9702 s15 ms_all

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CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the May/June 2015 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 2015 series for most
Cambridge IGCSE®
, Cambridge International A and AS Level components and some
Cambridge O Level components.

Mark Scheme Syllabus Paper
Cambridge International AS/A Level – May/June 2015 9702 11
© Cambridge International Examinations 2015
Question
Number
Key
Question
Number
Key
1 C 21 C
2 B 22 A
3 A 23 B
4 D 24 D
5 A 25 B
6 B 26 D
7 D 27 A
8 C 28 B
9 D 29 D
10 D 30 B
11 C 31 D
12 C 32 B
13 A 33 B
14 C 34 D
15 C 35 C
16 C 36 D
17 C 37 A
18 B 38 C
19 D 39 C
20 A 40 B

9702 PHYSICS
Cambridge IGCSE®

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

9702 PHYSICS
Cambridge IGCSE®

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

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.
Cambridge IGCSE®

1 (a) power = work/time or energy/time or (force × distance)/time B1
= kgms–2
× ms–1
= kgm2
s–3
A1 [2]
(b) power = VI [or V2
/R and V = IR orI2
R and V = IR] B1
(units of V:) kgm2
s–3
A–1
B1 [2]
2 (a) speed = distance/time and velocity = displacement/time B1
speed is a scalar as distance has no direction and
velocity is a vector as displacement has direction B1 [2]
(b) (i) constant acceleration or linear/uniform increase in velocity until 1.1s B1
rebounds or bounces or changes direction B1
decelerates to zero velocity at the same acceleration as initial value B1 [3]
(ii) a = (v – u)/t or use of gradient implied C1
= (8.8 + 8.8)/1.8 or appropriate values from line or = (8.6 + 8.6)/1.8 B1
= 9.8 (9.78)ms–2
or = 9.6ms–2
A1 [3]
(iii) 1. distance = first area above graph + second area below graph C1
= (1.1 × 10.8)/2 + (0.9 × 8.8)/2 (= 5.94 + 3.96) C1
= 9.9m A1 [3]
2. displacement = first area above graph – second area below graph C1
= (1.1 × 10.8)/2 – (0.9 × 8.8)/2
= 2.0 (1.98)m A1 [2]
(iv) correct shape with straight lines and all lines above the time axis or all below M1
correct times for zero speeds (0.0, 1.15s, 2.1s) and peak speeds
(10.8ms–1
at 1.1s and 8.8ms–1
at 1.2s and 3.0s) A1 [2]
3 (a) 4.5 × 50 – 2.8 × M ( = ...) C1
(...) = –1.8 × 50 + 1.4 × M C1
(M = ) 75g A1 [3]

(b) total initial kinetic energy/KE not equal to the total final kinetic energy/KE
or relative speed of approach is not equal to relative speed of separation
so not elastic or is inelastic B1 [1]
(c) force on X is equal and opposite to force on Y (Newton III) M1
force equals/is proportional to rate of change of momentum (Newton II) M1
time of collision same for both balls hence change in momentum is the same A1 [3]
4 (a) (i) two sets of co-ordinates taken to determine a constant value (F/x) M1
F/x constant hence obeys Hooke’s law A1 [2]
or
gradient calculated and one point on line used (M1)
to show no intercept hence obeys Hooke’s law (A1)
(ii) gradient or one point on line used e.g. 4.5/1.8 × 10–2
C1
(k =) 250Nm–1
A1 [2]
(iii) work done or EP = area under graph or ½Fx or ½kx2
C1
= 0.5 × 4.5 × 1.8 × 10–2
or 0.5 × 250 × (1.8 × 10–2
)2
C1
= 0.041 (0.0405)J A1 [3]
(b) KE = ½mv2
½mv2
= 0.0405 or KE = 0.0405 (J) C1
(v = [2 × 0.0405/1.7]1/2
=) 0.22 (0.218)ms–1
A1 [2]
5 (a) very high/infinite resistance for negative voltages up to about 0.4V B1
resistance decreases from 0.4V B1 [2]
(b) initial straight line from (0,0) into curve with decreasing gradient but not to
horizontal M1
repeated in negative quadrant A1 [2]
(c) (i) R = 122
/36 = 4.0Ω A1
or
I = P/V = 36/12 = 3.0 A and R = 12/3.0 = 4.0Ω (A1) [1]

(ii) lost volts = 0.5 × 2.8 = 1.4(V) or E = 12 = 2.8 × (R + r) C1
R = V/I = (12 – 1.4)/2.8 or (R + r) = 4.29 Ω C1
= 3.8 (3.79)Ω or R = 3.8Ω A1 [3]
(d) resistance of the lamp increases with increase of V or I B1 [1]
6 (a) diffraction is the spreading of a wave as it passes through a slit or past an edge B1
when two (or more) waves superpose/meet/overlap M1
resultant displacement is the sum of the displacement of each wave A1 [3]
(b) nλ = d sinθ and v = fλ C1
max order number for θ = 90°
hence n (= f/vN) = 7.06 × 1014
/(3 × 108
× 650 × 103
) M1
n = 3.6
hence number of orders = 3 A1 [3]
(c) greater wavelength so fewer orders seen A1 [1]
7 (a) a region/space/area where a (stationary) charge experiences an (electric) force B1 [1]
(b) (i) at least four parallel equally spaced straight lines perpendicular to plates B1
consistent direction of an arrow on line(s) from left to right B1 [2]
(ii) electric field strength E = V/d C1
E = (450/16 × 10–3
)
= 28 × 103
(28125)Vm–1
A1 [2]
(iii) W = Eqd or Vq C1
q = 3.2 × 10–19
(C) C1
W = 28125 × 3.2 × 10–19
× 16 × 10–3
or 450 × 3.2 × 10–19
= 1.4(4) × 10–16
J A1 [3]
(iv) 19
19
101.6450
103.2450
ratio −
−
×−×
××
= (evidence of working required)
= (–) 2 A1 [1]

9702 PHYSICS
Cambridge IGCSE®

1 (a) (work =) force × distance or force × displacement or (W =) F × d M1
units of work: kgms–2
× m = kgm2
s–2
A1 [2]
(b)
charge
forms)othertoelectrical(fromed)(transformenergyor(done)work
)(p.d. = B1 [1]
(c) R = V/I B1
units of V: kgm2
s–2
/As and units of I: A C1
or
R = P/I2
[or P = VI and V = IR] (B1)
units of P: kgm2
s–3
and units of I: A (C1)
or
R = V2
/P (B1)
units of V: kgm2
s–2
/As and units of P: kgm2
s–3
(C1)
units of R: (kgm2
s–2
/A2
s =)kgm2
s–3
A–2
A1 [3]
2 (a) speed decreases/stone decelerates to rest/zero at 1.25s B1
speed then increases/stone accelerates (in opposite direction) B1 [2]
(b) (i) v = u + at (or s = ut + ½at2
and v2
= u2
+ 2as) C1
= 0 + (3.00 – 1.25) × 9.81 C1
= 17.2 (17.17)ms–1
A1 [3]
(ii) s = ut + ½at2
s = ½ × 9.81 × (1.25)2
[= 7.66] C1
s = ½ × 9.81 × (1.75)2
[= 15.02] C1
(distance = 7.66 + 15.02)
[v = u + at = 0 + 9.81 × (2.50 – 1.25) = 12.26ms–1
]
or
s = ½ × 9.81 × (1.25)2
[= 7.66] (C1)
s = 12.26 × 0.50 + ½ × 9.81 × (3.00 – 2.50)2
[= 7.36] (C1)
(distance = 2 × 7.66 + 7.36)
Example alternative method:
s = (v2
– u2
)/2a = (12.262
– 0)/2 × 9.81 [= 7.66] (C1)
s = (v2
– u2
)/2a = (17.172
– 12.262
)/2 × 9.81 [= 7.36] (C1)
(distance = 2 × 7.66 + 7.36)

