GCE
Physics A
H156/01: Breadth in physics
Advanced Subsidiary GCE
Mark Scheme for Autumn 2021
Oxford Cambridge and RSA Examinations
GCE
Physics A
H156/01: Breadth in physics
Advanced Subsidiary GCE
Mark Scheme
...
GCE
Physics A
H156/01: Breadth in physics
Advanced Subsidiary GCE
Mark Scheme for Autumn 2021
Oxford Cambridge and RSA Examinations
GCE
Physics A
H156/01: Breadth in physics
Advanced Subsidiary GCE
Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of
qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications
include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals,
Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in
areas such as IT, business, languages, teaching/training, administration and secretarial skills.
It is also responsible for developing new specifications to meet national requirements and the
needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is
invested back into the establishment to help towards the development of qualifications and
support, which keep pace with the changing needs of today’s society.
This mark scheme is published as an aid to teachers and students, to indicate the requirements
of the examination. It shows the basis on which marks were awarded by examiners. It does not
indicate the details of the discussions which took place at an examiners’ meeting before marking
commenced.
All examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.
Mark schemes should be read in conjunction with the published question papers and the report
on the examination.
© OCR 2021H156/01 Mark Scheme October 2021
2
RM ASSESSOR
Annotations available in RM Assessor
Annotation Meaning
Correct response Used to indicate the point at which a mark has been awarded (one tick per mark awarded).
Incorrect response Used to indicate an incorrect answer or a point where a mark is lost.
AE 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.
BOD Benefit of doubt given
Used to indicate a mark awarded where the candidate provides an answer that is not totally satisfactory, but the
examiner feels that sufficient work has been done.
BP Blank page Use BP on additional page(s) to show that there is no additional work provided by the candidates.
CON Contradiction No mark can be awarded if the candidate contradicts himself or herself in the same response.
ECF Error carried forward
Used in numerical answers only, unless specified otherwise in the mark scheme. Answers to later sections of
numerical questions may be awarded up to full credit provided they are consistent with earlier incorrect answers.
Within a question, ECF can be given for AE, TE and POT errors but not for XP.
L1 Level 1 L1 is used to show 2 marks awarded and L1^ is used to show 1 mark awarded.
L2 Level 2 L2 is used to show 4 marks awarded and L2^ is used to show 3 marks awarded.
L3 Level 3 L3 is used to show 6 marks awarded and L3^ is used to show 5 marks awarded.
POT Power of 10 error
This is usually linked to conversion of SI prefixes. Do not allow the mark where the error occurs. Then follow
through the working/calculation giving ECF for subsequent marks if there are no further errors.
SEEN Seen To indicate working/text has been seen by the examiner.
SF Error in number of
significant figures
Where more SFs are given than is justified by the question, do not penalise. Fewer significant figures than
necessary will be considered within the mark scheme. Penalised only once in the paper.
TE Transcription error
This error is when there is incorrect transcription of the correct data from the question, graphical read-off, formulae
booklet or a previous answer. Do not allow the relevant mark and then follow through the working giving ECF for
subsequent marks.H156/01 Mark Scheme October 2021
3
XP Wrong physics or
equation
Used in numerical answers only, unless otherwise specified in the mark scheme. Use of an incorrect equation is
wrong physics even if it happens to lead to the correct answer.
^ Omission Used to indicate where more is needed for a mark to be awarded (what is written is not wrong but not enough).
Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific conventions).
Annotation Meaning
/ alternative and acceptable answers for the same marking point
Reject Answers which are not worthy of credit
Not Answers which are not worthy of credit
Ignore Statements which are irrelevant
Allow Answers that can be accepted
( ) Words which are not essential to gain credit
___
Underlined words must be present in answer to score a mark
ECF Error carried forward
AW Alternative wording
ORA Or reverse argumentH156/01 Mark Scheme October 2021
4
SECTION A
Question Answer Marks Guidance
1 D 1
2 B 1
3 D 1
4 C 1
5 B 1
6 C 1
7 C 1
8 D 1
9 A 1
10 D 1
11 A 1
12 A 1
13 B 1
14 B 1
15 D 1
16 C 1
17 D 1
18 A 1
19 D 1
20 C 1
Total 20H156/01 Mark Scheme October 2021
5
SECTION B
General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance
Question Answer Marks Guidance
21 (a) Separation between droplets increases (further down) B1
(b)
‘No motion’ explained either in terms of the first law or
second law
There is no / negligible resultant force
B1
B1
Allow F = ma, since F = 0, a is zero (hence at rest)
Allow an object continues in a state of rest or uniform
motion unless acted upon by a (resultant) force.
