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Cambridge International AS & A Level
PHYSICS 9702/11
Paper 1 Multiple Choice May/June 2021
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 May/June 2021 series for most Cambridge
IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9702/11 Cambridge International AS & A Level – Mark Scheme
PUBLISHED
May/June 2021
© UCLES 2021 Page 2 of 3
Question Answer Marks
1 A 1
2 D 1
3 C 1
4 B 1
5 D 1
6 C 1
7 C 1
8 D 1
9 A 1
10 C 1
11 A 1
12 D 1
13 C 1
14 D 1
15 A 1
16 D 1
17 A 1
18 B 1
19 B 1
20 C 1
21 D 1
22 B 1
23 A 1
24 A 1
25 A 1
26 B 1
27 A 1
28 B 1

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May/June 2021
29 B 1
30 D 1
31 C 1
32 A 1
33 C 1
34 B 1
35 D 1
36 C 1
37 B 1
38 B 1
39 B 1
40 A 1

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

PUBLISHED
May/June 2021
1 D 1
2 B 1
3 D 1
4 D 1
5 D 1
6 D 1
7 B 1
8 C 1
9 B 1
10 D 1
11 A 1
12 C 1
13 B 1
14 C 1
15 A 1
16 A 1
17 D 1
18 D 1
19 A 1
20 D 1
21 B 1
22 A 1
23 B 1
24 A 1
25 B 1
26 C 1
27 B 1
28 B 1

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May/June 2021
29 C 1
30 A 1
31 D 1
32 A 1
33 A 1
34 B 1
35 A 1
36 C 1
37 A 1
38 C 1
39 C 1
40 A 1

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

PUBLISHED
May/June 2021
1 C 1
2 C 1
3 A 1
4 B 1
5 B 1
6 D 1
7 C 1
8 C 1
9 D 1
10 B 1
11 A 1
12 B 1
13 A 1
14 A 1
15 B 1
16 D 1
17 C 1
18 B 1
19 D 1
20 D 1
21 C 1
22 B 1
23 A 1
24 A 1
25 D 1
26 B 1
27 C 1
28 B 1

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May/June 2021
29 A 1
30 B 1
31 D 1
32 A 1
33 C 1
34 A 1
35 A 1
36 C 1
37 B 1
38 C 1
39 C 1
40 D 1

PHYSICS 9702/21
Paper 2 AS Level Structured Questions May/June 2021
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
May/June 2021
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.

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May/June 2021
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.
Science-Specific Marking Principles
1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.
2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any
correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored.
3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus
terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).
4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where
necessary and any exceptions to this general principle will be noted.
5 ‘List rule’ guidance
For questions that require n responses (e.g. State two reasons …):
• The response should be read as continuous prose, even when numbered answer spaces are provided.
• Any response marked ignore in the mark scheme should not count towards n.
• Incorrect responses should not be awarded credit but will still count towards n.
• Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this
should be treated as a single incorrect response.
• Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

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May/June 2021
6 Calculation specific guidance
Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.
For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by
the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.
For answers given in standard form (e.g. a × 10n) in which the convention of restricting the value of the coefficient (a) to a value between 1
and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme.
Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded.
Exceptions to this general principle will be noted in the mark scheme.
7 Guidance for chemical equations
Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.
State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

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May/June 2021
Abbreviations
/ Alternative and acceptable answers for the same marking point.
( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an answer.
The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.
___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the
same technical meaning.
Mark categories
B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must be
seen specifically in the candidate’s answer.
M marks These are method marks upon which A marks later depend. For an M mark to be awarded, the point to which it refers must be seen
specifically in the candidate’s answer. If a candidate is not awarded an M mark, then the later A mark cannot be awarded either.
C marks These are compensatory marks which can be awarded even if the points to which they refer are not written down by the candidate,
providing subsequent working gives evidence that they must have known them. For example, if an equation carries a C mark and the
candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the
C mark is awarded.
If a correct answer is given to a numerical question, all of the preceding C marks are awarded automatically. It is only necessary to
consider each of the C marks in turn when the numerical answer is not correct.
A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.
Annotations
 Indicates the point at which a mark has been awarded.
X Indicates an incorrect answer or a point at which a decision is made not to award a mark.
XP Indicates a physically incorrect equation (‘incorrect physics’). No credit is given for substitution, or subsequent arithmetic, in a
physically incorrect equation.

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May/June 2021
ECF Indicates ‘error carried forward’. Answers to later numerical questions can always be awarded up to full credit provided they are
consistent with earlier incorrect answers. Within a section of a numerical question, ECF can be given after AE, TE and POT errors,
but not after XP.
AE Indicates an arithmetic error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
subsequent ECF if there are no further errors.
POT Indicates a power of ten error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
TE Indicates incorrect transcription of the correct data from the question, a graph, data sheet or a previous answer. For example, the
value of 1.6 × 10–19 has been written down as 6.1 × 10–19 or 1.6 × 1019.
Do not allow the mark where the error occurs. Then follow through the working/calculation giving full subsequent ECF if there are no
further errors.
SF Indicates that the correct answer is seen in the working but the final answer is incorrect as it is expressed to too few significant
figures.
BOD Indicates that a mark is awarded where the candidate provides an answer that is not totally satisfactory, but the examiner feels that
sufficient work has been done (‘benefit of doubt’).
CON Indicates that a response is contradictory.
I Indicates parts of a response that have been seen but disregarded as irrelevant.
M0 Indicates where an A category mark has not been awarded due to the M category mark upon which it depends not having previously
been awarded.
^ Indicates where more is needed for a mark to be awarded (what is written is not wrong, but not enough). May also be used to
annotate a response space that has been left completely blank.
SEEN Indicates that a page has been seen.

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May/June 2021
Question Answer Mark
1(a) mass / volume B1
1(b)(i) absolute uncertainty = 4.0 × (5 / 100)
= (±) 0.2 cm
B1
1(b)(ii) percentage uncertainty = 2 + 4 + (5 × 2) C1
= (±) 16% A1
1(c) p = F / A or p = W / A C1
p = (19.5 × 10–3 × 9.81) / (4.0 × 10–2)2 C1
= 120 Pa A1

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May/June 2021
2(a)(i) P = Fv C1
= 18 × 1.4
= 25 W
A1
2(a)(ii) a = F / m C1
a = 18 / 72 = 0.25 (ms–2)
t = 1.4 / 0.25
C1
= 5.6 s A1
2(b)(i) a = (54 – 18) / 72 or 36 / 72 (= 0.50 ms–2) C1
v2 = 2 × 0.50 × 9.5 C1
v = 3.1 m s–1 A1
2(b)(ii) W = 54 × 9.5
= 510 J
A1
2(b)(iii) curved line from the origin M1
gradient of line increases A1
2(c) (force due to) air resistance increases/changes/not constant or air resistance increases with speed B1

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May/June 2021
3(a) solid straight line drawn between centre of sphere at X and at Y B1
3(b) p = mv or 0.72 = mv C1
E = ½mv2 or 0.86 = ½mv2 C1
(m =) 0.722 / (2 × 0.86) = 0.30 (kg)
or
v = 2EK / p
v = (0.86 × 2) / 0.72 = 2.4 (to 2 s.f.)
m = 0.72 / 2.4 = 0.30 (kg)
A1
3(c) (Δ)E = mg(Δ)h C1
h = 0.86 / (0.30 × 9.81)
= 0.29 m
A1
3(d) cosθ = (0.93 – 0.29) / 0.93 so θ = 47° A1
3(e) moment = (0.30 × 9.81) × 0.93 × (sin 47° or cos 43°)
or
moment = (0.30 × 9.81) × [0.932 – (0.93 – 0.29)2]0.5
C1
= 2.0 N m A1

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May/June 2021
3(f) there is a resultant force (acting on sphere)
or
there is a resultant moment (about P acting on pendulum)
(so) not in equilibrium
B1
4(a) distance moved by wavefront/energy during one cycle/oscillation/period (of source)
or
minimum distance between two wavefronts
or
distance between two adjacent wavefronts
B1
4(b) v = λ / T
or
v = fλ and f = 1 / T
C1
T = 460 × 10–9 / 3.00 × 108 C1
= 1.5 × 10–15 s A1
4(c) waves pass through/enter the slit(s) B1
waves spread (into geometric shadow) B1
4(d)(i) nλ = d sinθ C1
G = sinθ / λ
d = 4 / G
A1
4(d)(ii) straight line from 400 nm to 700 nm that is always below printed line M1
straight line has smaller gradient than printed line and is 5 small squares high at wavelength of 700 nm A1

