9702 s17 ms_all

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice May/June 2017
MARK SCHEME
Maximum Mark: 40
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.
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 2017 series for most Cambridge IGCSE®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/11 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 3
Question Answer Marks
1 D 1
2 D 1
3 A 1
4 D 1
5 C 1
6 B 1
7 B 1
8 C 1
9 D 1
10 A 1
11 A 1
12 D 1
13 C 1
14 A 1
15 A 1
16 C 1
17 C 1
18 D 1
19 A 1
20 D 1
21 A 1
22 B 1
23 B 1
24 B 1
25 C 1
26 B 1
27 C 1
28 C 1

PUBLISHED
May/June 2017
29 A 1
30 A 1
31 B 1
32 A 1
33 C 1
34 A 1
35 C 1
36 B 1
37 C 1
38 A 1
39 B 1
40 C 1

PHYSICS 9702/12
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.
,
components.

PUBLISHED
May/June 2017
1 B 1
2 A 1
3 D 1
4 C 1
5 B 1
6 A 1
7 C 1
8 B 1
9 C 1
10 A 1
11 C 1
12 D 1
13 D 1
14 C 1
15 B 1
16 C 1
17 C 1
18 A 1
19 A 1
20 D 1
21 B 1
22 C 1
23 B 1
24 A 1
25 D 1
26 B 1
27 D 1
28 B 1

PUBLISHED
May/June 2017
29 B 1
30 C 1
31 C 1
32 A 1
33 A 1
34 C 1
35 A 1
36 B 1
37 B 1
38 A 1
39 B 1
40 A 1

PHYSICS 9702/13
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.
,
components.

PUBLISHED
May/June 2017
1 B 1
2 B 1
3 D 1
4 C 1
5 B 1
6 A 1
7 D 1
8 C 1
9 D 1
10 A 1
11 D 1
12 A 1
13 A 1
14 A 1
15 D 1
16 C 1
17 B 1
18 D 1
19 B 1
20 A 1
21 B 1
22 A 1
23 C 1
24 D 1
25 D 1
26 A 1
27 B 1
28 C 1

PUBLISHED
May/June 2017
29 D 1
30 D 1
31 B 1
32 B 1
33 C 1
34 B 1
35 C 1
36 A 1
37 A 1
38 D 1
39 A 1
40 A 1

PHYSICS 9702/21
Paper 2 AS Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 60
Published
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.
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a) (stress =) force / area or kg m s–2
/ m2
B1
= kg m–1
s–2
A1
1(b)(i) 0.58 = 2π × [(4 × 0.500 × 0.6003
) / (E × 0.0300 × 0.005003
)]0.5
C1
E = [4π2
× 4 × 0.500 × (0.600)3
] / [(0.58)2
× 0.0300 × (0.00500)3
]
= 1.35 × 1010
(Pa)
C1
= 14 (13.5) GPa A1
1(b)(ii)1. (accuracy determined by) the closeness of the value(s)/measurement(s) to the true value B1
(precision determined by) the range of the values/measurements B1
1(b)(ii)2. l is (cubed so) 3 × (percentage/fractional) uncertainty
and T is (squared so) 2 × (percentage / fractional) uncertainty
and (so) l contributes more
B1

PUBLISHED
May/June 2017
2(a) resultant force (in any direction) is zero B1
resultant torque/moment (about any point) is zero B1
2(b)(i) a = (v − u) / t or gradient or ∆v / (∆)t C1
e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s–2
A1
2(b)(ii) s = 4.6 × 4 + [(8.8 + 4.6) / 2] × 3 C1
= 18.4 + 20.1
= 39 (38.5) m
A1
2(b)(iii) ∆E = ½ × 95 [(8.8)2
− (4.6)2
] C1
= 3678 – 1005
= 2700 (2673) J
A1
2(b)(iv)1. weight = 95 × 9.81 (= 932 N) C1
vertical tension force = 280 sin 25° or 280 cos 65° (=118.3 N) C1
F = 932 + 118
= 1100 (1050) N
A1
2(b)(iv)2. horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N) C1
resultant force = 95 × 1.4 (= 133 N) C1
133 = 253.8 – R
R = 120 (120.8) N
A1

PUBLISHED
May/June 2017
3(a) ρ = m / V C1
V = πd2
L / 4 or πr2
L C1
weight = 2.7 × 103
× π (1.2 × 10–2
)2
× 5.0 × 10–2
× 9.81 = 0.60 N A1
3(b)(i) the point from where (all) the weight (of a body) seems to act B1
3(b)(ii) W × 12 C1
(0.25 × 8) + (0.6 × 38) C1
W = (2 + 22.8) / 12
= 2.1 (2.07) N
A1
3(c)(i) pressure changes with depth (in water)
or
pressure on bottom (of cylinder) different from pressure on top
B1
pressure on bottom of cylinder greater than pressure on top
or
force (up) on bottom of cylinder greater than force (down) on top
B1
3(c)(ii) anticlockwise moment reduced and reducing the weight of X reduces clockwise moment
or
anticlockwise moment reduced so clockwise moment now greater than (total) anticlockwise moment
B1

PUBLISHED
May/June 2017
4(a) (two) waves travelling (at same speed) in opposite directions overlap B1
waves (are same type and) have same frequency/wavelength B1
4(b)(i) λ = 12 / 250 (= 0.048 m) C1
distance = 1.5 × 0.048
= 0.072 m
A1
4(b)(ii) T = 1 / 250
= 0.004 (s) or 4 (ms)
C1
1. curve drawn is mirror image of that in Fig. 4.2 and labelled P A1
2. horizontal line drawn between A and B and labelled Q A1
5(a) observed frequency is different to source frequency when source moves relative to observer B1
5(b) 360 = (400 × 340) / (340 ± v) C1
v = 38 (37.8) m s–1
A1
away (from the observer) B1

PUBLISHED
May/June 2017
6(a) volt / ampere B1
6(b)(i) RT = [1 / 3.0 + 1 / 6.0]–1
+ 4.0 (= 6.0 Ω) C1
Ι = 1.5 / 6.0 C1
= 0.25 A A1
6(b)(ii) VB = 0.5 V
I = 0.5 / 3.0
= 0.17 (0.167) A
A1
6(b)(iii) P = I 2
R or VI or V 2
/ R C1
ratio = (0.1672
× 3.0) /(0.252
× 4.0)
= 0.33
A1
6(c)(i) vary/change/different radius/diameter/cross-sectional area (of wire) B1
6(c)(ii) v = I / Ane
( )
( )
= B
C
/
ratio
/
A
A
B
C
I
I
or × C
B
A
A
B
C
I
I
C1
(R ∝ 1 / A so) ratio = × B
C
R
R
B
C
I
I
= ×
0.167 3.0
0.25 4.0
= 0.50
A1
6(d)(i) 0.25 A to 0.13 (0.125) A or halved A1
6(d)(ii) no change A1

PUBLISHED
May/June 2017
7(a)(i) (proton is uud so) (2 / 3)e + (2 / 3)e – (1 / 3)e = e B1
7(a)(ii) (neutron is udd so) (2 / 3)e – (1 / 3)e –(1 / 3)e = 0 B1
7(b)(i) β–
β+
nucleon number 90 64
proton number 39 28
all correct
B1
7(b)(ii) weak (nuclear force/interaction) B1
7(b)(iii) β–
decay: electron and (electron) antineutrino
β+
decay: positron and (electron) neutrino
all correct
B1

PHYSICS 9702/22
Paper 2 AS Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
,
Cambridge International A and AS Level and Cambridge Pre–U components, and some Cambridge O Level
components.

