow labelled T on the diagram to represent the tension in the rod.
[1]

(ii) Show that the radius of the circular path followed by the sandbag is about 10 m.

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 [2]

(iii) Calculate the tension T in the rod.

T = ......................................................N [3]

(iv) Show that the angular velocity of the ride is about 0.8 radians per second.

[2]

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 (b) When the seat is at its highest point the sandbag is 17 m above the ground. The sandbag is released
from the seat to model an object being dropped by a rider.

(i) Calculate t, the time taken for the sandbag to reach the ground.

t = ........................................................ s [2]

(ii) Using your answer to (a)(iv), determine the horizontal displacement s travelled by the sandbag
before hitting the ground.

s = ......................................................m [3]

(iii) Determine, with reasons, the effect on the horizontal displacement travelled if the object released
from the ride was a shoe from a rider.

[3]

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 7(a) A particle-accelerator uses a ring of electromagnets to keep protons moving continuously in a circle.

The speed v of the protons depends on the frequency f of rotation of the protons in the circular orbit.

Fig. 22 shows data points plotted on a v against f grid.

Fig. 22

(i) Show that the gradient of the graph of v against f is equal to 2πr, where r is the radius of the
circular path of the protons.

[2]

(ii) Show that r is about 10 m by determining the gradient of the line of best fit through the data points
in Fig. 22.

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 [3]

(iii) The maximum speed of the protons from this accelerator is 2.0 × 107 m s–1.

Calculate the maximum centripetal force F acting on a proton at this speed.

mass of proton = 1.7 × 10–27 kg.

F = ...................................................... N [3]

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 (b) A new particle-accelerator is now built for moving the protons in a circle of a radius 20 m.

The ring of electromagnets for this new accelerator provides the same maximum centripetal force as
the accelerator in (a).

Calculate the maximum speed of the protons in this new accelerator.

maximum speed = ................................................ m s–1 [2]

END OF QUESTION PAPER

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

1 i 22.1 ± 0.9 B1 value plus uncertainty both required for

the mark allow ± 1.0

ii two points plotted correctly, including B1 ecf value and error bar of first point
error bars;

ii line of best fit B1 allow ecf from points plotted incorrectly

worst acceptable straight line. steepest or shallowest possible line
that passes through all the error bars;
should pass from top of top error bar to
bottom of bottom error bar or bottom of
top error bar to top of bottom error bar

ii gradient (= 4d / E) = 2.4 ± 0.4; B1 allow 2.4 ± 0.5

ii E = 4 × 2.0 × 10–2 / 2.4 × 10–6 = 3.3 × B1

104

ii (3.3) ± 0.6 × 104 B1 0.1/4 + 0.4/2.4 = 0.192 × 3.3 = 0.63

ii V m–1 or N C–1 B1 0.1/4 + 0.5/2.4 = 0.233 × 3.3 = 0.77

allow 3.3 ± 0.8 × 104

Total 7

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

2 i C1 E = 5.0 × 104 (V m-1)

C1 F = 8.0 × 10-15 (N)

C1 Allow this mark if the working is shown.

If only value is given, then the answer
or 8.78 × 1015 A0 must be 3SF or more

a = 8.8 × 1015 Examiner’s Comments

This question asks for a calculation to

show the value of the vertical
acceleration in an electric field. The
magnitude of the electric field strength
first needs to be calculated, followed
by the acceleration from Newton’s
second law. Candidates are reminded
that a show question needs to be
answered in detail and that each stage
should be clear. Roughly equal
numbers of candidates scored full
marks or zero on this question.

ii M1 Examiner’s Comments

A0 As with the previous question, there is

-9
(t = 2.0 × 10 s) the need to make sure that the
calculation leading to the given answer
is clearly set out.

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

iii (x = ) ½ × 8.78 × 1015 × (2.0 × 10-9)2 C1 Allow a = 8.8 × 1015

x = 1.8 × 10-2 (m) A1 Examiner’s Comments

Most candidates appreciated the need

to use an equation of motion in their
solution, but a significant number of
candidates used an initial horizontal
velocity in the expression, leading to
an incorrect answer. There were also
an unusually large number who gave
no response. Candidates should
appreciate that if they have been given
show questions, then it is likely that
these values will be used in alter
questions.

Misconception

Many candidates included an initial

vertical velocity – it may be helpful to
think of this process as analogous to
that of projectile motion.

