GCE
Physics A
H556/02: Exploring physics
Advanced GCE
Mark Scheme for November 2020
Oxford Cambridge and RSA Examinations
GCE
Physics A
H556/02: Exploring physics
Advanced GCE
Mark Scheme for November 2020Oxfor
...
GCE
Physics A
H556/02: Exploring physics
Advanced GCE
Mark Scheme for November 2020
Oxford Cambridge and RSA Examinations
GCE
Physics A
H556/02: Exploring physics
Advanced GCE
Mark Scheme for November 2020Oxford 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 2020H556/02 Mark Scheme November 2020
2
Here are the subject specific instructions for this question paper.
CATEGORISATION OF MARKS
The marking schemes categorise marks on the MACB scheme.
M marks These are method marks upon which A-marks (accuracy marks) later depend. For an M-mark to be scored, the point to which
it refers must be seen in the candidate’s answers. If a candidate fails to score a particular M-mark, then none of the
dependent A-marks can be scored.
A marks These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark to be scored.
C marks These are compensatory method marks which can be scored 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 it. 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 given.
B marks These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to
which it refers must be seen specifically in the candidate’s answers.
SIGNIFICANT FIGURES
If the data given in a question is to 2 sf, then allow an answer to 2 or more significant figures.
If an answer is given to fewer than 2 sf, then penalise once only in the entire paper.
Any exception to this rule will be mentioned in the Guidance.H556/02 Mark Scheme November 2020
3
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.
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).H556/02 Mark Scheme November 2020
4
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 argumentH556/02 Mark Scheme November 2020
5
SECTION A
Question Answer Marks Guidance
1 D 1
2 C 1
3 A 1
4 B 1
5 C 1
6 C 1
7 D 1
8 B 1
9 C 1
10 B 1
11 B 1
12 A 1
13 A 1
14 D 1
15 D 1
Total 15H556/02 Mark Scheme November 2020
6
SECTION B
General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance
Question Answer Marks Guidance
16 (a) (special coupling) gel is used that has the same
/’matching’ (acoustic) impedance as skin / body
Reduced / less / zero reflection (at the skin)
B1
B1
Allow Z
Allow gel and impedance is the same / matching for two
materials / mediums
Allow more transmission
(b) (Pulses of) ultrasound sent into the eye
Reflections from front and back of lens (and pulses
displayed on oscilloscope)
(Thickness of lens) determined from speed (of
ultrasound) and time (difference)
B1
B1
B1
Allow ultrasound reflected by any part of the eye
Allow ‘sound’ / wave (since ultrasound is in the question)
Ignore transducer placed close / next to eye
Allow thickness =
��
2
with c = speed (of ultrasound) and
t = time (difference)
Allow this mark even when the reflections are from
incorrect boundaries
Total 5H556/02 Mark Scheme November 2020
7
Question Answer Marks Guidance
17 (a) (E =) 1.8 × 1.6 × 10-19 or 2.88 × 10-19 (J)
1.8 × 1.6 × 10-19 = 6.63 × 10−34 × 3.0 × 108
�
λ = 6.9 × 10-7 (m)
C1
C1
A1
(b) (VR =) 2.7 (V) or (current =) 0.018 (A)
(ratio = 0.018 ×1.8
0.018 ×2.7
)
ratio = 0.67
C1
A1
Note the mark can be scored on circuit diagram
Note values of powers are: 0.0324 W and 0.0486 W
Allow 2/3; Not 0.66 (rounding error)
(c) (i) In darkness LDR has more resistance / p.d. across LDR is
large
or
In light LDR has less resistance / p.d. across LDR is small
Clear idea that when the LED is on, this will force the p.d.
across LED / LDR to decrease, forcing the LED to switch off
(ORA)
(The cycle of LED switching on and off is repeated)
B1
B1 Note the explanation must be in terms of p.d. / potential
divider. Ignore current
(ii) A sensible suggestion, e.g.
Point the LED away from the LDR / increase distance
(between LED and LDR) / insert a card between (LED and
LDR)
B1
Total 8H556/02 Mark Scheme November 2020
8
Question Answer Marks Guidance
18* Level 3 (5–6 marks)
Clear description and clear analysis of data
There is a well-developed line of reasoning which is clear
and logically structured. The information presented is
relevant and substantiated.