PUBLISHED
May/June 2017
1(a) kelvin, mole, ampere, candela
any two
B1
1(b) use of resistivity = RA / l and V = IR (to give ρ = VA/ Il) C1
units of V: (work done / charge) kg m2
s–2
(A s)–1
C1
units of resistivity: (kg m2
s–3
A–1
A–1
m)
= kg m3
s–3
A–2
A1
or
use of R = ρL / A and P = I2
R (gives ρ = PA / I2
L) (C1)
units of P: kg m2
s–3
(C1)
units of resistivity: (kg m2
s–3
× m2
) / (A2
× m)
= kg m3
s–3
A–2
(A1)
1(c)(i) ρ = (RA/l) C1
= (0.03 × 1.5 × 10–6
) / 2.5 (= 1.8 × 10–8
) C1
= 18 nΩ m A1
1(c)(ii) 1. precision is determined by the range in the measurements/values/readings/data/results B1
2. metre rule measures to ± 1 mm and micrometer to ± 0.01 mm (so there is less (percentage) uncertainty/random error) B1

PUBLISHED
May/June 2017
2(a) rate of change of displacement or change in displacement/time taken B1
2(b)(i) s = ut + ½at2
C1
t = [(2 × 1.25) / 9.81]1/2
(= 0.5048s) C1
or
v2
= u2
+ 2as
vvert = (2 × 9.81 × 1.25)1/2
(= 4.95)
(C1)
t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048s) (C1)
v = d / t = 1.5 / 0.50(48)
= 3.0 (2.97)ms–1
A1
2(b)(ii) vertical velocity = at
= 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9]
C1
velocity = [(vh)2
+ (vv)2
]1/2
C1
= [(2.97)2
+ (4.95)2
]1/2
= 5.8 (5.79) [using t = 0.50 leads to 5.7]
A1
direction (= tan–1
4.95/2.97) = 59° A1
2(b)(iii) kinetic energy = ½mv2
C1
= ½ × 0.45 × (5.8)2
= 7.6 (7.57) J [using t = 0.50 leads to 7.3 J]
A1

PUBLISHED
May/June 2017
2(b)(iv) potential energy = mgh C1
= (0.45 × 9.81 × 1.25)
= 5.5 (5.52) J
A1
2(c) there is KE of the ball at the start/leaving table
or
the ball has an initial/constant horizontal velocity
or
the ball has velocity at start/leaving table
B1
3(a) E = stress / strain or (F / A) / (e / l) C1
= [gradient × 3.5] / [π × (0.19 × 10–3
)2
]
e.g. E = [{(40 – 5) / ([11.6 – 3.2] × 10–3
)} × 3.5] / [π × (0.19 × 10–3
)2
]
or
[4170 × 3.5] / [π × (0.19 × 10–3
)2
]
C1
E (= 1.3 × 1011
) = 0.13 TPa (allow answers in range 0.120–0.136TPa) A1
3(b) a larger range of F required or range greater than 35 N B1

PUBLISHED
May/June 2017
4(a) a body/mass/object continues (at rest or) at constant/uniform velocity unless acted on by a resultant force B1
4(b)(i) initial momentum = final momentum
m1u1 + m2u2 = m1v1 + m2v2
C1
0.60 × 100 − 0.80 × 200 = −0.40 × 100 + v × 200
v = (−) 0.3(0) m s–1
A1
4(b)(ii) kinetic energy is not conserved/is lost (but) total energy is conserved/constant
or
some of the (initial) kinetic energy is transformed into other forms of energy
B1
5(a) frequency is the number of vibrations/oscillations per unit time or the number of wavefronts passing a point per unit time B1
5(b) vibrations/oscillation of the air particles are parallel to the direction of it (the direction of travel of the sound wave) B1
5(c)(i) T = 2(.0) (ms) C1
f = 500 Hz A1
5(c)(ii) 1. amplitude increases
(time) period decreases
2. amplitude decreases
(time) period increases
any 3 points
B3

PUBLISHED
May/June 2017
6(a)(i) waves at (each) slit/aperture spread B1
(into the geometric shadow) wave(s) overlap/superpose/sum/meet/intersect B1
6(a)(ii) there is not a constant phase difference/coherence (for two separate light source(s))
or
waves/light from the double slit are coherent/have a constant phase difference
B1
6(b) x = λD / a C1
λ = (36 × 10–3
× 0.48 × 10–3
) / (16 × 2.4) C1
= 4.5 × 10–7
m A1
6(c)(i) no movement of the water/water is flat/no ripples/disturbance B1
the path difference is 2.5λ or the phase difference is 900° or 5π rad B1
6(c)(ii) 1. surface/water/P vibrates/ripples
and
as (waves from the two dippers) arrive in phase
B1
2. surface/water/P vibrates/ripples
and
as amplitudes/displacements are no longer equal/do not cancel
B1

PUBLISHED
May/June 2017
7(a) energy transformed from chemical to electrical /unit charge (driven around a complete circuit) B1
7(b)(i) the current decreases (as resistance of Y increases) M1
lost volts go down (as resistance of Y increases) M1
p.d. AB increases (as resistance of Y increases) A1
7(b)(ii)1. 1.50 = 0.180 × (6.00 + 0.200 + RX) C1
RX = 2.1(3) Ω A1
7(b)(ii)2. p.d. AB = 1.5 − (0.180 × 0.200) or 0.18 × (2.13 + 6.00) C1
= 1.46(4) V A1
7(b)(ii)3. efficiency = (useful) power output / (total) power input or IV / IE C1
( = 1.46 / 1.5) = 0.97 [0.98 if full figures used] A1
8(a) β–
emission: neutron changes to proton (+ beta–
/electron)
and
β+
emission: proton changes to neutron (+ beta+
/positron)
B1
β–
emission: (electron) antineutrino also emitted
and
β+
emission: (electron) neutrino also emitted
B1
8(b) proton: up up down (and zero strange)
neutron: up down down (and zero strange)
B1

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 May/June 2017
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
Question Answer Mark
1(b)(iii) Value(s) of I1 with unit in the range 30–200 mA. 1
1(c)(iii) I2 > I1 (ECF unit from 1(b)(iii)). 1
1(d) Six sets of readings of x (different values), I1, I2 with correct trend and without help from Supervisor scores 5 marks, five sets
scores 4 marks etc.
5
Range: xmax ⩾ 90 cm. 1
Column headings:
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. x / m, I2 / mA, I2 / I1 no unit.
1
Consistency:
All values of x must be given to the nearest mm.
1
Significant figures:
All values of I2 / I1 must be given to the same number of s.f. as (or one more than) the s.f. in raw I1 and I2 (using the s.f. of the
I value with the lowest number of s.f.).
1
Calculation:
Values of I2 / I1 are correct.
1
1(e)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
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.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
1

PUBLISHED
May/June 2017
Quality:
All values of I2 / I1 must be greater than 1.
All points in the table must be plotted for this mark to be awarded.
It must be possible to draw a straight line that is within ± 0.05 on the I2 / I1 axis (normally y axis) of all plotted points.
1
1(e)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5). 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. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1
1(e)(iii) Gradient:
Gradient sign on answer line matches graph drawn.
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Method of calculation must be correct. Do not allow ∆x / ∆y.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
Read-off accurate to half a small square in both 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.
1
1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
The values must not be fractions.
1
Unit for P is correct (e.g. m–1
) and Q stated without a unit. 1