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May/June 2021
5(a) sum of e.m.f.(s) = sum of p.d.(s)
or
(algebraic) sum of e.m.f.(s) and p.d.(s) is zero
M1
around a loop/around a closed circuit A1
5(b)(i) I = 1.8 / 0.90
= 2.0 A
A1
5(b)(ii) Q = It C1
number = (2.0 × 45) / 1.60 × 10–19
= 5.6 × 1020
A1
5(b)(iii) 4.0 = 1.8 + [2.0 × (0.35 + R)]
or
4.0 = 2.0 × (0.90 + 0.35 + R)
C1
R = 0.75 Ω A1
5(c)(i) 1.2 / 1.8 = 0.30 / L C1
L = 0.45 m A1
5(c)(ii) p.d. across XY decreases/p.d. across XP decreases B1
(so) P is moved towards Y/away from X/to the right B1

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May/June 2021
6(a)(i) (electron) neutrino B1
6(a)(ii) weak (nuclear force/interaction) B1
6(a)(iii) β+ (particle) B1
6(a)(iv) (quark structure is) up up down or uud B1
(2 / 3)e + (2 / 3)e – (1 / 3)e = (+)e B1
6(a)(v) up up down changes to up down down or uud → udd
or
up changes to down or u → d
B1
6(b)(i) F = Eq C1
ratio = 6 / 8
= 0.75
A1
6(b)(ii) ratio = 0.75 × (16 / 12)
= 1.0
A1
6(b)(iii) horizontal straight line at a non-zero value of a B1

PHYSICS 9702/22
MARK SCHEME
Maximum Mark: 60
Published
Teachers.

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should
be treated as a single incorrect response.

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
Abbreviations
( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an answer.
The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.
___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the same
technical meaning.
Mark categories
candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the C
mark is awarded.
Annotations
XP Indicates a physically incorrect equation (‘incorrect physics’). No credit is given for substitution, or subsequent arithmetic, in a
physically incorrect equation.

PUBLISHED
May/June 2021
consistent with earlier incorrect answers. Within a section of a numerical question, ECF can be given after AE, TE and POT errors, but
not after XP.
TE Indicates incorrect transcription of the correct data from the question, a graph, data sheet or a previous answer. For example, the
value of 1.6 × 10–19 has been written down as 6.1 × 10–19 or 1.6 × 1019.
further errors.
SF Indicates that the correct answer is seen in the working but the final answer is incorrect as it is expressed to too few significant figures.
been awarded.
^ Indicates where more is needed for a mark to be awarded (what is written is not wrong, but not enough). May also be used to annotate
a response space that has been left completely blank.

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May/June 2021
1(a) acceleration: vector
electrical resistance: scalar
momentum: vector
1 mark for two correct, 2 marks for all three correct
B2
1(b) resultant force (in any direction) is zero B1
resultant torque/moment (about any point) is zero B1
1(c)(i) upthrust = ρg(∆)h× A C1
= (1.00 × 103 × 9.81 × 0.190) × 0.0230
= 42.9 N
A1
1(c)(ii) (T =) 43 – 28 = 15 (N)
or
(T =) 42.9 – 28 = 14.9 or 15 (N)
A1
1(c)(iii) σ = F / A or T / A C1
= 15 / (3.2 × 10–6)
= 4.7 × 106 Pa
A1
1(c)(iv) upthrust (on cylinder) increases (and weight constant) B1
tension/stress increases and (so) strain energy increases B1

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May/June 2021
2(a) v2 = u2 + 2as
u2 = 8.72 – (2 × 9.81 × 1.5)
C1
u = 6.8 m s–1 A1
2(b) (magnitude of) force on ball (by ground) equal to force on ground (by ball) B1
(direction of) force on ball (by ground) opposite to force on ground (by ball) B1
2(c)(i) (p = ) 0.059 × 8.7 or 0.059 × 5.4 C1
change in momentum = 0.059 (8.7 + 5.4)
= 0.83 N s
A1
2(c)(ii) resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4) / 0.091]
= 9.1 N
A1
2(c)(iii) (W =) 0.059 × 9.81 C1
(W =) 0.58 (N)
force = 9.1 + 0.58
= 9.7 N
A1
2(d) straight line with a positive gradient and starting from a non-zero value of speed at t = 0 and ending when t = T B1
2(e) air resistance increases B1
resultant force/acceleration decreases so gradient (of curve) decreases B1

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May/June 2021
3(a) (Δ)E= mg(Δ)h or W(∆)h C1
= 330 × (4.0 – 1.1)
= 960 J
A1
3(b) (work =) 960 – 540
(= 420 J)
C1
distance moved = (960 – 540) / 52
= 8.1 m
A1
3(c)(i) E = ½mv2 C1
540 = ½ × (330 / 9.81) × v2
v = 5.7 m s–1
A1
3(c)(ii) speed = horizontal component of velocity
= 5.7 × cos41°
C1
= 4.3 m s–1 A1

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May/June 2021
4(a) time for one oscillation/vibration/cycle
or
time between adjacent wavefronts (passing the same point)
or
shortest time between two wavefronts (passing the same point)
B1
4(b) (when two or more) waves meet/overlap (at a point) B1
(resultant) displacement is sum of the individual displacements B1
4(c)(i) microwave(s) B1
4(c)(ii) v = λ / T
or
v = fλ and f = 1/T
C1
T= 0.040 / 3.00 × 108 C1
= 1.33 × 10–10 (s)
= 1.33 × 10–10 / 10–12 (ps)
= 130 ps
A1
4(c)(iii) (1.380 – 1.240) / 0.040 = 3.5
or
1.380 / 0.040 – 1.240 / 0.040 = 3.5
A1
4(c)(iv) phase difference = 1260° or 180° A1
4(c)(v) (always) zero A1
4(c)(vi) increase in distance between (adjacent intensity) maxima/minima A1

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5(a) volt / ampere B1
5(b) R = ρL / A B1
(A = V / L)
(so) R = ρL2 / V (with ρ and V constant so R ∝ L2)
B1
5(c)(i) E = 2.4 V A1
5(c)(ii) P= VI or I2R or V2 / R C1
= 1.3 × 5.0 or 5.02 × 0.26 or 1.32 / 0.26 C1
= 6.5 W A1
5(c)(iii) (–) internal resistance or (–) r B1

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6(a) the nucleus is charged B1
the majority of the mass (of atom) is in the nucleus B1
6(b) made up of quarks (so) not a fundamental particle B1
6(c) (Q =) 6.9 × 10–9 × 60 C1
number = (6.9 × 10–9 × 60) / (2 × 1.60 × 10–19) C1
= 1.3 × 1012 A1
6(d) (magnitude of electric) force is constant B1
(so magnitude of) acceleration is constant B1
6(e) (nuclei have) same charge/same number of protons B1
(so) same (magnitude of) force B1

PHYSICS 9702/23
MARK SCHEME
Maximum Mark: 60
Published
Teachers.

PUBLISHED
May/June 2021
marking principles.
descriptors.

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May/June 2021
2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for
any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored.
3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other
syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme
where necessary and any exceptions to this general principle will be noted.

PUBLISHED
May/June 2021
your working’.
For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded
by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.

PUBLISHED
May/June 2021
Abbreviations
( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an answer. The
context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.
___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the same
technical meaning.
Mark categories
candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the
C mark is awarded.
Annotations
XP Indicates a physically incorrect equation (‘incorrect physics’). No credit is given for substitution, or subsequent arithmetic, in a physically
incorrect equation.

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May/June 2021
not after XP.
TE Indicates incorrect transcription of the correct data from the question, a graph, data sheet or a previous answer. For example, the value
of 1.6 × 10–19 has been written down as 6.1 × 10–19 or 1.6 × 1019.
further errors.
been awarded.
^ Indicates where more is needed for a mark to be awarded (what is written is not wrong, but not enough). May also be used to annotate
a response space that has been left completely blank.