22.7 (22.69 or 23)m A1 [3]
(iii) (s = 15.02 – 7.66 =) 7.4 (7.36)m (ignore sign in answer) A1
down A1 [2]
(c) straight line from positive value of v to t axis M1
same straight line crosses t axis at t = 1.25s A1
same straight line continues with same gradient to t = 3.0s A1 [3]
3 (a) (i) (vertical component = 44 sin 30° =) 22N A1 [1]
(ii) (horizontal component = 44 cos 30° =) 38(.1)N A1 [1]
(b) W × 0.64 = 22 × 1.60 C1
(W =) 55N A1 [2]
(c) F has a horizontal component (not balanced by W)
or F has 38N acting horizontally
or 38N acts on wall
or vertical component of F does not balance W
or F and W do not make a closed triangle of forces B1 [1]
(d) line from P in direction towards point on wire vertically above W and direction up B1 [1]
4 (a) (p =) mv C1
∆p (= – 6.64 × 10–27
× 1250 – 6.64 × 10–27
× 1250) = 1.66 × 10–23
Ns A1 [2]
(b) (i) molecule collides with wall/container and there is a change in momentum B1
change in momentum / time is force or ∆p = Ft B1
many/all/sum of molecular collisions over surface/area of container produces
pressure B1 [3]
(ii) more collisions per unit time so greater pressure B1 [1]
5 (a) curved line showing decreasing gradient with temperature rise M1
smooth line not touching temperature axis, not horizontal or vertical anywhere A1 [2]
(b) (i) (no energy lost in battery because) no/negligible internal resistance B1 [1]

(ii) I = V/R
= 8/15 × 103
or 1.6/3.0 × 103
or 2.4/4.5 × 103
or 12/22.5 × 103
C1
= 0.53 × 10–3
A A1 [2]
(iii) p.d. across X = 12 – 8.0 – 3.0 × 103
× 0.53 × 10–3
(= 2.4V) C1
RX = 2.4/(0.53 × 10–3
) C1
or
Rtot = 12/0.53 × 10–3
(= 22.5 × 103
Ω) (C1)
RX = (22.5 – 15.0 – 3.0) × 103
(C1)
4.5(2) × 103
Ω A1 [3]
(iv) resistance decreases hence current (in circuit) is greater M1
p.d. across X and Y is greater hence p.d across Z decreases A1
or explanation in terms of potential divider:
RZ decreases so RZ /(RX + RY + RZ) is less (M1)
therefore p.d. across Z decreases (A1) [2]
6 (a) progressive waves transfer/propagate energy and stationary waves do not B1
amplitude constant for progressive wave and varies (from max/antinode to
min/zero/node) for stationary wave B1
adjacent particles in phase for stationary wave and out of phase for progressive
wave (B1) [2]
(b) (i) wave/microwave from source/S reflects at reflector/R B1
reflected and (further) incident waves overlap/meet/superpose B1
waves have same frequency/wavelength/period and speed (so stationary
waves formed) B1 [3]
(ii) detector/D is moved between reflector/R and source/S (or v.v.) B1
maximum, minimum/zero, (maximum… etc.) observed on
meter/deflections/readings/measurements/recordings B1 [2]
(iii) determine/measure the distance between adjacent minima/nodes or
maxima/antinodes or across specific number of nodes/antinodes B1
wavelength is twice distance between adjacent nodes/minima or maxima/
antinodes (or other correct method of calculation of wavelength from
measurement) B1 [2]

(c) v = fλ C1
f = 3.0 × 108
/(2.8 × 10–2
) [= 1.07 × 1010
Hz] C1
11 (10.7)GHz A1 [3]
7 (a) 92 protons and 143 neutrons B1 [1]
(b)
value
a 1
b 0 (a and b both required) B1
c 141 B1
d 55 B1 [3]
(c) kinetic energy (of products) or gamma/γ (radiation or photon) B1 [1]
(d) (total) mass on left-hand side/reactants is greater than (total) mass on right-hand
side/products M1
difference in mass is (converted to) energy A1 [2]

9702 PHYSICS
Cambridge IGCSE®

1 (a) 150 or 1.5 × 102
Gm A1 [1]
(b) distance = 2 × (42.3 – 6.38) × 106
(= 7.184 × 107
m) C1
(time =) 7.184 × 107
/ (3.0 × 108
) = 0.24 (0.239)s A1 [2]
(c) units of pressure P: kgms–2
/m2
= kgm–1
s–2
M1
units of density ρ: kgm–3
and speed v: ms–1
M1
simplification for units of C: C = v2
ρ/P units: (m2
s–2
kgm–3
)/kgm–1
s–2
and cancelling to give no units for C A1 [3]
(d) energy and power (both underlined and no others) A1 [1]
(e) (i) vector triangle of correct orientation M1
three arrows for the velocities in the correct directions A1 [2]
(ii) length measured from scale diagram 5.2 ± 0.2cm or components of
boat speed determined parallel and perpendicular to river flow C1
velocity = 2.6 ms–1
(allow ± 0.1ms–1
) A1 [2]
2 (a) constant rate of increase in velocity/acceleration from t = 0 to t = 8s B1
constant deceleration from t = 8s to t = 16s or constant rate of increase in
velocity in the opposite direction from t = 10s to t = 16s B1 [2]
(b) (i) area under lines to 10s C1
(displacement =) (5.0 × 8.0) / 2 + (5.0 × 2.0) / 2 = 25m
or ½ (10.0 × 5.0) = 25m A1 [2]
(ii) a = (v – u)/t or gradient of line C1
= (–15.0 –5.0) / 8.0
= (–) 2.5ms–2
A1 [2]
(iii) KE = ½mv2
C1
= 0.5 × 0.4 × (15.0)2
= 45J A1 [2]
(c) (distance =) 25(m) (= ut + ½at2
) = 0 + ½ × 2.5 × t2
C1
(t = 4.5 (4.47)s therefore) time to return = 14.5s A1 [2]

3 (a) (power =) work done / time (taken) or rate of work done A1 [1]
(b) (i) F – R = ma C1
F = 1500 × 0.82 + 1200 C1
= 2400 (2430)N A1 [3]
(ii) P = Fv C1
= (2430 × 22) = 53000 (53500) W A1 [2]
(c) (there is maximum power from car and) resistive force = force produced by
car hence no acceleration
or
suggestion in terms of power produced by car and power
wasted to overcome resistive force B1 [1]
4 (a) (i) diameter and extension: micrometer (screw gauge) or digital calipers B1
length: tape measure or metre rule B1
load: spring balance or Newton meter B1 [3]
(ii) to reduce the effect of random errors or to plot a graph to check for zero
error in measurement of extension or to see if limit of proportionality is
exceeded B1 [1]
(b) plot a graph of F against e and determine the gradient B1
E = (gradient × l)/[πd2
/4] B1 [2]
5 (a) R = ρl / A C1
= (5.1 × 10−7
× 0.50) / π(0.18 × 10−3
)2
= 2.5 (2.51)Ω M1 [2]
(b) (i) resistance of CD = 8 × resistance of AB = 20(Ω) C1
circuit resistance = [1/5.0 + 1/20]−1
= 4.0(Ω) C1
current = V/R = 6.0/4.0 C1
= 1.5A A1 [4]
(ii) power in AB = I2
R or power = V2
/R C1
= (1.2)2
× 2.5 = 3.6W = (3.0)2
/2.5 = 3.6W A1 [2]

(iii) potential drop A to M = 1.25 × 1.2 = 1.5V M1
potential drop C to N = 3.0V
p.d. MN = 1.5V A1 [2]
6 (a) (i) coherent: constant phase difference B1
interference is the (overlapping of waves and the) sum of/addition of
displacement of two waves B1 [2]
(ii) wavelength = 3.2m (allow ± 0.05m) M1
f (= v/λ = 240 / 3.2) = 75Hz A1 [2]
(iii) 90° (allow ± 2°) or π/2 rad A1 [1]
(iv) sketch has amplitude 3.0 ± 0.1cm M1
correct displacement values at previous peaks to produce correct shape A1 [2]
(b) (i) λ = ax/D C1
x = (546 × 10–9
× 0.85) / 0.13 × 10–3
(= 3.57 × 10–3
m) C1
AB = 8.9 (8.93) × 10–3
m A1 [3]
(ii) shorter wavelength for blue light so separation is less B1 [1]
7 (a) (i) (rate of decay) not affected by any external factors or changes in
temperature and pressure etc. B1 [1]
(ii) two protons and two neutrons B1 [1]
(b) (i) (total) mass before decay/on left-hand side is greater than (total) mass M1
on right-hand side/after the decay
the difference in mass is released as kinetic energy of the products A1 [2]
(may also be some γ radiation) (to conserve mass-energy)
(ii) (6.2 × 106
× 1.6 × 10−19
=) 9.9(2) × 10−13
J A1 [1]