PUBLISHED
May/June 2017
2(b)(ii) Values of time in the range 3–7s and within 0.4s of each other.
All raw readings stated to same precision and to at least 0.1s.
1
2(c)(ii) x on the answer line in the range 35.0–40.0 cm and all value(s) of raw x to nearest mm with unit. 1
2(c)(iii) Absolute uncertainty in x in range 2–5 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
2(e)(i) Value of n on answer line in the range 10 to 20. No unit. 1
Evidence of repeat readings. 1
2(e)(ii) Correct calculation of (n + 1)2
/n2
. 1
2(f) Second value of x. 1
Second value of n. 1
Quality: second value of n < first value of n (if x2 < x1 allow n2 > n1). 1
2(g)(i) Two values of k calculated correctly. 1
2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1
2(h) Correct calculation of L written to 3 s.f. and correctly rounded to 3 s.f. 1

PUBLISHED
May/June 2017
2(i)(i) A Two readings are not enough to draw a valid conclusion (not “not enough for accurate results”, “few readings”).
B Difficult to judge when the balls are back in phase.
C Balls hit each other/blown by moving air/balls move sideways/irregular movement.
D Difficult to ensure both lengths 35 cm or difficult to measure x with a reason e.g. string kinked/parallax error.
E Damping/oscillations die out quickly.
F Difficulty linked to fixing block/wood with reason e.g. longer block requires more tape (therefore an unfair test)/block
comes unstuck/blocks attached at an angle (x not constant).
G For two/both balls the release point is different or release instant is different.
1 mark for each point up to a maximum of 4.
4
2(i)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).
B Video/film/record and perpendicular to plane of oscillation/from the side.
C Turn off air conditioning/close windows/use windshield/use a longer rod.
D Improved method of measuring 35 cm or x e.g. use clamps to fix loop and ball.
F Improved method of attachment e.g. Blu-Tack/glue/Velcro/stickier tape/stronger adhesive on the tape.
G Improved method of release e.g. bar to hold both balls then drop away like a gate.
4

PHYSICS 9702/32
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a)(v) Value of p in the range 20.0–30.0 cm, with unit. 1
Value of q less than p. 1
1(b) Six sets of readings of p, q and V (with correct trend and without help from Supervisor) scores 5 marks, five sets scores 4
marks etc.
5
Range:
pmax ⩾ 40.0 cm and pmin ⩽ 10.0 cm.
1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. 1/p/cm–1
.
1
Consistency:
All values of p and q must be given to the nearest mm.
1
Significant figures for every value of 1 /q must be same as, or one greater than, the s.f. of q as recorded in table.
1
Calculation:
Values of 1 /q calculated correctly.
1
1(c)(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:
Plots must be accurate to within half a small square in both x and y directions.
1

PUBLISHED
May/June 2017
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded.
It must be possible to draw a straight line that is within ± 1.0 m–1
(± 0.01 cm–1
) of all the plotted points in the 1/q direction.
1
1(c)(ii) Line of best fit:
Judge by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate.
1
1(c)(iii) Gradient:
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression.
or
Intercept read directly from the graph, with read-off at x = 0 accurate to half a small square in y direction.
1
1(d) Value of a = candidate’s gradient and value of b = candidate’s intercept.
Values must not be fractions.
1
There must be no unit for a.
Unit for b correct e.g. m–1
or cm–1
.
1

PUBLISHED
May/June 2017
2(a)(ii) Value of h1 to nearest 0.1 cm. 1
h2 < h1. 1
2(a)(iv) Absolute uncertainty in h2 of 2–5 mm 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
2(b)(i) Correct calculation of k. 1
2(b)(ii) Justification for s.f. in k linked to s.f. in m, g and (h1–h2) or m, g, h1 and h2. 1
2(c)(iii) Value of T in range 1.00–3.00 s. 1
Evidence of repeat readings of nT with n ⩾ 5. 1
2(d) Second values of h1 and h2. 1
Second value of T. 1
Quality: T for three springs > T for two springs. 1
2(e)(i) Two values of C calculated correctly. 1
2(e)(ii) Valid comment consistent with the calculated values of C, testing against a criterion specified by the candidate. 1

PUBLISHED
May/June 2017
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B (Difficulty) measuring h because of parallax/metre rule not vertical.
C Oscillation dies away quickly/oscillation damped.
D Difficult to judge/tell end of oscillation/decide when to operate stopwatch.
E Other modes of oscillation occur.
F Some movement at joints still occurs.
4
2(f)(ii) A Take more readings and plot a graph/take more readings and compare C values (not “repeat readings” on its own).
B Use set square against rule (with detail)/use pointer attached to bottom of mass/use set square from rule to bottom of
mass.
C Twist through larger angle to get more rotations.
D View (video) recording/film of motion with timer in view or put mark on mass to make it easier to see motion.
E Workable method of restricting vertical/swinging movement (e.g. enclose in transparent tube).
F Better method of fixing joints/use two complete springs.
4

PHYSICS 9702/33
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a) Value of L with unit to the nearest mm in the range 2.5–3.5 cm. 1
1(c) Value of T with unit in the range 0.7s to 1.5s. 1
Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1
1(d) Six sets of readings of n (different values) and time with correct trend and without help from Supervisor scores 4 marks, five
sets scores 3 marks etc.
4
Range of n ⩾ 9. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. T / s. No unit for n or √n.
1
Consistency:
All values of raw time must be given to either 0.1 s or 0.01 s.
1
All values of √n must be given to 3 significant figures.
1
1(e)(i) Axes:
1
Plotting of points:
1
Quality:
It must be possible to draw a straight line that is within ± 0.10 on the √n axis of all plotted points.
1

PUBLISHED
May/June 2017
1(e)(ii) Line of best fit:
1
1(e)(iii) Gradient:
Both read-offs must be accurate to half a small square in both x and y directions.
1
y-intercept:
or
Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction.
1
1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
1
Units for P and Q both s. 1
1(g) Correct calculation of g = Lπ2
/ P2
with consistent unit. 1

PUBLISHED
May/June 2017
2(a) Value(s) of A with unit in the range 97.5–99.5 cm. 1
2(b)(ii) Values of all raw x to nearest mm. 1
2(c)(iii) Value of y ˃ x. 1
2(c)(iv) Correct calculation of (y – x). 1
2(c)(v) Percentage uncertainty in (y – x) based on absolute uncertainty of 2–5 mm.
1
2(d)(i) Correct calculation of m(A – 2y). 1
2(d)(ii) Justification for s.f. in m(A – 2y) linked to s.f. in A, y and m or (A – 2y) and m. 1
2(e)(i) Second value of y. 1
Quality: second value of y > first value of y. 1
2(f)(i) Two values of k calculated correctly. 1
2(f)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1
2(g) Correct calculation of M = 1/k – B. 1

PUBLISHED
May/June 2017
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficult to measure x or y with reason e.g. markings on rule obscured because of thickness of string/twist in string/ruler
oscillating/ruler swinging/rule slides in loop/string not vertical.
C Large (%) uncertainty in (y – x) or (y – x) is small.
D Little difference in y values.
E Difficult to balance rule.
4
2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).
B Use thinner string/tie a knot in loop/use two loops/use a hook/use longer loop/string.
C Heavier (added/slotted) masses (not heavier hangers).
D Use wider range of (added) masses.
E Suspend rule from its top edge/balance on fulcrum/use thicker string with reason such as to make easier to
balance/prevent rule from slipping.
4