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May/June 2021
1(a)(i) two correct scalar quantities e.g. time, mass, distance, temperature B1
two correct vector quantities e.g. force, acceleration, velocity, displacement B1
1(a)(ii) magnitude B1
unit B1
1(b)(i) north component of velocity = 11 m s–1 A1
east component of velocity = 7.5 m s–1 A1
1(b)(ii) velocity = 7.5 – 2.7
= 4.8 m s–1
A1
1(b)(iii) velocity = √(112 + 4.82) C1
= 12 m s–1 A1
1(b)(iv) angle = tan–1 (4.8 / 11) C1
= 24° A1

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May/June 2021
2(a) change in velocity / time (taken) B1
2(b)(i) air resistance increases (with speed/with time) B1
resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time) B1
when air resistance equals the weight the speed/velocity/v becomes constant B1
2(b)(ii) speed = 36 m s–1 A1
2(b)(iii) height given by area under the curve C1
height = 950 m
Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range
900–910 m or 990–1000 m.
A2
2(b)(iv) line starting at (0, 9.8) B1
curve with negative gradient between t = 0 and t = 20 s B1
line showing zero acceleration between t = 20 s and t = 30 s B1

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May/June 2021
3(a) force × distance M1
perpendicular distance of (line of action of) force from the point A1
3(b)(i) distance moved by pointer = 123 – 86 (= 37 mm) C1
(extension =) 37 × (1.8 / 52.6) = 1.3 (mm)
or
sin or tan θ = 37 / 526 (so θ = 4.0° so extension =) sin or tan θ × 18 = 1.3 (mm)
A1
3(b)(ii) moment = 0.472 × 9.81 × 6.2 × 10–2 C1
= 0.29 N m A1
3(b)(iii) (Δ)F × 1.8 × 10–2 = 0.29 C1
ΔF = 16 N A1
3(b)(iv) k = F / x C1
= 16 / (1.3 × 10–3)
= 1.2 × 104 N m–1
A1

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May/June 2021
4(a) (when two or more) waves meet/overlap (at a point) B1
(resultant) displacement is sum of the individual displacements B1
4(b) intensity ∝ amplitude2 C1
maximum intensity = 9I A1
4(c)(i) x = λD / a C1
= (550 × 10–9 × 1.2) / (0.35 × 10–3) C1
= 1.9 × 10–3 m A1
4(c)(ii) red light has longer wavelength (than 550 nm) so distance (between fringes) increases B1

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May/June 2021
5(a) energy per unit charge B1
energy transferred by source driving charge around the complete circuit
or
energy transferred from other forms to electrical energy
B1
5(b) there is a p.d. across the internal resistance/r B1
change in current/I results in a change in p.d. across the internal resistance B1
V = E – p.d. across internal resistance
or
change in p.d. across r causes a change in V (as e.m.f. is constant)
B1
5(c)(i) E = 7.4 V A1
5(c)(ii) maximum current = 0.92 A A1
5(c)(iii) r = E / IMAX or (–)gradient C1
e.g. r = 7.4 / 0.92
= 8.0 Ω
A1
5(d) straight line with negative gradient that is smaller in magnitude than the original line B1
line which would have intercept on V-axis below the original line B1

PUBLISHED
May/June 2021
6(a)(i) up up down B1
6(a)(ii) up down down B1
6(a)(iii) (alpha-particle is) 2 protons and 2 neutrons C1
6 up, 6 down A1
6(b)(i) most of an atom is empty space
or
the nucleus (volume) is (very) small compared with the atom
B1
6(b)(ii) the nucleus is charged B1
the majority of the mass of atom is in the nucleus B1
6(c) F = Eq and a = F / m C1
a = Eq / m
ratio = (e / m) / (2e / 4m)
= 2
A1

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 May/June 2021
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
1(a) Final value of L with unit and in the range 28.0–32.0 cm. 1
1(b) Raw value(s) of x and b to nearest mm and with units. 1
1(c) Six (or more) sets of readings of x and b (different values) with correct trend (as m decreases, x decreases and b increases)
and without help from Supervisor scores 5 marks, five sets scores 4 marks, etc.
5
Range: (mmax – mmin) ⩾ 70 g. 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. 1/b / cm–1 or 1/b (1/cm).
1
Significant figures:
All values of 1 / b must be given to the same number of significant figures as, or one greater than, the number of significant
figures in raw b.
1
Calculation: Values of 1 / b are correct. 1
1(d)(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
Axes 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 in the table must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square.
Points must be plotted to an accuracy of half a small square.
1
Quality:
All points in the table must be plotted (at least 5) on the grid.
Trend of points must be correct.
It must be possible to draw a straight line that is within ± 0.2 m–1 (± 0.002 cm–1) on the 1 / b axis (normally y-axis) of all
plotted points.
1

PUBLISHED
May/June 2021
1(d)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of
points either side of the line along the full length.
Allow one anomalous point only if clearly indicated 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(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, i.e. Δy /Δx.
Gradient sign on answer line matches graph drawn.
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 and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
or
Intercept read directly from the graph at x = 0, accurate to half a small square.
1
1(e) Value of P equal to candidate’s gradient and value of Q equal to candidate’s intercept.
Values must not be written as fractions.
1
Unit for P (e.g. cm–2) and unit for Q (e.g. cm–1) correct. 1
1(f) Line W shown with greater gradient. 1

PUBLISHED
May/June 2021
2(a) Value of d to the nearest mm with unit. 1
2(b) Correct calculation of temperature change Δθ. 1
2(c)(i) Value of h with unit. 1
D > d. 1
2(c)(ii) Percentage uncertainty based on absolute uncertainty Δh in the range 2–6 mm.
If repeat readings have been taken, then the absolute 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(c)(iii) Correct calculation of C. 1
2(c)(iv) Justification for significant figures in C linked to s.f. in D and d or linked to s.f. in (D – d). 1
2(d) Second values of θ0 and θ. 1
Second values of h and D. 1
Temperature decreases in both experiments and second value of Δθ < first value of Δθ. 1
2(e)(i) Two values of k calculated correctly. The final k values must not be written as fractions. 1
2(e)(ii) Valid comment consistent with the calculated values of k, testing against a criterion stated by the candidate. 1

PUBLISHED
May/June 2021
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 measure h with reason, e.g. scale of ruler does not start at the end/opaque cup/water too hot/rule in water
changes water level.
C Difficult to measure D with reason, e.g. parallax.
D Large % uncertainty (error) in Δθ
or
thermometer is not precise enough to measure a small change in temperature.
E Difficulty with taking temperature at the end of the two-minute period with reason, e.g. cannot look at stop-watch and
thermometer simultaneously.
1 mark for each point up to a maximum of 4.
4
2(f)(ii) A Take more readings and plot a graph or take more readings and compare k values (not “repeat readings” on its own).
B Use a transparent cup/ruler without a space at end/measure (empty depth – depth to water surface) with detailed
description e.g. use rod at right angles to a rule across top of cup to reach water surface or cup bottom.
C To measure D or inside of cup, use calipers/dividers/vertical pointers/travelling microscope.
D Improved method to measure Δθ e.g. more precise thermometer/thermometer reading to 0.1 °C/thinner capillary in
thermometer/more sensitive thermometer
or
allow experiment to go on for more than two minutes.
E Method for simultaneous measurement of temperature and time e.g. sound to mark two-minute period/temperature
probe linked to data logger system/video with thermometer and timer in view.
4

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
1(a) Final value of sA to at least two significant figures and in the range 1.05–1.10mm. 1
Evidence that sA has been correctly calculated from a measurement of at least 10sA. 1
1(b) Value of G in range 0°–45°. 1
1(c) Six (or more) sets of readings of G and F (different values) with correct trend (F increases as G increases) and without help
from the Supervisor scores 3 marks, five sets scores 2 marks, four or fewer sets scores 1 mark.
3
Range: Gmin ⩽ 3° and Gmax ⩾ 17°. 1
Column headings:
Headings for sin F and sin (F–G) must have no unit.
The presentation of quantity and unit must conform to accepted scientific convention e.g. F / °.
1
Consistency: All values of raw G and raw F must be given to the nearest degree. 1
Values of sin F should be to the same number of significant figures as, or one greater than, the number of significant
figures in the corresponding value(s) of raw F.
1
Calculation: Values of sin (F–G) calculated correctly. 1

PUBLISHED
May/June 2021
1(d)(i) Axes:
Scales are chosen so that the plotted points occupy at least half the graph grid in both x and y directions
1
Plotting of points:
Diameter of plotted points are ⩽ half a small square.
Points must be plotted to an accuracy of half a small square.
1
Quality:
Trend of points on graph must be correct.
It must be possible to draw a straight line that is within ± 0.02 on the sin F axis of all plotted points.
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.
1
1(d)(iii) Gradient:
1
y-intercept:
or
1

PUBLISHED
May/June 2021
1(e) Value of p equal to candidate’s gradient and value of q equal to candidate’s intercept. Values must not be written as
fractions.
1
Values for p and q both given without a unit. 1
1(f) Correct calculation of sB using sB = psA. 1

PUBLISHED
May/June 2021
2(a)(i) Value of TV with unit and in range 0.20–0.40 s. 1
At least two measurements of nTV where n ⩾ 5. 1
2(a)(ii) Value for TS larger than TV. 1
2(b) Second values of TV and TS. 1
Second TS > first TS. 1
2(c)(i) Two values of TS
2 – TV
2 calculated correctly. 1
2(c)(ii) Justification based on significant figures in TS and TV. 1
2(c)(iii) Valid comment consistent with the calculated values of TS
2 – TV
2, testing against a criterion stated by the candidate. 1
2(d)(i) Value for x1 in range 4.0–6.0 cm. 1
2(d)(ii) Percentage uncertainty based on an absolute uncertainty in the range 2–3 mm.
If repeat readings have been taken, then the absolute uncertainty can be half the range (but not zero) if the working is
clearly shown.
1
2(d)(iii) Raw value(s) for x2 to nearest 0.1 cm. 1
2(d)(iv) Correct calculation of g with consistent unit. 1

