9702 m19 ms_all

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Cambridge Assessment International Education
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/12
Paper 1 Multiple Choice March 2019
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 International will not enter into discussions about these mark schemes.
Cambridge International is publishing the mark schemes for the March 2019 series for most Cambridge
IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9702/12 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
March 2019
© UCLES 2019 Page 2 of 3
Question Answer Marks
1 A 1
2 B 1
3 D 1
4 A 1
5 B 1
6 A 1
7 C 1
8 B 1
9 A 1
10 C 1
11 C 1
12 D 1
13 D 1
14 D 1
15 C 1
16 D 1
17 B 1
18 D 1
19 B 1
20 D 1
21 C 1
22 D 1
23 C 1
24 D 1
25 D 1
26 C 1
27 C 1
28 A 1

PUBLISHED
March 2019
29 B 1
30 A 1
31 B 1
32 D 1
33 C 1
34 A 1
35 D 1
36 B 1
37 A 1
38 D 1
39 D 1
40 D 1

PHYSICS 9702/22
Paper 2 AS Level Structured Questions March 2019
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.

PUBLISHED
March 2019
Generic Marking Principles
These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the
specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these
marking principles.
GENERIC MARKING PRINCIPLE 1:
Marks must be awarded in line with:
• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.
Marks awarded are always whole marks (not half marks, or other fractions).
Marks must be awarded positively:
• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the
scope of the syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the
question as indicated by the mark scheme. The meaning, however, should be unambiguous.
Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

PUBLISHED
March 2019
Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may
be limited according to the quality of the candidate responses seen).
Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

PUBLISHED
March 2019
1(a) kilogram / kg B1
kelvin / K B1
1(b) units for v: m s–1 and units for F: kg m s–2
C1
units for e: A s C1
units for µ: ms–1 A s / kg m s–2
= A kg–1 s2
A1
2(a)(i) distance in a specified direction (from a point) B1
2(a)(ii) change in velocity / time (taken) B1
2(b)(i) constant velocity so no resultant force B1
no resultant force so in equilibrium B1
2(b)(ii) (difference in height =) 47 × 2.8 × 60 × sin24° = 3200 m A1

PUBLISHED
March 2019
2(b)(iii) 1 (∆)E = mg(∆)h
= 85 × 9.81 × 3200
C1
= 2.7 × 106 J A1
2 In terms of energy:
work done = 2.7 × 106 J
force = 2.7 × 106 / (47 × 2.8 × 60)
C1
= 340 N A1
In terms of forces:
component of weight along path = force due to air resistance
force = 85 × 9.81 × sin24°
(C1)
= 340 N (A1)
2(b)(iv) (∆)p = ρg(∆)h
(92 – 63) × 103 = ρ × 9.81 × 3200
C1
ρ = 0.92 kg m–3 A1
3(a) (m × 3.0) or (2.5 × 9.6 × cos 60°) C1
(m × 3.0) – (2.5 × 9.6 × cos 60°) = 0 so m = 4.0 (kg) A1

PUBLISHED
March 2019
3(b) 2.5 × 9.6 × sin60° = (4.0 + 2.5) × V C1
V = 3.2 m s–1
A1
or
use of momentum vector triangle:
(4.0 × 3.0)2 + [(4.0 + 2.5) × V]2 = (2.5 × 9.6)2
(C1)
V = 3.2 m s–1
(A1)
3(c) E = ½mv 2
difference in EK = ½ × 2.5 × (9.6)2 – ½ × 4.0 × (3.0)2
C1
= 97 J A1
4(a) force per unit positive charge B1
4(b)(i) 1 E = V / d or E = ∆V / ∆d
d = 4.0 × 103 / 5.0 × 104
C1
= 8.0 × 10–2 m A1
2 plates are (in) horizontal (plane) (above and below the rod) B1
top (plate) negative and bottom (plate) positive B1
4(b)(ii) magnitude = 5.0 × 104 × 3 × 1.6 × 10–19
= 2.4 × 10–14 N
A1
direction is (vertically) downwards / down B1

