Cambridge International Examinations physics mark schemes for multiple choice papers

CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2014 series
9702 PHYSICS
9702/11 Paper 1 (Multiple Choice), maximum raw mark 40
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Mark Scheme Syllabus Paper
GCE AS/A LEVEL – May/June 2014 9702 11
© Cambridge International Examinations 2014
Question
Number
Key
Question
Number
Key
1 A 21 C
2 B 22 D
3 A 23 B
4 D 24 D
5 B 25 B
6 D 26 B
7 D 27 D
8 A 28 D
9 B 29 A
10 A 30 A
11 B 31 D
12 C 32 B
13 B 33 A
14 D 34 D
15 C 35 B
16 B 36 A
17 D 37 C
18 D 38 B
19 B 39 B
20 B 40 A

9702 PHYSICS

Question
Number
Key
Question
Number
Key
1 C 21 C
2 B 22 C
3 B 23 C
4 C 24 C
5 D 25 B
6 B 26 D
7 C 27 D
8 C 28 A
9 C 29 D
10 B 30 B
11 A 31 A
12 D 32 A
13 A 33 A
14 C 34 D
15 B 35 B
16 B 36 B
17 C 37 B
18 B 38 C
19 D 39 C
20 A 40 C

9702 PHYSICS

Question
Number
Key
Question
Number
Key
1 C 21 D
2 D 22 D
3 A 23 A
4 B 24 D
5 C 25 C
6 C 26 C
7 C 27 B
8 D 28 A
9 C 29 B
10 C 30 B
11 B 31 A
12 C 32 C
13 C 33 D
14 A 34 D
15 D 35 A
16 D 36 A
17 C 37 A
18 B 38 C
19 C 39 C
20 C 40 D

9702 PHYSICS
9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.

1 (a) (i) either rate of change of displacement
or (change in) displacement/time (taken) B1 [1]
(ii) speed has magnitude only B1
velocity has magnitude and direction B1 [2]
(b) (i) idea of area under graph/use of t
vu
s ×
+
=
2
)(
C1
2.5
2
32)(18
×
+
=s C1
= 62.5 m A1 [3]
(ii) a = (18 – 32)/2.5 (= –5.6) C1
F = ma C1
F = 1500 × (–) 5.6 = (–) 8400N A1 [3]
(c) arrow labelled A and arrow labelled F both to the left B1 [1]
2 (a) (i) work (done)/time (taken) B1 [1]
(ii) work = force × displacement (in direction of force) B1
power = force × displacement/time (taken) = force × velocity B1 [2]
(b) (i) weight = mg C1
P = Fv = 2500 × 9.81 × sin 9° × 8.5 (or use cos81°) C1
= 33 (32.6)kW A1 [3]
(ii) no gain or loss of KE B1
no work (done) against air resistance B1 [2]
3 (a) (i) resultant force is zero B1
weight of plank + weight of man = FA + FB
or 200(N) + 880(N) or 1080 = FA + FB B1 [2]
(ii) principle of moments used C1
(anticlockwise moments) FB × 5.0 C1
(clockwise moments) 880 × 0.5 + 200 × 2.5 C1
FB = (440 + 500)/5.0 = 188N A1 [4]
(b) straight line with positive gradient (allow freehand) M1
start point (0, 100) A1
finish point (5, 980) A1 [3]

4 (a) kinetic energy = ½ mv2
C1
= ½ × 0.040 × (2.8)2
= 0.157J or 0.16J A1 [2]
(b) (i) k = F/x or F = kx C1
XB = 14/800
= 0.0175m A1 [2]
(ii) area under graph = elastic potential energy stored C1
or ½ kx2
or ½ Fx
(energy stored =) 0.1225J less than KE (of 0.16J) A1 [2]
5 (a) (i) displacement is the distance from the
equilibrium position/undisturbed position/midpoint/rest position B1
amplitude is the maximum displacement B1 [2]
(ii) frequency is the number of wavefronts/crests passing a point
per unit time/number of oscillations per unit time B1
time period is the time between adjacent wavefronts
or time for one oscillation B1 [2]
(b) (i) 1. amplitude = 1.5mm A1 [1]
2. wavelength = 25/6 C1
= 4.2cm or 4.2 × 10–2
m A1 [2]
(ii) v = λ/T or v = f λ and T= 1/f C1
T = 4.2/7.5 = 0.56s A1 [2]
(c) (i) progressive M0
wavefront/crests moving/energy is transferred by the waves A1 [1]
(ii) transverse M0
the vibration is perpendicular to the direction of energy transfer/wave velocity
or travel of the wave/wavefronts A1 [1]
6 (a) e.m.f.: energy converted from chemical/other forms to electrical
per unit charge B1
p.d.: energy converted from electrical to other forms per unit charge B1 [2]
(b) (i) the p.d. across the lamp is less than 12V
or there are lost volts/power/energy in the battery/internal resistance B1 [1]
(ii) R = V2
/P (or V = RI and P = VI) C1
= 144/48
= 3.0 Ω A1 [2]

(iii) I = E/(RT + r) C1
= 12/2.0
= 6.0A A1 [2]
(iv) power of each lamp = I 2
R
= (3.0)2
× 3.0 C1
= 27W A1 [2]
(c) less resistance (in circuit)/more current M1
more lost volts/less p.d. across battery A1 [2]
7 (a) α: helium nucleus
β: electron
γ: electromagnetic radiation/wave/ray or photon
three correct 2/2, two correct 1/2 B2 [2]
(b) (i) atomic number/proton number/Z –2, nucleon/mass number/A –4 B1 [1]
(ii) atomic number/proton number/Z +1
nucleon/mass number/A no change B1 [1]
(iii) no change in proton or mass number
or “no change” B1 [1]

9702 PHYSICS

1 (a) power = energy/time or work done/time B1
force: kgms–2
(including from mg in mgh or Fv)
or kinetic energy (
2
1
mv2
): kg (ms–1
)2
B1
(distance: m and (time) –1
: s–1
) and hence power: kgms–2
ms–1
= kgm2
s–3
B1 [3]
(b) Q/t : kgm2
s–3
C1
A: m2
and x: m and T: K C1
correct substitution into C = (Qx) /tAT or equivalent, or with cancellation C1
units of C : kgms–3
K–1
A1 [4]
2 (a) ρ = m/V C1
V = (πd2
/4) × t = 7.67 × 10–7
m3
ρ = (9.6 × 10–3
)/[π(22.1/2 × 10–3
)2
× 2.00 × 10–3
] C1
ρ = 12513kgm–3
(allow 2 or more s.f.) A1 [3]
(b) (i) ∆ρ/ρ = ∆m/m + ∆t/t + 2∆d/d C1
= 5.21% + 0.50% + 0.905% [or correct fractional uncertainties] C1
= 6.6% (6.61%) A1 [3]
(ii) ρ = 12500 ± 800kgm–3
A1 [1]
3 (a) a body/mass/object continues (at rest or) at constant/uniform velocity unless
acted on by a resultant force B1 [1]
(b) (i) weight vertically down B1
normal/reaction/contact (force) perpendicular/normal to the slope B1 [2]
(ii) 1. acceleration = gradient or (v – u)/t or ∆v/t C1
= (6.0 – 0.8)/(2.0 – 0.0) = 2.6 ms–2
M1 [2]
2. F = ma
= 65 × 2.6
= 169 N (allow to 2 or 3 s.f.) A1 [1]
3. weight component seen: mgsinθ (218 N) C1
218 – R = 169 C1
R = 49 N (require 2 s.f.) A1 [3]

4 (a) GPE: energy of a mass due to its position in a gravitational field B1
KE: energy (a mass has) due to its motion/speed/velocity B1 [2]
(b) (i) 1. KE =
2
1
mv2
C1
=
2
1
× 0.4 × (30)2
C1
= 180J A1 [3]
2. s = 0 +
2
1
× 9.81 × (2.16)2
or s = (30sin45°)2
/(2 × 9.81) C1
= 22.88 (22.9)m = 22.94 (22.9)m A1 [2]
3. GPE = mgh C1
= 0.4 × 9.81 × 22.88 = 89.8 (90) J A1 [2]
(ii) 1. KE = initial KE – GPE = 180 – 90 = 90J A1 [1]
2. (horizontal) velocity is not zero/(object) is still moving/answer explained
in terms of conservation of energy B1 [1]
5 (a) (Young modulus/E =) stress/strain B1 [1]
(b) (i) stress = F/A
or = F/(πd2
/4)
or = F/ (πd2
) M1
ratio = 4 (or 4:1) A1 [2]
(ii) E is the same for both wires (as same material) [e.g. EP = EQ] M1
strain = stress/E
ratio = 4 (or 4:1) [must be same as (i)] A1 [2]
6 (a) there are no lost volts/energy lost in the battery
or there are no lost volts/energy lost in the internal resistance B1 [1]
(b) the current/I decreases (as R increases) M1
p.d. decreases (as R increases) A1
or
the parallel resistance (of X and R) increases M1
p.d. across parallel resistors increases, so p.d. (across Y) decreases A1 [2]