PUBLISHED
May/June 2021
2(e)(i) A Two TS
2 – TV
2 values are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few
readings”).
B Difficult to maintain single mode of oscillation e.g. spring swings when measuring vertical oscillations/spring bounces
when measuring swinging oscillations/spring swings in more than one plane.
C Spring slides along rod during the oscillation.
D Difficult to judge/determine/decide when an oscillation starts/ends/is complete.
E Large % uncertainty in Tv
or
Tv is small so large uncertainty.
F Difficult to measure x1 or x2 with reason e.g. parallax error
or
difficult to measure x2 due to space at end of ruler.
4
2(e)(ii) A Take more readings and plot a graph or take more readings and compare (not “repeat readings” on its own).
B Method to help maintain single mode of oscillation e.g. restrict sideways motion with tube/use parallel guides.
C Method to attach spring to rod/stop spring sliding on rod e.g. adhesive putty/glue spring to rod/cut notch in rod/use rod
with diameter same as diameter of spring loop/rougher rod.
D Video/record/film with timer in view/frame by frame
or
use fiducial marker at centre of oscillation.
E Use larger masses/use spring with lower spring constant/stiffness.
F Use calipers/travelling microscope/use ruler starting at zero/use blocks with detail.
4

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
1(b) Raw values of I with unit and to the nearest 0.1 mA and final value of I < 1 A. 1
1(c) Six sets of readings of x and I (different values) without help from the Supervisor and showing the correct trend
(I decreases as x increases) scores 4 marks, five sets scores 3 marks, etc.
4
Range: xmin ⩽ 10.0 cm and xmax ⩾ 60.0 cm. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention, e.g. 1/I / A–1 or 1/I (A–1) and x/m.
1
Consistency: All values of x must be given to the nearest mm. 1
All values of 1 / I must be given to the same number of significant figures as, or one greater than, the number of
significant figures in raw I.
1
Calculation: Values of 1 / I are correct. 1

PUBLISHED
May/June 2021
1(d)(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:
Points must be plotted to an accuracy of half a small square in both x and y directions.
1
Quality:
All points in the table (at least 5) must be plotted on the grid.
Trend of points on graph must be correct.
It must be possible to draw a straight line that is within 2.0 cm (to scale) on the x-axis of all plotted points.
1
1
1(d)(iii) Gradient:
1
y-intercept:
or
1

PUBLISHED
May/June 2021
1(e) Value of P = candidate’s gradient and value of Q =candidate’s intercept.
1
Unit for P is correct (e.g. A–1 m–1 or mA–1 cm–1) and unit for Q is correct (e.g. A–1 or mA–1). 1
1(f) Correct calculation of ρA /ρB using PL/Q + 1. 1

PUBLISHED
May/June 2021
2(a) Final value for d with unit and in the range 28.0–40.0 cm. 1
2(b) Percentage uncertainty based on absolute uncertainty in the range 2–6 mm.
clearly shown.
1
2(c)(i) Measurement of raw b to the nearest mm. Final value with unit and in the range 9.0–11.0 cm. 1
2(c)(ii) Correct calculation of α. 1
2(c)(iii) Justification for the number of significant figures in α linked to s.f. in b and d. 1
2(d) Final value for T with unit and in the range 1.50–2.50 s. 1
At least two measurements of nT where n ⩾ 5. 1
2(e) Second values of b and T. 1
Second value of T < first value of T. 1
2(f)(i) Two values of C calculated correctly. The final values must not be written as fractions. 1
2(f)(ii) Valid comment consistent with calculated values of C, testing against a criterion stated by the candidate. 1
2(g) Value of k correctly calculated from the second value of C and with consistent unit, i.e. N m–1 or kg s–2. 1

PUBLISHED
May/June 2021
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Masses falling off/masses stick to a small surface area/not enough adhesive putty for the masses/adhesive putty not
strong enough to hold masses.
C Difficult to judge when the wooden strip is horizontal/difficult to set wooden strip horizontal
or
difficult to judge or set the spring or string vertical.
D Difficulty measuring d with a reason, e.g. finding/determining the centre of the mass/hole
or
difficulty measuring b with a reason, e.g. finding/determining the centre of the nail/holding the ruler parallel to the
strip/set-up wobbly/holding ruler in mid-air.
E Difficult to judge/determine/decide when an oscillation starts/finishes/is complete.
F Different modes of oscillation/string moves during oscillation
or
strip bends/twists/flexes (during oscillation).
4

PUBLISHED
May/June 2021
2(h)(ii) A Take more readings and plot a graph or take more readings and compare C values (not “repeat readings” on its
own).
B Wider strip/named method for improving the adhesion of the masses, e.g. glue/tape with reference to masses.
C Method of ensuring strip is horizontal, e.g. use a spirit level
or
method to ensure spring/string is vertical, e.g. use a plumb-line.
D Improved method to measure b or d, e.g. add a scale to the strip/mark on the strip/clamp ruler.
E Fiducial marker at the centre of the oscillation
or
video/record/film with timer in view/play back frame by frame.
F Use a thicker/stiffer/laminated strip
or
sand to make rougher/add notch to wooden strip or rod/stick sandpaper to wooden strip or rod.
4

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
1(a)(i) Value of C to the nearest mm. 1
1(a)(ii) Value of T with unit and in the range 0.50–1.50 s. 1
At least two measurements of nT where n ⩾ 5. 1
1(b) Six sets of readings of C and T (different values) with correct trend (T increases as C increases) and without help from the
Supervisor scores 4 marks, five sets scores 3 marks etc.
4
Range: Cmin ⩽ 18.0 cm and Cmax ⩾ 35.0 cm. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. 1/√C / cm–½, 1/T (s–1).
1
Consistency:
All values of raw times must be given to the nearest 0.1 s or all to the nearest 0.01 s.
1
Values of 1 / √C must be given to the same number of significant figures as, or one greater than, the number of significant
figures in C.
1
Calculation: Values of 1 / √C calculated correctly. 1

PUBLISHED
May/June 2021
1(c)(i) Axes:
1
Plotting of points:
Points must be plotted to an accuracy of half a small square in both x and y directions.
1
Quality:
It must be possible to draw a straight line that is within ± 5.0 × 10–3 cm–½ (± 5.0 × 10–2 m–½) on the 1 / √C axis (normally x-
axis) of all plotted points.
1
1(c)(ii) Line of best fit:
1
1(c)(iii) Gradient:
1
y-intercept:
or
1

PUBLISHED
May/June 2021
1(d) Value of a equal to candidate’s gradient and value of b equal to candidate’s intercept. Values must not be written as
fractions.
1
Units for a and b correct, e.g. cm½ s–1 for a and s–1 for b. 1

PUBLISHED
May/June 2021
2(a)(i) Values of D1 and D2 both to nearest mm. 1
Evidence of repeat readings for D1 and D2. 1
2(a)(ii) Percentage uncertainty based on an absolute uncertainty of 2–5 mm.
clearly shown.
1
2(b) All values of h1 and h2 to nearest 0.1 mm or all to nearest 0.01 mm. 1
Correct calculation of y. 1
2(c)(i) Values of A and B recorded and A < B. 1
2(c)(ii) Correct calculation of F. 1
2(c)(iii) Justification for significant figures in F linked to s.f. in B and A. 1
2(d) Second values of D1, D2, h1 and h2. 1
y smaller for smaller D2. 1
2(e)(i) Two values of k calculated correctly. 1
2(e)(ii) Valid comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1

PUBLISHED
May/June 2021
B Difficult to judge/determine whether rod is horizontal.
C Difficult to measure h with reason e.g. parallax error/wooden block or ring getting in way.
D Large percentage uncertainty in y.
E Difficult to measure A or B with reason e.g. judging centre of slotted mass when measuring B/judging centre of nail for
A or B/parallax error in A or B/difficult to hold ruler steady when measuring A or B. (Allow parallax error linked to h or
A or B only once.)
4
B Improved method to determine if rod horizontal e.g. use a spirit level/use set square(s) with description of method.
C Method to reduce error in measuring h e.g. use wider rod/move ring to edge of block/use travelling microscope.
D Method to reduce percentage uncertainty in y e.g. measure change in height at end of rod/use larger mass/increase B.
E Improved method of measuring A or B e.g. clamp ruler/method to locate and mark centre of slotted mass e.g. measure
diameter and halve to find radius.
4