PHYSICS 9702/34
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a)(v) Value of T in the range 0.10–0.90 s. 1
Evidence of repeated readings. Must see nT repeated where n ⩾ 5. 1
1(b) Six sets of readings of M, h and T showing the correct trend and without help from the Supervisor scores 5 marks, five sets
scores 4 marks etc.
5
Range: Mmax ⩾ 450g and Mmin ⩽ 200g. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. T 3
/ s3
.
1
Consistency:
All values of h must be given to the nearest mm.
1
Significant figures for every value of T 3
must be the same as (or one greater than) the s.f. of raw times as recorded in table.
1
Calculation:
Values of T 3
calculated correctly.
1
1(c)(i) Axes:
1
Plotting of points:
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded.
It must be possible to draw a straight line that is within ± 1.0 cm (to scale) of all the plotted points in the h direction.
1

PUBLISHED
May/June 2017
1(c)(ii) Line of best fit:
Judge by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
Lines must not be kinked or thicker than half a square.
1
1(c)(iii) Gradient:
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression.
or
Intercept read directly from the graph, with read-off at h = 0, accurate to half a small square in y direction.
1
1(d) Value of a = candidate’s gradient and value of b = candidate’s intercept.
1
Unit for a correct (e.g. s3
cm–1
).
Unit for b is s3
.
1

PUBLISHED
May/June 2017
2(a) Value for x0 to nearest 0.1cm. 1
2(b)(ii) x greater than x0. 1
2(b)(iii) Raw θ in range 91–110° and recorded to nearest degree. 1
2(b)(v) Absolute uncertainty in (φ – 90°) of 2–5° and correct method of calculation to obtain percentage uncertainty.
If repeated readings of φ have been taken, then the absolute uncertainty can be half the range (but not zero) only if working
shown clearly.
1
2(c)(ii) Second value of x. 1
Second value of φ. 1
Quality: second value of φ ⩾ first value of φ. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Justification for s.f. in k linked to s.f. in θ, x and x0. 1
2(d)(iii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1
2(e)(ii) Raw value(s) of D and d recorded to the nearest 0.1 cm. 1
2(e)(iv) Value of ρ calculated correctly. 1

PUBLISHED
May/June 2017
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficult to get/adjust wooden strip so that it is horizontal/parallel to bench.
C Difficulty when measuring x (or x0), with reason e.g. parallax error/thickness of string/difficult to judge centre of nail.
D Difficult to measure angle with reason, e.g. parallax error/hard to hold protractor steady/hard not to touch wooden
strip/difficult to align protractor correctly.
(Do not credit parallax error twice for both C and D.)
E Large (percentage) uncertainty in (φ – 90°) or (φ – 90°) is small.
F Difficult to measure D (or d) with reason linked to use of ruler.
4
2(f)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).
B Use thinner nail/fulcrum/prism.
C Mark a scale on the wooden strip/use a thinner string.
D Hold protractor in a clamp/
take photograph and measure angle on photo/
trigonometric method with detail e.g. measure height(s) of end(s) of wooden strip
E Method to increase (φ – 90°) e.g. more paper clips/larger values of x/move pipe and pivot closer.
F Use vernier/digital calipers/travelling microscope.
4

PHYSICS 9702/35
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a)(iii) Value of x with unit to the nearest mm in the range 10.0–15.0 cm. 1
1(b)(ii) Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1
1(c) Six sets of readings of x (different values) and time with correct trend and without help from Supervisor scores 5 marks, five
sets scores 4 marks etc.
5
Range: xmin ⩽ 5.0cm and xmax ⩾ 20.0cm. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. x2
/ m2
.
1
Consistency:
All values of raw x must be given to the nearest mm.
1
All values of x2
must be given to the same number of s.f. as (or one more than) the s.f. in raw x.
1
1(d)(i) Axes:
1
Plotting of points:
1
Quality:
It must be possible to draw a straight line that is within ± 0.025s on the T axis of all plotted points.
1

PUBLISHED
May/June 2017
1(d)(ii) Line of best fit:
1
1(d)(iii) Gradient:
Gradient sign must match graph.
Method of calculation must be correct. Do not allow ∆x/∆y.
1
y-intercept:
or
Intercept read directly from the graph, with read-off at x2
= 0, accurate to half a small square in the y direction.
1
1(e) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
1
Unit for P dimensionally correct (e.g. sm–2
or scm–2
or smm–2
).
Unit for Q correct (s).
1
1(f)(iii) Correct calculation of x. 1

PUBLISHED
May/June 2017
2(a) Value of raw L with unit in range 1.5–2.5cm. 1
2(c)(ii) Value of y ⩾ L and to the nearest mm. 1
2(c)(iii) Value of raw θ to the nearest degree. 1
2(d) Percentage uncertainty in θ based on absolute uncertainty of 2–10°.
1
2(e)(i) Correct calculation of (y – L). 1
2(e)(ii) Correct calculation of cos(θ / 2). 1
2(e)(iii) Justification for s.f. in cos(θ / 2) linked to s.f. in θ. 1
2(f) Second value of y. 1
Second value of θ. 1
Quality: second value of θ > first value of θ. 1
2(g)(i) Two values of k calculated correctly. 1
2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1

PUBLISHED
May/June 2017
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Moving hands affecting keeping 15cm the same or measuring y or θ.
C Parallax error affecting the measurement of y or θ.
D Reason for d not remaining constant e.g. loops slip on stand.
E Difficult to know where to measure from for the angle or y.
4
2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).
B Use another stand for valid purpose (e.g. instead of hand to hold spring).
C Photo or still from video to measure θ or including scale in frame to measure y.
D Improved method of fixing springs onto stand e.g. use Blu-Tack/tape/vertically clamped hacksaw blade/bulldog clips or
use sandpaper to make stand rough.
E Use a grid behind the springs/markers on spring.
4

PHYSICS 9702/41
Paper 4 A Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a) gravitational force (of attraction between satellite and planet) B1
provides / is centripetal force (on satellite about the planet) B1
1(b) M = (4/3) × πR3
ρ B1
ω = 2π / T or v = 2πnR / T B1
GM / (nR)2
= nRω2
or v 2
/ nR M1
substitution clear to give ρ = 3πn3
/ GT 2
A1
1(c) n = (3.84 × 105
) / (6.38 × 103
) = 60.19 or 60.2 C1
ρ = 3π × 60.193
/ [(6.67 × 10–11
) × (27.3 × 24 × 3600)2
] C1
ρ = 5.54 × 103
kg m–3
A1

PUBLISHED
May/June 2017
2(a) e.g. period = 3 / 2.5 C1
frequency = 0.83 Hz A1
2(b) light (damping) B1
2(c) at 2.7 s, A0 = 1.5 (cm) B1
energy = ½ m × 4π2
f 2
A0
2
B1
= ½ × 0.18 × 4π2
× 0.832
× (1.5 × 10–2
)2
= 5.51 × 10–4
(J)
C1
at 7.5 s, A0 = 0.75 (cm) B1
energy = ¼ × 5.51 × 10–4
or
energy = ½ × 0.18 × 4π2
× 0.832
× (0.75 × 10–2
)2
C1
energy = 1.38 × 10–4
(J)
change = (5.51 × 10–4
– 1.38 × 10–4
) = 4.13
J
A1