(c) (i) current = 2.4 (A) C1
p.d. across AB = 24 – 2.4 × 6 = 9.6V M1
or
total resistance = 10Ω (= 24V/2.4A) C1
(parallel resistance = 4Ω), p.d. = 24 × (4/10) = 9.6 V M1 [2]
(ii) R (AB) = 9.6/2.4 = 4.0Ω C1
1/6 + 1/X = 1/4 [must correctly substitute for R] C1
X = 12Ω A1
or
IR = 9.6/6.0 = 1.6 (A) (C1)
IX = 2.4 – 1.6 = 0.8 (A) (C1)
X (= 9.6/0.8) = 12Ω (A1) [3]
(iii) power = VI or EI or V2
/R or E2
/R or І 2
R C1
= 24 × 2.4 or (24)2
/10 or (2.4)2
× 10
= 57.6W (allow 2 or more s.f.) A1 [2]
(d) power decreases M0
e.m.f. constant or power = 24 × current, and current decreases
or e.m.f. constant or power = 242
/resistance, and resistance increases A1 [1]
7 (a) waves from the double slit are coherent/constant phase difference B1
waves (from each slit) overlap/superpose/meet (not interfere) B1
maximum/bright fringe where path difference is nλ
or phase difference is n360U/2πn rad
or minimum/dark fringe where path difference is (n +
2
1
)λ
or phase difference is (2n + 1) 180U/(2n + 1)π rad B1 [3]
(b) v = fλ C1
λ = (3 × 108
) / 670 × 1012
= 448 (or 450) (nm) M1 [2]
(c) w = 12 / 9 C1
a (= Dλ/w) = (2.8 × 450 × 10–9
) /(12 / 9 × 10–3
) [allow nm, mm] C1
= 9.5 × 10–4
m [9.4 × 10–4
m using λ = 448nm] A1 [3]
(d) (red light has) larger/higher/longer wavelength (must be comparison) M1
fringes further apart/larger separation A1 [2]

9702 PHYSICS

1 (a) current, mass and temperature
two correct 2/2, one omission or error 1/2 A2 [2]
(b) σ : no units, V: m3
C1
EP: kgm2
s–2
C1
C: kgm2
s–2
× m–3
= kgm–1
s–2
A1 [3]
2 (a) scalar has magnitude only B1
vector has magnitude and direction B1 [2]
(b) (i) v2
= 0 + 2 × 9.81 × 25 (or using
2
1
mv2
= mgh) C1
v = 22(.1) ms–1
A1 [2]
(ii) 22.1 = 0 + 9.81 × t (or 25 =
2
1
× 9.81 × t2
) M1
t (=22.1/9.81) = 2.26s or t [=(5.097)1/2
] =2.26s A0 [1]
(iii) horizontal distance = 15 × t
= 15 × 2.257 = 33.86 (allow 15 × 2.3 = 34.5) C1
(displacement)2
= (horizontal distance)2
+ (vertical distance)2
C1
= (25)2
+ (33.86)2
C1
displacement = 42 (42.08) m (allow 43 (42.6) m, allow 2 or more s.f.) A1 [4]
(iv) distance is the actual (curved) path followed by ball B1
displacement is the straight line/minimum distance P to Q B1 [2]
3 (a) work done is the product of force and the distance moved in the direction of the
force
or product of force and displacement in the direction of the force B1 [1]

(b) (i) work done equals the decrease in GPE – gain in KE B1 [1]
(ii) 1. distance = area under line C1
= (7.4 × 2.5) /2 = 9.3m (9.25m) M1 [2]
or
acceleration from graph a = 7.4/2.5 (= 2.96) (C1)
and equation of motion (7.4)2
= 2 × 2.96 × s gives s = 9.3 (9.25) m (A1)
2. kinetic energy =
2
1
mv2
C1
=
2
1
× 75 × (7.4)2
C1
= 2100J A1 [3]
3. potential energy = mgh C1
h = 9.3sin30° C1
PE = 75 × 9.81 × 9.3sin30° = 3400J A1 [3]
4. work done = energy loss C1
R = (3421 – 2054) /9.3 C1
= 150 (147)N A1 [3]
4 (a) add small mass to cause extension then remove mass to see if spring returns to
original length M1
repeat for larger masses and note maximum mass for which, when load is
removed, the spring does return to original length A1 [2]
(b) Hooke’s law requires force proportional to extension B1
graph shows a straight line, hence obeys Hooke’s law M1 [2]
(c) k = force/extension C1
= (0.42 × 9.81) /[(30 – 21.2) × 10–2
] C1
= 47 (46.8) Nm–1
A1 [3]
5 (a) lost volts/energy used within the cell/internal resistance B1
when cell supplies a current B1 [2]

(b) (i) E = І(R + r) C1
4.5 = 0.65 (6.0 + r)
r = 0.92Ω A1 [2]
(ii) І = 0.65 (A) and V = ІR C1
V = 0.65 × 6 = 3.9V A1 [2]
(iii) P = V2
/R or P = І2
R and P = ІV C1
= (3.9)2
/6 = 2.5W A1 [2]
(iv) efficiency = power out/power in C1
= І 2
R/І 2
(R + r) = R/(R + r) = 6.0/( 6.0 + 0.92 ) = 0.87 A1 [2]
(c) (circuit) resistance decreases B1
current increases M1
more heating effect A1 [3]
6 (a) (i) progressive wave transfers energy, stationary wave no transfer of energy/
keeps energy within wave B1 [1]
(ii) (progressive) wave/wave from loudspeaker reflects at end of tube B1
reflected wave overlaps (another) progressive wave B1
same frequency and speed hence stationary wave formed B1 [3]
(iii) (side to side) along length of tube/along axis of tube B1 [1]
(b) all three nodes clearly marked with N/clearly labelled at cross-over points B1 [1]
(c) phase difference = 0 A1 [1]
(d) (i) v = fλ C1
λ = 330/440 = 0.75m A1 [2]
(ii) L = 5/4 λ C1
= 5/4 × 0.75 = 0.94m A1 [2]

9702 PHYSICS
9702/31 Paper 3 (Advanced Practical Skills 1),
maximum raw mark 40

1 (a) Value of L0 in range 0.045m–0.070m (4.5 to 7.0cm). [1]
If out of range, compare to Supervisor’s value ± 20%.
(b) (iii) Value of L > L0. [1]
(c) Six sets of readings of m and L scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend then –1. Correct trend is L decreases as m increases for all m
values.
Major help from Supervisor –2. Minor help from Supervisor –1.
Range: at least one value of m less than 200g and one value more than 200g. [1]
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit. The
presentation of quantity and unit must conform to accepted scientific convention,
e.g. θ/o
, L/m, m2
/kg2
, e/m, e2
(m2
).
Consistency: [1]
All values of L must be given to the nearest mm only.
Significant figures:
Significant figures for every row of m2
same as (or one greater than) the s.f. in m
as recorded in table. [1]
Calculation: [1]
Values of e2
calculated correctly to the number of significant figures given by the
candidate.
(d) (i) Axes: [1]
Sensible scales must be used, no awkward scales (e.g. 3:10).
Scales must be chosen so that the plotted points occupy at least half the
graph grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
Plotting of points: [1]
All observations must be plotted.
Diameter of plots must be Y half a small square (no “blobs”).
Work to an accuracy of half a small square.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
Scatter of points must be less than ± 0.0005m2
of e2
from a straight line.
(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidates’ line (at least
5 points). There must be an even distribution of points either side of the line
along the full length.
Allow one anomalous point only if clearly indicated by the candidate.
Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be at least half the length of the drawn
line.
Both read-offs must be accurate to half a small square in both the x and y
directions. The method of calculation must be correct.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into
y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph.
(e) P = – value of the gradient and Q = value of the y-intercept. [1]
Do not allow fractions. Do not allow substitution methods.
(f) Check substitution and value of M in range 0.100–0.500kg with unit. [1]
[Total: 20]
2 (b) (ii) Value of x in the range 25.0–35.0cm with unit. [1]
x to the nearest mm [1]
(iii) Absolute uncertainty in x in range 2mm–5mm. [1]
If repeated readings have been taken, then the uncertainty can be half the
range (not zero) only if working shown.
Correct method of calculation to obtain percentage uncertainty.
(c) (ii) Value of T with unit. [1]
Evidence of repeats. [1]
(iii) Correct calculation of f. [1]
(d) (ii) Second value of x. [1]
Second value of T. [1]
Second value of T < first value of T. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Justification of significant figures in k linked to significant figures in x and
T/time. (Do not allow “raw readings”.) [1]
(iii) Valid comment relating to the calculated values of k, testing against a
criterion specified by the candidate. [1]