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
1(b) Final value of T in the range 0.80–1.20 s. 1
1(c) Five sets of readings of d and time (different values) without help from the Supervisor and with the correct trend
(d increases, T increases) scores 5 marks, four sets scores 4 marks etc.
5
Range: Includes d ⩽ 25.0 cm and d ⩾ 40.0 cm. 1
Column headings:
The presentation of the quantity and the unit must conform to accepted scientific convention e.g. T / √d / s m–½.
1
Consistency: Raw values of d must all be given to the nearest mm. 1
All values of
d L
d
−
√ must be given to the same number of significant figures as, or one greater than, the least number of
number of significant figures in either (d – L) or d.
1
Calculation: Correct calculation of T / √d and
d L
d
−
√ .
1

PUBLISHED
May/June 2021
1(d)(i) Axes:
Axes must be labelled with the quantity which is being plotted.
1
Plotting of points:
Points must be plotted to an accuracy of half a small square in both the x and y directions.
1
Quality:
All points in the table (at least 4) must be plotted on the grid.
It must be possible to draw a straight line that is within 0.01 on the
d L
d
−
√ axis of all plotted points.
1
Allow one anomalous point only if clearly indicated by the candidate. There must be at least four points left after the
1
1(d)(iii) Gradient:
1
y-intercept:
Check correct read-off from a point on the line and substituted into y = mx + c.
or
1

PUBLISHED
May/June 2021
1(e) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
1
Unit for P is correct (e.g. s m–½) and unit for Q is correct (s m–½). 1

PUBLISHED
May/June 2021
2(a) Final value of H in the range 38.0–42.0 cm. 1
2(b)(i) Value of raw x to the nearest mm. 1
Value of rawθ to the nearest degree and in the range 60°–80°. 1
2(b)(ii) Percentage uncertainty in θ based on absolute uncertainty of 2°–5°.
clearly shown.
1
2(b)(iii) Correct calculation of x tan θ. 1
2(b)(iv) Justification for significant figures in x tan θ linked to s.f. in x and θ. 1
2(c) Second value of x. 1
Second value of θ. 1
Second value of θ < first value of θ. 1
2(d)(i) Two values of k calculated correctly. Final values must not be written as fractions. 1
2(d)(ii) Valid comment consistent with calculated values of k, testing against a criterion stated by the candidate. 1
2(e) Correct calculation of M with consistent unit. 1

PUBLISHED
May/June 2021
B Difficult to measure H with a reason e.g. locating centre of modelling clay/parallax error/locating centre of the hole.
C Difficult to measure H because the ruler is not long enough.
D Difficult to determine if the string is horizontal
or
difficult to set up the string horizontally.
E Difficulty measuring θ with reason e.g. parallax error/holding protractor in air/wooden strip moves when knocked by
protractor. (Allow parallax error linked to H or θ only once.)
F The modelling clay falls off.
4
B Method of locating and marking the centre.
C Use a half-metre rule or a metre rule.
D Use a spirit level/metre rule and set square with detail.
E Clamp protractor
or
take a photo and measure angle
or
attach protractor to wooden rod.
F Use glue to stick a sphere of clay to end/use a regular shape of mass (instead of the modelling clay).
4

PHYSICS 9702/41
Paper 4 A Level Structured Questions May/June 2021
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
Abbreviations
( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an
answer. The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.
Mark categories
B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must
be seen specifically in the candidate’s answer.
providing subsequent working gives evidence that they must have known them. For example, if an equation carries a C mark and
the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then
the C mark is awarded.

PUBLISHED
May/June 2021
Annotations
incorrect equation.
not after XP.
further errors.
been awarded.

PUBLISHED
May/June 2021
^ Indicates where more is needed for a mark to be awarded (what is written is not wrong, but not enough). May also be used to annotate a
response space that has been left completely blank.

PUBLISHED
May/June 2021
1(a) force per unit mass B1
1(b) GMm / r 2 = mrω 2 and ω = 2π/T
or
GMm / r 2 = mv2 / r and v = 2πr / T
C1
6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1
r = 6.9 × 106 m A1
1(c)(i) r3ω2 = constant or r3 / T2 = constant C1
r3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1
or
GMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1)
6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2
so r = 9.4 × 106 m
(A1)
1(c)(ii) separation increases so (potential energy) increases
or
movement is against gravitational force so (potential energy) increases
B1
1(c)(iii) potential energy = (–)GMm / r C1
ΔEP = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1
= 1.9 × 1010 J A1

PUBLISHED
May/June 2021
2(a) pV = NkT C1
N = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1
or
pV = nRT and nNA = N (C1)
N = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1)
2(b) speed of molecule decreases on impact with moving piston B1
mean square speed (directly) proportional to (thermodynamic) temperature
or
mean square speed (directly) proportional to kinetic energy (of molecules)
or
kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature
B1
kinetic energy (of molecules) decreases (so temperature decreases) B1
2(c)(i) ΔU = 3/2 × k × ΔT × N C1
= 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1
= – 64 J A1
2(c)(ii) decrease in internal energy is less than work done by gas M1
(thermal energy is) transferred to the gas (during the expansion) A1

PUBLISHED
May/June 2021
3(a) acceleration (directly) proportional to displacement B1
acceleration is in opposite direction to displacement B1
3(b) ω2 = 2k / m and ω = 2πf C1
(2πf)2 = (2 × 130) / 0.84 C1
f = 2.8 Hz A1
3(c)(i) resonance B1
3(c)(ii) oscillator supplies energy (continuously) B1
energy of trolley constant so energy must be dissipated
or
without loss of energy the amplitude would continuously increase
B1
4 (ultrasound) pulse B1
reflected at boundaries B1
gel is used to minimise reflection at skin
or
generated and detected by quartz crystal
B1
time delay between generation and detection gives information about depth B1
intensity (of reflected wave) gives information about nature of boundary B1

PUBLISHED
May/June 2021
5(a) amplitude of the carrier wave varies M1
in synchrony with the displacement of the (information) signal A1
5(b)(i) wavelength = (3.0 × 108) / (300 × 103)
= 1000 m
A1
5(b)(ii) bandwidth = 16 kHz A1
5(b)(iii) frequency = 8 kHz A1
5(c) attenuation = 10 lg (P1 / P2) C1
73 = 10 lg (PT / PR)
73 = 10 lg (PT x2 / 0.082 PT) or x2 / 0.082 = 107.3
C1
x = 1300 m A1

PUBLISHED
May/June 2021
6(a) from x = 0 to x = r : E = 0 B1
from x = r to x = 3r : curve with negative gradient of decreasing magnitude passing through (r, E0) B1
line passing through (2r, E0 / 4) and (3r, E0 / 9) B1
6(b) from p = p0 / 2 to p = p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0) B1
line passing through (½p0, 2λ0) B1
6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1
line passing through (15, ½N0) B1
line passing through (30, 0.75N0) and (45, 0.88N0) B1

PUBLISHED
May/June 2021
7(a) charge / potential M1
charge is on one plate, potential is p.d. between the plates A1
7(b)(i) I = Q / t M1
charge = CV and time = 1 / f leading to I = fCV A1
7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1
C = 530 pF A1
7(c) (total) capacitance is halved B1
charge (for each cycle/discharge) is halved
or
since f and V are constant, current is proportional to capacitance
B1
current = 2.4 μA B1

PUBLISHED
May/June 2021
8(a) V+ = 3.0 × 3.0 / (2.5 + 3.0) C1
= 1.6 V A1
8(b) V – is +2.0 V
or
V – > V +
B1
output is negative so (LED) does not emit light B1
8(c) at 0 °C, V – = 1.7 V
or
for all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ
B1
V – always greater than V + (so no switching) B1
8(d) (at 20 °C,) RT = 1.8 kΩ C1
2.5 / 3.0 = 1.8 / R
or
[R / (R + 1.8)] × 3.0 = 1.6
C1
R = 2.2 kΩ A1

PUBLISHED
May/June 2021
9(a) region where there is a force exerted on M1
a current-carrying conductor
or
a moving charge
or
a magnetic material/magnetic pole
A1
9(b)(i) face PSWV shaded B1
9(b)(ii) accumulating electrons cause an electric field (between the faces) B1
force due to electric field opposes force due to magnetic field B1
accumulation stops when magnetic force equals electric force B1
9(c)(i) number density of charge carriers B1
9(c)(ii) PV or QT or SW B1
9(d) (for semiconductor,) n is (much) smaller so VH (much) larger B1