PUBLISHED
May/June 2017
3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1
3(a)(ii) component X: parallel-to-serial converter B1
component Y: DAC/digital-to-analogue converter B1
3(a)(iii) sample the (analogue) signal M1
at regular intervals and converts the analogue number to a digital number A1
3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1
ratio = 16 + 28
= 44 dB
A1
3(b)(ii) ratio / dB = 10 lg (P2 / P1) C1
44 = 10 lg ({9.7 × 10–3
} / P)
or
–44 = 10 lg (P / {9.7 × 10–3
})
C1
power = 3.9 × 10–7
W A1

PUBLISHED
May/June 2017
4(a) random/haphazard B1
constant velocity or speed in a straight line between collisions
or
distribution of speeds/different directions
B1
4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1
moving haphazardly/randomly/jerky/in a zigzag fashion A1
4(c)(i) pV = ⅓ Nm〈c2
〉
1.05 × 105
× 0.0240 = ⅓ × 4.00 × 10–3
× 〈c2
〉
C1
〈c2
〉 = 1.89 × 106
C1
or
½ m〈c2
〉 = (3 / 2) kT
0.5 × (4.00 × 10–3
/ 6.02 × 1023
) × 〈c2
〉 = 1.5 × 1.38 × 10–23
× 300
(C1)
〈c2
〉 = 1.87 × 106
(C1)
or
nRT = ⅓ Nm〈c2
〉
1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3
× 〈c2
〉
(C1)
〈c2
〉 = 1.87 × 106
(C1)
cr.m.s. = 1.37 × 103
m s–1
A1

PUBLISHED
May/June 2017
4(c)(ii) 〈c2
〉 ∝ T C1
〈c2
〉 at 177 °C = 1.89 × 106
× (450 / 300) C1
cr.m.s. at 177 °C = 1.68 × 103
m s–1
A1
5(a) (loss in) kinetic energy of α-particle = Qq / 4πε0r
or
7.7 × 10–13
= Qq / 4πε0r
C1
7.7 × 10–13
= 8.99 × 109
× 79 × 2 × (1.60 × 10–19
)2
/ r M1
r = 4.7 × 10–14
m
r is closest distance of approach so radius less than this
A1
5(b) force = Qq / 4πε0r 2
= 4u × a C1
8.99 × 109
× 79 × 2 × (1.60 × 10–19
)2
/ (4.7 × 10–14
)2
= 4 × 1.66 × 10–27
× a C1
a = 2.5 × 1027
m s–2
A1
5(c) so that single interactions between nucleus and α-particle can be studied
or
so that multiple deflections with nucleus do not occur
B1

PUBLISHED
May/June 2017
6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1
op-amp can deliver only a small current/small voltage B1
6(a)(ii) correct symbol for relay coil connected between output and earth B1
switch between mains supply and lamp B1
6(b)(i) vary light intensity at which lamp is switched on/off B1
6(b)(ii) so that relay operates for only one current/voltage direction
or
so that relay/lamp operates for either dark or light conditions
B1
6(c) when light level increases, LDR resistance decreases B1
(RLDR low,) so V –
> V+
, so VOUT negative/–5 V (must be consistent with B1 mark) M1
or
when light level decreases, LDR resistance increases (B1)
(RLDR high,) so V –
< V+
, so VOUT is positive/+5 V (must be consistent with B1 mark) (M1)
lamp comes on as light level decreases
or
lamp goes off as light level increases
A1

PUBLISHED
May/June 2017
7(a) (magnetic) force (always) normal to velocity/direction of motion M1
(magnitude of magnetic) force constant
or
speed is constant/kinetic energy is constant
M1
so provides the centripetal force A1
7(b) increase in KE = loss in PE or ½ mv 2
= qV M1
p = mv with algebra leading to p = √(2mqV) A1
7(c) Bqv = mv2
/ r
mv = Bqr or p = Bqr
C1
(2 × 9.11 × 10–31
× 1.60 × 10–19
× 120)1/2
= B × 1.60 × 10–19
× 0.074 C1
B = 5.0 × 10–4
T A1
7(d) greater momentum M1
(p = Bqr and) so r increased A1

PUBLISHED
May/June 2017
8 strong (uniform) magnetic field B1
* nuclei precess/rotate about field (direction)
radio frequency pulse/RF pulse (applied) B1
* RF or pulse is at Larmor frequency / frequency of precession
causes resonance / excitation (of nuclei)/nuclei to absorb energy B1
on relaxation/de-excitation, nuclei emit RF/pulse B1
* (emitted) RF/pulse detected and processed
non-uniform field (superposed on uniform field) B1
allows positions of (resonating) nuclei to be determined B1
* allows for position of detection to be changed/different slices to be studied
max. 2 of additional detail points marked * B2

PUBLISHED
May/June 2017
9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1
9(a)(ii) reduces (size of eddy) currents in core B1
(so that) heating of core is reduced B1
9(b) alternating voltage gives rise to changing magnetic flux in core M1
(changing) flux links the secondary coil A1
induced e.m.f. (in secondary) only when flux is changing/cut B1
10(a)(i) penetration of beam M1
greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1
10(a)(ii) greater accelerating potential difference
or
greater p.d. between anode and cathode
B1
10(b) I = I0 exp(–µx)
ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55})
C1
= 0.58 A1

PUBLISHED
May/June 2017
11(a) electrons (in gas atoms/molecules) interact with photons B1
photon energy causes electron to move to higher energy level/to be excited B1
photon energy = difference in energy of (electron) energy levels B1
when electrons de-excite, photons emitted in all directions (so dark line) B1
11(b)(i) photon energy ∝ 1 / λ C1
energy = 1.68 eV A1
or
E = hc / λ
E = 6.63 × 10–34
× 3.0 × 108
/ (740 × 10–9
)
= 2.688 × 10–19
J
(C1)
energy = 1.68 eV (A1)
11(b)(ii) 3.4 eV → 1.5 eV
3.4 eV → 0.85 eV
3.4 eV → 0.54 eV
all correct and none incorrect 2/2
2 correct and 1 incorrect or only 2 correctly drawn 1/2
B2

PUBLISHED
May/June 2017
12(a) x = 7 A1
12(b)(i) E = mc2
C1
= 1.66 × 10–27
× (3.0 × 108
)2
= 1.494 × 10–10
J
C1
division by 1.6 × 10–13
clear to give 934 MeV A1
12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4
)
or
∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4
)
C1
= 0.21053 u C1
energy = 0.21053 × 934
= 197 MeV
A1
12(c) kinetic energy of nuclei/particles/products/fragments B1
γ–ray photon energy B1

PHYSICS 9702/42
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a) force per unit mass B1
1(b)(i) g = GM / r 2
= (6.67 ×10–11
× 1.0 × 1013
) / (3.6 × 103
)2
C1
= 5.1 × 10–5
N kg–1
A1
1(b)(ii) mass = (960 / 9.81) kg
weight on comet = (960 / 9.81) × 5.1 ×10–5
C1
= 5.0 × 10–3
N A1
1(c) similarity: e.g. both attractive/pointed towards the comet
e.g. same order of magnitude
B1
difference: e.g. radial/non-radial
e.g. same (over surface)/varies (over surface)
B1