Level 2 (3–4 marks)
Some description and some analysis of data
OR
Clear description
OR
Clear analysis of data
There is a line of reasoning presented with some
structure. The information presented is in the most-part
relevant and supported by some evidence.
Level 1 (1–2 marks)
Limited description and limited analysis
OR
Some description
OR
Some analysis of data
There is an attempt at a logical structure with a line of
reasoning. The information is in the most part relevant.
0 marks
No response or no response worthy of credit
B1×6
Indicative scientific points may include:
Description
• Circuit showing supply, ammeter, voltmeter and
resistance wire /coil
• Measure I (in coil) with ammeter
• Measure V (across coil) with voltmeter
• Power (for coil) calculated: P = VI
• Resistance of thermistor either calculated using
R = V/I or measured with ohmmeter
• Change P / change V / use variable power supply / use
variable resistor (to change I)
• Keep the number of turns of coil constant throughout /
no draughts / wait until the resistance stabilises
Analysis
• lgP = lgk +nlgR (or natural logs ln)
• Plot a graph of lgP against lgR
• If expression is correct, then a straight line with nonzero intercept
• gradient = n
• intercept = lgk
• k = 10intercept (or k = eintercept for natural logs)
Total 6H556/02 Mark Scheme November 2020
9
Question Answer Marks Guidance
19 (a) h → J s / h → N m s / J → kg m2 s-2
base unit = kg m2 s-1
C1
A1
(b) (i) Vq = ½ mv2 and λ = ℎ
��
Clear algebra leading to �2 = ℎ2
2��
×
1 �
M1
A1
Allow p for mv
Allow e for q in (b)(i) – this is to be treated as a ‘slip’
(ii)1 (% uncertainty in λ2 =) 10%
(% uncertainty in λ =) 5%
C1
A1 Note 10 (%) on answer line will score the C1 mark
(ii)2 Straight line of best fit passes through all error bars B1
(ii)3 gradient = 1.0 (× 10-22)
ℎ2
2�� = gradient
(6.63×10−34)2
2 × � × 3.2 × 10−19 = gradient
m = 6.9 × 10 -27 (kg) (hence about 10-26 kg)
C1
C1
C1
A1
Ignore POT for this mark; Allow ± 0.20 (× 10-22)
Possible ECF for incorrect value of gradient
Note check for AE (condone rounding error here) and
answer must be about 10-26 (kg) for any incorrect gradient
value for this A1 mark
Special case: 1.37 × 10-26 kg scores 3 marks for
q = 1.6 × 10-19 C because answer is about 10-26 kg
Total 11H556/02 Mark Scheme November 2020
10
Question Answer Marks Guidance
20 (a) (i) sensible diameter, e.g. 7 (mm)
(power = 4.8 × 10-7 × π × (0.0035)2 )
power = 1.8 × 10-11 (W)
C1
A1
Allow 2 – 16 (mm)
Not πd2; this is XP
Note check for AE (condone rounding error here)
Possible ECF for diameter outside the range 2 – 16 (mm)
Allow 1 SF answer here
(ii) (I ∝ A2; intensity doubles)
A = √2 × 7.8 (or equivalent)
A = 11 (nm)
C1
A1
Allow the C1 mark for 4.8 (× 10-7) = k × [7.8 × (10-9)]2
(b) (When two or more waves meet at a point)
the resultant displacement is (equal to)
the sum of the (individual) displacements (of the waves)
B1 Allow sum / total / net for resultant
Ignore vector sum
(c) (i) phase difference = n × 360(°) for bright (fringes) /
constructive (interference)
phase difference = (n + ½) × 360(°) for dark (fringes) /
destructive (interference)
B1
B1
Allow zero or n × 2π(rad) or even number of π (rad) or
even number of 180(°)
Allow 180(°) or (n + ½) × 2π(rad) or odd number of π (rad)
or odd number of 180(°)
Special case: 1 mark for ‘completely in phase for bright
fringes/constructive (interference) and in anti-phase /
completely out of phase for dark fringes /destructive
(interference)’
(ii) λ = 3.0 × 108
4.75 × 1014 or λ = 6.316 × 10-7 (m)
x =
6.316 × 10−7 × 8.2
0.20 × 10−3 or x = 0.0259 (m)