PUBLISHED
March 2019
4(b)(iii) 6.2 × 10–16 = 2.4 × 10–14 × 72 × 10–3 × cosθ C1
θ = 69° A1
5(a)(i) (two) waves meet/overlap (at a point) B1
(resultant) displacement is sum of the displacement of each wave B1
5(a)(ii) constant phase difference (between the waves) B1
5(b) I ∝ A 2
3I / I = (A + 1.5)2 / 1.52
C1
A = 1.1 cm A1
5(c)(i) λ = ax / D C1
e.g. a = 680 × 10–9 × 2.0 / 4.0 × 10–3 C1
a = 3.4 × 10–4 m A1
5(c)(ii) straight line from positive value on x-axis and always below ‘old’ line B1
straight line with a smaller positive gradient than ‘old’ line B1
6(a) e.m.f.: energy transferred from chemical to electrical (per unit charge) B1
p.d.: energy transferred from electrical to thermal (per unit charge) B1

PUBLISHED
March 2019
6(b)(i) 1 I = 4.8 / 32
= 0.15 A
A1
2 P = EI or P = VI or P = I 2R or P = V 2 / R
= 6.0 × 0.15 or 0.152 × 40 or 6.02 / 40
C1
= 0.90 W A1
3 number = It / e
= [0.15 × 25] / 1.6 × 10–19
C1
= 2.3 × 1019 A1
or
Q = 0.15 × 25 (= 3.75)
number = 3.75 / 1.6 × 10–19
(C1)
= 2.3 × 1019 (A1)
4 4.8 / 6.0 = 32 / (RXY + 32) or 1.2 / 6.0 = RXY / (RXY + 32)
or 4.8 / 1.2 = 32 / RXY
C1
RXY = 8.0 Ω A1
Alternative methods:
RXY = (6.0 – 4.8) / 0.15 or (C1)
= 8.0 Ω (A1)
or
6.0 = 0.15 (32 + RXY)
(C1)
RXY = 40 – 32
= 8.0 Ω
(A1)

PUBLISHED
March 2019
6(b)(i) 5 1 / 8.0 = 1 / RX + 1 / 24 C1
Rx = 12 Ω A1
Alternative method:
IZ = 4.8 / 32 = 0.15 and IY = 1.2 / 24 = 0.05
IX = 0.15 – 0.05 (= 0.10)
(C1)
RX = 1.2 / 0.10 = 12 Ω (A1)
6(b)(ii) total resistance decreases M1
(so voltmeter) reading increases A1
7(a)(i) alpha, neutron and proton B1
7(a)(ii) neutron B1
7(a)(iii) beta plus or β+ B1
7(b) d has charge (+)⅓ e C1
(so) other quark has charge = e – ⅓ e
= (+)⅔ e
M1
other quark is an up / u A1

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 March 2019
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
March 2019
marking principles.
descriptors.

PUBLISHED
March 2019

PUBLISHED
March 2019
1(a) value of x to nearest mm and in range 4.5 to 5.5 cm 1
1(b) value of T in range 1.50 s to 2.50 s 1
repeat readings – at least two values of at least 5T 1
1(c) six sets of readings of x and T with correct trend and without help from supervisor scores 4 marks, five sets scores 3 marks
etc.
4
range:
xmin ⩽ 2.5 cm and xmax ⩾ 13.5 cm
1
column headings:
each column heading must contain a quantity and a unit
the presentation of quantity and unit must conform to accepted scientific convention. e.g. T 2 / s2
1
consistency:
all values of raw times must be given to the nearest 0.1 s or all values to the nearest 0.01 s
1
significant figures:
significant figures for every value of 1 / x same as, or one greater than, the s.f. of x as recorded in table
1
calculation:
values of 1 / x calculated correctly
1

PUBLISHED
March 2019
1(d)(i) axes:
sensible scales must be used, no awkward scales (e.g. 3:10)
scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
scales must be labelled with the quantity which is being plotted
scale markings should be no more than 3 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)
plots must be accurate to within half a small square in both x and y directions
1
quality:
all points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of plots must be no more than
± 0.05 cm–1 from a straight line in the 1 / x direction
1
1(d)(ii) line of best fit:
judged by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
either side of the line along the full length
one anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the candidate
lines must not be kinked or thicker than half a small square
1
1(d)(iii) gradient:
the hypotenuse of the triangle used must be greater than half the length of the drawn line
method of calculation must be correct
both read-offs must be accurate to half a small square in both the x and y directions
1
y-intercept:
Either
correct read-off from a point on the line substituted into y = mx + c or an equivalent expression, with read-off accurate to half a s
square in both x and y directions
Or
intercept read directly from the graph, with read-off at x = zero accurate to half a small square in y direction
1