PUBLISHED
May/June 2021
10(a) direction of (induced) e.m.f. M1
is such as to oppose the change causing it A1
10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1
(induced) e.m.f. causes (eddy/induced) currents (in ring) B1
currents (in ring) cause magnetic field (around ring) M1
two fields interact to cause resistive/opposing force A1
or
current (in ring) is in a magnetic field (M1)
which causes resistive force (A1)
or
currents (in ring) dissipate thermal energy (M1)
(thermal) energy comes from energy of oscillations (A1)
10(c) current cannot pass all the way around the ring B1
(induced) currents smaller B1
smaller resistive force (so more oscillations)
or
smaller rate of dissipation of energy (so more oscillations)
B1

PUBLISHED
May/June 2021
11(a) intensity: vary filament current/p.d. across filament B1
hardness: vary accelerating potential difference B1
11(b)(i) I = I0e –μx C1
IS = I0 exp(–0.92 × 9.0)
= 2.5 × 10–4 I0
A1
11(b)(ii) IC = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I0 C1
= 6.7 × 10–7 I0 A1
11(c) conclusion consistent with values in (b)(i) and (b)(ii)
e.g. IS ≫ IC so good contrast
B1

PUBLISHED
May/June 2021
12(a) • frequency determines energy of photon
• intensity determines number of photons (per unit time)
• intensity does not determine energy of a photon
Any two points, 1 mark each
B2
kinetic energy (of the electron) depends on the energy of one photon B1
12(b)(i) E = hc / λ
or
E = hf and c = fλ
C1
E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1
(= 7.96 × 10–19 J)
= 5.0 eV
A1
12(b)(ii) EMAX = photon energy – work function C1
work function = 5.0 – 1.4
= 3.6 eV
A1

PHYSICS 9702/42
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

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marking principles.
descriptors.

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PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
Abbreviations
Mark categories

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May/June 2021
Annotations
incorrect equation.
not after XP.
further errors.
been awarded.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
1(a) (gravitational) force per unit mass B1
1(b)(i) g = GM / r2 C1
= (6.67 × 10–11 × 6.42 × 1023) / (3.39 × 106)2
= 3.73 N kg–1
A1
1(b)(ii) a = rω2 and ω = 2π / T
or
a = v2 / r and v = 2πr / T
C1
a = 3.39 × 106 × (2π / (24.6 × 3600))2
= 0.0171 m s–2
A1
1(b)(iii) force per unit mass = 3.73 – 0.0171
= 3.71 N kg–1
A1

PUBLISHED
May/June 2021
2(a) pV = nRT C1
pV = nRT and N = nNA
or
pV = NkT
C1
3.1 × 10–3 × 8.5 × 105 = (N × 290 × 8.31) / (6.02 × 1023)
so N = 6.6 × 1023
or
3.1 × 10–3 × 8.5 × 105 = N × 1.38 × 10–23 × 290
so N = 6.6 × 1023
A1
2(b)(i) (3.1 × 10–3 × 8.5 × 105) / 290 = (6.3 × 10–3 × 2.7 × 105) / T
so T = 190 K
or
6.3 × 10–3 × 2.7 × 105 = 6.6 × 1023 × 1.38 × 10–23 × T
so T = 190 K
A1
2(b)(ii) ΔU = 3/2 × k × ΔT × N C1
= 3/2 × 1.38 × 10–23 × (190 – 290) × 6.6 × 1023 C1
= –1400 J A1
2(c) ΔU = q + w M1
q = 0 so ΔU = w A1

PUBLISHED
May/June 2021
3(a) acceleration in opposite direction to displacement shown by – sign B1
g / L is constant M1
(so) acceleration is (directly) proportional to displacement A1
3(b) ω2 = g / L C1
ω = 2π / T
or
ω = 2πf and f = 1 / T
C1
(2π / T)2 = 9.81 / 0.18
T = 0.85 s
A1
3(c) energy ∝ x0
2 C1
(after 3 cycles,) amplitude = (0.94)3x0
= 0.83x0
C1
ratio final energy / initial energy = 0.832
= 0.69
A1

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May/June 2021
4(a)(i) frequency (modulation) B1
4(a)(ii) 1. zero B1
2. frequency (of 1.2 MHz) varies by ±50 kHz B1
frequency varies (by ±50 kHz) at a rate of 8000 times per second B1
4(b)(i) wavelength = (3.00 × 108) / (240 × 103) C1
(= 1250 m)
= 1.25 km
A1

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May/June 2021
5(a) from x = 0 to x = r: horizontal line at V = 1.0V0 B1
from x = r to x = 3r: curve with negative gradient of decreasing magnitude starting at (r, 1.0V0) B1
line passing through (2r, ½V0) and (3r, ⅓V0) B1
5(b) line with negative gradient from λ = ⅓λ0 to λ = λ0 B1
line passing through (λ0, 0) B1
curve with negative gradient of decreasing magnitude passing through (½λ0, EMAX) and (⅓λ0, 2EMAX) B1
5(c) 1.0T½ shown at ½N0 and 2.0T½ shown at ¼N0 B1
line starting at (0, 0) and reaching (T, N0–N) B1
line starting at (0, 0) and reaching original curve at (1.0T½, ½N0) B1

PUBLISHED
May/June 2021
6(a) potential difference applied between the plates M1
causes charge separation (between the plates)
or
causes energy to be stored (between the plates)
A1
6(b)(i) I = Q / t M1
clear substitution of Q = CV and f = 1 / t, leading to I = fCV A1
6(b)(ii) 2.5 × 10–6 = 50 × C × 180 C1
C = 280 pF A1
6(c) (total) capacitance increases B1
greater charge (for each cycle/discharge) so greater (average) current
or
V and f are constant so (average) current increases
or
I is (directly) proportional to C so (average) current increases
B1

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May/June 2021
7(a)(i) no current enters/leaves the input B1
7(a)(ii) gain is the same for all frequencies B1
7(b)(i) VIN = 1.5 × 400 / (400 + 1100) = 0.40 V
or
VIN = 1.5 – (1.5 × 1100 / 1500) = 0.40 V
or
(1.5 – VIN) / 1100 = VIN / 400 so VIN = 0.40 V
A1
7(b)(ii) gain = (–) Rf / Ri C1
VOUT / 0.40 = (360 + 100) / 96 C1
VOUT = 1.9 V A1
7(b)(iii) resistance of thermistor decreases B1
(magnitude of) gain decreases so reading decreases B1
7(b)(iv) (at gain 12.5) VOUT is 5.0 V, so (above gain 12.5) output becomes saturated B1

PUBLISHED
May/June 2021
8(a) • force per unit length
• force per unit current
• length/current perpendicular to field
1 mark for any two points, 2 marks for all three points
B2
8(b) change in potential energy = change in kinetic energy
or
qV = ½mv2
B1
v = √(2qV / m) A1
8(c)(i) magnetic force = centripetal force
or
Bqv = mv2 / r
M1
clear substitution of expression for v and correct algebra leading to q / m = 2V / B2r2 A1
8(c)(ii) q / m = (2 × 230) / [(0.38 × 10–3)2 × 0.142] C1
= 1.6 × 1011 C kg–1 A1
8(c)(iii) (for α-particle,) q / m is (much) smaller B1
r would be much larger B1

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May/June 2021
9(a) (particle is) stationary/not moving B1
(particle is) moving parallel to the (magnetic) field B1
9(b) magnetic field around each coil is circular
or
each coil is normal to magnetic field due to adjacent coils
B1
current in coil interacts with (magnetic) field to exert force (on coil) B1
force is normal to both coil and magnetic field
or
force parallel to axis (of coil)
B1
forces between coils are attractive so spring contracts B1
9(c) (oscillating) coils cut magnetic flux
or
as separation of coils changes, magnetic flux changes
B1
cutting flux causes induced e.m.f. in coils B1
changing (induced) e.m.f. causes changing current (in coil) B1

PUBLISHED
May/June 2021
10(a) the steady current
or
the direct current
M1
that produces the same heating effect (as the alternating current) A1
10(b)(i) peak current = 2.6 A and r.m.s. current = 1.8 A A1
10(b)(ii) peak current = 2.0 A and r.m.s. current = 2.0 A A1
10(c)(i) k = 2πf C1
= 2π × 50
= 310 rad s–1
A1
10(c)(ii) power = VRMS
2 / R or power = V0
2 / 2R C1
R = (240 / √2)2 / 3200 or R = 2402 / (2 × 3200)
R = 9.0 Ω
A1

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May/June 2021
11(a) to produce a 3-dimensional image of structure/body B1
11(b) X-rays (are used) B1
scanning in sections B1
scanning from many angles B1
image of each section is 2-dimensional B1
scanning repeated for many sections
or
images of many sections combined together
B1
12(a) quantum of energy M1
of electromagnetic radiation A1
12(b)(i) energy = hc / λ
or
energy = hf and f = c / λ
C1
0.57 × 106 × 1.60 × 10–19 = (6.63 × 10–34 × 3.00 × 108) / λ
λ = 2.2 × 10–12 m
A1

PUBLISHED
May/June 2021
12(b)(ii) p = h / λ C1
= (6.63 × 10–34) / (2.2 × 10–12)
= 3.0 × 10–22 N s
A1
or
p = E / c (C1)
= (0.57 × 106 × 1.60 × 10–19) / (3.00 × 108)
= 3.0 × 10–22 N s
(A1)
12(c)(i) mass (of Sm-157 nucleus) = 157 × 1.66 × 10–27
or
mass (of Sm-157 nucleus) = 0.157 / (6.02 × 1023)
C1
recoil speed = (3.00 × 10–22) / (157 × 1.66 × 10–27)
= 1.2 × 103 m s–1
A1
12(c)(ii) (1.2 ×) 103 m s–1 is much less than (3.0 ×) 108 m s–1 B1

PHYSICS 9702/43
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Maximum Mark: 100
Published
Teachers.