(f) (i) Limitations (4 max) (ii) Improvements (4 max) Do not credit
A Two readings not enough to
draw a conclusion.
Take more readings for
different lengths and plot a
graph or take more readings
and compare k values.
Not enough repeat
readings.
Few readings.
Idea of repeats.
“Too few readings/two
readings” on its own.
B Difficulty linked to timing with
reason, e.g. time small/short/
vibrates fast/high frequency/
oscillates fast/swings fast.
or
Large uncertainty in time with
reason.
or
Human reaction time with
reason, e.g. short time.
Improved method of timing,
e.g. video with timer/video and
view frame by frame/light gate
placed at the centre/motion
sensor at side of blade (to
timer/datalogger display).
“Human errors/reaction
time” on their own.
“Light gate” on its own.
Moves fast.
Video and playback.
Fans.
Longer blade.
Simultaneous release of
blade and start timer.
Amplitude.
High speed camera or slow
motion cameras.
C Unevenness of oscillation or
G-clamp moving.
Improved method of smoother
oscillation, e.g. use of wooden
block either side of hacksaw
blade.
Method of fixing G-clamp, e.g.
clamp G-clamp (to table).
D Difficulty judging centre of
masses (due to slots).
Use masses without slots/
measure to the top and bottom
and average.
Blu Tack.
Masses different heights.
E Difficult to measure x with
reason, e.g. difficult to know
where to start x as jaws of
clamp are rounded/blade may
not be vertical (due to clamp).
Improved method to measure
x, e.g. description of use of set
square
Uncertainty in metre rule.
Parallax error.
F Small range for x for
measurable times
[Total: 20]

GCE Advanced Level
9702 PHYSICS
maximum raw mark 40

GCE A LEVEL – May/June 2014 9702 32
1 (a) (i) Value of l0 in range 4.0cm Y l0 Y 8.0cm. [1]
(b) (iii) Value of h to nearest mm, in the range 40.0cm Y h Y 50.0cm. [1]
(c) Six sets of values for h and l scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend –1. Help from Supervisor –1.
Range: [1]
h values must include 20cm or less.
Each column heading must contain a quantity and an appropriate unit.
The presentation of quantity and unit must conform to accepted scientific
convention, e.g. 1/h2
/cm–2
or 1/h2
(1/cm2
) but not 1/h2
(cm2
).
Consistency: [1]
All values of h and l must be given to the nearest mm only.
Significant figures: [1]
Every value of 1/h2
must be given to the same s.f. as (or one greater than) the
s.f. in the corresponding h.
Calculation: [1]
Values of (l–l0)2
calculated correctly.
(d) (i) Axes: [1]
Scale markings must be no more than 3 large squares apart.
Plotting: [1]
All observations in the table must be plotted.
Diameter of plotted points must be Y half a small square (no “blobs”).
Quality: [1]
All points must be plotted (at least 5) for this mark to be scored. Scatter of
points must be within ±10cm2
of (l–l0)2
from a straight line.
Judge by balance of all points about the candidate’s line (at least 5 points).
There must be an even distribution of points either side of the line along the
full length.
Allow one anomalous plot only if clearly indicated (i.e. circled or labelled) by
the candidate.

(iii) Gradient: [1]
The hypotenuse must be at least half the length of the drawn line.
Both read-offs must be accurate to half a small square in both x and y
directions.
The method of calculation must be correct.
y-intercept: [1]
Either:
Read-off from a point on the line is substituted into y = mx + c. Read-off must
be accurate to half a small square in both x and y directions.
Or:
(e) p = value of the gradient, and q = value of the intercept. [1]
Dimensionally correct units for p and q. [1]
[Total: 20]
2 (a) (i) Value for d to nearest mm, in range 1.0cm Y d Y 2.0cm. [1]
(ii) Correct calculation of l. [1]
(b) (ii) Value for t in range 4.0sY t Y 10.0s, with unit. [1]
Evidence of repeat readings of t. [1]
(c) (iii) Value for A to nearest mm, with unit. [1]
(d) Absolute uncertainty in A in range 2 to 5mm. [1]
If repeated readings have been taken, then absolute uncertainty could be half the
range (but not zero) only if working shown.
(e) Second value of t. [1]
Second value of A. [1]
(f) (i) Two values of k calculated correctly. [1]
kl in range 0.20 to 0.30cms–2
. [1]
(ii) Justification based on the number of s.f. in d, n, t and A (not just “raw readings”). [1]
(iii) Valid comment relating to the calculated values of k, testing against a criterion
specified by the candidate. [1]

(g) (i) Limitations (4 max) (ii) Improvements (4 max) Do not credit
A Two readings are not
enough to draw a valid
conclusion
Take more readings (for
different masses) and plot a
graph, or take more readings
and compare k values
Not enough repeat readings
Few readings
Idea of repeats
readings” on its own
B Difficult to measure d, with
reason, e.g. parallax error/
measuring outside
diameter/loop gets in the
way
Use vernier calipers/micrometer
Measure inside and outside
diameter and find average
“Parallax error” on its own
“Calipers” on its own
C Difficult to judge exactly
when 10 oscillations
completed
Video + timer/video and view
frame-by-frame.
Use distance sensor in stated
and correct position
Use a (fiducial) marker at centre
of oscillation/equilibrium position
Light gate at equilibrium
position/centre of oscillation
Human/reaction time error
High speed cameras/slow
motion cameras
Oscillations too fast
D Mass swings as it
oscillates/non-uniform
oscillation/spring moves
along bolt
Use tube to act as a guide
Use deeper groove on bolt
Fix top of spring to bolt with,
e.g. Blu-tack/Sellotape
E Difficult to judge when
contact is lost/hard to see
gap
Use pressure sensor on bolt
Use video close-up/zoom lens/
magnifying glass
Video + slow-motion, linked to
observing gap
Better/more sensitive method of
adjusting height of bolt, e.g. lab
jack
“Video + slow motion” on its
own
F n is not a whole number Measure n to nearest ¼ turn
Do not credit use of an assistant, fans, air conditioning, or use of computers/data loggers on its own.
[Total: 20]

9702 PHYSICS
maximum raw mark 40

Quality:
Scatter of points must be less than ± 0.25 A–1
of 1/I from a straight line.
[1]
1 (a) Value of L with unit in range 90.0cm ≤ L ≤ 110.0cm. [1]
(d) (ii) Value of I with unit in the range 50 mA ≤ I ≤ 150 mA.
Allow Supervisor’s value ± 20%.
[1]
(e) Six sets of readings of x and I scores 5 marks, five sets scores 4 marks etc.
Incorrect trend –1 (correct trend is I increases as x increases for all values of
x).
Major help from Supervisor –2. Minor help from Supervisor –1.
[5]
Range: ∆x ≥ 70 cm. [1]
Column headings:
Each column heading must contain a quantity and a unit.
convention e.g. x/cm or x(cm), x2
/(x+L)/m but not x2
/(x+L)/m2
/m, 1/I (A–1
), 1/I
(1/A) but not 1/I (A) or 1/I(A)–1
.
[1]
Consistency:
All values of x must be given to the nearest mm only.
[1]
Significant figures:
Significant figures for every row of x2
/(x+L) same as (or one greater than) the
s.f. in x.
Calculation:
Values of x2
/(x + L) calculated correctly to the number of significant figures
given by the candidate.
[1]
[1]
(f) (i) Axes:
Scales must be chosen so that the plotted points occupy at least half the graph
grid in both x and y directions.
[1]
Plotting of points:
Diameter of points must be ≤ half a small square (no “blobs”).
[1]

(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5
points). There must be an even distribution of points either side of the line
[1]
(iii) Gradient:
line.
directions. The method of calculation must be correct.
[1]
y-intercept:
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Or:
[1]
(g) P = – the value of the gradient and Q = the value of the y-intercept.
Do not allow fractions. Do not allow a substitution method.
Unit for P (A–1
m–1
) and Q (A–1
) consistent with value.
[1]
[1]
[Total: 20]
2 (b) (i) Value of y in range 8.0 cm ≤ y ≤ 12.0 cm with unit. [1]
(ii) Absolute uncertainty in y in range 2 mm – 5 mm.
If repeated readings have been taken, then allow uncertainty to be half the
range (but not zero) only if working shown.
[1]
(iii) Value of x to the nearest mm with unit. [1]
(iv) Correct calculation of R. [1]
(c) (iii) Value of T with unit in range 1.5 s ≤ T ≤ 3.0 s.
Evidence of repeats.
[1]
[1]
(d) Second value of R.
Second value of T.
Correct trend for T with respect to R (T decreases as R decreases).
[1]
[1]
[1]