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

1 (a) (ii) Value of w with unit, in range 45.0 cm to 55.0 cm. [1]
(c) (iii) Value of IB, with unit, to nearest 0.1mA, in range 70.0 ≤ IB ≤ 100.0mA. [1]
(d) Six sets of readings of w, IA and IB, different values, scores 5 marks,
five sets scores 4 marks, etc. [5]
Incorrect trend –1.
Major help from Supervisor –2. Minor help from Supervisor –1.
Range: [1]
Range of w ≥ 60.0cm.
Column headings: [1]
Each column heading must contain a quantity and a unit.
The presentation of quantity and unit must conform to accepted scientific
convention e.g. w/cm, w (cm), (IA+IB)/(IAIB)/A–1
, (IA+IB)/(IAIB)/(1/A).
Do not allow (IA+IB)/(IAIB)/(A/A2
).
Consistency: [1]
All values of w must be given to the nearest mm only.
Significant figures: [1]
Every value of (IA+IB)/(IAIB) must be given to the same number of s.f. as (or one
more than) the least s.f. in the corresponding values of IA and IB.
Calculated values: [1]
Values of (IA+IB)/(IAIB) calculated correctly to the number of significant figures
given by the candidate.
(e) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
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 in the table must be plotted.
Diameter of points must be ≤ half a small square (no “blobs”).
Plotted points must be accurate to within half a small square.
Quality: [1]
All points in the table must be plotted on the grid (at least 5) for this mark to
be awarded.
All points must be within ± 5cm (± 0.05m) on the w-axis from a straight line.
(ii) Line of best fit: [1]
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 along the
full length.
Allow one anomalous point only if clearly indicated by the candidate.
Lines must not be kinked or thicker than half a square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the
drawn line.
The method of calculation must be correct.
Both read-offs must be accurate to half a small square in both the x and y directions.
y-intercept: [1]
Either:
Correct read-offs 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
(accurate to half a small square).
(f) M = value of the candidate’s gradient and N = value of the candidate’s y-intercept. [1]
Do not allow substitution methods. Do not allow fractions.
Unit for M correct (e.g. A–1
m–1
or A–1
cm–1
or A–1
mm–1
or mA–1
m–1
or mA–1
cm–1
or
mA–1
mm–1
)
and unit for N correct (e.g. mA–1
or A–1
). [1]
2 (a) (i) Value of L with unit, in range 55.0cm ≤ L ≤ 65.0cm. [1]
(ii) Value of m to nearest gram or better, in range 10.0 g ≤ m ≤ 100.0g. [1]
(iv) Correct justification of significant figures in p linked to significant figures in L and m. [1]
(b) (i) Value of M to the nearest gram or better, in range 90.0g ≤ M ≤ 110.0g. [1]
(iii) Correct calculation of C. [1]
(c) (ii) Value of x to the nearest mm, with unit, in range 5.0cm ≤ x ≤ 20.0cm. [1]
(iii) Absolute uncertainty in x in range 2 – 5mm.
If repeated readings have been taken, then the uncertainty can be half the range
(but not zero) if the working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
(d) Second value of L. [1]
Second value of x. [1]
Correct trend for x with respect to L (x decreases as L decreases). [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Valid comment consistent with 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 valid conclusion.
Take more readings (for
different L) and plot a graph/
take more readings and
compare k values.
“repeat readings”/
“too few readings”
B Difficult to measure x with
reason, e.g. parallax/ruler not
in line with wood/strip moves
as touched while taking
measurement/mass obscures
end of rule/strip
oscillates/balance achieved for
a short time
Improved method to measure x
e.g. attach mass to bottom of
strip/mark scale on strip/mark
strip at balance point/measure
(L–x)/clamp ruler horizontally
Travelling microscope
Video
C Difficult to balance with reason,
e.g. wind/air conditioning or
pivot moves
Method to remove wind, e.g.
turn off fans/close windows or
method of fixing pivot to bench
i.e. tape/heavier pivot
Sliding rule
Pivot size
D Problem with Blu-Tack, e.g.
mass of Blu-Tack not taken
into account
Method to overcome problem
with Blu-Tack, e.g. measure
mass of Blu-Tack and add to
value of M or fix mass with
named adhesive, e.g.
tape/glue because this has
less mass
E Difficult to know where centre
of mass is with reason, e.g.
slot in mass
or
Difficult to place centre of mass
at end of strip
Detailed method of finding
centre of mass
Method to attach mass on strip
to ensure centre of mass is at
the end of strip, e.g. hang
mass from strip with thread
Mark centre of mass
Measure diameter
F Two strips have different
density/p
Find mass or p of second strip Different thickness/cross-
sectional area

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (c) (ii) Value of h in the range 45.0 to 55.0cm. [1]
(iii) Value of x less than 50.0cm. [1]
(d) Six sets of readings of x and h scores 5 marks, five sets scores 4 marks etc. [5]
Incorrect trend –1. Help from Supervisor –1.
Range: [1]
xmax – xmin ≥ 60.0cm.
Each column heading must contain a quantity and a unit where appropriate. The
presentation of quantity and unit must conform to accepted scientific convention.
e.g. 1/h/cm–1
. x/h must have no unit.
Consistency: [1]
All values of h and all values of x must be given to the nearest mm.
Every value of x/h must be given to the same number of s.f. (or one more than) the least
number of s.f. in the corresponding values of x and h as recorded in table.
Calculation: [1]
Values of x/h calculated correctly.
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph
grid in both x and y directions.
Plotting: [1]
All observations must be plotted on the grid. Diameter of plotted points must
be ≤ half a small square (no “blobs”).
Plotted points must be accurate to within half a small square in both x and y
directions.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
Scatter of points must be no more than ± 0.1 from a straight line in the x/h
direction.
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 (i.e. circled or labelled) by the
candidate.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the drawn line.
y-intercept: [1]
Either:
Correct read-offs from a point on the line and substituted into y = mx + c or an
equivalent expression.
Both read-offs accurate to half a small square in both the x and y directions.
Or:
Intercept read directly from the graph, with read-off at x = 0 accurate to half a small
square in y direction.
(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]
Units for a and b both correct and consistent with values. [1]
2 (a) (ii) All values of D to nearest 0.1cm and in range 2.0cm to 4.0cm. [1]
Evidence of repeat readings of D. [1]
(iii) Absolute uncertainty in D in range 0.2 to 0.5cm and correct method of calculation
to obtain percentage uncertainty. If repeated readings have been taken, then the
absolute uncertainty can be half the range (but not zero) if the working is clearly
shown. [1]
(iv) Correct calculation of C with consistent unit. [1]
(b) Justification for significant figures in C linked to significant figures in D only. [1]
(d) (ii) r1 in range 5.0cm to 25.0cm, with unit, to nearest mm. [1]
(v) r2 in range 5.0cm to 25.0cm. [1]
(e) Second value of D. [1]
Second values of r1 and r2. [1]
Second value of |r1 – r2| > first value of |r1 – r2|. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Sensible 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 and plot a
graph/
obtain more k values and
compare
“few readings”/
only one reading/
take more readings
and (calculate)
average k
B Difficult to measure D (or there is
uncertainty in D or C) because loop
is not circular/not flat/deforms
Workable method of making a
more circular loop, e.g. wrap
loop around tube
Use micrometer
Use vernier calipers
Material weak
Material flexible
C Parallax error with pointer/
pointer moves away from scale/
pointer (or spring) vibrates
Use shadow method
D Ruler not vertical Use set square to ensure ruler
vertical/clamp ruler
E Difficult to judge reading when loop
breaks away/
loop breaks away suddenly
Video with scale/
use maximum marker
Slow motion camera
High speed camera
Difficult to determine
point (or moment)
loop breaks away
F Difficult to lower beaker steadily Use adjustable-height stand
G Reading affected by contact
between loop and beaker/
impurities in water
Use larger diameter container/
wider container
Use distilled water
Larger beaker

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (b) (iv) Value of y in the range 10.0cm to 11.0cm with unit. [1]
(c) (ii) Value of y > value in (b)(iv). [1]
(d) Six sets of readings of m and y scores 5 marks, five sets scores 4 marks etc. [5]
Help from Supervisor –1.
Range: [1]
Range of m to include m = 0g and m = 50 g or 60g.
Each column heading must contain a quantity and a unit where appropriate.
The unit must conform to accepted scientific convention e.g. y(C+m)/cmg.
Consistency: [1]
All values of raw y must be given to the nearest mm.
Every value of value of y(C+m) must be given to the same number of s.f. as (or one
more than) the least s.f. in the corresponding values of y, C and m as stated in the
candidate’s table and (a)(ii).
Calculation: [1]
Values of y(C+m) calculated correctly to the number of s.f. given by the candidate.
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph grid
in both x and y directions.
Plotting: [1]
All observations must be plotted.
Diameter of plotted points must be ≤ half a small square (no “blobs”).
Quality: [1]
All points in the table (at least 5) must be plotted on the grid for this mark to be
awarded.
All points must be within ±40gcm of a straight line in the y(C+m) direction.
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.