PUBLISHED
March 2019
1(e) a equal to candidate’s gradient, and b equal to candidate’s intercept, and values are not written as fractions 1
unit for a is correct and consistent with value
unit for b is correct
1
2(a) value for B, with unit 1
2(b) measured value(s) for d, with unit, to nearest mm 1
evidence of repeat readings 1
2(c) absolute uncertainty in d value ⩾ 0.2 cm and correct method of calculation to obtain percentage uncertainty.
if several readings have been taken, then the absolute uncertainty can be half the range if the working is clearly shown (but
not zero if values are equal)
1
2(d) calculation of P correct 1
POT of P value matches unit 1
2(e) justification based on s.f. in d, m and g 1
2(f) value for second B 1
value for second d 1
Quality:
d smaller for smaller B
1
2(g)(i) two values of k calculated correctly 1
2(g)(ii) sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate 1

PUBLISHED
March 2019
2(h)(i) two k values are not enough to draw a valid conclusion 4 max
difficult to measure B with reason, e.g. pressing on balloon / ruler scale doesn’t start at zero / parallax
difficult to balance Perspex block on balloon / balloon kept moving
difficult to measure d with reason, e.g. not circular / changes when block touched / parallax / refraction changes apparent size
difficult to see (edges of) contact patch (clearly)
2(h)(ii) take more readings and plot a graph / calculate more k values and compare 4 max
workable method of reducing parallax for B e.g. callipers
fix balloon in place with tape / glue / Blu-tack / trap between blocks at sides
grid on block / travelling microscope / thinner block
use contrasting colours for balloon and liquid

PHYSICS 9702/42
Paper 4 A Level Structured Questions March 2019
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
March 2019
marking principles.
descriptors.

PUBLISHED
March 2019

PUBLISHED
March 2019
1(a)(i) work done per unit mass B1
idea of work done moving mass from infinity (to the point) B1
1(a)(ii) (gravitational) force is attractive B1
(gravitational) potential at infinity is zero B1
decrease in potential energy as masses approach
or displacement and force in opposite directions
B1
1(b)(i) Either mv2 / R = GMm / R2
Or v = √( GM / R)
v2 = (6.67 × 10–11 × 6.00 × 1024) / (7.30 × 106)
C1
giving v = 7.4 × 103 m s–1 A1
1(b)(ii) VP = – GMm / R C1
= – (6.67 × 10–11 × 6.00 × 1024 × 340) / (7.30 × 106) C1
VP = – 1.9 × 1010 J A1
1(c) v2 ∝ 1 / r, (r smaller) so v greater M1
and EK greater A1

PUBLISHED
March 2019
2(a)(i) gas obeys formula pV / T = constant M1
symbols V and T explained A1
2(a)(ii) mean-square-speed (of atoms / molecules) B1
2(b)(i) use of T = 393 C1
pV = nRT C1
2.4 × 105 × 6.8 × 10–3 = n × 8.31 × 393
and N = n × 6.02 × 1023 = 3.0 × 1023
A1
or
pV = NkT
(C1)
2.4 × 105 × 6.8 × 10–3 = N × 1.38 x 10–23 × 393
hence N = 3.0 × 1023
(A1)
2(b)(ii) volume of one atom = 4 / 3πr3 C1
volume occupied = 3.0 × 1023 × 4 / 3 × π × (3.2 × 10–11 )3
= 4 × 10–8 m3
A1
2(b)(iii) assumption: volume of atoms negligible compared to volume of container / cylinder B1
4 × 10–8 (m3) << 6.8 × 10–3 (m3) so yes B1