PUBLISHED
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marking principles.
descriptors.

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your working’.

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May/June 2021
Abbreviations
Mark categories

PUBLISHED
May/June 2021
Annotations
incorrect equation.
not after XP.
further errors.
been awarded.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
1(a) force per unit mass B1
1(b) GMm / r 2 = mrω 2 and ω = 2π/T
or
GMm / r 2 = mv2 / r and v = 2πr / T
C1
6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1
r = 6.9 × 106 m A1
1(c)(i) r3ω2 = constant or r3 / T2 = constant C1
r3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1
or
GMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1)
6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2
so r = 9.4 × 106 m
(A1)
1(c)(ii) separation increases so (potential energy) increases
or
movement is against gravitational force so (potential energy) increases
B1
1(c)(iii) potential energy = (–)GMm / r C1
ΔEP = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1
= 1.9 × 1010 J A1

PUBLISHED
May/June 2021
2(a) pV = NkT C1
N = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1
or
pV = nRT and nNA = N (C1)
N = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1)
2(b) speed of molecule decreases on impact with moving piston B1
mean square speed (directly) proportional to (thermodynamic) temperature
or
mean square speed (directly) proportional to kinetic energy (of molecules)
or
kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature
B1
kinetic energy (of molecules) decreases (so temperature decreases) B1
2(c)(i) ΔU = 3/2 × k × ΔT × N C1
= 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1
= – 64 J A1
2(c)(ii) decrease in internal energy is less than work done by gas M1
(thermal energy is) transferred to the gas (during the expansion) A1

PUBLISHED
May/June 2021
3(a) acceleration (directly) proportional to displacement B1
acceleration is in opposite direction to displacement B1
3(b) ω2 = 2k / m and ω = 2πf C1
(2πf)2 = (2 × 130) / 0.84 C1
f = 2.8 Hz A1
3(c)(i) resonance B1
3(c)(ii) oscillator supplies energy (continuously) B1
energy of trolley constant so energy must be dissipated
or
without loss of energy the amplitude would continuously increase
B1
4 (ultrasound) pulse B1
reflected at boundaries B1
gel is used to minimise reflection at skin
or
generated and detected by quartz crystal
B1
time delay between generation and detection gives information about depth B1
intensity (of reflected wave) gives information about nature of boundary B1

PUBLISHED
May/June 2021
5(a) amplitude of the carrier wave varies M1
in synchrony with the displacement of the (information) signal A1
5(b)(i) wavelength = (3.0 × 108) / (300 × 103)
= 1000 m
A1
5(c) attenuation = 10 lg (P1 / P2) C1
73 = 10 lg (PT / PR)
73 = 10 lg (PT x2 / 0.082 PT) or x2 / 0.082 = 107.3
C1
x = 1300 m A1

PUBLISHED
May/June 2021
6(a) from x = 0 to x = r : E = 0 B1
from x = r to x = 3r : curve with negative gradient of decreasing magnitude passing through (r, E0) B1
line passing through (2r, E0 / 4) and (3r, E0 / 9) B1
6(b) from p = p0 / 2 to p = p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0) B1
line passing through (½p0, 2λ0) B1
6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1
line passing through (15, ½N0) B1
line passing through (30, 0.75N0) and (45, 0.88N0) B1

PUBLISHED
May/June 2021
7(a) charge / potential M1
charge is on one plate, potential is p.d. between the plates A1
7(b)(i) I = Q / t M1
charge = CV and time = 1 / f leading to I = fCV A1
7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1
C = 530 pF A1
7(c) (total) capacitance is halved B1
charge (for each cycle/discharge) is halved
or
since f and V are constant, current is proportional to capacitance
B1
current = 2.4 μA B1

PUBLISHED
May/June 2021
8(a) V+ = 3.0 × 3.0 / (2.5 + 3.0) C1
= 1.6 V A1
8(b) V – is +2.0 V
or
V – > V +
B1
output is negative so (LED) does not emit light B1
8(c) at 0 °C, V – = 1.7 V
or
for all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ
B1
V – always greater than V + (so no switching) B1
8(d) (at 20 °C,) RT = 1.8 kΩ C1
2.5 / 3.0 = 1.8 / R
or
[R / (R + 1.8)] × 3.0 = 1.6
C1
R = 2.2 kΩ A1

PUBLISHED
May/June 2021
9(a) region where there is a force exerted on M1
a current-carrying conductor
or
a moving charge
or
a magnetic material/magnetic pole
A1
9(b)(i) face PSWV shaded B1
9(b)(ii) accumulating electrons cause an electric field (between the faces) B1
force due to electric field opposes force due to magnetic field B1
accumulation stops when magnetic force equals electric force B1
9(c)(i) number density of charge carriers B1
9(c)(ii) PV or QT or SW B1
9(d) (for semiconductor,) n is (much) smaller so VH (much) larger B1

PUBLISHED
May/June 2021
10(a) direction of (induced) e.m.f. M1
is such as to oppose the change causing it A1
10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1
(induced) e.m.f. causes (eddy/induced) currents (in ring) B1
currents (in ring) cause magnetic field (around ring) M1
two fields interact to cause resistive/opposing force A1
or
current (in ring) is in a magnetic field (M1)
which causes resistive force (A1)
or
currents (in ring) dissipate thermal energy (M1)
(thermal) energy comes from energy of oscillations (A1)
10(c) current cannot pass all the way around the ring B1
(induced) currents smaller B1
smaller resistive force (so more oscillations)
or
smaller rate of dissipation of energy (so more oscillations)
B1

PUBLISHED
May/June 2021
11(a) intensity: vary filament current/p.d. across filament B1
hardness: vary accelerating potential difference B1
11(b)(i) I = I0e –μx C1
IS = I0 exp(–0.92 × 9.0)
= 2.5 × 10–4 I0
A1
11(b)(ii) IC = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I0 C1
= 6.7 × 10–7 I0 A1
11(c) conclusion consistent with values in (b)(i) and (b)(ii)
e.g. IS ≫ IC so good contrast
B1

PUBLISHED
May/June 2021
12(a) • frequency determines energy of photon
• intensity determines number of photons (per unit time)
• intensity does not determine energy of a photon
Any two points, 1 mark each
B2
kinetic energy (of the electron) depends on the energy of one photon B1
12(b)(i) E = hc / λ
or
E = hf and c = fλ
C1
E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1
(= 7.96 × 10–19 J)
= 5.0 eV
A1
12(b)(ii) EMAX = photon energy – work function C1
work function = 5.0 – 1.4
= 3.6 eV
A1

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation May/June 2021
MARK SCHEME
Maximum Mark: 30
Published
Teachers.

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marking principles.
descriptors.

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May/June 2021
your working’.

