GCE Physics A H556/02: Exploring physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H556/02: Exploring physics Advanced GCE Mark Scheme fo... r 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 2021H556/02 Mark Scheme October 2021 2 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.H556/02 Mark Scheme October 2021 3 ^ 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 argumentH556/02 Mark Scheme October 2021 4 SECTION A Question Answer Marks Guidance 1 D 1 2 A 1 3 D 1 4 C 1 5 D 1 6 C 1 7 B 1 8 A 1 9 B 1 10 B 1 11 C 1 12 D 1 13 A 1 14 B 1 15 C 1 Total 15H556/02 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 16 (a) (i) (v = fλ) 340 = 20 × 103 × λ wavelength = 1.7 × 10-2 (m) C1 A1 Allow 1 mark for 17 (m); 20 Hz used (ii) Loudspeaker and signal generator Frequency increased until limit of hearing frequency determined using f = 1/T B1 B1 B1 Allow this mark for a labelled diagram Do not allow t for time period (b) Difference: (stationary waves) has nodes / antinodes Similarity: Oscillations are longitudinal B1 B1 Differences and/or similarities can be described in terms of net energy transfer, phase or amplitude variations (c) Diagram showing angle within the block measured relative to the normal Increase the (incident) angle until the ray of light runs along the boundary / suffers total internal reflection (ORA) or angle measured using a protractor n determined using n = 1/sinC B1 B1 B1 Allow i / θ / C as the angle to be measured. Must be clear which angle is being measured. Expect the normal as a line perpendicular to straight edge of block, and emergent ray. No labels expected for the rays or the normal. Formula in this arrangement Total 10H556/02 Mark Scheme October 2021 6 Question Answer Marks Guidance 17 (a) The minimum energy needed to remove an electron (from the surface of a metal) B1 Allow work done for energy Allow photoelectron for electron (b) (i) energy of blue light / photon of blue light > 2.3 eV / work function or energy of red light / photon of red light < 2.3 eV / work function Energy of photon is independent of intensity (energy of photon given by equation) E = hf / E=hc/λ One photon interacts with one electron B1 B1 B1 B1 Not blue light has frequency > threshold frequency Or red light has frequency < threshold frequency Allow intensity linked to rate of photons / rate of electrons emitted per second Allow E proportional f / E proportional to 1/λ (ii) (φ =) 2.3 × 1.6 × 10-19 or (E =) 6.63×10−34×3.0×108 320×10−9 (KEmax =) 6.63×10−34×3.0×108 320×10−9 − 2.3 × 1.6 × 10−19 (v =) �2×92..11 5356 ×10 ×−1031−19 (wavelength =) 6.63×10−34 9.11×10−31×7.46×105 wavelength = 9.8 × 10-10 (m) C1 C1 C1 A1 φ = 3.68 × 10-19 (J); E = 6.2156 × 10-19 (J) KEmax = 2.5356 × 10-19 (J) v = 7.46 × 105 (m s-1) Total 9H556/02 Mark Scheme October 2021 7 Question Answer Marks Guidance 18 (a) (i) Arrow in anticlockwise direction B1 Allow this mark for correct direction shown on diagram either on or off connecting wires (ii) (E =) 4.5 – 2.4 or (RT =) 0.80 +0.50 + 1.2 4.5 – 2.4 = I × (0.80 +0.50 + 1.2) I = 0.84 (A) C1 C1 A1 E = 2.1 (V); RT = 2.5 (Ω) Treat missing 1.2 resistance as TE Allow 2 marks for 2.8 (A); E = 6.9 V used (iii) (I = Anev) 0.84 = π × (2.3 × 10-4)2 × 4.2 × 1028 × 1.60 × 10-19 × v v = 7.5 × 10-4 (m s-1) C1 A1 Possible ECF from (ii) Note answer is 2.5 × 10-3 (m s-1) for I = 2.76 (A) Allow 1 mark for 1.9 × 10-4; diameter used as radius (iv) Sensible suggestion, e.g. use a water bath / fan / only switch on when taking readings Need to lower the