PUBLISHED
March 2019
3(a)(i) mention of upthrust and weight B1
3(a)(ii) upthrust is greater than the weight B1
(resultant force is) upwards B1
3(b) A, ρ, g and M are constant B1
either acceleration ∝ – displacement
or acceleration ∝ displacement and (– sign indicates) a and x in opposite directions
B1
3(c)(i) either ω = 2π / T or ω = 2πf and f = 1 / T C1
ω = 2π / 1.3
= 4.8 rad s–1
A1
3(c)(ii) ω2 = Aρg / m C1
4.832 = (4.5 × 10–4 × ρ × 9.81) / 0.17 C1
ρ = 900 kg m–3 A1

PUBLISHED
March 2019
4(a) Any three from:
above the Equator
period 24 hours
orbits west to east
one particular orbital radius
B3
4(b) attenuation = 10 lg(P1 / P2)
194 = 10 lg (3.2 × 103 / P2)
C1
P2 = 1.3 × 10–16 W A1
4(c) advantage: e.g. no tracking required B1
disadvantage: e.g. longer time delay B1
5(a) region where charge experiences an (electric) force B1
5(b) graph: field strength zero from x = 0 to x = R B1
curve with negative gradient, decreasing from x = R to x = 3R B1
line passes through field strength E at x = R, B1
line passes through field strength 0.25E at x = 2R and field strength 0.11E at x = 3R B1

PUBLISHED
March 2019
5(c) field strength = q / 4πϵ0x2 C1
2.0 × 106 = q / (4 × π × 8.85 × 10–12 × 0.262) C1
q = 1.5 × 10–5 C A1
6(a) charge / potential (difference) M1
charge on one plate, p.d. between the plates A1
6(b)(i) all three capacitors connected in series B1
6(b)(ii) 8 ( µF) in parallel with the two 4 (µF) capacitors connected in series B1
6(c) discharge from 7.0 V to 4.0 V C1
Either energy = ½CV2 or energy = ½ QV and C = Q / V C1
energy = ½ × 47 × 10–6 × (72 – 42)
= 7.8 × 10–4 J
A1
7(a)(i) output voltage / input voltage B1
7(a)(ii) no time delay between input and output B1
clear reference to change(s) in input and / or output B1
7(b)(i) VIN only connected to non-inverting input B1
midpoint between R1 and R2 only connected to inverting input B1

PUBLISHED
March 2019
7(b)(ii) gain = 1 + (R1 / R2)
25 = 1 + (12 × 103) / R2
C1
R2 = 500 Ω A1
7(b)(iii) VMAX = 9/25
= 0.36 V
C1
range is –0.36 V to + 0.36 V A1
8(a)(i) Either Newton’s third law or equal and opposite forces B1
force on magnet is upwards B1
so force on wire downwards B1
8(a)(ii) using (Fleming’s) left-hand rule M1
current from B to A A1
8(b) sinusoidal wave with at least 1 cycle B1
peaks at +6.4 mN and –6.4 mN B1
time period 25 ms B1

PUBLISHED
March 2019
9 X-rays (are used) B1
(object is) scanned in sections / slices B1
either: scans taken at many angles / directions
or images of each section / slice are 2-dimensional
B1
scans of many sections / slices are combined B1
(to give) 3-dimensional image (of whole structure) B1
10(a) single straight line along full length of solenoid B1
at least two more parallel lines along full length of solenoid B1
correct direction – right to left B1
10(b) (induced) e.m.f. proportional / equal to rate M1
of change of (magnetic) flux (linkage) A1
10(c) increasing current causes increasing flux B1
increasing flux induces e.m.f. in coil B1
(induced) e.m.f. opposes growth of current B1

PUBLISHED
March 2019
11(a) quantum / packet / discrete amount of energy M1
of electromagnetic radiation A1
11(b) E = hc / λ C1
= (6.63 × 10–34 × 3.0 × 108) / (540 × 10–9) C1
= (3.68 × 10–19) / (1.6 × 10–19)
= 2.3 eV
A1
11(c) Any 4 from:
photon absorbed by electron in valence band (1)
photon energy > energy of forbidden band (1)
electron promoted to conduction band (1)
hole left in valence band (1)
more charge carriers so lower resistance (1)
B4
12(a)(i) fission B1
12(a)(ii) either 0
–1e or 0
–1β M1
7 A1
12(b)(i) energy = c2 ∆m
= 0.223 × 1.66 × 10–27 × (3.00 × 108)2
C1
= 3.33 × 10–11 J A1

PUBLISHED
March 2019
12(b)(ii) Any 2 from:
kinetic energy of products
gamma photons
neutrinos
B2

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation March 2019
MARK SCHEME
Maximum Mark: 30
Published
Teachers.