PUBLISHED
May/June 2021
Annotations
 Correct point
Method of analysis marks in Question 1
1–10 Additional detail marks in Question 1
X Incorrect point
^ Omission
BOD Benefit of the doubt
NBOD No benefit of the doubt given
ECF Error carried forward
P Defining the problem marks in Question 1
Power of ten error in Question 2
M0 Methods of data collection marks in Question 1
SF Incorrect number of significant figures

PUBLISHED
May/June 2021
1 Defining the problem
R is the independent variable and t is the dependent variable or vary R and measure t 1
keep the number of turns on the coil/N constant 1
Methods of data collection
labelled diagram or correct symbols including:
• labelled (d.c.) power supply
• switch in series with power supply, resistor and coil
• complete workable circuit
1
circuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other
connections (not ohmmeter in main circuit)
1
method to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to
datalogger/computer
1
method to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2 /4 1
Method of analysis
plot a graph of t against 1 / R
(allow log t against log R)
1
relationship valid if a straight line passing through the origin is produced
(allow gradient = –1 for graph of log t against log R)
1
2
gradient
.
L
K
AN
×
=
1

PUBLISHED
May/June 2021
1 Additional detail including safety considerations 6
D1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or
ensure no bare wires/use shrouded connectors
D2 wear (insulating) gloves to prevent electric shock/electrocution
D3 keep A and L constant
D4 use ruler/calipers to measure L
D5 repeat measurements of diameter in different directions/at points along the coil and average
D6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure
resistance using ohmmeter
D7 repeat experiment for each value of R and average t
D8 method to determine t:
use of time-base from oscilloscope explained
or
use of time axis of output from data logger/computer explained
D9 use smaller values of R to increase I
D10 reduce L or increase N or increase A to increase t

PUBLISHED
May/June 2021
2(a)
gradient =
1
uA
y-intercept =
1
u
1
2(b)
(M + m) / g
1
v
/ s cm–1
380 0.226 or 0.2262
480 0.255 or 0.2551
580 0.294 or 0.2941
680 0.331 or 0.3311
830 0.388 or 0.3876
930 0.429 or 0.4292
Values of (M + m) and
1
v
as shown above.
1
Absolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 1
2(c)(i) Six points plotted correctly.
Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square.
1
Error bars in (M + m) plotted correctly.
All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
1

PUBLISHED
May/June 2021
2(c)(ii) Line of best fit drawn covers all points.
Points must be balanced. Do not allow line from top point to bottom point.
Line must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400).
1
Worst acceptable line drawn (steepest or shallowest possible line that passes through all error bars).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx.
Distance between data points must be at least half the length of the drawn line.
1
Gradient of worst acceptable line determined.
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)
Do not allow ECF from false origin method.
1
2(d)(i) u determined using y-intercept and u and A given to two or three significant figures.
1
-intercept
u
y
=
1
A determined using gradient with correct substitution and units with correct power of ten for u and A.
-intercept 1
gradient gradient
y
A A
u
= =
×
or
1

PUBLISHED
May/June 2021
2(d)(ii) Percentage uncertainty in A determined, e.g.
gradient -intercept
percentage uncertainty in
gradient -intercept
y
A
y
 
Δ Δ
= +
 
 
or
Δu clearly determined using the value of u and
gradient
percentage uncertainty in 100
gradient
u
A
u
 
Δ Δ
= + ×
 
 
or
correct substitution for max/min methods e.g.
1
max
min min gradient
A
u
=
×
1
min
max maxgradient
A
u
=
×
1
2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten.
( )
330
2
A u
m A
×
= − +
or
0.5 -intercept
330
gradient
y
m
−
= −
1

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
Annotations
 Correct point
X Incorrect point
^ Omission

PUBLISHED
May/June 2021
A is the independent variable and t is the dependent variable or vary A and measure t 1
keep Δθ constant 1
labelled diagram of workable experiment including:
• beaker of water
• cylinder in water
• electrical heater in water
• thermometer in water
• minimum of three labels from heater, thermometer, cylinder, water, beaker
1
circuit diagram to determine power of the heater e.g. ammeter and voltmeter correctly positioned with a power supply or
wattmeter correctly connected to power supply and heater
1
method to determine time for temperature of water to increase or t, e.g. use a stopwatch/timer 1
method to determine A, e.g. micrometer/calipers to determine diameter of cylinder and A = πd2 /4 1
Method of analysis
plot a graph of t against A (not logarithmic graphs) 1
gradient P
W
h θ
×
=
Δ
1
-intercept
y P
Z
θ
×
=
Δ
1

PUBLISHED
May/June 2021
D1 wear (heat proof) gloves to prevent burns from hot beaker/cylinder/heater/water
D2 keep P and h constant
D3 check that/ensure/keep initial temperature of the water constant or volume/mass of water constant
D4 use calipers/ruler to measure h
D5 repeat measurements of diameter in different directions/at different positions along cylinder and average
D6 method to calculate power of heater e.g. P = VI linked to correct circuit diagram for ammeter/voltmeter method
D7 repeat measurements of t for same A and average t
D8 ensure heater and cylinder are (totally) submerged/immersed
or
stir water (using a glass rod/stirrer)
D9 relationship valid if a straight line (not passing through the origin)
D10 method to insulate beaker, e.g. use of a lid on the beaker or foam/insulation around outside of beaker

PUBLISHED
May/June 2021
2(a)
gradient =
1
E
y-intercept =
r
E
1
2(b)
(R1 + R2) / Ω
1
I
/ A–1
55 58.1 or 58.14
69 70.4 or 70.42
78 78.1 or 78.13
80 80.6 or 80.65
89 87.7 or 87.72
103 99.0 or 99.01
Values of (R1 + R2) and
1
I
as shown above.
1
Absolute uncertainties in (R1 + R2) from ± (2.75 or 2.8 or 3) to ± (5.15 or 5.2 or 5). 1
1
Error bars in (R1 + R2) plotted correctly.
1

PUBLISHED
May/June 2021
Line must pass between (61.0, 65.0) and (63.5, 65.0) and between (96.5, 95.0) and (98.5, 95.0).
1
1
1
or
1
or
1
2(d)(i) E determined using gradient and E and r given to two or three significant figures.
1
gradient
E =
1
r determined using y-intercept with correct substitution and units with correct power of ten for E and r.
r = y-intercept/gradient or r = E × y-intercept
1

PUBLISHED
May/June 2021
2(d)(ii) Absolute uncertainty in E determined with method shown e.g.
gradient
gradient
E E
Δ
Δ = ×
or
1
min gradient
E E
Δ = −
1
maxgradient
E E
Δ = −
1
2(e) Value of R2 determined from (d)(i) or (c)(iii) and (c)(iv), with correct substitution and correct power of ten.
( )
2 22
0.0075
E
R r
= − +
or
( )
2
1
22
0.0075 gradient
R r
= − +
×
1

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

PUBLISHED
May/June 2021
marking principles.
descriptors.

PUBLISHED
May/June 2021

PUBLISHED
May/June 2021
your working’.

PUBLISHED
May/June 2021
Annotations
 Correct point
X Incorrect point
^ Omission

PUBLISHED
May/June 2021
R is the independent variable and t is the dependent variable or vary R and measure t 1
keep the number of turns on the coil/N constant 1
labelled diagram or correct symbols including:
• labelled (d.c.) power supply
• switch in series with power supply, resistor and coil
• complete workable circuit
1
circuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other
connections (not ohmmeter in main circuit)
1
method to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to
datalogger/computer
1
method to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2 /4 1
Method of analysis
plot a graph of t against 1 / R
(allow log t against log R)
1
relationship valid if a straight line passing through the origin is produced
(allow gradient = –1 for graph of log t against log R)
1
2
gradient
.
L
K
AN
×
=
1

PUBLISHED
May/June 2021
D1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or
ensure no bare wires/use shrouded connectors
D2 wear (insulating) gloves to prevent electric shock/electrocution
D3 keep A and L constant
D4 use ruler/calipers to measure L
D5 repeat measurements of diameter in different directions/at points along the coil and average
D6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure
resistance using ohmmeter
D7 repeat experiment for each value of R and average t
D8 method to determine t:
use of time-base from oscilloscope explained
or
use of time axis of output from data logger/computer explained
D9 use smaller values of R to increase I
D10 reduce L or increase N or increase A to increase t

PUBLISHED
May/June 2021
2(a)
gradient =
1
uA
y-intercept =
1
u
1
2(b)
(M + m) / g
1
v
/ s cm–1
380 0.226 or 0.2262
480 0.255 or 0.2551
580 0.294 or 0.2941
680 0.331 or 0.3311
830 0.388 or 0.3876
930 0.429 or 0.4292
Values of (M + m) and
1
v
as shown above.
1
Absolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 1
1
Error bars in (M + m) plotted correctly.
1

PUBLISHED
May/June 2021
Line must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400).
1
1
1
or
1
or
1
2(d)(i) u determined using y-intercept and u and A given to two or three significant figures.
1
-intercept
u
y
=
1
A determined using gradient with correct substitution and units with correct power of ten for u and A.
-intercept 1
gradient gradient
y
A A
u
= =
×
or
1

PUBLISHED
May/June 2021
2(d)(ii) Percentage uncertainty in A determined, e.g.
gradient -intercept
percentage uncertainty in
gradient -intercept
y
A
y
 
Δ Δ
= +
 
 
or
Δu clearly determined using the value of u and
gradient
percentage uncertainty in 100
gradient
u
A
u
 
Δ Δ
= + ×
 
 
or
1
max
min min gradient
A
u
=
×
1
min
max maxgradient
A
u
=
×
1
2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten.
( )
330
2
A u
m A
×
= − +
or
0.5 -intercept
330
gradient
y
m
−
= −
1

9702_s21_ms_all.pdf

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9702_s21_ms_all.pdf