temperature / reduce resistance of R M1 A1 Allow keep the surroundings cold Allow to keep the temperature / resistance constant OR allow increase in temperature increases resistanceH556/02 Mark Scheme October 2021 8 (b)* Level 3 (5–6 marks) E and r calculated correctly and table completed correctly and clear description of P and R 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) Table completed correctly and some description of P and R / some attempt at E and r OR E and r calculated correctly OR Some attempt at calculating E and r and some description of P and R 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 calculation of E and r OR Table completed correctly OR Limited description of relationship between P and R 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: Calculating E and r • E = Ir + V • gradient = (-) r • y-intercept = E • Line extrapolated to y-axis • E = 1.2 (V) • r = 0.8(0 Ω) Table and description • Table completed (ignore SF) – see below • R increases as V increases (or I decreases) • P increases and decreases • Maximum power is when internal resistance is equal to R (0.8 Ω) V / V I / A R / Ω P / W 0.20 1.25 0.16 0.25 0.40 1.00 0.40 0.40 0.60 0.75 0.80 0.45 0.80 0.50 1.60 0.40 1.00 0.25 4.00 0.25 Total 14H556/02 Mark Scheme October 2021 9 Question Answer Marks Guidance 19 (a) Direction of field shown as clockwise Three field lines shown as concentric circles and distance between adjacent field lines increasing as distance from wire increases B1 B1 Expect at least one field line with an arrow Allow more than three lines, but distance between adjacent field lines increasing distance from wire must increase for all (b) (force =) 2.2 × 10-3 × 9.81 2.2 × 10-3 × 9.81 = B × 5.0 × 0.060 (= 0.072 T) (absolute uncertainty =) 0.2 6.0 + 0.1 5.0 (× 0.072= 0.0038 T) B = 0.072 ± 0.004 C1 C1 C1 A1 Allow calculation of percentage uncertainty = 5.3% Allow calculation of max B (=0.0759 T) and min B (=0.0683 T) Note B must be given to 2 SF and the uncertainty given to 1 SF. Special case: allow follow through from incorrect B calculation. Total 6H556/02 Mark Scheme October 2021 10 Question Answer Marks Guidance 20 (a) (CR =) 2000 × 10-6 × 120 × 103 1.00 = 1.48 × [1 – e-t/240] or 0.48 = 1.48e-t/240 (t =) - 240 × ln(0.48/1.48) t = 270 (s) C1 C1 C1 A1 CR = 240 (s) Special case: 94 (s) for use of discharging equation. Max 2 marks (b) Line of best fit drawn through the data points Gradient = 38 (Ckln2 = gradient) 1.2 × 10-3 × k × ln2 = 38 k = 4.6 × 104 (Ω m-1) B1 C1 C1 A1 Allow ± 2. Not calculated through use of a single point. Possible ECF from incorrect gradient Note: gradient of 40 gives 4.8 × 104 and gradient of 36 gives 4.3 × 104 Total 8H556/02 Mark Scheme October 2021 11 Question Answer Marks Guidance 21 (a) Electron removed / ejected (from atom) Photon (scattered with) increased wavelength / lower frequency / lower energy B1 B1 Needs a comparative statement (b) (intensity I = I0e-µx) = 4.6 × 103 × e-0.85 × 2.1 Either: (power =) 4.6 × 103 × e-0.85 × 2.1 × 3.4 × 10-4 Or (energy per unit area =) 4.6 × 103 × e-0.85 × 2.1 × 30 energy = 4.6 × 103 × e-0.85 × 2.1 × 3.4 × 10-4 × 30 energy = 7.9 (J) C1 C1 C1 A1 intensity = 772 (W m-2) power = 0.262 (W) energy per unit area = 23160 J m-2 energy at surface = 47 (J) 2 marks (c) CAT (CT) scan Any one from • A CAT scan will give 3D image • A CAT scan gives better contrast M1 A1 Insufficient: more detail / clearer image Total 8H556/02 Mark Scheme October 2021 12 Question Answer Marks Guidance 22 The positrons / beta-plus particles annihilate electrons (within the patient) Two gamma-photons are produced these (photons / rays) travel in opposite directions The difference in the arrival times at the detectors is used to locate the point of annihilation / nuclei B1 B1 B1 B1 Allow ‘two gamma rays’ instead of ‘two gamma-photons’ Allow gamma symbol Allow delay time Total 4H556/02 Mark Scheme October 2021 13 Question Answer Marks Guidance 23 (a) Control rods: absorb the neutrons (without further fission) Moderator: Slow down the neutrons / decrease KE of neutrons B1 B1 Not collide for absorb (b)* Level 3 (5–6 marks) Clear description and clear calculations of energy per kg 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) Clear description OR Clear calculations of energy per kg OR Some description and some calculations 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 OR Limited calculations 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 • Energy is produced in both reactions • More energy produced (per reaction) in fission • The (total) binding energy of ‘products’ is greater • In fusion, nuclei repel (each other) • Fusion requires high temperatures / high KE • Fission reactions are triggered by (slow-)neutrons • Chain reaction possible in fission Calculations • 1 kg of uranium has 4.26 mols / 2.56 × 1024 nuclei • 1 kg of deuterium has 500 mol / 3.01 × 1026 nuclei / 1.50 × 1026 ‘reactions’ • 200 MeV = 3.2 × 10-11 J • 4 MeV = 6.4 × 10-13 J • Uranium: ~ 1014 (J kg-1) (actual value 8.2 × 1013) • Deuterium: ~ 1014 (J kg-1) (actual value 9.6 × 1013) • The energy per kg is roughly the same Total 8H556/02 Mark Scheme October 2021 14 Question Answer Marks Guidance 24 (a) (i) (E = ) 4000 0.080 (F = ) 4000 0.080 × 1.6 × 10−19 (a = ) 8.0×10−15 9.11×10−31 or 8.78 × 1015 a = 8.8 × 1015 C1 C1 C1 A0 E = 5.0 × 104 (V m-1) F = 8.0 × 10-15 (N) Allow this mark if the working is shown. If only value is given, then the answer must be 3SF or more (ii) (t = ) 0.12 6.0×107 (t = 2.0 × 10-9 s) M1 A0 (iii) (x = ) ½ × 8.78 × 1015 × (2.0 × 10-9)2 x = 1.8 × 10-2 (m) C1 A1 Allow a = 8.8 × 1015 (b) Downward curved path Same x B1 B1 Ignore any line outside of the plates Expect same x by eye (c) Apply a magnetic field at right angles to electric field electric force = magnetic force No resultant vertical force, so only beta-particles with a specific speed will travel horizontally B1 B1 B1 Note this mark is for the idea that E and B are perpendicular even if direction of B is incorrect Allow ‘apply horizontal magnetic field’ Allow Eq = Bqv Allow v = E/B in this arrangement Total 11H556/02 Mark Scheme October 2021 15 Question Answer Marks Guidance 25 (a) (i) Material X because of the shorter half-life B1 Must be comparative Allow explanation in terms of decay constant (ii) (Alpha particles are stopped by the glass but) the betaparticles are not (AW) B1 Allow symbols (b) (i) 1 B1 (ii) Either: mass of nucleus 14.000 × 1.66 × 10-27 ( = 2.324 × 10-26 kg) Or: mass of nucleons = 8 × 1.675 ×10-27 + 6 × 1.673 × 10-27 (= 2.3438 × 10-26 kg) (∆m =) 2.3438 × 10-26 - 2.324 × 10-26 = (1.98 × 10-28 kg) (∆E =) 1.98 × 10-28 × (3.00 × 108)2 (BE per nucleon =) 1.782 × 10-11/14 binding energy per nucleon = 1.27 × 10-12 (J per nucleon) C1 C1 C1 A1 ∆m = 1.9262 × 10-28 kg Ignore sign throughout ∆E = 1.782 × 10-11 J Allow for any mass difference × (3.00 × 108)2 Note A mark for correct answer to 3sf only Total 7OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]
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