Total 6

3 i Parallel and equidistant field lines. B1 Note: Field lines must be right angle to
the plates.

i Field direction is correct (from left to B1

right).

ii work done = 1500 × 1.6 × 10–19 × 1.2 × C1

10–2 = 2.88 × 10–18 (J)

ii C1 Correct use of: final KE = initial KE –

work done.

ii speed = 4.3 × 106 (m s–1) A1

Total 5

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

4 i E = 60x103 ÷0.25 / E = 2.4 × 105 (V C1

m–1)

i F = 2.4 × 105 × 1.5 × 10–13 Allow: F = [1.5 × 10–13 × 60 × 103]/0.25

for the first C1 mark

i force = 3.6 × 10–8 (N) A1 Allow: 1 mark for 7.2 × 10–8 (N); d =
12.5 cm used

Examiner's Comments

Most candidates scored two marks in

this question. Answers were generally
well-structured and prefixes correctly
identified. The majority of the answers
were written in scientific notation. A
pleasingly number of candidates
derived an expression for the force and
then substituted the values. A small
number of candidates used a
separation of 12.5 cm to determine the
electric field strength. This gave an
answer of 7.2 × 10–8 N. On the basis
of error-carried-forward rules,
Examiners awarded one mark for this
incorrect answer.

ii C1 Possible ecf from (d)(i)

t = 1.8/1.2 (= 1.5 s) or (=
4.5 × 10–2 m s–2)

ii

ii C1 Note: No ecf within calculation if t ≠

1.8/1.2

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

ii displacement = 5.1 × 10–2 (m) A1 Note: Answer to 3 sf is 5.06 × 10–2 (m)

Examiner's Comments

This was a challenging question

requiring synoptic knowledge of
equations of motion and dynamics.
Many candidates scored a mark for
either calculating the horizontal
acceleration of the charged particle or
the time it took to fall between the
parallel plates. Further success in the
question hinged on how effectively the
question was scrutinised. Top-end
candidates effortlessly calculated the
horizontal displacement to be 0.051 m.
However, many candidates ignored
that the charged particle was falling
vertically at a constant speed and
attempted to determine the time of fall
using an equation of motion with 9.81
m s–2. A very small number of
candidates in the lower quartile
attempted to use either Coulomb's law
or triangle of forces to determine the
displacement.

Total 5

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

5 a F (= EQ) = 0.90 × 1.60 × 10–19 = 1.4(4) B1 Working and answer must both be
× 10–19 (N) shown
Answer must be given to 2sf or more
Unit need not be given but, if given,
must be correct

Examiner’s Comments

This was an easy introduction to the

question, which used the definition of
electric field strength; E = FE / Q. Being
a ‘show that’ question, candidates
needed to show their working in full,
including writing the value for the
electronic charge (rather than simply
‘e’) and giving the answer to at least 2
s.f.

b i (F = BQv but B and Q are constant, so) B1 Allow speed

Ignore the direction is different
(the magnitude of) the velocity is
different /changes Examiner’s Comments

The force on a charged particle moving

at right angles to a magnetic field is
given by the formula Fmag = BQv. Since
B and Q are constants in this case, the
reason for the different magnitude of F
must be that the proton has a different
velocity, v.

Common problems in 6(b)(i)

using the formula F = BIlsinθ and

suggesting that the proton might be
travelling at a different angle to the
field, not realising that the proton is
always travelling at right angles to
the magnetic field in this question
suggesting that the proton may be
in a weaker (or stronger) field at X
than at Y, not realising that the
magnetic field is uniform and so its
field strength is constant throughout

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

ii C1 v = 7.0 × 104 (m s–1) implies first C1

C1
C1 Allow 10–19 for 1.4×10–19 (giving FR=
resultant force FR= (5.6 – 1.4) × 10–19 A1 4.6 × 10–19) FR = 4.2 × 10–19 implies
(C1) second C1
(C1) Do not credit if used as Fmag in Fmag in
(C1) Fmag = BQv
(A1)
Third C1 is for correct substitution into
r = 20 (m)
formula
Allow mp = 1.67 × 10–27 kg given to 3
Alternative all-in-one method:
s.f.
Not mp = 1.661 × 10–27 kg or mp =
1.675 × 10–27 kg
Allow ECF for incorrect v
Use of FR = 5.6 × 10–19 or = 1.4 × 10–19
resultant force FR= (5.6 – 1.4) × 10–19 is XP

Allow r = 19 (m)

FR = 4.2 × 10–19 (4.16 × 10–19 to 3sf)

r = 20 (m) implies second C1
An incorrect value of FR is XP from this
point

Third C1 is for correct substitution into

formula

Allow r = 19 (m)

Examiner’s Comments

This question could not be done in one

step, by equating the magnetic force to
the centripetal force. This is because,
at X, the centripetal force is being
provided by a combination of forces
from both the electric and the magnetic
field.