PUBLISHED
May/June 2017
2(a)(i) mean/average square speed/velocity B1
2(a)(ii) pV = NkT or pV = nRT B1
ρ = Nm /V
or
ρ = nNAm / V and k = nR / N
B1
EK = ½ m〈c2
〉 with algebra to (3 / 2)kT B1
2(b)(i) no (external) work done or ∆U = q or w = 0 B1
q = NA × (3 / 2)k × 1.0 M1
NAk = R so q = (3 / 2)R A1
2(b)(ii) specific heat capacity = {(3 / 2) × R} / 0.028 C1
= 450 J kg–1
K–1
A1

PUBLISHED
May/June 2017
3(a)(i) e.g. period = 6 / 2.5 C1
3(a)(ii) energy = ½ m × 4π2
f 2
y0
2
C1
= ½ × 0.25 × 4π2
× 0.422
× (1.5 × 10–2
)2
C1
= 2.0 × 10–4
J A1
3(b)(i) (induced) e.m.f. proportional to rate of M1
change of magnetic flux (linkage)
or
cutting of magnetic flux
A1
3(b)(ii) coil cuts flux/field (of moving magnet) inducing e.m.f. in coil B1
(induced) current in resistor causes heating (effect) M1
thermal energy/heat derived from energy of oscillations (of magnet) A1

PUBLISHED
May/June 2017
4(a) pulse (of ultrasound) B1
* produced by quartz crystal/piezo-electric crystal
* gel/coupling medium (on skin) used to reduce reflection at skin
reflected from boundaries (between media) B1
reflected pulse/wave detected by (ultrasound) transmitter B1
reflected wave processed and displayed B1
* intensity of reflected pulse/wave gives information about boundary
* time delay gives information about depth of boundary
4(b) IT = I0 exp (–µx) C1
2.9 = exp (4.6µ) C1
µ = 0.23 cm–1
A1

PUBLISHED
May/June 2017
5(a) any two reasonable suggestions e.g.
• signal can be regenerated/noise removed (not “no noise”)
• circuits more reliable
• circuits cheaper to produce
• multiplexing (is possible)
• error correction/checking
• easier encryption/better security
B2
5(b)(i) samples the analogue signal M1
at regular intervals and converts it (to a digital number) A1
5(b)(ii) 1. smaller step depth B1
2. smaller step height B1

PUBLISHED
May/June 2017
6(a) force proportional to product of charges and inversely proportional to the square of the separation M1
reference to point charges A1
6(b)(i) (near to each sphere,) fields are in opposite directions
or
point (between spheres) where fields are equal and opposite
or
point (between spheres) where field strength is zero
M1
so same (sign of charge) A1
6(b)(ii) (at x = 5.0 cm,) E = 3.0 × 103
V m–1
and a = qE / m C1
E = (1.60 × 10–19
× 3.0 × 103
) / (1.67 × 10–27
) C1
= 2.9 × 1011
ms–2
A1
6(c) field strength or E is potential gradient
or
field strength is rate of change of (electric) potential
M1
(field strength) maximum at x = 6 cm A1

PUBLISHED
May/June 2017
7(a) equal and opposite charges on the plates so no resultant charge B1
+ve and –ve charges separated so energy stored B1
7(b) charge / potential difference M1
reference to charge on one plate and p.d. between plates A1
7(c) energy = ½ CV2
or
energy = ½ QV and C = Q / V
C1
(1 / 16) × ½ CV0
2
= ½ CV2
V = ¼ V0
A1

PUBLISHED
May/June 2017
8(a)(i) circle around both diodes B1
8(a)(ii) indicates (whether) temperature M1
(is) above or below a set value A1
8(b)(i) (when resistance of C > RV,) V–
> V+
or V+
< 3V
or
p.d. across RV < p.d. across R/Y/3V
or
p.d. across C > p.d. across R/ X/3V
M1
op-amp output is negative M1
(only) green A1
8(b)(ii) resistance of C becomes less than RV
or
V–
< V+
B1
green (LED) goes out A1
blue (LED) comes on A1
8(c) changes/determines temperature at which LEDs switch B1

PUBLISHED
May/June 2017
9(a)(i) Hall voltage depends on thickness of slice C1
thinner slice, larger Hall voltage A1
9(a)(ii) Hall voltage depends on current in slice B1
9(b) sinusoidal wave, one cycle B1
at θ = 0 and at θ = 360°, VH = VMAX B1
at θ = 180°, VH = –VMAX B1

PUBLISHED
May/June 2017
10(a) two from:
• frequency below which electrons not ejected
• maximum energy of electron depends on frequency
• maximum energy of electrons does not depend on intensity
• instantaneous emission of electrons
B2
10(b)(i) (λ0 is the) threshold wavelength
or
wavelength corresponding to threshold frequency
or
maximum wavelength for emission of electrons
B1
10(b)(ii)1. intercept = 1 /λ0 = 2.2 × 106
m–1
λ0 = 4.5 × 10–7
m or 450 nm
A1
10(b)(ii)2. gradient = hc C1
gradient = 2.0 × 10–25
or correct substitution into gradient formula C1
h = (2.0 × 10–25
) / (3.0 × 108
) = 6.7 × 10–34
J s A1
10(c) line: same gradient B1
straight line, positive gradient, intercept at greater than 2.2 × 106
when candidate’s line extrapolated B1

PUBLISHED
May/June 2017
11(a) loss of (electric) potential energy = gain in kinetic energy
or
qV = ½ mv2
or
EK = p2
/ 2m = qV
B1
p = mv with algebra leading to p = √(2mqV) B1
11(b)(i) particle/electron has a wavelength (associated with it) B1
dependent on its momentum
or
when/because particle is moving
B1
11(b)(ii) p = (2 × 9.11 × 10–31
× 1.60 × 10–19
× 120)1/2
C1
λ = (6.63 × 10–34
) / (5.91 × 10–24
) C1
= 1.12 × 10–10
m A1
11(c) wavelength is similar to separation of atoms M1
so diffraction observed A1

PUBLISHED
May/June 2017
12(a) 7 e0
1−
A1
12(b)(i) E = mc2
C1
= 1.66 × 10–27
× (3.00 × 108
)2
M1
= 1.494 × 10–10
J
clear to give 934 MeV
A1
12(b)(ii) ∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u
= (–) 1.16762 (u)
C1
energy = 1.16762 × 934 C1
energy per nucleon = (1.16762 × 934) / 139
= 7.85 MeV
A1
12(c) above A = 56, binding energy per nucleon decreases as A increases B1
U-235 has larger nucleon number M1
so less (binding energy per nucleon) A1
or
fission takes place with uranium (B1)
fission reaction releases energy (M1)
binding energy per nucleon less (for uranium than for products) (A1)

PHYSICS 9702/43
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1(a) gravitational force (of attraction between satellite and planet) B1
provides / is centripetal force (on satellite about the planet) B1
1(b) M = (4/3) × πR3
ρ B1
ω = 2π / T or v = 2πnR / T B1
GM / (nR)2
= nRω2
or v 2
/ nR M1
substitution clear to give ρ = 3πn3
/ GT 2
A1
1(c) n = (3.84 × 105
) / (6.38 × 103
) = 60.19 or 60.2 C1
ρ = 3π × 60.193
/ [(6.67 × 10–11
) × (27.3 × 24 × 3600)2
] C1
ρ = 5.54 × 103
kg m–3
A1