ALLOW no frictional/extra/new force
(c) (i)
Tangent drawn at t = 4.0 s
Attempt at calculating the gradient
v calculated from gradient and between 9.50 - 10.50 (m s-1)
OR
s = 20 (m) and s = ½ at2
20 = ½ a × 4.02 or a = 2.5 (m s-2)
v = 2.5 × 4.0 or v2 = 2 × 2.5 × 20
v = 10 (m s-1)
C1
C1
A1
C1
C1
C1
A0
Allow other correct methods
Note working required for this mark
(c) (ii) change in momentum = 1200 × 10 or 12000 (kg m s-1)
rate of change of momentum = 3000
unit: kg m s-2 or N
OR
F = 1200 × 2.5
rate of change of momentum = 3000
unit: kg m s-2 or N
C1
A1
B1
C1
A1
B1
Allow ECF from (c)(i)
Allow 2850 - 3150
Allow newton
Allow ECF from (c)(i)
Allow newton
Total 9H156/01 Mark Scheme October 2021
6
Question Answer Marks Guidance
22 (a) Ruler (with mm scale) B1 Allow any instruments with a mm scale e.g. tape measure
NOT calipers and micrometre
(b) 0.002
0.1000
(× 100) or 0.1
1.4
(× 100) or g = 1.42
2 ×0.100
(2 × 0.071… + 0.02) or 0.1628 …. or 16.3 %
absolute uncertainty = 1.6 (m s-2)
OR
gmax =
1.52
2 × 0.098
(= 11.48) or gmin = 1.32
2 ×0.102
(= 8.28)
range = 3.2 (m s-2)
absolute uncertainty = 1.6 (m s-2)
C1
C1
A1
C1
C1
A1
Allow 1SF answers here for uncertainties
Not g = 9.8 for this C1 mark; must see working
Allow 0.16 or 16%
Note: The answer must be given to 2 SF
Ignore value of g given on the answer line, e.g. 9.8 ± 1.6
Note: The answer must be given to 2 SF
Total 4H156/01 Mark Scheme October 2021
7
Question Answer Marks Guidance
23 (a) Circuit diagram with LED, voltmeter and ammeter
connected to a supply with correct polarity and some
means of adjusting the p.d.
Increase the p.d. until the LED is just lit / current is shown in
the circuit ORA
To observe the light from LED use a tube / turn lights off
B1
B1
B1
Allow variable (power) supply / variable resistor in circuit
Allow this on a labelled diagram e.g. tube drawn around
the diode
(b) (i) 0.045
1.6 × 10−19
number of electrons = 2.8 × 1017
C1
A1
(ii) A = π × (0.12 × 10-3)2 or 4.5(2) × 10-8 (m2)
0.045 = π × (0.12 × 10-3)2 × 6.3 × 1028 × 1.6 × 10-19 × v
v = 9.9 × 10-5 (m s-1)
C1
C1
A1 Allow 2 marks for 2.5 × 10-5 (m s-1); 0.24 mm and POT
error
(c) (i) (Current causes) increase in temperature of thermistor
Resistance of thermistor decreases (and hence V
decreases)
or
Current in the circuit increases, p.d. across resistor
increases (and hence V decreases)
B1
B1
Allow warms up/heat ups
Ignore increase temperature of the circuit
(ii) V = 2.4 (V) or VR = 3.6 (V)
I = 0.30 (A)
resistance = 8.0 (Ω)
OR
V = 2.4 (V) and a potential divider equation / idea
2.4 = �
� +12
× 6.0 or �
2.4
=
12
3.6
resistance = 8.0 (Ω)
C1
C1
A1
C1
C1
A1
Not V = 2.2 (V); misreading
Allow ECF if V = 2.2 (V) is used
Allow 8 (1 SF answer)
Not V = 2.2 (V); misreading
Allow ECF if V = 2.2 (V) is used
Allow 8 (1 SF answer)
Total 13H156/01 Mark Scheme October 2021
8
Question Answer Marks Guidance
24 (a) Both travel at the same speed / speed of light/ 3.0 × 108
m s-1
B1 Ignore travels at ‘c’
(b) 3.0 × 108 = f × 2.5 × 10-11
f = 1.2 × 1019 (Hz)
C1
A1
(c) Use of E = hf or E = hc/λ
EITHER
(λ) number = (500 × 10-9)/(2.5 × 10-11)
number = 2.0 × 104
OR
(E) number = (7.96 × 10-15)/ (3.98 × 10-19)
number = 2.0 × 104
OR
(f) number = (1.2 x 1019) / (6 x 1014)
number = 2.0 × 104
C1
C1
A1
C1
A1
C1
A1
Allow ECF from (b)
Allow ECF from (b)
Total 6H156/01 Mark Scheme October 2021
9
Question Answer Marks Guidance
25 (a) (i) It is longitudinal B1
(ii) Loudspeaker, microphone/ear and slit
Sound spreads from the slit AW
Size of slit comparable to the wavelength (of sound)
B1
B1
B1
Allow doorway for a slit/gap – receiver for microphone
Not diffraction (since it is in the stem of the question)
(b) (i) (refraction index) = speed of light in vacuum ÷ speed of
light in material
B1 Note light must be mentioned at least once
Allow n = c/v if all terms defined
Allow ration of speed of light in vacuum to speed of light
in material
NOT speed of light in air for c
(ii)1 Frequency (of light) is the same (in A and B)
(Light travels) slower in B or vB = 0.77vA ORA
v = fλ and λB < λA
B1
B1
B1
Allow f for frequency
Allow v directly proportional to λ
(ii)2 sin60° = 1.3 × sinθ
θ = 42 (°)
C1
A1
(ii)3 (No total internal reflection)
Internal reflection / critical angle can only occur for light
travelling from B to A AW
B1 Allow TIR can only occur for light entering an optically
less dense material/lower refractive index ORA
Not θ < φ
Total 11H156/01 Mark Scheme October 2021
10
Question Answer Marks Guidance
26 (a) weight × y = Fx
(ALρg) × y = Fx
� = ����� � ��
M1
M1
A0
Allow W or mg Wy = Fx or mgy = Fx
(b) (i) Straight line of best fit drawn through the data points
Gradient = 1.5
B1
B1 Allow gradient in the range 1.40 - 1.60
(ii) ����� � � = 1.5
6.8
6.4 × 10−5 × 0.90 × � × 9.81 = 1.5
ρ = 8.0 × 103 (kg m-3)
C1
C1
A1
Allow ECF from (b)(i)
Allow 8 × 103 (1 SF answer)
Note must be consistent with gradient value from (b)(i)
Total 8OCR (Oxford Cambridge and RSA Examinations)
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