The easiest approach is to find the

velocity of the proton using Fmag = BQv
(the value for Fmag is given in the
diagram as 5.6 × 1019 N). This velocity
v can then be used in the formula F =

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

mv2/r in order to calculate the radius, r.
F here is the resultant force towards
the centre of the circle, which is found
from magnetic force downwards -
electric force upwards (the electric
force having been calculated in part
(a)).

Exemplar 3 is an example of a correct

answer, clearly written to show each
stage in the calculation:

Exemplar 3

iii C1 Ignore attempt to calculate weight of

A1 proton
|resultant force| = 4.1 × 10–19 (N) Allow FE = 10–19

Allow |F| = 4.0 × 10–19 (N) using FE =

1.0 × 10–19
Allow |F| = 4.2 × 10–19 (N) using FE =
1.44 × 10–19

Examiner’s Comments

There are two forces acting on the

proton at Y: an electric force upwards
(given in (a)) and a magnetic force to
the left (shown on the diagram). These
two forces act at right angles to each
other, and so the magnitude of their
resultant can be found using
Pythagoras’s Theorem.

Credit was given for using a value for

the electric force to 1, 2 or more
significant figures.

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

iv resultant / net force is not B1 Allow direction of motion / path but not
perpendicular to velocity B1 speed for velocity
Allow acceleration / resultant force is
work is done on proton (therefore not (always) towards centre (of circle)
kinetic energy changes so speed is not Allow electric force is not perpendicular
constant) to velocity / is in the same direction as
velocity
Ignore references to centripetal

Ignore references to centripetal

Examiner’s Comments

At Y, the proton is moving downwards,

with a resultant force being the
combination of an electric force
upwards and a magnetic force to the
left (calculated in part 1). The resultant
force cannot be at right angles to the
velocity, so we cannot have circular
motion.

The component of the resultant force

acting in the direction of the proton’s
motion will do work on the proton and
change its speed. So, the proton
cannot be travelling at a constant
speed.

Total 10

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

6 a i Arrow along the line of the support rod B1 Allow unlabelled single arrow along
labelled tension or T. either rod
Allow unlabelled arrows along both
rods
Allow arrow(s) up, down or both

NOT any contradictory arrows

Examiner’s Comments

In Question 17 (a) (i) of this question,

the phrase ‘tension in the rod’ can
mean several different things, all of
which were given in the mark scheme.

ii 11.1 sin 35 or 11.1 cos 55 seen M1 NOT use of tan 35 or tan 55

M1
addition of 3.9 (half the diameter of the A0 allow 7.8/2 for 3.9
support disc) to candidate’s horizontal
component of rod length
10.27 to 2 dp
Total= 10.3m
NB use of 11.1 cos 35 or 11.1 sin 55
arriving at 12.99 scores 1 (wrong trig)
NB reject use of radians (scores 0)

Examiner’s Comments

Many candidates approached part (ii)

with some confidence, spotting that the
horizontal portion of the rod was 11.1
sin(35) and that it should be added to
the radius of the disc.

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

iii mg=T cos 35 C1 Allow use of sin 55

C1 NOT use of tan 35 or tan 55
T= mg ÷ cos 35 A1
= 140 N Answer is 143.7 N to 4 sf

Examiner’s Comments

Parts (a) (iii) and (iv) were more

challenging, requiring good knowledge
of both circular motion and how to
calculate components of forces. Again
there were several legitimate routes to
the right answer, all of which were
mentioned in the mark scheme. Very
logical approaches were in part (a) (iii),
to equate the vertical component of the
tension with the weight of the sandbag.

iv T sin 35 = mrω2 M1 Allow use of W tan 35 or W tan 55

A1 Allow use of cos 55 and/or mv2/r
A0 Allow use of Pythogoras to find
centripetal force (82.4...)
NOT use of T tan 35 or T tan 55
=0.8(17) radian s–1
Allow ω2 subject.