(iii) Gradient: [1]
y-intercept: [1]
Either:
Correct read-offs from a point on the line and substituted into y = mx + c.
Read-offs 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
(accurate to half a small square).
(f) Value of A = candidate’s gradient and value of B = 2 × candidate’s intercept / A. [1]
Do not allow fractions or final answer to 1 s.f.
Unit for A (m, cm or mm) and B (g or kg) correct and consistent with value, with
correct power of ten. [1]
2 (a) (i) Raw values of d and D to nearest 0.1mm and with consistent SI unit, in ranges:
10.0 mm ≤=d ≤ 25.0 mm
20.0 mm ≤ D ≤ 40.0 mm. [1]
(ii) Value of h with consistent unit in range 40.0 mm ≤ h ≤ 60.0 mm. [1]
(iii) Percentage uncertainty in d based on absolute uncertainty of 0.1 or 0.2 mm.
If repeated readings have been taken, then the uncertainty can be half the range
(but not zero) if the working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
(b) (iii) Correct calculation of x. Answer must be correct when rounded to 2 s.f. [1]
(c) Correct justification of s.f. in x linked to s.f. in D, d and h. [1]
(d) (ii) Value of average t ≥ 0.5s with unit. [1]
Evidence of repeated readings (here or in (e)). [1]
(e) Second value of x. [1]
Second value of t. [1]
Second value of t < first value of t. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Sensible 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 not enough to draw
a valid conclusion
Take many readings and plot
a graph/
take more readings and
compare k values.
“too few readings”
B Difficulty in release of cylinder
from same position every time
with reason, e.g. placing fingers in
water, level of water surface
changing, difficult to judge start
point
Better method of holding and
releasing cylinder e.g. stop
gate/
use mark to ensure the water
level is the same for each
release
Clamps
Force on release
C Cylinder doesn’t always fall
vertically (i.e. path at an angle or
cylinder tilted)/
hits sides on descent
Method of attaching string
symmetrically/
method of symmetrical
distribution of mass
e.g. use glass beads or sand/
modelling clay distributed
evenly/
glue marbles in symmetrically
Ignore string effects
Sand on its own
Narrow cylinder
D Times short/
large uncertainty in time
Use longer tube/
video with timer (or video and
view frame by frame)
Reaction time is short
“too fast/quick”
High speed camera
Light gate(s)
Slow motion camera
Terminal velocity
E Difficulty in identifying end point
with reason e.g. refraction, glass
curvature, tray in the way, bottom
of cylinder not flat
Method to identify end point
e.g. time to a mark on
cylinder/listening to impact
Flat bottomed cylinder
Sensors
F Limited number of marbles to fit in
container/
different diameter marbles/
bubbles or air in container/
container deforms when
measuring D
Use different shapes e.g.
cubes/smaller spheres to
occupy more space/
use sand/modelling clay to fill
more space/
measure and account for
different diameter of marbles
in equation for x
Sand on its own without
explanation

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (b) (i) Value of r in the range 28.0cm to 32.0cm, with unit. [1]
(c) (ii) Value of T in range 2.0s to 4.0s. If out of range, allow Supervisor’s value ±20%. [1]
Evidence of repeat measurements for T. [1]
(d) Six sets of readings of r and T scores 4 marks, five sets scores 3 marks etc. [4]
Incorrect trend –1. Help from Supervisor –1.
Range: [1]
rmax – rmin ≥ 30cm.
Each column heading must contain a quantity and a unit. The presentation of
quantity and unit must conform to accepted scientific convention e.g. r2
/m2
.
Consistency: [1]
All values of r must be given to the nearest mm.
The number of significant figures for every value of T3
must be the same as, or
one more than, the number of significant figures in the corresponding time.
Calculation: [1]
Values of T3
calculated correctly to the number of significant figures given by the
candidate.
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the
graph grid in both x and y directions.
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”).
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
All points must be within ± 2s3
of a straight line in the T3
direction.
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 (i.e. circled or labelled) by the
candidate.

(iii) Gradient: [1]
y-intercept: [1]
Either:
Correct read-offs from a point on the line substituted into y = mx + c or an
equivalent expression.
Read-offs must be accurate to half a small square in both x and y directions.
Or:
Intercept read directly from the graph, with read-off accurate to half a small square.
(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]
Units for a and b are correct (e.g. s3
m–2
for a ands3
for b). [1]
2 (a) (ii) Value for t in range 0.10cm to 0.90cm and given to nearest 0.01cm. [1]
Value for D in range 3.0cm to 6.0cm. [1]
Value for h less than t. [1]
(b) Correct calculation of R. [1]
Value of R given to 2 or 3 significant figures. [1]
(c) (ii) Value for f in range 13.0cm to 17.0cm or 28.0 to 32.0cm. [1]
(iii) Absolute uncertainty in f in range 0.2cm to 0.5cm and correct method of
calculation to obtain percentage uncertainty. If repeated readings have been
taken, then the absolute uncertainty can be half the range (but not zero) if
the working is clearly shown. [1]
(d) Second values for t, D and h. [1]
Second value for f. [1]
(e) (i) Two values of k calculated correctly. [1]
Quality: Both k values in range 0.50 to 1.50. [1]
(ii) Sensible 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 are not enough to
draw a valid conclusion
Take more readings and plot a
graph/
obtain more k values and
compare
“few readings”/
only one reading/
take more readings
and (calculate)
average k
B Reason for difficulty in measuring t,
h or D e.g. jaws of calipers slip off
ends of lens/jaws too short and
cannot reach centre of lens
Use a travelling microscope References to
parallax
C h is small/large uncertainty in h Use micrometer/travelling
microscope
D Difficult to obtain sharp image/hard
to focus/blurred image
Use a dark(ened) room/
turn off lights/
use point/more compact source
of light
E Difficult to measure f/take
measurement with ruler/measure
distance, with reason e.g. difficult
to keep lens steady/screen not
vertical/lens not vertical/ruler not
perpendicular to lens or screen
Mount lens in holder/clamp/
fix lens to bench with e.g.
Blu-Tack/
use optical bench
Flexible/bendy
screens

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

Section A
1 (a) (gravitational) force proportional to product of masses and inversely proportional
to square of separation M1
reference to either point masses or particles or ‘size’ much less than separation A1 [2]
(b) gravitational force provides/is the centripetal force B1
GMNm/r2
= mrω2
(or mv2
/r) M1
2π/T (or v = 2πr/T) leading to GMN = 4π2
r3
/T2
A1 [3]
(c) MN /MU = (3.55/5.83)3
× (13.5/5.9)2
x3
factor correct C1
T2
factor correct C1
ratio = 1.18 (allow 1.2) A1
alternative method: mass of Neptune = 1.019 × 1026
kg (C1)
mass of Uranus = 8.621 × 1025
kg (C1)
ratio = 1.18 (A1) [3]
2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1
mention of random motion/distribution A1 [2]
(b) (i) pV = nRT
either at A, 1.2 × 105
× 4.0 × 10−3
= n × 8.31 × 290
or at B, 3.6 × 105
× 4.0 × 10−3
= n × 8.31 × 870 C1
n = 0.20mol A1 [2]
(ii) 1.2 × 105
× 7.75 × 10–3
= 0.20 × 8.31 × T or T = (7.75/4.0) × 290 C1
T = 560 K A1 [2]
(Allow tolerance from graph: 7.7–7.8 × 10–3
m3
)
(c) temperature changes/decreases so internal energy changes/decreases B1
volume changes (at constant pressure) so work is done B1 [2]
3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of
unit mass M1
at constant temperature A1 [2]
(allow 1/2 for definition restricted to fusion or vaporisation)
(b) (i) at 70W, mass s–1
= 0.26gs–1
A1
at 110W, mass s–1
= 0.38gs–1
A1 [2]