(ii) Correct justification of significant figures in k linked to significant figures in x
and y and time (do not allow “raw readings”).
[1]
(iii) Valid comment consistent with the calculated values of k, testing against a
criterion specified by the candidate.
[1]
A Two readings not
enough to draw a
conclusion.
Take more readings (for
different radii) and plot a graph
or take more readings and
compare k values.
Not enough repeat
readings.
Few readings.
Idea of repeats.
readings” on its own.
B Only a few cycles
possible/(heavily)
damped/ball does not
return to the original
height.
Use a longer track/heavier ball. Smoother or lubricated
track.
C Difficult to judge exactly
when an oscillation is
complete.
Improved method of timing e.g.
video with timer/video and view
frame by frame/measure time
from centre of oscillation/put
marker at the centre of
oscillation/light gate at centre.
“Ball too fast” on its own.
Human reaction time.
Video and play back.
High speed cameras or
slow motion cameras.
Use of motion sensor.
D Difficult to measure y
with reason e.g. not
sure where horizontal is
at the top/ruler not
vertical/allowing for the
thickness/lips of the
track/can’t hold ruler
steady.
Improved method of measuring
y e.g. use a ruler across the
top/tie a string across the top/
set square on bench/use of
plumbline/measure two heights
and find difference/use of
micrometer for thickness of
track/clamp ruler to measure
y/clamp ruler vertically.
Parallax error.
Not clear where the
lowest point of track is.
E Limitation of track e.g.
is not entirely
circular/portion of track
in clamps is straight/too
stiff to bend/track
twisted/skewed.
Improved method of
supporting track e.g. attach
track to board/clamp in more
places.
Changing track e.g. more
flexible or rubber track.
Difficult to clamp.
“Friction on track” on its
own.
F Difficulty in placing ball
at the same point each
time/difficulty in
releasing the ball with
no force.
Method of marking the same
point on the track/method of
releasing the ball e.g. use card
as a gate/electromagnet
release and steel ball
If electromagnetic
release do not allow
‘metal’ ball.
Clamp or robotic arm.
Burning through string.
Do not credit measurement of x, use of an assistant, fans, air conditioning, use of
computers/dataloggers on its own.
[Total: 20]

9702 PHYSICS
maximum raw mark 40

1 (a) (i) Value for θ in range 80° to 100°, with unit. [1]
(b) (ii) Value for t in range 10 to 40s, with unit.. [1]
Evidence of repeat readings of t. [1]
(c) Five sets of values for θ and t scores 4 marks, four sets scores 3 marks etc. [4]
Incorrect trend –1. Help from Supervisor –1.
Range: [1]
θ values must include 75° or less and 105° or more.
Each column heading must contain a quantity and an appropriate unit.
convention
e.g. t2
/s2
or t2
(s2
), θ (°) or θ (deg) etc. sin2
(θ /2) must have no unit.
Consistency: [1]
All values of t must be given to the nearest 0.1s, or all to the nearest 0.01s.
Every value of t2
must be given to the same s.f. as (or one greater than) the s.f.
in the corresponding t.
Calculation: [1]
Values of sin2
(θ /2) calculated correctly.
(d) (i) Axes: [1]
Scale markings must be no more than three large squares apart.
Plotting: [1]
All observations in the table must be plotted on the grid.
Diameter of plotted points must be ≤ half a small square (no “blobs”).
Plotting must be accurate to half a small square.
Quality: [1]
Scatter of points must be within ± 20s2
of a straight line in the y (t2
) direction.

Judge by balance of all points on the grid about the candidate's line (at least
4 points).
There must be an even distribution of points either side of the line along the
full length.
Allow one anomalous plot only if clearly indicated by the candidate.
(iii) Gradient: [1]
line.
Both read-offs must be accurate to half a small square in both x and y
directions.
y-intercept: [1]
Either:
Correct read-off from a point on the line substituted into y = mx + c. Read-off
must be accurate to half a small square in both x and y directions.
Or:
Correct read-off of the intercept directly from the graph.
(e) q = candidate's gradient and p = candidate's intercept. [1]
Correct units for q and p (s2
for q and s2
for p). [1]
[Total: 20]

2 (a) r in range 5.0mm to 12.0mm and to nearest 0.1mm or better. [1]
(b) (i) Value for l in range 61.0mm to 65.0mm. [1]
(ii) Value for e in range 6.0mm to 8.0mm. [1]
(c) (ii) Value for x. [1]
Evidence of repeat readings of x. [1]
(iii) Absolute uncertainty in x in range 2 to 9mm. [1]
If repeated readings have been taken, then absolute uncertainty could be
half the range (but not zero) only if working is shown.
(iv) Calculated value of θ correct. [1]
θ given to 2 or 3 significant figures. [1]
(d) Second values of e and x. [1]
Both values for k in range 0.80 to 1.20. [1]
(ii) Valid comment consistent with the calculated values of k, testing against a

(f) Limitations (4 max) Improvements (4 max) Do not credit
A Two readings are not enough
to draw a valid conclusion
Take more readings and
plot graph/
take more readings and
compare k values
Repeat readings/
too few readings/
two readings
B Difficult to align groove with
line/
difficult to estimate centre of
groove
Mark centre line of groove/
method of aligning groove
(e.g. use lines on
paper/use graph paper/
transparent ramp)
C e is small so uncertainty in e
is large/
small change in e gives large
change in θ
Use larger spheres to
enable larger e
Just “use larger
spheres”
D Parallax error when
measuring x
Use set square (with detail
of workable method)
E Difficult to locate centre of
sphere when measuring x
Measure to edge of sphere
and add r
Difficult to locate
centre of sphere
when measuring r
F Sphere rolls slightly after
hitting tape/
sphere does not stick
Use video with scale (in
view)/
Description of workable
improvement (e.g. powder
on strip/plasticine surface)
High speed
camera/
slow motion
camera/
video camera/
stickier surface/
magnets
[Total: 20]

9702 PHYSICS
maximum raw mark 40

1 (b) (iii) Values of a in range 53.0cm – 57.0cm and b < a with unit. [1]
(iv) Value of L in the range 12.0cm – 16.0cm with unit. [1]
(d) Six sets of readings of a and b scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend –1 (correct trend is b increases as a increases).
Help from Supervisor –1.
Range: ∆a [ 39cm. [1]
Each column heading must contain a quantity and unit.
convention, e.g. (1/b)/m–1
.
Consistency: [1]
All values of a and b must be given to the nearest mm.
Significant figures for every row of values of 1/b same as (or one greater than) b
as recorded in table.
Calculation: [1]
Values of a/b calculated correctly.
(e) (i) Axes: [1]
Plotting of points: [1]
Diameter of plotted points must be Y half a small square (no “blobs”).
Quality: [1]
Scatter of points must be less than ± 0.001cm–1
(0.1m–1
) of 1/b from a
straight line.
Judge by balance of all points on the grid about the candidate’s line (at least
5 points). There must be an even distribution of points either side of the line

(iii) Gradient: [1]
line.
directions.
The method of calculation must be correct.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into
y = mx + c.
Or:
(f) Value of P = –gradient and value of Q = intercept. [1]
(g) Value of M in range 40–200g with unit. No POT error allowed. [1]
[Total: 20]
2 (a) (ii) Values of d to the nearest 0.01mm with unit. [1]
6.00mm Y d Y 10.00mm. If out of range allow Supervisor’s value ± 2.00mm. [1]
Evidence of repeat readings. [1]
(iii) Absolute uncertainty in d in range 0.05mm–2.00mm. [1]
If repeated readings have been taken, then the uncertainty could be half the
range (but not zero) only if the working is shown.
Correct method of calculation to get percentage uncertainty.
(c) (iv) Value of x1 > x with unit to the nearest mm. [1]
(v) Correct calculation of e. [1]
(d) (ii) Second value of d. [1]
Second value of x1. [1]
Second value of e < first value of e. [1]
(ii) Justification based on the number of significant figures in e (or x1 – x) and d. [1]
(iii) Valid comment relating to the calculated values of k, testing against a

A Two readings not enough to draw a
conclusion
Take many readings for different
diameters and plot a graph
Repeat readings
Few readings
Too few readings/only
two readings
B Large uncertainty in extension
because extension small
Use longer/thinner cylinders or
thinner central portion/time for
hanging longer/greater mass to
give greater extension
C Difficult to roll uniform cylinder/
cylinder not symmetrical/not
uniform diameter/density or
consistency not the same
Viable suggestion for
improvement, e.g. spacers,
mould, force through hole
D Reading of x1 is imprecise because
marks have widened
Improved method of marking
without a dent
E Difficulty with clamping the
plasticine, e.g. breaks prematurely/
twists in clamp
Improved method to attach
weight to plasticine,
e.g string loop through handles/
place clamp lengthways
F Micrometer digs into plasticine and
may weaken it/gives incorrect
diameter reading
Improved method to measure d,
e.g. travelling microscope/work
out diameter from volume or
circumference
G Difficulties relating to the properties
of the plasticine over time, e.g. high
temperature making too
soft/picking up impurities/re-
breaking at fractured points/as roll
temperature increases and affects
Use new piece of plasticine
each time/roll and leave until
reaches room temperature
[Total: 20]