t = 0.14 (s)
C1
C1
A1 Note the answer must be given to 2 SF for this mark
Special case: allow 1 mark for 8.6 × 10-11 s on the answer
line; incorrect physics using 0.18 = 4.75 × 1014 λ
Total 10H556/02 Mark Scheme November 2020
11
Question Answer Marks Guidance
21 (a) Electrons and quarks identified as fundamental particles
There are 6 electrons, 6 protons and 8 neutrons
Composition of proton → u ud
Composition of neutron → u d d
B1
B1
B1
B1
Allow e for electron, p for proton, and n for neutron
throughout
Allow 6 electrons, 20 u and 22 d
Do not award this mark if electron has quark-composition
Allow ‘2 up and 1 down’
Allow ‘2 down and 1 up’
(b) (i) (decay constant =) ln2
5700
decay constant = 1.2(2) × 10-4 (y-1)
C1
A0
(ii) 0.78 = e− � �
ln0.78 = (-) 1.2 × 10-4 × t
age = 2100 (y)
C1
C1
A1
Note 1 = 0.78e− � � is XP; answer is negative (- 2100 y)
There is no ECF from (b)(i)
Note 1.22 × 10-4 gives an answer of 2040 y or 2000 y
(iii) The ratio (of carbon-14 to carbon-12) has remained
constant
B1H556/02 Mark Scheme November 2020
12
(c)* Level 3 (5–6 marks)
Some description and clear analysis for r ∝ A1/3 and
correct calculation of mean density
There is a well-developed line of reasoning which is clear
and logically structured. The information presented is
relevant and substantiated.
Level 2 (3–4 marks)
Some description and some analysis for r ∝ A1/3 or some
calculation of mean density
OR
Some description and clear analysis for r ∝ A1/3
OR
Some description and correct calculation of mean density
OR
Clear analysis for r ∝ A1/3 and correct calculation of mean
density
There is a line of reasoning presented with some structure.
The information presented is in the most-part relevant and
supported by some evidence.
Level 1 (1–2 marks)
Some description
OR
Limited analysis for r ∝ A1/3
OR
Limited calculation of mean density
There is an attempt at a logical structure with a line of
reasoning. The information is in the most part relevant.
0 marks
No response or no response worthy of credit
B1×6 Indicative scientific points may include:
Description
• The density is independent of A
• The density is constant for most of d
• Nucleus with bigger A is larger (d / volume / mass)
Analysis for r ∝ A1/3
• r ≈ 3.6 (× 10-15 m) for Al-27 / r ≈ 5.5 (× 10-15 m) for Mo-
96 / r ≈ 7.0 (× 10-15 m) for Hg-200
• r/A1/3 = constant (or equivalent)
• Evidence for r ∝ A1/3 with at least 2 nuclei
(Note: 3.6 (×10-15)/271/3 ≈ 5.5 (×10-15)/961/3 ≈
7.0 (×10-15)/2001/3 ≈ 1.2 (×10-15)
or
• r3/A = constant (or equivalent)
• Evidence for r3 ∝ A with at least 2 nuclei
(Note. 3.63 (×10-45)/27 ≈ 5.53 (×10-45)/96 ≈
7.03 (×10-45)/200 ≈ 1.7 (×10-45)
Calculation for density
• ρ = M/V
• ρ = Amn ÷ 4
3
π r3 or ρ ≈ Amn ÷ diameter3
• mn≈ 1.7 × 10-27 (kg); ρ = 2.3 × 1017 (kg m-3) for at least
one of the nuclei given in the figure or table
Total 15H556/02 Mark Scheme November 2020
13
Question Answer Marks Guidance
22 (a) Magnetic field (around current-carrying wire)
(Fleming’s) left-hand rule mentioned
(Magnetic) field into page, (current is up the page) and
force is to the left / towards X
B1
B1
B1
Not magnetic force
Allow ‘field into page and wires attract’
Note the field direction and force direction can be shown
on the figure
(b) (i) (induced) e.m.f. is (directly) proportional / equal to the rate
of change of (magnetic) flux linkage
B1 Not current
Allow ‘rate of cutting’ for ‘rate of change’
(ii) Connect the primary (coil) to an alternating voltage / current
Oscilloscope connected across secondary coil / to measure
E
A graph of E against N will be a straight line through the
origin.