(ii) 1. P + h = mL or substitution of one set of values C1
(110 – 70) = (0.38 – 0.26)L C1
L = 330Jg–1
A1 [3]
2. either 70 + h = 0.26 × 330
or 110 + h = 0.38 × 330 C1
h = 17/16/15W A1 [2]
4 (a) (i) frequency at which object is made to vibrate/oscillate B1 [1]
(ii) frequency at which object vibrates when free to do so B1 [1]
(iii) maximum amplitude of vibration of oscillating body B1
when forced frequency equals natural frequency (of vibration) B1 [2]
(b) e.g. vibration of quartz/piezoelectric crystal (what is vibrating) M1
either for accurate timing
or maximise amplitude of ultrasound waves (why it is useful) A1 [2]
(c) e.g. vibrating metal panels (what is vibrating) M1
either place strengthening struts across the panel
or change shape/area of panel (how it is reduced) A1 [2]
5 (a) (magnitude of electric field strength is the potential gradient B1
use of gradient at x = 4.0cm M1
gradient = 4.5 × 104
NC–1
(allow ± 0.3 × 104
) A1
or
x
Q
V
04πε
= and 2
04 x
Q
E
πε
= leading to
x
V
E = (B1)
E = 1.8 × 103
/0.04 (M1)
= 4.5 × 104
NC–1
(A1) [3]
(b) (i) 3.6 × 103
V A1 [1]
(ii) capacitance = Q/V C1
= (8.0 × 10–9
)/(3.6 × 103
)
= 2.2 × 10–12
F A1 [2]
6 (a) (i) gravitational B1 [1]
(ii) gravitational and electric B1 [1]
(iii) magnetic and one other field given B1
magnetic, graviational and electric B1 [2]

(b) (i) out of (plane of) paper/page (not “upwards”) B1 [1]
(ii) B = mv/qr C1
= (3.32 × 10–26
× 7.6 × 104
)/(1.6 × 10–19
× 6.1 × 10–2
) C1
= 0.26T A1 [3]
(c) sketch: semicircle with diameter < 12.2cm B1 [1]
7 (a) can change (output) voltage efficiently or to suit different consumers/appliances B1
by using transformers B1 [2]
(b) for same power, current is smaller B1
less heating in cables/wires
or thinner cables possible
or less voltage loss in cables B1 [2]
8 (a) (i) p = h/λ
= (6.63 × 10–34
)/(6.50 × 10–12
) C1
= 1.02 × 10–22
Ns A1 [2]
(ii) E = hc/λ or E = pc
= (6.63 × 10–34
× 3.00 × 108
)/(6.50 × 10–12
) C1
= 3.06 × 10–14
J A1 [2]
(b) (i) 0.34 × 10–12
= (6.63 × 10–34
)/(9.11 × 10–31
× 3.0 × 108
) × (1 – cos θ) C1
θ = 30.7° A1 [2]
(ii) deflected electron has energy M1
this energy is derived from the incident photon A1
deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3]
9 (a) nucleus/nuclei emits M1
spontaneously/randomly A1
α-particles, β-particles, γ-ray photons A1 [3]
(b) (i) N – ∆N A1 [1]
(ii) ∆N/∆t A1 [1]
(iii) ∆N/N A1 [1]
(iv) ∆N/N∆t A1 [1]
(c) graph: smooth curve in correct direction starting at (0,0) M1
n at 2t½ is 1.5 times that at t½ (± 2mm) A1 [2]

Section B
10 (a) (i) (potential =) 1.2/(1.2 + 4.2) × 4.5 = +1.0V A1 [1]
(ii) (for VIN > 1.0V) V+
> V–
B1
output (of op-amp) is +5V or positive M1
diode conducts giving +5V across R or Vout is +5V A1
(for VIN < 1.0V) output of op-amp –5V/negative so diode does not conduct,
giving Vout = 0 or 0V across R A1 [4]
(b) (i) square wave with maximum value +5 V and minimum value 0 M1
vertical sides in correct positions and correct phase A1 [2]
(ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1]
11 (a) change/increase/decrease anode/tube voltage B1
electrons striking anode have changed (kinetic) energy/speed B1
X-ray/photons/beam have different wavelength/frequency B1 [3]
(b) (i) I = I0 e–µx
B1 [1]
(ii) contrast is difference in degree of blackening (of regions of the image) B1
µ (very) similar so similar absorption of radiation (for same thickness) so little
contrast A1 [2]
12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1]
(ii) television set/audio amplifier etc. B1 [1]
(iii) satellite/satellite dish/mobile phone etc. B1 [1]
(b) e.g. lower attenuation/fewer repeaters
more secure
less prone to noise/interference
physically smaller/less weight
lower cost
greater bandwidth
(any two sensible suggestions, 1 each) B2 [2]
(c) (i) ratio = 25 + (62 × 0.21) C1
= 38dB A1 [2]
(ii) ratio/dB= 10 lg(P2 /P1) C1
38 = 10 lg(P/{9.2 × 10–6
})
P = 58mW or 5.8 × 10–2
W A1 [2]
(allow 1/2 for missing 10 in equation)

13 (a) (i) to align nuclei/protons B1
to cause Larmor/precessional frequency to be in r.f. region B1 [2]
(ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1
knowing field strength enables (region of precessing) nuclei to be located M1
by knowing the frequency A1 [3]
(b) E = 2.82 × 10–26
× B
6.63 × 10–34
× 42 × 106
= 2.82 × 10–26
× B C1
B = 0.99T A1 [2]

9702 PHYSICS
Cambridge IGCSE®

1 (a) (i) 1. F = Gm1m2 /x2
= (6.67 × 10–11
× 2.50 × 5.98 × 1024
)/(6.37 × 106
)2
M1
= 24.6N (accept 2 s.f. or more) A1 [2]
2. F = mxω2
or F = mv2
/x and v = ωx (accept x or r for distance) C1
= 2.50 × 6.37 × 106
× (2π/24 × 3600)2
= 0.0842N (accept 2 s.f. or more) A1 [2]
(ii) reading = 24.575 – 0.0842 B1
= 24.5N (accept only 3 s.f.) A1 [2]
(b) gravitational force provides the centripetal force M1
gravitational force is ‘equal’ to the centripetal force
(accept Gm1m2 /x2
= mxω2
or FC = FG) M1
‘weight’/sensation of weight/contact force/reaction force is difference between FG
and FC which is zero A1 [3]
2 (a) mean speed = 1.44 × 103
ms–1
A1 [1]
(b) evidence of summing of individual squared speeds C1
mean square speed = 2.09 × 106
m2
s–2
A1 [2]
(c) root-mean-square speed = 1.45 × 103
ms–1
A1 [1]
(allow ECF from (b) but only if arithmetic error)
3 (a) (numerically equal to) quantity of heat/(thermal) energy to change state/phase of
unit mass M1
(b) (i) constant gradient/straight line (allow linear/constant slope) B1 [1]
(ii) Pt = mL or power = gradient × L C1
use of gradient of graph
(or two points separated by at least 3.5 minutes) M1
110 × 60 = L × (372 – 325) × 10–3
/7.0
L = 9.80 × 105
Jkg–1
(accept 2 s.f.) (allow 9.8 to 9.9 rounded to 2 s.f.) A1 [3]
(iii) some energy/heat is lost to the surroundings or vapour condenses on sides M1
so value is an overestimate A1 [2]
4 (a) displacement (directly) proportional to acceleration/force M1
either displacement and acceleration in opposite directions
or acceleration (always) towards a (fixed) point A1 [2]