9702 PHYSICS
9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100

Section A
1 (a) work done bringing unit mass M1
from infinity (to the point) A1 [2]
(b) EP = –mφ B1 [1]
(c) φ ∝ 1/x C1
either at 6R from centre, potential is (6.3 × 107
)/6 (= 1.05 × 107
J kg–1
)
and at 5R from centre, potential is (6.3 × 107
)/5 (= 1.26 × 107
J kg–1
) C1
change in energy = (1.26 – 1.05) × 107
× 1.3 C1
= 2.7 × 106
J A1
or change in potential = (1/5 – 1/6) × (6.3 × 107
) (C1)
change in energy = (1/5 – 1/6) × (6.3 × 107
) × 1.3 (C1)
= 2.7 × 106
J (A1) [4]
2 (a) the number of atoms M1
in 12 g of carbon-12 A1 [2]
(b) (i) amount = 3.2/40
= 0.080 mol A1 [1]
(ii) pV = nRT
p × 210 × 10–6
= 0.080 × 8.31 × 310 C1
p = 9.8 × 105
Pa A1 [2]
(do not credit if T in °C not K)
(iii) either pV = 1/3 × Nm <c2
>
N = 0.080 × 6.02 × 1023
(= 4.82 × 1022
)
and m = 40 × 1.66 × 10–27
(= 6.64 × 10–26
) C1
9.8 × 105
× 210 × 10–6
= 1/3 × 4.82 × 1022
× 6.64 × 10–26
× <c2
> C1
<c2
> = 1.93 × 105
cRMS = 440 m s–1
A1 [3]
or Nm = 3.2 × 10–3
(C1)
9.8 × 105
× 210 × 10–6
= 1/3 × 3.2 × 10–3
× <c2
> (C1)
<c2
> = 1.93 × 105
cRMS = 440 m s–1
(A1)
or 1/2 m<c2
> = 3/2 kT (C1)
1/2 × 40 × 1.66 × 10–27
<c2
> = 3/2 × 1.38 × 10–23
× 310 (C1)
<c2
> = 1.93 × 105
cRMS = 440 m s–1
(A1)
(if T in °C not K award max 1/3, unless already penalised in (b)(ii))

3 (a) either change in volume = (1.69 – 1.00 × 10–3
)
or liquid volume << volume of vapour M1
work done = 1.01 × 105
× 1.69 = 1.71 × 105
(J) A1 [2]
(b) (i) 1. heating of system/thermal energy supplied to the system B1 [1]
2. work done on the system B1 [1]
(ii) ∆U = (2.26 × 106
) – (1.71 × 105
) C1
= 2.09 × 106
J (3 s.f. needed) A1 [2]
4 (a) kinetic (energy)/KE/EK B1 [1]
(b) either change in energy = 0.60 mJ
or max E proportional to (amplitude)2
/equivalent numerical working B1
new amplitude is 1.3 cm B1
change in amplitude = 0.2 cm B1 [3]
5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1
curve with decreasing gradient M1
passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3]
(b) graph: straight line at E = 0 from x = 0 to x = r B1
curve with decreasing gradient from (r, E0) M1
passing through (2r, ¼E0) A1 [3]
(for 3rd mark line must be drawn to x = 4r and must not touch x-axis)
6 (a) (i) energy = EQ C1
= 9.0 × 22 × 10–3
= 0.20 J A1 [2]
(ii) 1. C = Q/V
V = (22 × 10–3
)/(4700 × 10–6
) C1
= 4.7 V A1 [2]
2. either E = ½CV2
C1
= ½ × 4700 × 10–6
× 4.72
= 5.1 × 10–2
J A1 [2]
or E = ½QV (C1)
= ½ × 22 × 10–3
× 4.7
= 5.1 × 10–2
J (A1)
or E = ½Q2
/C (C1)
= ½ × (22 × 10–3
)2
/4700 ×10–6
= 5.1 × 10–2
J (A1)

(b) energy lost (as thermal energy) in resistance/wires/battery/resistor B1 [1]
(award only if answer in (a)(i) > answer in (a)(ii)2)
7 (a) graph: VH increases from zero when current switched on B1
VH then non-zero constant B1
VH returns to zero when current switched off B1 [3]
(b) (i) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(ii) pulse as current is being switched on B1
zero e.m.f. when current in coil B1
pulse in opposite direction when switching off B1 [3]
8 (a) discrete and equal amounts (of charge) B1 [1]
allow: discrete amounts of 1.6 × 10–19
C/elementary charge/e
integral multiples of 1.6 × 10–19
(b) weight = qV/d
4.8 × 10–14
= (q × 680)/(7.0 × 10–3
) C1
q = 4.9 × 10–19
C A1 [2]
(c) elementary charge = 1.6 × 10–19
C (allow 1.6 × 10–19
C to 1.7 × 10–19
C ) M0
either the values are (approximately) multiples of this
or it is a common factor C1
it is the highest common factor A1 [2]
9 (a) e.g. no time delay between illumination and emission
max. (kinetic) energy of electron dependent on frequency
max. (kinetic) energy of electron independent of intensity
rate of emission of electrons dependent on/proportional to intensity
(any three separate statements, one mark each, maximum 3) B3 [3]
(b) (i) (photon) interaction with electron may be below surface B1
energy required to bring electron to surface B1 [2]

(ii) 1. threshold frequency = 5.8 × 1014
Hz A1 [1]
2. Φ = hf0 C1
= 6.63 × 10–34
× 5.8 × 1014
= 3.84 × 10–19
(J) C1
= (3.84 × 10–19
)/(1.6 × 10–19
)
= 2.4 eV A1 [3]
or
hf = Φ + EMAX (C1)
chooses point on line and substitutes values EMAX, f and h into
equation with the units of the hf term converted from J to eV (C1)
Φ = 2.4 eV (A1)
10 (a) energy required to separate the nucleons (in a nucleus) M1
to infinity A1 [2]
(allow reverse statement)
(b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1
= 9.11 × 10–3
u C1
binding energy = 9.11 × 10–3
× 930
= 8.47 MeV A1 [3]
(allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV)
(allow 2 s.f.)
(ii) ∆m = 211.70394 – 209.93722
= 1.76672 u C1
binding energy per nucleon = (1.76672 × 930)/210 C1
= 7.82 MeV A1 [3]
(allow 2 s.f.)
(c) total binding energy of barium and krypton M1
is greater than binding energy of uranium A1 [2]
Section B
11 (a) (i) inverting amplifier B1 [1]
(ii) gain is very large/infinite B1
V+
is earthed/zero B1
for amplifier not to saturate, P must be (almost) earth/zero B1 [3]
(b) (i) RA = 100 kΩ A1
RB = 10 kΩ A1
VIN = 1000 mV A1 [3]
(ii) variable range meter B1 [1]

12 (a) series of X-ray images (for one section/slice) M1
taken from different angles M1
to give image of the section/slice A1
repeated for many slices M1
to build up three-dimensional image (of whole object) A1 [5]
(b) deduction of background from readings C1
division by three C1
P = 5 Q = 9 R = 7 S =13
(four correct 2/2, three correct 1/2) A2 [4]
13 (a) e.g. noise can be eliminated/waveform can be regenerated
extra bits of data can be added to check for errors
cheaper/more reliable
greater rate of transfer of data
(1 each, max 2) B2 [2]
(b) receives bits all at one time B1
transmits the bits one after another B1 [2]
(c) sampling frequency must be higher than/(at least) twice frequency to be sampled M1
either higher (range of) frequencies reproduced on the disc
or lower (range of) frequencies on phone A1
either higher quality (of sound) on disc
or high quality (of sound) not required for phone B1 [3]
14 (a) reduction in power (allow intensity/amplitude) B1 [1]
(b) (i) attenuation = 2.4 × 30
= 72 dB A1 [1]
(ii) gain/attenuation/dB = 10 lg(P2/P1) C1
72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1
ratio = 1.6 × 107
A1 [3]
(c) e.g. enables smaller/more manageable numbers to be used
e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]

GCE Advanced Level
9702 PHYSICS
9702/42 Paper 4 (A2 Structured Questions),
maximum raw mark 100

Section A
1 (a) gravitational force provides/is the centripetal force B1
GMm/r2
R mv2
/r M1
v R √(GM/r) A0 [2]
allow gravitational field strength provides/is the centripetal acceleration (B1)
GM/r2
R v2
/r (M1)
(b) (i) kinetic energy increase/change R loss/change in (gravitational) potential
energy B1
½mV0
2
R GMm/x C1
V0
2
R 2GM/x
V0 R √(2GM/x) A1 [3]
(max. 2 for use of r not x)
(ii) V0 is (always) greater than v (for x = r) M1
so stone could not enter into orbit A1 [2]
(expressions in (a) and (b)(i) must be dimensionally correct)
2 (a) use of kelvin temperatures B1
both values of (V/T) correct (11.87), V/T is constant so pressure is constant M1 [2]
(allow use of n R1. Do not allow other values of n.)
(b) (i) work done R p∆V
R 4.2 × 105
× (3.87 – 3.49) × 103
× 10–6
C1
= = = = = R 160 J A1 [2]
(do not allow use of V instead of ∆V)
(ii) increase/change in internal energy R heating of system
N work done on system C1
R 565 – 160
= = = = R 405J A1 [2]
(c) internal energy R sum of kinetic energy and potential energy/EK N EP B1
no intermolecular forces M1
no potential energy (so ∆U R ∆EK) A1 [3]
3 (a) resonance B1 [1]
(b) Pt R mc ∆θ C1
750 × 2 × 60 R 0.28 × c × (98 – 25) C1
c R 4400Jkg–1
K–1
A1 [3]
(use of ∆θ R 73 N 273 max. 1/3)
(use of t R 2s not 120s max. 2/3)