B1
B1
B1
Allow AC (can be on the figure)
Not changing / variable for alternating
Allow voltmeter (can be on the figure)
Allow p.d. / voltage for e.m.f. / E throughout
Ignore any component (e.g. lamp or resistor) connected
across the secondary coil
Allow (E ÷ N) = constant
Total 7H556/02 Mark Scheme November 2020
14
Question Answer Marks Guidance
23 (a) (i) (force =) (1.6 × 10−19)2
4π�0×(1.0 × 10−15)2
(F =) 230 (N)
F2 = 2302 + 2302 – 2 × 230 × 230 × cos120°
or
F = 2 × 230cos30°
F = 400 (N)
C1
C1
C1
A1
Special case: � = ��
4π�0�2 =
2 × 1.6 × 10−19
4π�0×(1.0 × 10−15)2 loses this
C1 mark, then ECF for the rest of the marks
Not the first two C1 marks for incorrect charge, then allow
ECF for the final C1A1 marks
Note force to 4 SF is 230.2 N
Allow sine rule / scale drawing
Allow this mark for 230cos30° or 200 (N)
Allow ± 10 (N) if scale drawing used
(ii) F / arrow vertical up the page B1 Allow correct arrow direction anywhere on the figure
(iii) Strong (nuclear) force (acts on the protons)
The strong (nuclear) force is attractive
B1
B1
Ignore gravitational force
Allow pulls / holds (the protons) / binds (the protons) for
‘attractive’
(b) (i) 12 000 = �
4π�0�
12 000 = �
4π�0 × 0.19
Q = 2.5(4) × 10-7 (C)
C1
C1
A0
Allow E = (V/d =) 6.316 × 104 C1
and
E = 6.316 × 104 = �
4π�0 × 0.192 C1
(ii)1 t = 78 × 3600
(I =) 2.5 × 10−7
78 ×3600
I = 8.9 × 10-13 (A)
C1
C1
A0
There is no ECF from (b)(i)
Note 2.54 × 10-7 gives an answer 9.0 × 10-13 A
(ii)2 (R =) 6000
9.0 × 10−13 or 6.7 × 1015 (Ω) or V = IR and � = �� �
6000
9.0 × 10−13 =
� × 0.38
1.1 × 10−4
ρ = 1.9 × 1012 (Ω m)
C1
C1
A1
There is no ECF from (b)(ii)1
Take 12000 V as TE for this C1 mark, then ECF
Note 8.9 × 10-13 (A) gives an answer 2.0 × 1012 (Ω m)
Total 14H556/02 Mark Scheme November 2020
15
Question Answer Marks Guidance
24 (a) Emits gamma (photons / radiation / waves / rays)
Any one from:
(Diagnosing the) function of organ
Detecting tumour
Small half-life
(Gamma rays) can be detected outside body / passes
through patient / least ionising
Position of tracer located
B1
B1 Not injected into a patient / non-invasive
Allow for half-life is a few hours
(b)
Collimator: Allows gamma (photons) parallel to the axis of
the tubes to pass through
Scintillator: gamma (photons) produces (many) light
(photons)
Photomultiplier (tubes): light (photons) produces electrons
/ current / electrical pulse / p.d. / signal
B1
B1
B1
Ignore any other components named / described
Allow photon / waves / rays
Allow idea of tubes allowing the gamma (photons) to
travel in the same direction
Allow crystal (or named crystal) for scintillator
Allow high-energy photons produce (many) low-energy
photons
Total 5H556/02 Mark Scheme November 2020
16
Question Answer Marks Guidance
25 (a) All except pair production / PP B1 Allow PE, S and C
(b) (energy =) 9.11 × 10-31 × (3.0 × 108)2
(energy =) 2 × 9.11 × 10-31 × (3.0 × 108)2 / 1.60 × 10-19
lg1.0(2) × 106 = 6 (as on graph)
OR
(energy =) 1.0 × 106 (eV) or lg1.0 × 106 = 6 (from graph)
(energy =) 1.6 × 10-13 J and evidence of mc2
2 × 9.11 × 10-31 × (3.0 × 108)2 ≈ 1.6 × 10-13
B1
B1
B1
B1
B1
B1
Note this is 8.2 × 10-14 (J)
Note this is 1.0(2) × 106 eV
Note this can be shown in a variety of ways
Total 4OCR (Oxford Cambridge and RSA Examinations)
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