(b) (i) ⅓π rad or 1.05 rad (allow 60° if unit clear) A1 [1]
(ii) a0 = –ω2
x0
= (–) (2π/1.2)2
× 0.030 C1
= (–) 0.82ms–2
A1 [2]
(special case: using oscillator P gives x0 = 1.7cm and a0 = 0.47ms–1
for 1/2)
(iii) max. energy ∝ x0
2
ratio = 3.02
/1.72
C1
= 3.1 (at least 2 s.f.) A1 [2]
(if has inverse ratio but has stated max. energy ∝ x0
2
then allow 1/2)
(c) graph: straight line through (0,0) with negative gradient M1
correct end-points (–3.0, +0.82) and (+3.0, –0.82) A1 [2]
5 (a) work done bringing/moving per unit positive charge M1
from infinity (to the point) A1 [2]
(b) (i) slope/gradient (of the line/graph/tangent) B1 [1]
(allow dV/dx, but not ∆V/∆x or V/x)
(allow potential gradient)
(negative sign not required)
(ii) maximum at surface of sphere A or at x = 0(cm) B1
zero at x = 6(cm) B1
then increases but in opposite direction B1 [3]
(any mention of attraction max. 2/3)
(c) (i) M shown between x = 5.5cm and x = 6.5cm B1 [1]
(ii) 1. ∆V = (570 – 230) = 340V (allow 330V to 340V) A1 [1]
2. q(∆)V = ½mv2
or change/loss in PE = change/gain in KE or ∆EK = ∆EP B1
4.8 × 107
× 340 = ½v2
C1
v2
= 3.26 × 1010
v = 1.8 × 105
ms–1
(not 1 s.f.) A1 [3]
6 (a) packet/quantum/discrete amount of energy M1
of electromagnetic energy/radiation/waves A1 [2]
(b) (i) arrow below axis and pointing to right B1 [1]

(ii) 1. E = hc/λ
= (6.63 × 10–34
× 3.0 × 108
)/(6.80 × 10–12
) C1
= 2.93 × 10–14
J (accept 2 s.f.) A1 [2]
2. energy of electron = (3.06 – 2.93) × 10–14
= 1.3 × 10–15
J C1
speed = )/(2 mE C1
= 5.4 × 107
ms–1
A1 [3]
(c) momentum is a vector quantity B1
either must consider momentum in two directions
or direction changes so cannot just consider magnitude B1 [2]
7 (a) moving magnet gives rise to/causes/induces e.m.f./current in solenoid/coil B1
(induced current) creates field/flux in solenoid that opposes (motion of) magnet B1
work is done/energy is needed to move magnet (into solenoid) B1
(induced) current gives heating effect (in resistor) which comes from the work done B1 [4]
(b) current in primary coil give rise to (magnetic) flux/field B1
(magnetic) flux/field (in core) is in phase with current (in primary coil) B1
(magnetic) flux threads/links/cuts secondary coil inducing e.m.f. in secondary coil B1
(there must be a mention of secondary coil)
e.m.f. induced proportional to rate of change/cutting of flux/field so not in phase B1 [4]
8 (a) (i) energy = 5.75 × 1.6 × 10–13
= 9.2 × 10–13
J A1 [1]
(ii) number = 1900/(9.2 × 10–13
× 0.24) C1
= 8.6 × 1015
s–1
A1 [2]
(b) (i) decay constant = 0.693/(2.8 × 365 × 24 × 3600) C1
= 7.85 × 10–9
s–1
(allow 7.8 or 7.9 to 2 s.f.) A1 [2]
(ii) A = λN
8.6 × 1015
= 7.85 × 10–9
× N C1
N = 1.096 × 1024
C1
mass = (1.096 × 1024
× 236)/(6.02 × 1023
) M1
= 430g A1 [4]
(c) 0.84 = 1.9 exp(–7.85 × 10–9
t) C1
t = 1.04 × 108
s
= 3.3 years A1 [2]

Section B
9 (a) VB = 1000mV C1
when strained, VA = 2000 × 121.5/(121.5 +120.0)
= 1006.2mV M1
change = 6.2mV (allow 6mV) A1 [3]
(b) (i) 1. resistor between VIN and V–
and V+
connected to earth B1
resistor between V–
and VOUT B1 [2]
2. P/+ sign shown on earth side of voltmeter B1 [1]
(ii) ratio of RF /RIN = 40 M1
RIN between 100Ω and 10kΩ A1 [2]
(any values must link to the correct resistors on the diagram)
10 (a) product of density (of medium) and speed (of ultrasound) M1
in the medium A1 [2]
(b) (i) 7.0 × 106
= 1.7 × 103
× speed C1
speed = 4.12 × 103
ms–1
wavelength = (4.12 × 103
)/(9.0 × 105
)m C1
= 4.6mm (2 s.f. minimum) A1 [3]
(ii) for air/tissue boundary, IR /I ≈ 1 M1
for air/tissue boundary, (almost) complete reflection/no transmission A1
for gel/tissue boundary, IR /I = 0.12
/3.12
= 1.04 × 10–3
(accept 1 s.f.) M1
gel enables (almost) complete transmission (into the tissue) A1 [4]
11 (a) (i) metal (allow specific example of a metal) B1 [1]
(ii) e.g. provides ‘return’ for the signal
shields inner core from interference/reduces cross-talk/reduces noise
increased security
(b) (i) (gradual) loss of power/intensity/amplitude B1 [1]
(ii) dB is a log scale B1
either large (range of) numbers are easier to handle (on a log scale)
or compounding attenuations/amplifications is easier B1 [2]
(c) attenuation = 190 × 11 × 10–3
= 2.09dB C1
–2.09 = 10 lg(POUT /PIN) C1
ratio = 0.62 A1 [3]

12 handset transmits (identification) signal to number of base stations B1
base stations transfers (signal) to cellular exchange B1
(idea of stations needed at least once in first two marking points)
computer at cellular exchange selects base station with strongest signal B1
computer at cellular exchange selects a carrier frequency for mobile phone B1 [4]
(idea of computer needed at least once in these two marking points)

9702 PHYSICS
Cambridge IGCSE®

Section A
1 (a) (gravitational) force proportional to product of masses and inversely proportional
to square of separation M1
reference to either point masses or particles or ‘size’ much less than separation A1 [2]
(b) gravitational force provides/is the centripetal force B1
GMNm/r2
= mrω2
(or mv2
/r) M1
2π/T (or v = 2πr/T) leading to GMN = 4π2
r3
/T2
A1 [3]
(c) MN /MU = (3.55/5.83)3
× (13.5/5.9)2
x3
factor correct C1
T2
factor correct C1
ratio = 1.18 (allow 1.2) A1
alternative method: mass of Neptune = 1.019 × 1026
kg (C1)
mass of Uranus = 8.621 × 1025
kg (C1)
ratio = 1.18 (A1) [3]
2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1
mention of random motion/distribution A1 [2]
(b) (i) pV = nRT
either at A, 1.2 × 105
× 4.0 × 10−3
= n × 8.31 × 290
or at B, 3.6 × 105
× 4.0 × 10−3
= n × 8.31 × 870 C1
n = 0.20mol A1 [2]
(ii) 1.2 × 105
× 7.75 × 10–3
= 0.20 × 8.31 × T or T = (7.75/4.0) × 290 C1
T = 560 K A1 [2]
(Allow tolerance from graph: 7.7–7.8 × 10–3
m3
)
(c) temperature changes/decreases so internal energy changes/decreases B1
volume changes (at constant pressure) so work is done B1 [2]
3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of
unit mass M1
(b) (i) at 70W, mass s–1
= 0.26gs–1
A1
at 110W, mass s–1
= 0.38gs–1
A1 [2]

(ii) 1. P + h = mL or substitution of one set of values C1
(110 – 70) = (0.38 – 0.26)L C1
L = 330Jg–1
A1 [3]
2. either 70 + h = 0.26 × 330
or 110 + h = 0.38 × 330 C1
h = 17/16/15W A1 [2]
4 (a) (i) frequency at which object is made to vibrate/oscillate B1 [1]
(ii) frequency at which object vibrates when free to do so B1 [1]
(iii) maximum amplitude of vibration of oscillating body B1
when forced frequency equals natural frequency (of vibration) B1 [2]
(b) e.g. vibration of quartz/piezoelectric crystal (what is vibrating) M1
either for accurate timing
or maximise amplitude of ultrasound waves (why it is useful) A1 [2]
(c) e.g. vibrating metal panels (what is vibrating) M1
either place strengthening struts across the panel
or change shape/area of panel (how it is reduced) A1 [2]
5 (a) (magnitude of electric field strength is the potential gradient B1
use of gradient at x = 4.0cm M1
gradient = 4.5 × 104
NC–1
(allow ± 0.3 × 104
) A1
or
x
Q
V
04πε
= and 2
04 x
Q
E
πε
= leading to
x
V
E = (B1)
E = 1.8 × 103
/0.04 (M1)
= 4.5 × 104
NC–1
(A1) [3]
(b) (i) 3.6 × 103
V A1 [1]
(ii) capacitance = Q/V C1
= (8.0 × 10–9
)/(3.6 × 103
)
= 2.2 × 10–12
F A1 [2]
6 (a) (i) gravitational B1 [1]
(ii) gravitational and electric B1 [1]
(iii) magnetic and one other field given B1
magnetic, graviational and electric B1 [2]