PUBLISHED
May/June 2017
2(a) e.g. period = 3 / 2.5 C1
2(b) light (damping) B1
2(c) at 2.7 s, A0 = 1.5 (cm) B1
energy = ½ m × 4π2
f 2
A0
2
B1
= ½ × 0.18 × 4π2
× 0.832
× (1.5 × 10–2
)2
= 5.51 × 10–4
(J)
C1
at 7.5 s, A0 = 0.75 (cm) B1
energy = ¼ × 5.51 × 10–4
or
energy = ½ × 0.18 × 4π2
× 0.832
× (0.75 × 10–2
)2
C1
energy = 1.38 × 10–4
(J)
change = (5.51 × 10–4
– 1.38 × 10–4
) = 4.13
J
A1

PUBLISHED
May/June 2017
3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1
3(a)(ii) component X: parallel-to-serial converter B1
component Y: DAC/digital-to-analogue converter B1
3(a)(iii) sample the (analogue) signal M1
at regular intervals and converts the analogue number to a digital number A1
3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1
ratio = 16 + 28
= 44 dB
A1
3(b)(ii) ratio / dB = 10 lg (P2 / P1) C1
44 = 10 lg ({9.7 × 10–3
} / P)
or
–44 = 10 lg (P / {9.7 × 10–3
})
C1
power = 3.9 × 10–7
W A1

PUBLISHED
May/June 2017
4(a) random/haphazard B1
constant velocity or speed in a straight line between collisions
or
distribution of speeds/different directions
B1
4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1
moving haphazardly/randomly/jerky/in a zigzag fashion A1
4(c)(i) pV = ⅓ Nm〈c2
〉
1.05 × 105
× 0.0240 = ⅓ × 4.00 × 10–3
× 〈c2
〉
C1
〈c2
〉 = 1.89 × 106
C1
or
½ m〈c2
〉 = (3 / 2) kT
0.5 × (4.00 × 10–3
/ 6.02 × 1023
) × 〈c2
〉 = 1.5 × 1.38 × 10–23
× 300
(C1)
〈c2
〉 = 1.87 × 106
(C1)
or
nRT = ⅓ Nm〈c2
〉
1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3
× 〈c2
〉
(C1)
〈c2
〉 = 1.87 × 106
(C1)
cr.m.s. = 1.37 × 103
m s–1
A1

PUBLISHED
May/June 2017
4(c)(ii) 〈c2
〉 ∝ T C1
〈c2
〉 at 177 °C = 1.89 × 106
× (450 / 300) C1
cr.m.s. at 177 °C = 1.68 × 103
m s–1
A1
5(a) (loss in) kinetic energy of α-particle = Qq / 4πε0r
or
7.7 × 10–13
= Qq / 4πε0r
C1
7.7 × 10–13
= 8.99 × 109
× 79 × 2 × (1.60 × 10–19
)2
/ r M1
r = 4.7 × 10–14
m
r is closest distance of approach so radius less than this
A1
5(b) force = Qq / 4πε0r 2
= 4u × a C1
8.99 × 109
× 79 × 2 × (1.60 × 10–19
)2
/ (4.7 × 10–14
)2
= 4 × 1.66 × 10–27
× a C1
a = 2.5 × 1027
m s–2
A1
5(c) so that single interactions between nucleus and α-particle can be studied
or
so that multiple deflections with nucleus do not occur
B1

PUBLISHED
May/June 2017
6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1
op-amp can deliver only a small current/small voltage B1
6(a)(ii) correct symbol for relay coil connected between output and earth B1
switch between mains supply and lamp B1
6(b)(i) vary light intensity at which lamp is switched on/off B1
6(b)(ii) so that relay operates for only one current/voltage direction
or
so that relay/lamp operates for either dark or light conditions
B1
6(c) when light level increases, LDR resistance decreases B1
(RLDR low,) so V –
> V+
, so VOUT negative/–5 V (must be consistent with B1 mark) M1
or
when light level decreases, LDR resistance increases (B1)
(RLDR high,) so V –
< V+
, so VOUT is positive/+5 V (must be consistent with B1 mark) (M1)
lamp comes on as light level decreases
or
lamp goes off as light level increases
A1

PUBLISHED
May/June 2017
7(a) (magnetic) force (always) normal to velocity/direction of motion M1
(magnitude of magnetic) force constant
or
speed is constant/kinetic energy is constant
M1
so provides the centripetal force A1
7(b) increase in KE = loss in PE or ½ mv 2
= qV M1
p = mv with algebra leading to p = √(2mqV) A1
7(c) Bqv = mv2
/ r
mv = Bqr or p = Bqr
C1
(2 × 9.11 × 10–31
× 1.60 × 10–19
× 120)1/2
= B × 1.60 × 10–19
× 0.074 C1
B = 5.0 × 10–4
T A1
7(d) greater momentum M1
(p = Bqr and) so r increased A1

PUBLISHED
May/June 2017
8 strong (uniform) magnetic field B1
* nuclei precess/rotate about field (direction)
radio frequency pulse/RF pulse (applied) B1
* RF or pulse is at Larmor frequency / frequency of precession
causes resonance / excitation (of nuclei)/nuclei to absorb energy B1
on relaxation/de-excitation, nuclei emit RF/pulse B1
* (emitted) RF/pulse detected and processed
non-uniform field (superposed on uniform field) B1
allows positions of (resonating) nuclei to be determined B1
* allows for position of detection to be changed/different slices to be studied

PUBLISHED
May/June 2017
9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1
9(a)(ii) reduces (size of eddy) currents in core B1
(so that) heating of core is reduced B1
9(b) alternating voltage gives rise to changing magnetic flux in core M1
(changing) flux links the secondary coil A1
induced e.m.f. (in secondary) only when flux is changing/cut B1
10(a)(i) penetration of beam M1
greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1
10(a)(ii) greater accelerating potential difference
or
greater p.d. between anode and cathode
B1
10(b) I = I0 exp(–µx)
ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55})
C1
= 0.58 A1

PUBLISHED
May/June 2017
11(a) electrons (in gas atoms/molecules) interact with photons B1
photon energy causes electron to move to higher energy level/to be excited B1
photon energy = difference in energy of (electron) energy levels B1
when electrons de-excite, photons emitted in all directions (so dark line) B1
11(b)(i) photon energy ∝ 1 / λ C1
energy = 1.68 eV A1
or
E = hc / λ
E = 6.63 × 10–34
× 3.0 × 108
/ (740 × 10–9
)
= 2.688 × 10–19
J
(C1)
energy = 1.68 eV (A1)
11(b)(ii) 3.4 eV → 1.5 eV
3.4 eV → 0.85 eV
3.4 eV → 0.54 eV
all correct and none incorrect 2/2
2 correct and 1 incorrect or only 2 correctly drawn 1/2
B2

PUBLISHED
May/June 2017
12(a) x = 7 A1
12(b)(i) E = mc2
C1
= 1.66 × 10–27
× (3.0 × 108
)2
= 1.494 × 10–10
J
C1
clear to give 934 MeV A1
12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4
)
or
∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4
)
C1
= 0.21053 u C1
energy = 0.21053 × 934
= 197 MeV
A1
12(c) kinetic energy of nuclei/particles/products/fragments B1
γ–ray photon energy B1

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation May/June 2017
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
1 Defining the problem
(sin) θ is the independent variable and v is the dependent variable or vary (sin) θ and measure v 1
keep s (PQ) constant 1
Methods of data collection
labelled diagram showing inclined plane with labelled support and P and Q marked 1
method to measure angle e.g. use a protractor to measure θ or use a ruler to measure marked distances from which sin θ or θ
may be determined
1
method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light
gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger
1
measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card
to interrupt light beam or distance from motion sensor to Q
1
Method of analysis
plot a graph of v2
against sin θ 1
relationship valid if a straight line produced (not passing through the origin) 1
+
= − ×gradient
2
B m
g
ms
or
=g
+
×-intercept
2
B m
y
Bs
1