PUBLISHED
March 2019
marking principles.
descriptors.

PUBLISHED
March 2019

PUBLISHED
March 2019
1 Defining the problem
R is the independent variable and T is the dependent variable, or vary R and measure T 1
keep C constant 1
Methods of data collection
labelled diagram or correct symbols of workable circuit including:
• (d.c.) power supply correctly positioned
• (neon) lamp correctly positioned
do not accept ohmmeter in circuit
1
circuit diagram to determine resistance of resistors e.g. using ammeter and voltmeter OR ohmmeter 1
method to determine period or T, e.g. use a stopwatch / timer / oscilloscope
do not accept counting the flashes in a specified time
1
circuit diagram showing voltmeter(s) or oscilloscope(s) to determine Vi and VF 1
Method of Analysis
plots a graph of T against R 1
gradient
K
C
=
1
( ) ( )
gradient
K C
L i i F i i FV V V V e V V V e= − − = − −
1

PUBLISHED
March 2019
1 Additional detail including safety considerations Max 6
switch off (high voltage) circuit (before changing the resistor) / wear insulating gloves to prevent electrocution / shock D1
resistance of resistors linked to diagram is V / I for ammeter / voltmeter method or gradient of appropriate graph or resistance
from ohmmeter
D2
input voltage or Vi is constant D3
repeat experiment for each value of R and average T D4
90 V (or larger) power supply
do not accept a.c. or signal generator
D5
for stopwatch method:
time 10 or more flashes and divide by number of flashes
for oscilloscope method:
length of wave × timebase
D6
record value of capacitance from the capacitor or method to determine capacitance D7
appropriate circuit to enable capacitance to be determined D8
relationship valid if a straight line passing through the origin is produced D9
method to obtain a measurable time period e.g. do a preliminary experiment to choose appropriate resistors, use large values
of R or C
D10

PUBLISHED
March 2019
2(a) gradient = q
y-intercept = lg p
1
2(b)
1.58 or 1.580 1.61 or 1.613
1.66 or 1.663 1.51 or 1.505
1.74 or 1.740 1.40 or 1.398
1.81 or 1.806 1.30 or 1.301
1.86 or 1.857 1.23 or 1.230
1.90 or 1.898 1.15 or 1.146
1
absolute uncertainties in lg η: ± 0.01 to ± 0.03 1
2(c)(i) six points plotted correctly
must be accurate to the nearest half small square
diameter of points must be less than half a small square
1
error bars in lg η plotted correctly
all error bars to be plotted
total length of bar must be accurate to less than half a small square and symmetrical
1
2(c)(ii) line of best fit drawn
points must be balanced
do not allow line from top plot to bottom plot
if points are plotted correctly then lower end of line should pass between (1.820, 1.275) and (1.835, 1.275) and upper end of
line should pass between (1.640, 1.525) and (1.650, 1.525)
1
worst acceptable line drawn
steepest or shallowest possible line
mark scored only if all error bars are plotted
1

PUBLISHED
March 2019
2(c)(iii) gradient determined with clear substitution of data points into ∆y / ∆x; distance between data points must be at least half the
length of the drawn line
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 determined by substitution of correct point into y = mx + c 1
y-intercept of worst acceptable line determined by substitution into y = mx + c
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)
no ECF from false origin method
1
2(d) -intercept
10y
p =
and
given to 2 or 3 sf
1
q = gradient
and
q and p have correct power of ten from (c)(iii) and (c)(iv)
1
absolute uncertainty in p = intercept of WAL
10y
p−
−
absolute uncertainty in q = uncertainty in gradient
correct substitution of numbers must be seen
1
2(e) 100
p
qθ = or
( )lg 100 lg 2 intercept
lg
gradient
p y
q
θ
− − −
= =
correct substitution of numbers must be seen
1

9702 m19 ms_all

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9702 m19 ms_all