(b) (i) out of (plane of) paper/page (not “upwards”) B1 [1]
(ii) B = mv/qr C1
= (3.32 × 10–26
× 7.6 × 104
)/(1.6 × 10–19
× 6.1 × 10–2
) C1
= 0.26T A1 [3]
(c) sketch: semicircle with diameter < 12.2cm B1 [1]
7 (a) can change (output) voltage efficiently or to suit different consumers/appliances B1
by using transformers B1 [2]
(b) for same power, current is smaller B1
less heating in cables/wires
or thinner cables possible
or less voltage loss in cables B1 [2]
8 (a) (i) p = h/λ
= (6.63 × 10–34
)/(6.50 × 10–12
) C1
= 1.02 × 10–22
Ns A1 [2]
(ii) E = hc/λ or E = pc
= (6.63 × 10–34
× 3.00 × 108
)/(6.50 × 10–12
) C1
= 3.06 × 10–14
J A1 [2]
(b) (i) 0.34 × 10–12
= (6.63 × 10–34
)/(9.11 × 10–31
× 3.0 × 108
) × (1 – cos θ) C1
θ = 30.7° A1 [2]
(ii) deflected electron has energy M1
this energy is derived from the incident photon A1
deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3]
9 (a) nucleus/nuclei emits M1
spontaneously/randomly A1
α-particles, β-particles, γ-ray photons A1 [3]
(b) (i) N – ∆N A1 [1]
(ii) ∆N/∆t A1 [1]
(iii) ∆N/N A1 [1]
(iv) ∆N/N∆t A1 [1]
(c) graph: smooth curve in correct direction starting at (0,0) M1
n at 2t½ is 1.5 times that at t½ (± 2mm) A1 [2]

Section B
10 (a) (i) (potential =) 1.2/(1.2 + 4.2) × 4.5 = +1.0V A1 [1]
(ii) (for VIN > 1.0V) V+
> V–
B1
output (of op-amp) is +5V or positive M1
diode conducts giving +5V across R or Vout is +5V A1
(for VIN < 1.0V) output of op-amp –5V/negative so diode does not conduct,
giving Vout = 0 or 0V across R A1 [4]
(b) (i) square wave with maximum value +5 V and minimum value 0 M1
vertical sides in correct positions and correct phase A1 [2]
(ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1]
11 (a) change/increase/decrease anode/tube voltage B1
electrons striking anode have changed (kinetic) energy/speed B1
X-ray/photons/beam have different wavelength/frequency B1 [3]
(b) (i) I = I0 e–µx
B1 [1]
(ii) contrast is difference in degree of blackening (of regions of the image) B1
µ (very) similar so similar absorption of radiation (for same thickness) so little
contrast A1 [2]
12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1]
(ii) television set/audio amplifier etc. B1 [1]
(iii) satellite/satellite dish/mobile phone etc. B1 [1]
(b) e.g. lower attenuation/fewer repeaters
more secure
less prone to noise/interference
physically smaller/less weight
lower cost
greater bandwidth
(c) (i) ratio = 25 + (62 × 0.21) C1
= 38dB A1 [2]
(ii) ratio/dB= 10 lg(P2 /P1) C1
38 = 10 lg(P/{9.2 × 10–6
})
P = 58mW or 5.8 × 10–2
W A1 [2]
(allow 1/2 for missing 10 in equation)

13 (a) (i) to align nuclei/protons B1
to cause Larmor/precessional frequency to be in r.f. region B1 [2]
(ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1
knowing field strength enables (region of precessing) nuclei to be located M1
by knowing the frequency A1 [3]
(b) E = 2.82 × 10–26
× B
6.63 × 10–34
× 42 × 106
= 2.82 × 10–26
× B C1
B = 0.99T A1 [2]

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

1 Planning (15 marks)
Defining the problem (3 marks)
P V is the independent variable, or vary V and f is the dependent variable, or
measure f.
Or f is the independent variable, or vary f and V is the dependent variable, or
measure V. [1]
P Change f (allow V) until the mass leaves/gap between plate. [1]
P Keep the position of the mass constant. (Do not allow keep mass constant.) [1]
Methods of data collection (5 marks)
M Labelled diagram showing signal generator/a.c. supply connected to vibrator with
two wires with mass on plate. At least two labels needed. [1]
M Voltmeter/c.r.o. connected in parallel with vibrator in a workable circuit. [1]
M Measure f or T from signal generator/c.r.o. (Allow detailed use of motion
sensor/stroboscope.) [1]
M Detail regarding mass leaving the plate: listen to noise, look for gap. [1]
M Repeat each experiment for the same value of V (allow f if consistent with above)
and average. [1]
Method of analysis (2 marks)
Plot a graph of:
A
f2
against
1/V
1/V
against
f2
f
against
1/ V
1/ V
against
f
lg V
against
lg f
lg f
against
lg V
or or or or
V
against
1/f2
1/f2
against
V
V
against
1/f
1/f
against
V
[1]
A
k =
2gradient π× gradient
2
π
=k
k =
22
gradient π× 2
2
gradient
π
=k k = c102 ×π k = c22 10×π [1]
Safety considerations (1 mark)
S Precaution linked to mass leaving vibrating plate, e.g. use safety
screen/goggles/sand tray. [1]

Additional detail (4 marks)
D Relevant points might include [4]
1 Wait for vibrator to oscillate evenly
2 Method to determine period of oscillation from c.r.o., i.e. one time period × time-base
3 Method to determine f from c.r.o. having determined T, i.e. f = 1/T
4 Method to determine V from c.r.o, i.e. amplitude (height) × y-gain
5 Relationship is valid if the graph is a straight line passing through the origin
[For lg – lg graph the gradient must be correct (–2 or –0.5)]
6 Determine f (allow V if consistent with above) by increasing and decreasing V or f
7 Clean surfaces of metal plate/small mass
8 Spirit level to keep plate horizontal/eye level to look for gap
Do not allow vague computer methods.

2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a) A1 gradient = m
y-intercept = lg k
(b) T1
T2 1.70 or 1.699 1.312 or 1.3118
1.79 or 1.785 1.204 or 1.2041
1.85 or 1.851 1.114 or 1.1139
1.90 or 1.903 1.041 or 1.0414
1.95 or 1.954 0.98 or 0.978
2.00 or 1.996 0.90 or 0.903
Allow a mixture of significant figures.
T1 (first column) and T2 (second column)
must be values in table.
U1 From ±0.01 to ±0.03 Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Do not allow “blobs”.
Ecf allowed from table.
U2 Error bars in lg P plotted correctly All error bars to be plotted. Must be
accurate to less than half a small square.
(ii) G2 Line of best fit Upper end of line must pass between
(1.75, 1.24) and (1.75, 1.255) and lower
end of line must pass between (2.00, 0.900)
and (2.00, 0.915).
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
error bars are plotted.
(iii) C1 Gradient of line of best fit 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 –1.35.)
U3 Uncertainty in gradient Method of determining absolute uncertainty:
difference in worst gradient and gradient.
(iv) C2 y-intercept Check substitution into y = mx + c.
Allow ecf from (c)(iii).
(Should be about 4.)
Do not allow read-off of false origin.