(c) e.g. some microwave leakage from the cooker
e.g. container for the water is also heated
(any sensible suggestion) B1 [1]
4 (a) (i) FE R Q1Q2/4πε0r2
C1
= = = = R 8.99 × 109
× (1.6 × 10–19
)2
/(2.0 × 10–15
)2
R 58 N A1 [2]
(ii) FG R Gm1m2/r2
C1
= = = R 6.67 × 10–11
× (1.67 × 10–27
)2
/(2.0 × 10–15
)2
= = = R 4.7 × 10–35
N A1 [2]
(b) (i) force of repulsion (much) greater than force of attraction B1
must be some other force of attraction M1
to hold nucleus together A1 [3]
(Do not allow if FG > FE in (a) or one of the forces not calculated in (a))
(ii) outside nucleus there is repulsion between protons B1
either attractive force must act only in nucleus
or if not short range, all nuclei would stick together B1 [2]
5 (a) only curve with decreasing gradient M1
acceptable value near xR 0 and does not reach zero A1 [2]
(if graph line less than 4.0 cm do not allow A1 mark)
(no credit if graph line has positive and negative values of VH)
(b) graph: from 0 to 2T, two cycles of a sinusoidal wave M1
all peaks above 3.5mV C1
peaks at 4.95/5.0mV (allow 4.8mV to 5.2mV) A1 [3]
(c) e.m.f. induced in coil when magnetic field/flux is changing/cutting B1
either at each position, magnetic field does not vary
so no e.m.f. is induced in the coil/no reading on the millivoltmeter
or at each position, switch off current and take millivoltmeter reading
or at each position, rapidly remove coil from field and take meter reading B1 [2]
6 (a) electric and magnetic fields normal to each other B1
either charged particle enters region normal to both fields
or correct B direction w.r.t. E for zero deflection B1
for no deflection, v R E/B B1 [3]
(no credit if magnetic field region clearly not overlapping with electric field region)

(b) (i) m R Bqr/v C1
= = = R=(640 × 10–3
× 1.6 × 10–19
× 6.2 × 10–2
)/(9.6 × 104
) C1
= = = R=6.61 × 10–26
kg C1
= = = R=(6.61 × 10–26
)/(1.66 × 10–27
)u
= = = R=40u A1 [4]
(ii) q/m ∝ 1/r or m constant and q ∝ 1/r B1
q/m for A is twice that for B B1
ions in path A have (same mass but) twice the charge (of ions in path B) B1 [3]
7 (a) angle subtended at the centre of a circle B1
by an arc equal in length to the radius B1 [2]
(b) (i) arc R distance × angle C1
diameter R 3.8 × 105
× 9.7 × 10–6
= = = R 3.7km A1 [2]
(ii) Mars is (much) further from Earth/away (answer must be comparative) B1
angle (at telescope is much) smaller B1 [2]
8 (a) photon energy R hc/λ
R (6.63 × 10–34
× 3.0 × 108
)/(590 × 10–9
) C1
R 3.37 × 10–19
J C1
number R (3.2 × 10–3
)/(3.37 × 10–19
)
= = = R 9.5 × 1015
(allow 9.4 × 1015
) A1 [3]
(b) (i) p R h/λ C1
= = = R (6.63 × 10–34
)/(590 × 10–9
)
= = = R 1.12 × 10–27
kgms–1
C1
total momentum R 9.5 × 1015
× 1.12 × 10–27
= = = R 1.06 × 10–11
kgms–1
A1 [3]
(ii) force R 1.06 × 10–11
N A1 [1]
9 (a) time for number of atoms/nuclei/activity (of the isotope) M1
to be reduced to one half (of its initial value) A1 [2]
(b) (i) A R λN C1
460 R N × ln 2/(8.1 × 24 × 60 × 60) C1
N R 4.6 × 108
A1 [3]
(ii) number of water molecules in 1.0 kg R (6.02 × 1023
)/(18 × 10–3
) C1
R 3.3 × 1025
ratio R (3.3 × 1025
)/(4.6 × 108
)
R 7.2 (7.3) × 1016
A1 [2]

(c) A R A0 e–λt
and λt½ R ln 2 C1
170 R 460 exp (–{ln 2 t }/8.1) C1
t R 11.6 days (allow 2 s.f.) A1 [3]
Section B
10 (a) compares the potentials/voltages at the (inverting and non-inverting) inputs B1
either output (potential) dependent on which input is the larger
or V+
> V–
, then VOUT is positive B1
states the other condition B1 [3]
(b) (i) ring drawn around both the LEDs (and series resistors) B1 [1]
(ii) V–
R (1.5 × 2.4)/(1.2 N 2.4) R 1.0V B1 [1]
(allow 1.5 × 2.4/3.6 R 1.0V)
(iii) 1. VOUT switches at N1.0V B1
maximum VOUT is 5.0V B1
when curve is above N1.0V, VOUT is negative (or v.v.) B1 [3]
2. at time t1, diode R is emitting light, diode G is not emitting B1
at time t2, diode R is not emitting, diode G is emitting B1 [2]
(must be consistent with graph line. If no graph line then 0/2)
11 (a) X-ray: flat/shadow/2D image B1
regardless of depth of object/depth not indicated B1
CT scan: built up from (many) images at different angles B1
image is three-dimensional B1
image can be rotated/viewed at different angles B1 [5]
(b) (i) I R I0 e–µx
C1
0.25 R e–0.69x
x R2.0mm (allow 1 s.f.) A1 [2]
(ii) for aluminium, I/I0 R e–0.46 × 2.4
R 0.33 C1
fraction R 0.33 × 0.25
= = = R 0.083 A1 [2]
(iii) gain/dB R 10lg(I/I0) C1
R 10lg(0.083)
= = = R (–) 10.8dB (allow 2 s.f.) A1
with negative sign B1 [3]
12 (a) (i) satellite is in equatorial orbit B1
travelling from west to east B1
period of 24 hours/1 day B1 [3]

(ii) either uplink signal is highly attenuated
or signal is highly amplified (before transmission) as downlink signal B1
prevents downlink signal swamping the uplink signal B1 [2]
(b) speed of signal is same order of magnitude in both systems B1
optic fibre link (much) shorter than via satellite M1
time delay using optic fibre is less A1 [3]

9702 PHYSICS
9702/43 Paper 4 (A2 Structured Questions), maximum raw mark 100

Section A
1 (a) work done bringing unit mass M1
from infinity (to the point) A1 [2]
(b) EP = –mφ B1 [1]
(c) φ ∝ 1/x C1
either at 6R from centre, potential is (6.3 × 107
)/6 (= 1.05 × 107
J kg–1
)
and at 5R from centre, potential is (6.3 × 107
)/5 (= 1.26 × 107
J kg–1
) C1
change in energy = (1.26 – 1.05) × 107
× 1.3 C1
= 2.7 × 106
J A1
or change in potential = (1/5 – 1/6) × (6.3 × 107
) (C1)
change in energy = (1/5 – 1/6) × (6.3 × 107
) × 1.3 (C1)
= 2.7 × 106
J (A1) [4]
2 (a) the number of atoms M1
in 12 g of carbon-12 A1 [2]
(b) (i) amount = 3.2/40
= 0.080 mol A1 [1]
(ii) pV = nRT
p × 210 × 10–6
= 0.080 × 8.31 × 310 C1
p = 9.8 × 105
Pa A1 [2]
(do not credit if T in °C not K)
(iii) either pV = 1/3 × Nm <c2
>
N = 0.080 × 6.02 × 1023
(= 4.82 × 1022
)
and m = 40 × 1.66 × 10–27
(= 6.64 × 10–26
) C1
9.8 × 105
× 210 × 10–6
= 1/3 × 4.82 × 1022
× 6.64 × 10–26
× <c2
> C1
<c2
> = 1.93 × 105
cRMS = 440 m s–1
A1 [3]
or Nm = 3.2 × 10–3
(C1)
9.8 × 105
× 210 × 10–6
= 1/3 × 3.2 × 10–3
× <c2
> (C1)
<c2
> = 1.93 × 105
cRMS = 440 m s–1
(A1)
or 1/2 m<c2
> = 3/2 kT (C1)
1/2 × 40 × 1.66 × 10–27
<c2
> = 3/2 × 1.38 × 10–23
× 310 (C1)
<c2
> = 1.93 × 105
cRMS = 440 m s–1
(A1)
(if T in °C not K award max 1/3, unless already penalised in (b)(ii))