Allow any combination of

rearrangement and substitution

ECF allowed for T and r.

Use of 2 s.f. values for T and r gives
0.84 m

Examiner’s Comments

In part (a) (iv), the quickest approach

was to equate the horizontal
component of the tension with the
centripetal force. The data booklet
provides a convenient expression for
the centripetal force in terms of the
angular velocity, without the need for
finding the tangential velocity.

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

b i Use of 17=1/2 gt2 C1 i.e. substitution of 17 and g or 9.81 or

A1 9.8

e.g.
=1.9 (1.86) s s=(ut)+ ½ at2

Allow any subject

ii Horizontal speed = rω or Horizontal C1 Use of data in the question stem (0.8

distance = speed × time C1 and 10) allowed, which gives 15.2 m.
A1
= 0.82 radians s–1 × 10.26 m × 1.86 s
Ecf for use of candidate’s value of r
= 16 m (15.6 m) and ω, giving

values rounding to between 14.9 and

16.0 m

iii Relevant variable identified M1 e.g mass/weight, drag/air resistance,

M1 radius, height, starting condition (e.g.
A1 kicking shoe off) Assume “it” in
Effect on speed of shoe or time of flight response refers to the shoe.
of shoe correctly identified ignore velocity for first M1
allow correct explanation of "no effect"
Conclusion consistent with relevant on speed or time by change of mass
physics
Examiner’s Comments
e.g.
Question 17 (b) explored ideas about
Shoe is lower mass yet no change parabolic flight due to gravitation. The
in angular velocity or radius since only force acting on the sandbag and
independent of mass so no change the shoe after they have been released
in horizontal displacement. is the weight force, which acts vertically
Shoe is below seat so would be downwards.
travelling with larger radius/speed
so larger distance travelled Many candidates realised that the
horizontally vertical velocity of the sandbag when it
Shoe might have be kicked off left the swing was zero, enabling them
backwards so have lower speed so to calculate the time for the bag to fall
lower distance 17m vertically downwards (using s = ½

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Shoe would come from below the gt2)
seat/lower than the sandbag i.e.
vertical distance to fall less, thus To calculate the horizontal distance
time of flight and horizontal travelled required both the horizontal
distance less. velocity (from v = rω) and the time of
Effect of air resistance hadn’t been flight from part (b) (iii). There was lots
included so shoe suffers drag, of scope for applying error carried
decelerating horizontally so forward rules as mentioned in the mark
distance would be smaller scheme.

Misconception

In lots of questions, candidates make

assumptions when trying to use
formulae to justify their ideas. In this
case, it was that for the same radius,
the shoe must leave the seat faster
than the sandbag, purely because the
shoe had less mass.

Often, what is constant is as important

to consider as what is changing. Here,
if the radius for the sandbag and the
shoe are the same, then the horizontal
velocity at release must be the same,
since the radius and angular speed for
both are the same, using v = rω.

Total 16

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 Mark Scheme

Question Answer/Indicative content Marks Guidance

7 a i {v = ωr and ω = 2πf} or v = 2 π fr B1 Allow v/f = 2πr

Comparison with y = mx leading to B1 Examiner's Comments

gradient = 2 π r This part tested ideas about
or investigative experiments: there was a
Δv/Δf = 2πr solid focus on elements of data-taking
and instruments that should be used.
Typically at A Level, analysis should
include an appropriate graph and a
comparison between the line of best fit
and the equation under test. Putting
the general equation below the given
equation would make it much clearer
how the candidate linked the gradient
or y-intercept with the required
property.

ii Line of best fit drawn B1 Allow ± 3

Gradient = 62.5 (m) M1 Allow ECF on gradient

2πr = 62.5 M1

r = 9.9 (m) A0

iii C1 Allow use of candidate’s answer for (ii)

or use of ‘10’
C1
Expect answers of 6.8 or 6.9 × 10-14
A1 (N)

F = 6.8 × 10-14 (N)

b r ∝ v2 / speed increases by a factor of C1 Allow substitution into correct equation

√2 with r doubled
A1
maximum speed = 2.8 × 107 (m s-1) Allow recalculation from previous value
of force in (a)(iii)

Total 10

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