U4 Uncertainty in y-intercept Uses worst gradient and point on worst
acceptable line.
Do not check calculation. Do not allow if
false origin used.
(d) (i) C3 k = 10y-intercept
C4 m = gradient and given to 2 or 3 s.f.
and in the range –1.30 to –1.44
Must be negative.
Allow –1.3 or –1.4 (2 s.f.)
(ii) U5 Percentage uncertainty in k
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)
(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) (ii) [U5]
max k = 10max y-intercept
and min k = 10min y-intercept
percentage uncertainty = 100
max
×
−
k
kk
= 100
min
×
−
k
kk
=
( )
100
minmax
2
1
×
−
k
kk

9702 PHYSICS
maximum raw mark 30
Cambridge IGCSE®

P t is the independent variable and I (or amplitude of reflected signal) is the dependent
variable, or vary t and measure I (or amplitude of reflected signal). [1]
P Keep distance from the wall/foam to the speaker/microphone constant. [1]
P Keep the amplitude or intensity I0 of the sound before reflection constant. [1]
M Labelled diagram of workable experiment including speaker, microphone/sound detector,
foam and wall. [1]
M Signal generator/a.c. power supply connected to speaker. [1]
M Microphone connected to oscilloscope or sound (intensity) meter. [1]
M Measure the thickness with a rule/micrometer/vernier calipers. [1]
M Method to determine the density; ρ = m/V. [1]
A Plot a graph of ln I against t.
(Allow log I against t and lg I against t graphs.) [1]
A α = –gradient/ρ (must be consistent with graph plotted) [1]
S Precaution linked to loud sounds, e.g. use ear plugs/muffs/defenders.
Allow switch off sound source to prevent damage to ears. [1]
1 Keep the frequency constant
2 Carry out experiment in a quiet room/no other sources of sound
3 Method to keep angles constant/positions of speaker and microphone constant.
4 Method and explanation to detect reflected sound from foam only, e.g. barrier, tube or
method to avoid reflections
5 Method to determine mass, e.g. use scales/balance and method to determine volume
6 Relationship is valid if the graph is a straight line (ignore reference to y-intercept)
7 Method to check that emitted sound I0 is constant or method to check y-intercept is ln I0.
8 Intensity is proportional to the amplitude2
.

Expected Answer Additional Guidance
(a) A1 gradient =
d
Efε
(b) T1 X/10–2
m2
T2
4.80 or 4.800
5.40 or 5.400
6.30 or 6.300
7.20 or 7.200
8.10 or 8.100
9.00 or 9.000
Must be table values.
U2 Error bars in X plotted
correctly
All error bars to be plotted. Must be accurate to
less than half a small square.
(ii) G2 Line of best fit Lower end of line must pass between (5.1, 5.0)
and (5.3, 5.0) and upper end of line must pass
between (8.5, 8.5) and (8.8, 8.5).
Steepest or shallowest
possible line that passes
through all the error bars.
Examiner judgement on worst acceptable line.
Lines must cross. Mark scored only if error bars
are plotted.
(iii) C1 Gradient of best fit line 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 1×10–4
.)
(d) (i) C2 ε = 6.25 × 10–7
× gradient Do not penalise POT.
(Should be about 6 or 7 × 10–11
.)
C3 Fm–1
or CV–1
m–1
Allow Am–1
V–1
Hz–1
or Asm–1
V–1
or A2
s4
kg–1
m–3
.
Power of 10 must be correct.
(ii) U4 Percentage uncertainty in ε 10.83% + percentage uncertainty in gradient

(e) C4 f in the range 73.0 to 84.4 and
given to 2 or 3 s.f.
Allow 73 to 84 for 2 s.f.
f =
ε
9
100.5 −
×
U5 Absolute uncertainty in f Clear working needed.
Allow ecf from (d)(ii).
(d) (ii) [U4]
max ε =
fE
d
minmin
maxgradientmax
×
×
min ε =
fE
d
maxmax
mingradientmin
×
×
% uncertainty = 100
gradient
gradient
×




 ∆
+
∆
+
∆
+
∆
E
E
f
f
d
d
= 100
12.0
0.2
400
10
0.0030
0.0002
gradient
gradient
×





+++
∆
(e) [U5]
max f =
EX
d
minminmin
maxmax
××
×
ε
I
min f =
EX
d
maxmaxmax
minmin
××
×
ε
I
∆f = 




 ∆
+
∆
+
∆
+
∆
+
∆
ε
ε
E
E
d
d
l
l
2
I
I
f = 




 ∆
++++
ε
ε
12.0
0.2
0.500
0.001
2
0.0030
0.0002
5.0
0.1
f = 




 ∆
+
ε
ε
0.107 f
∆f = f




 +
100
10.7 (d)(ii)











 +
= correctisif
100
gradientinyuncertaint%21.5
(d)(ii)f

9702 PHYSICS
maximum raw mark 30
Cambridge IGCSE®

P V is the independent variable, or vary V and f is the dependent variable, or
measure f.
Or f is the independent variable, or vary f and V is the dependent variable, or
measure V. [1]
P Change f (allow V) until the mass leaves/gap between plate. [1]
P Keep the position of the mass constant. (Do not allow keep mass constant.) [1]
M Labelled diagram showing signal generator/a.c. supply connected to vibrator with
two wires with mass on plate. At least two labels needed. [1]
M Voltmeter/c.r.o. connected in parallel with vibrator in a workable circuit. [1]
M Measure f or T from signal generator/c.r.o. (Allow detailed use of motion
sensor/stroboscope.) [1]
M Detail regarding mass leaving the plate: listen to noise, look for gap. [1]
M Repeat each experiment for the same value of V (allow f if consistent with above)
and average. [1]
Plot a graph of:
A
f2
against
1/V
1/V
against
f2
f
against
1/ V
1/ V
against
f
lg V
against
lg f
lg f
against
lg V
or or or or
V
against
1/f2
1/f2
against
V
V
against
1/f
1/f
against
V
[1]
A
k =
2gradient π× gradient
2
π
=k
k =
22
gradient π× 2
2
gradient
π
=k k = c102 ×π k = c22 10×π [1]
S Precaution linked to mass leaving vibrating plate, e.g. use safety
screen/goggles/sand tray. [1]

1 Wait for vibrator to oscillate evenly
2 Method to determine period of oscillation from c.r.o., i.e. one time period × time-base
3 Method to determine f from c.r.o. having determined T, i.e. f = 1/T
4 Method to determine V from c.r.o, i.e. amplitude (height) × y-gain
5 Relationship is valid if the graph is a straight line passing through the origin
[For lg – lg graph the gradient must be correct (–2 or –0.5)]
6 Determine f (allow V if consistent with above) by increasing and decreasing V or f
7 Clean surfaces of metal plate/small mass
8 Spirit level to keep plate horizontal/eye level to look for gap

Mark Expected Answer Additional Guidance
(a) A1 gradient = m
y-intercept = lg k
(b) T1
T2 1.70 or 1.699 1.312 or 1.3118
1.79 or 1.785 1.204 or 1.2041
1.85 or 1.851 1.114 or 1.1139
1.90 or 1.903 1.041 or 1.0414
1.95 or 1.954 0.98 or 0.978
2.00 or 1.996 0.90 or 0.903
T1 (first column) and T2 (second column)
must be values in table.
U2 Error bars in lg P plotted correctly All error bars to be plotted. Must be
accurate to less than half a small square.
(ii) G2 Line of best fit Upper end of line must pass between
(1.75, 1.24) and (1.75, 1.255) and lower
end of line must pass between (2.00, 0.900)
and (2.00, 0.915).
Steepest or shallowest possible
line that passes through all the
error bars.
Examiner judgement on worst acceptable
line. Lines must cross. Mark scored only if
error bars are plotted.
(iii) C1 Gradient of line of best fit 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 –1.35.)
(iv) C2 y-intercept Check substitution into y = mx + c.
Allow ecf from (c)(iii).
(Should be about 4.)
Do not allow read-off of false origin.

U4 Uncertainty in y-intercept Uses worst gradient and point on worst
acceptable line.
Do not check calculation. Do not allow if
false origin used.
(d) (i) C3 k = 10y-intercept
C4 m = gradient and given to 2 or 3 s.f.
and in the range –1.30 to –1.44
Must be negative.
Allow –1.3 or –1.4 (2 s.f.)
(ii) U5 Percentage uncertainty in k
(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) (ii) [U5]
max k = 10max y-intercept
and min k = 10min y-intercept
percentage uncertainty = 100
max
×
−
k
kk
= 100
min
×
−
k
kk
=
( )
100
minmax
2
1
×
−
k
kk

9702 s15 ms_all

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9702 s15 ms_all