3 (a) either change in volume = (1.69 – 1.00 × 10–3
)
or liquid volume << volume of vapour M1
work done = 1.01 × 105
× 1.69 = 1.71 × 105
(J) A1 [2]
(b) (i) 1. heating of system/thermal energy supplied to the system B1 [1]
2. work done on the system B1 [1]
(ii) ∆U = (2.26 × 106
) – (1.71 × 105
) C1
= 2.09 × 106
J (3 s.f. needed) A1 [2]
4 (a) kinetic (energy)/KE/EK B1 [1]
(b) either change in energy = 0.60 mJ
or max E proportional to (amplitude)2
/equivalent numerical working B1
new amplitude is 1.3 cm B1
change in amplitude = 0.2 cm B1 [3]
5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1
curve with decreasing gradient M1
passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3]
(b) graph: straight line at E = 0 from x = 0 to x = r B1
curve with decreasing gradient from (r, E0) M1
passing through (2r, ¼E0) A1 [3]
(for 3rd mark line must be drawn to x = 4r and must not touch x-axis)
6 (a) (i) energy = EQ C1
= 9.0 × 22 × 10–3
= 0.20 J A1 [2]
(ii) 1. C = Q/V
V = (22 × 10–3
)/(4700 × 10–6
) C1
= 4.7 V A1 [2]
2. either E = ½CV2
C1
= ½ × 4700 × 10–6
× 4.72
= 5.1 × 10–2
J A1 [2]
or E = ½QV (C1)
= ½ × 22 × 10–3
× 4.7
= 5.1 × 10–2
J (A1)
or E = ½Q2
/C (C1)
= ½ × (22 × 10–3
)2
/4700 ×10–6
= 5.1 × 10–2
J (A1)

(b) energy lost (as thermal energy) in resistance/wires/battery/resistor B1 [1]
(award only if answer in (a)(i) > answer in (a)(ii)2)
7 (a) graph: VH increases from zero when current switched on B1
VH then non-zero constant B1
VH returns to zero when current switched off B1 [3]
(b) (i) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(ii) pulse as current is being switched on B1
zero e.m.f. when current in coil B1
pulse in opposite direction when switching off B1 [3]
8 (a) discrete and equal amounts (of charge) B1 [1]
allow: discrete amounts of 1.6 × 10–19
integral multiples of 1.6 × 10–19
(b) weight = qV/d
4.8 × 10–14
= (q × 680)/(7.0 × 10–3
) C1
q = 4.9 × 10–19
C A1 [2]
(c) elementary charge = 1.6 × 10–19
C (allow 1.6 × 10–19
C to 1.7 × 10–19
C ) M0
either the values are (approximately) multiples of this
or it is a common factor C1
it is the highest common factor A1 [2]
9 (a) e.g. no time delay between illumination and emission
max. (kinetic) energy of electron dependent on frequency
max. (kinetic) energy of electron independent of intensity
rate of emission of electrons dependent on/proportional to intensity
(any three separate statements, one mark each, maximum 3) B3 [3]
(b) (i) (photon) interaction with electron may be below surface B1
energy required to bring electron to surface B1 [2]

(ii) 1. threshold frequency = 5.8 × 1014
Hz A1 [1]
2. Φ = hf0 C1
= 6.63 × 10–34
× 5.8 × 1014
= 3.84 × 10–19
(J) C1
= (3.84 × 10–19
)/(1.6 × 10–19
)
= 2.4 eV A1 [3]
or
hf = Φ + EMAX (C1)
chooses point on line and substitutes values EMAX, f and h into
equation with the units of the hf term converted from J to eV (C1)
Φ = 2.4 eV (A1)
10 (a) energy required to separate the nucleons (in a nucleus) M1
to infinity A1 [2]
(allow reverse statement)
(b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1
= 9.11 × 10–3
u C1
binding energy = 9.11 × 10–3
× 930
= 8.47 MeV A1 [3]
(allow 2 s.f.)
(ii) ∆m = 211.70394 – 209.93722
= 1.76672 u C1
binding energy per nucleon = (1.76672 × 930)/210 C1
= 7.82 MeV A1 [3]
(allow 2 s.f.)
(c) total binding energy of barium and krypton M1
is greater than binding energy of uranium A1 [2]
Section B
11 (a) (i) inverting amplifier B1 [1]
(ii) gain is very large/infinite B1
V+
is earthed/zero B1
for amplifier not to saturate, P must be (almost) earth/zero B1 [3]
(b) (i) RA = 100 kΩ A1
RB = 10 kΩ A1
VIN = 1000 mV A1 [3]
(ii) variable range meter B1 [1]

12 (a) series of X-ray images (for one section/slice) M1
taken from different angles M1
to give image of the section/slice A1
repeated for many slices M1
to build up three-dimensional image (of whole object) A1 [5]
(b) deduction of background from readings C1
division by three C1
P = 5 Q = 9 R = 7 S =13
(four correct 2/2, three correct 1/2) A2 [4]
13 (a) e.g. noise can be eliminated/waveform can be regenerated
extra bits of data can be added to check for errors
cheaper/more reliable
greater rate of transfer of data
(1 each, max 2) B2 [2]
(b) receives bits all at one time B1
transmits the bits one after another B1 [2]
(c) sampling frequency must be higher than/(at least) twice frequency to be sampled M1
either higher (range of) frequencies reproduced on the disc
or lower (range of) frequencies on phone A1
either higher quality (of sound) on disc
or high quality (of sound) not required for phone B1 [3]
14 (a) reduction in power (allow intensity/amplitude) B1 [1]
(b) (i) attenuation = 2.4 × 30
= 72 dB A1 [1]
(ii) gain/attenuation/dB = 10 lg(P2/P1) C1
72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1
ratio = 1.6 × 107
A1 [3]
(c) e.g. enables smaller/more manageable numbers to be used
e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]

9702 PHYSICS
9702/51 Paper 5 (Planning, Analysis and Evaluation),
maximum raw mark 30

1 Planning (15 marks)
Defining the problem (3 marks)
P r is the independent variable or vary r. [1]
P T (or t) is the dependent variable or measure T (or t). [1]
P Keep the radius of curvature (of the track) or C constant (or radius of track
constant). Do not allow “use same track”. [1]
Methods of data collection (5 marks)
M Diagram showing ball in a (curved) track with supports for track, e.g. retort
stands. Minimum of two labels (from ball, track, supports; not stopwatch, bench,
micrometer).
Supports making contact with track higher than ball/at least half way up. [1]
M Measure time using stopwatch or light gates and timer or datalogger with motion
sensor.
Detail needed for video camera. [1]
M Use many oscillations (at least 10 or at least 10s of timing) and determine
T = t/n. [1]
M Measure diameter (radius) of ball with a micrometer/vernier calipers.
Do not allow travelling microscope. [1]
M radius = diameter/2. [1]
Method of analysis (2 marks)
A Plot a graph of T2
against r (or r against T2
)
Do not allow log graphs. [1]
A C = y-intercept ×
gradient
intercepty
28
5
2
−
=
π
g
(or for r against T2
, C = y-intercept) [1]
Safety considerations (1 mark)
S Precaution linked to ball escaping on to floor, e.g. use barrier/safety
screen/sand tray to prevent balls rolling on to floor. [1]

Additional detail (4 marks)
D Relevant points might include [4]
1 Add weights to/G-clamp retort stands
2 Keep the material/density of the ball constant
3 Use of fiducial marker near centre of track/mark on the track
4 Clean track/balls. Do not allow oil the track.
5 Repeat measurements of t (for each ball) and average
6 Repeat measurement for d (or r) and average
7 Relationship is valid if straight line, provided plotted graph is correct
8 Relationship is valid if straight line not passing through origin or has an intercept,
provided plotted graph is correct (any quoted expression must be correct, e.g.
y-intercept =
g
C
5
28 2
π )
Do not allow vague computer methods.
[Total: 15]

2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a) A1 gradient = –R
y-intercept = –R/P
Must be negative.
(b) T1
T2 6.7 or 6.67 –5.1 or –5.13
4.5 or 4.55 –3.8 or –3.75
3.0 or 3.03 –2.8 or –2.75
2.0 or 2.00 –2.1 or –2.06
1.5 or 1.52 –1.8 or –1.75
1.1 or 1.11 –1.5 or –1.50
Allow a mixture of significant figures.
T1 and T2 must be table values.
Ignore “–” omissions.
U1 From ± 0.3 or ± 0.4 to ± 0.1 or ± 0.2 Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Penalise “blobs”.
Ecf allowed from table.
U2 Error bars in V/E plotted correctly All error bars to be plotted. Must be
accurate to less than half a small
square.
(ii) G2 Line of best fit If points are plotted correctly then
lower end of line should pass between
(6.4,–5.0) and (6.6,–5.0) and upper
end of line should pass between
(1.0,–1.5) and (1.2,–1.5).
G3 Worst acceptable straight line.
Steepest or shallowest possible line
that passes through all the error bars.
Line should be clearly labelled or
dashed. 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. Mark scored only if all
error bars are plotted.
(iii) C1 Gradient of best fit line Must be negative. The triangle used
should be at least half the length of the
drawn line. Check the read offs. Work
to half a small square. Do not penalise
POT. (Should be about –650.)
U3 Uncertainty in gradient Method of determining absolute
uncertainty.
Difference in worst gradient and
gradient.