PUBLISHED
May/June 2017
Additional detail including safety considerations Max. 6
D1 use cushion/foam/sandbox for falling body (B)
D2 (sin) θ determined using trigonometry relationship using marked lengths
D3 appropriate equation to determine v (at Q) e.g. =
2s
v
t
D4 repeat experiment for each θ and average v or t
D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s
D6 =
+
2
-intercept .
Bsg
y
B m
D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth
D8 keep B and m constant or keep mass of block and mass of falling body constant
D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of
block
D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop

PUBLISHED
May/June 2017
2(a)
gradient = −
2
v
y-intercept =
2L
v
1
2(b)
0.80 ± 0.01
0.77 ± 0.01
0.73 ± 0.01
0.70 ± 0.01
0.66 ± 0.01
0.62 ± 0.01
First mark for all values of t correct.
Second mark for uncertainties correct.
2
2(c)(i) Six points plotted correctly.
Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square.
1
Error bars in t plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1

PUBLISHED
May/June 2017
2(c)(ii) Line of best fit drawn.
If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line
should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.
1
Worst acceptable line drawn (steepest or shallowest possible line).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line.
Gradient must be negative.
1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
1
2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.
= − = −
2 2
gradient 2(c)(iii)
v
1
L determined from y-intercept and v and L given to 2 or 3 significant figures.
Correct substitution of numbers must be seen.
= × = × = − = −
-intercept (c)(iv)
-intercept (c)(iv)
2 2 gradient (c)(iii)
v v y
L y
1

PUBLISHED
May/June 2017
2(d)(ii) % uncertainty in v = % uncertainty in gradient 1
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient
or
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in v
Maximum/minimum methods:
= ×
max -intercept
Max max -intercept max or
mingradient
y
L y v
= ×
min -intercept
Min min -intercept min or
max gradient
y
L y v
1

PHYSICS 9702/52
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
r is the independent variable and f (frequency of turntable) is the dependent variable or vary r and measure f (frequency of
turntable)
1
keep m constant 1
labelled diagram showing power supply connected to motor (two leads) within turntable; circuits must be workable 1
method to change frequency of rotation of the turntable, e.g. adjust output of (variable) power supply or adjust variable
resistor
1
increase frequency until the cube moves (relative to the turntable) 1
method to determine period of rotation of the turntable, e.g. stopwatch, light gate attached to a timer/data-logger or
stroboscope
1
Method of analysis
plots a graph of f against 1 / r (allow log f against log r) 1
relationship valid if a straight line produced passing through the origin
(for lg f vs. lg r straight line of gradient of –1)
1
K = gradient × 4π2
m
(for lg f vs. lg r, K = 10y-intercept
× 4π2
m)
1

PUBLISHED
May/June 2017
D1 use safety screen
D2 time at least 10 rotations of turntable or detailed use of stroboscope
D3 f = 1 / T for correct determination of period of rotation of turntable
D4 repeat experiment for each r and average f
D5 use balance to measure mass of cube
D6 wait for turntable to rotate steadily before increasing frequency
or
gradual/incremental/slowly increase in frequency
D7 use a spirit level to check that turntable is horizontal or clean cube/surface
D8 use a rule to measure r
D9 method to ensure r is measured to the centre of the cube, e.g. put a mark on the cube or align front or back of cube
by a set distance
D10 method to determine centre of the turntable e.g. measure two or more diameters/maximum distance ideas

PUBLISHED
May/June 2017
2(a)
gradient =
1
E
y-intercept =
Q
E
1
2(b)
P / Ω
1
I
/ A–1
± 9 29 or 29.4
± 11 36 or 35.7
± 16.5 53 or 52.6
± 23.5 71 or 71.4
± 28 83 or 83.3
± 34 100
First mark for uncertainties correct. Allow 1 s.f. e.g. 10, 10, 20, 20, 30, 30.
Second mark for all second column correct. Allow a mixture of significant figures.
2
1
Error bars in P plotted correctly.
1

PUBLISHED
May/June 2017
If points are plotted correctly then lower end of line should pass between (200, 32) and (200, 34) and upper end of line should
pass between (600, 88) and (600, 91).
1
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. 1
or
1
2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1
or
1

PUBLISHED
May/June 2017
2(d)(i) E determined using gradient and units for E and Q with correct power of ten.
= =
1 1
gradient 2(c)(iii)
E
1
Q determined using y-intercept and E and Q given to 2 or 3 significant figures.
= × = × = =
-intercept 2(c)(iv)
-intercept 2(c)(iv)
gradient 2(c)(iii)
y
Q E y E
1
2(d)(ii) % uncertainty in E = % uncertainty in gradient 1
% uncertainty in Q = % uncertainty in E + % uncertainty in y-intercept
or
% uncertainty in Q = % uncertainty in gradient + % uncertainty in y-intercept.
= ×
max -intercept
mingradient
y
Q y E
= ×
min -intercept
max gradient
y
Q y E
1

PHYSICS 9702/53
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
,
components.

PUBLISHED
May/June 2017
(sin) θ is the independent variable and v is the dependent variable or vary (sin) θ and measure v 1
keep s (PQ) constant 1
labelled diagram showing inclined plane with labelled support and P and Q marked 1
method to measure angle e.g. use a protractor to measure θ or use a ruler to measure marked distances from which sin θ or θ
may be determined
1
method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light
gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger
1
measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card
to interrupt light beam or distance from motion sensor to Q
1
Method of analysis
plot a graph of v2
against sin θ 1
relationship valid if a straight line produced (not passing through the origin) 1
+
= − ×gradient
2
B m
g
ms
or
=g
+
×-intercept
2
B m
y
Bs
1

PUBLISHED
May/June 2017
D1 use cushion/foam/sandbox for falling body (B)
D2 (sin) θ determined using trigonometry relationship using marked lengths
D3 appropriate equation to determine v (at Q) e.g. =
2s
v
t
D4 repeat experiment for each θ and average v or t
D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s
D6 =
+
2
-intercept .
Bsg
y
B m
D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth
D8 keep B and m constant or keep mass of block and mass of falling body constant
D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of
block
D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop

PUBLISHED
May/June 2017
2(a)
gradient = −
2
v
y-intercept =
2L
v
1
2(b)
0.80 ± 0.01
0.77 ± 0.01
0.73 ± 0.01
0.70 ± 0.01
0.66 ± 0.01
0.62 ± 0.01
First mark for all values of t correct.
Second mark for uncertainties correct.
2
1
Error bars in t plotted correctly.
1

PUBLISHED
May/June 2017
If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line
should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.
1
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line.
Gradient must be negative.
1
or
1
2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1
or
1
2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.
= − = −
2 2
gradient 2(c)(iii)
v
1
L determined from y-intercept and v and L given to 2 or 3 significant figures.
= × = × = − = −
-intercept (c)(iv)
-intercept (c)(iv)
2 2 gradient (c)(iii)
v v y
L y
1

PUBLISHED
May/June 2017
2(d)(ii) % uncertainty in v = % uncertainty in gradient 1
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient
or
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in v
= ×
max -intercept
mingradient
y
L y v
= ×
min -intercept
max gradient
y
L y v
1

9702 s17 ms_all

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