(iv) C2 Negative y-intercept Must be negative.
Check substitution into y = mx + c
Allow ecf from (c)(iii). (Should be
about –0.8.)
FOX does not score.
U4 Uncertainty in y-intercept Uses worst gradient and point on WAL.
Check method.
FOX does not score.
(d) (i) C3 P = R/y-intercept
= –gradient/y-intercept
Include unit [Ω] for P and R.
Do not penalise POT.
C4 R = gradient in the range 620 to 680
and given to 2 or 3 s.f.
(ii) U5 Percentage uncertainty in P Percentage uncertainty in gradient +
percentage uncertainty in y-intercept
[Total: 15]
Uncertainties in Question 2
(c) (iii) Gradient [U3]
Uncertainty = gradient of line of best fit – gradient of worst acceptable line
Uncertainty = ½ (steepest worst line gradient – shallowest worst line
gradient)
(c) (iv) [U4]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (i) [U5]
Percentage uncertainty in gradient + percentage uncertainty in y-intercept
max P =
interceptymin
gradientmax
interceptymin
max
−
=
−
R
min f =
interceptmax y
gradientmin
interceptmax y
min
−
=
−
R

9702 PHYSICS
maximum raw mark 30

P r is the independent variable, B is the dependent variable or vary r and measure B. [1]
P Keep the number of turns on the coil(s) constant. [1]
Do not accept “same coil”.
P Keep the current in the coil constant. [1]
M Diagram showing flat coils and labelled Hall probe positioned at X. Minimum two labels
needed.
Solenoids will not be credited. [1]
M Workable circuit diagram for coil connected to a (d.c.) power supply and ammeter.
Do not allow a.c. power supply or incorrect circuit diagrams. [1]
M Connect Hall probe to voltmeter/c.r.o.
Allow galvanometer but do not allow ammeter. [1]
M Measure diameter (radius) with a ruler/vernier calipers. Do not allow micrometer. [1]
M Calibrate Hall probe with a known magnetic flux density. [1]
A Plot a graph of B against 1/r [allow lg B against lg r or other valid graph] [1]
A µ0
IN0.72
gradient
= [1]
S Precaution linked to (large) heating of coil, e.g. switch off when not in use to avoid
overheating coil; do not touch coil because it is hot. [1]

1 Use large current/large number of turns to create a large magnetic field
2 Use rheostat (to adjust current in circuit) (with ammeter) to keep the current constant
3 Hall probe at right angles to direction of magnetic field/parallel to coils. Allow adjust to obtain
maximum reading
4 Reasoned method to keep Hall probe perpendicular to direction of magnetic field or at X (e.g.
use of set square, fix to rule, optical bench or equivalent)
5 Method to check coils are correctly aligned in parallel
6 Repeat experiment with Hall probe reversed and average
8 Relationship is valid if the graph is a straight line passing through the origin for appropriate
graph
[if lg–lg then straight line with gradient = –1 (ignore reference to y-intercept)]
[Total: 15]
(a) A1
gradient =
g
2
4π−
y-intercept = k
g
2
4π
Gradient must be negative.
Allow y-intercept = –gradient × k
(b) T1 (mean) t/s, T/s and T2
/s2
All column headings to be correct.
T2
31.8 or 31.81
30.8 or 30.80
29.6 or 29.59
28.7 or 28.73
27.8 or 27.77
26.8 or 26.83
Check all values of T2
.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Penalise “blobs”
U1 Error bars in d plotted
correctly
All error bars to be plotted. Must be accurate to less
than half a small square.
(c) (ii) G2 Line of best fit Lower end of line should pass between (1.60, 27.0)
and (1.64,27.0) and upper end of line should pass
between (0.44,31.8) and (0.48,31.8).

G3 Worst acceptable straight
line.
Steepest or shallowest
possible line that passes
through all the error bars.
Line should be clearly labelled or dashed.
Examiner judgement on worst acceptable line.
Lines must cross. Mark scored only if all error bars
are plotted.
(c) (iii) C1 Gradient of best fit line Must be negative.
The triangle used should be at least half the length
of the drawn line. Check the read offs. Work to half
a small square. Do not penalise POT. (Should be
about –4.)
U2 Uncertainty in gradient Method of determining absolute uncertainty:
difference in worst gradient and gradient.
(c) (iv) C2 y-intercept FOX does not score.
Allow ecf from (c)(iii). (Should be about 33.7.)
Do not check calculation.
FOX does not score.
(d) (i) C3 g between 9.20 and 9.90
given to 2 or 3 s.f. and
correct unit (ms–2
) having
used gradient.
m
g
2
4π
−= ; allow Nkg–1
C4 k determined correctly with
correct unit (m) m
cg
ck
−
==
2
4π
(k must be positive.)
(d) (ii) U4 Percentage uncertainty in g
U5 Percentage uncertainty in k Percentage uncertainty in k must be larger than the
percentage uncertainty in g.
[Total: 15]
Uncertainty =
2
1
(steepest worst line gradient – shallowest worst line gradient)
(c) (iv) [U3]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
Uncertainty =
2
1
(steepest y-intercept – shallowest y-intercept)

(d) (ii) [U4]
Percentage uncertainty in g = 100100 ×
∆
=×∆
g
g
m
m
[U5]
Percentage uncertainty in k= 100100100 ×∆+×
∆
=×∆
c
c
g
g
k
k
max k = 2
4
interceptmaxmax
π
× -yg
=
gradientmin
interceptmax −y
min k = 2
4
interceptminmin
π
× -g
=
gradientmax
interceptmin −y

9702 PHYSICS
maximum raw mark 30

P r is the independent variable or vary r. [1]
P T (or t) is the dependent variable or measure T (or t). [1]
P Keep the radius of curvature (of the track) or C constant (or radius of track
constant). Do not allow “use same track”. [1]
M Diagram showing ball in a (curved) track with supports for track, e.g. retort
stands. Minimum of two labels (from ball, track, supports; not stopwatch, bench,
micrometer).
Supports making contact with track higher than ball/at least half way up. [1]
M Measure time using stopwatch or light gates and timer or datalogger with motion
sensor.
Detail needed for video camera. [1]
M Use many oscillations (at least 10 or at least 10s of timing) and determine
T = t/n. [1]
M Measure diameter (radius) of ball with a micrometer/vernier calipers.
Do not allow travelling microscope. [1]
M radius = diameter/2. [1]
A Plot a graph of T2
against r (or r against T2
)
Do not allow log graphs. [1]
A C = y-intercept ×
gradient
intercepty
28
5
2
−
=
π
g
(or for r against T2
, C = y-intercept) [1]
S Precaution linked to ball escaping on to floor, e.g. use barrier/safety
screen/sand tray to prevent balls rolling on to floor. [1]

1 Add weights to/G-clamp retort stands
2 Keep the material/density of the ball constant
3 Use of fiducial marker near centre of track/mark on the track
4 Clean track/balls. Do not allow oil the track.
5 Repeat measurements of t (for each ball) and average
7 Relationship is valid if straight line, provided plotted graph is correct
8 Relationship is valid if straight line not passing through origin or has an intercept,
provided plotted graph is correct (any quoted expression must be correct, e.g.
y-intercept =
g
C
5
28 2
π )
[Total: 15]

(a) A1 gradient = –R
y-intercept = –R/P
Must be negative.
(b) T1
T2 6.7 or 6.67 –5.1 or –5.13
4.5 or 4.55 –3.8 or –3.75
3.0 or 3.03 –2.8 or –2.75
2.0 or 2.00 –2.1 or –2.06
1.5 or 1.52 –1.8 or –1.75
1.1 or 1.11 –1.5 or –1.50
T1 and T2 must be table values.
Ignore “–” omissions.
U1 From ± 0.3 or ± 0.4 to ± 0.1 or ± 0.2 Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Penalise “blobs”.
U2 Error bars in V/E plotted correctly All error bars to be plotted. Must be
accurate to less than half a small
square.
(ii) G2 Line of best fit If points are plotted correctly then
lower end of line should pass between
(6.4,–5.0) and (6.6,–5.0) and upper
end of line should pass between
(1.0,–1.5) and (1.2,–1.5).
G3 Worst acceptable straight line.
Steepest or shallowest possible line
that passes through all the error bars.
Line should be clearly labelled or
dashed. 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. Mark scored only if all
error bars are plotted.
(iii) C1 Gradient of best fit line Must be negative. The triangle used
should be at least half the length of the
drawn line. Check the read offs. Work
to half a small square. Do not penalise
POT. (Should be about –650.)
U3 Uncertainty in gradient Method of determining absolute
uncertainty.
Difference in worst gradient and
gradient.

(iv) C2 Negative y-intercept Must be negative.
Allow ecf from (c)(iii). (Should be
about –0.8.)
FOX does not score.
Check method.
FOX does not score.
(d) (i) C3 P = R/y-intercept
= –gradient/y-intercept
Include unit [Ω] for P and R.
Do not penalise POT.
C4 R = gradient in the range 620 to 680
and given to 2 or 3 s.f.
(ii) U5 Percentage uncertainty in P Percentage uncertainty in gradient +
percentage uncertainty in y-intercept
[Total: 15]
Uncertainty = ½ (steepest worst line gradient – shallowest worst line
gradient)
(c) (iv) [U4]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (i) [U5]
Percentage uncertainty in gradient + percentage uncertainty in y-intercept
max P =
interceptymin
gradientmax
interceptymin
max
−
=
−
R
min f =
interceptmax y
gradientmin
interceptmax y
min
−
=
−
R

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