GCE Physics A H556/01: Modelling physics Advanced GCE Mark Scheme for Autumn 2021

GCE Physics A H556/01: Modelling physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H556/01: Modelling physics Advanced GCE Mark Scheme for Autumn 2021Oxford Ca ...

GCE Physics A H556/01: Modelling physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H556/01: Modelling physics Advanced 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 2021H556/01 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. ^ Omission Used to indicate where more is needed for a mark to be awarded (what is written is not wrong but not enough).H556/01 Mark Scheme October 2021 3 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 argumentH556/01 Mark Scheme October 2021 4 SECTION A Question Answer Marks Guidance 1 C 1 2 A 1 3 C 1 4 B 1 5 A 1 6 B 1 7 B 1 8 D 1 9 D 1 10 C 1 11 C 1 12 B 1 13 C 1 14 B 1 15 A 1 Total 15H556/01 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 16 (a) (i) (g →) [m s-2] and (t →) [s] or (gt2 →) [m s-2 × s2] Clear evidence of working leading to m on both sides M1 A1 (ii) s / distance measured with a ruler / tape measure Timer mentioned for measuring t / time Measure distance from bottom of ball to (top of) trapdoor Any one from: • Take repeated readings (for t for same s) to determine average t • Avoid parallax error when using the ruler B1 B1 B1 B1 (b) (i) (p1 = 4.4 × 0.050 ) = 0.22 (kg m s-1) B1 (ii) (impulse =) ½ × 30 × 0.02 or 0.30 (kg m s-1) -0.30 = p2 – 0.22 p2 = (–) 0.08 (kg m s-1) C1 C1 A1 Allow any correct re-arrangement Possible ECF from (b)(i) Ignore sign Allow 0.52 for 2 marks (iii) (momentum of trapdoor =) 0.30 (kg m s-1) v = 3.0 (m s-1) C1 A1 Allow (KEtrapdoor =) ½ × 0.05 × (4.42 – 1.62) or 0.42 (J) Possible ECF from (b)(ii) Allow 1 SF answer here Allow alternate methods involving CoE (giving 2.9) and e(giving 2.8) Total 12H556/01 Mark Scheme October 2021 6 Question Answer Marks Guidance 17 (a) (mean) = 1.87(2) (mm) (range) = 0.04 mm (percentage uncertainty =) 0.02 1.872 × 100 percentage uncertainty = 1 (%) C1 C1 A1 Allow use of resolution of micrometer (gives percentage uncertainty of 0.5%) Allow use of maximum or minimum deviation from the mean Allow 2 or 3 SF answerH556/01 Mark Scheme October 2021 7 (b)* Level 3 (5–6 marks) Clear description and clear analysis 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 or Clear description or Clear analysis 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 analysis 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 Use level of response annotation in RM Assessor. Indicative scientific points may include: Description • Determine T by measuring several oscillations • Independent and dependent variables identified (e.g. L and T) • Variables kept constant (e.g. for L and T experiment, m is kept constant) • Repeating to determine average T • Measure length L and width w with ruler • Measure thickness t with a vernier (calliper) / micrometer • Use video/phone camera / stopwatch / data-logger and motion sensor / light gates and timer • Use top-pan balance / scales to measure m Analysis • Plot an appropriate graph, e.g. T2 against L3 or tabulate T2 ÷ L3 • Gradient of best line determined or average of T2 ÷ L3 • Use a large triangle to determine gradient • Gradient (or equivalent) related to E, e.g. gradient = 16π2m/wEt3 for T2 against L3 graph Total 9H556/01 Mark Scheme October 2021 8 Question Answer Marks Guidance 18 (a) (i) (energy =) 150 × 7 or 1050 (J) 1050 = 0.025 × c × 20 (c =) 2100 (J kg-1 K-1) C1 C1 A1 Allow any correct re-arrangement (ii) (energy=) 150 × (63 – 7) or 8400 (J) 8400 = L(f) × 0.025 (Lf = ) 3.4 × 105 (J kg-1) C1 A1 (iii) Longer time to heat water (through the same temperature) / shorter time to heat (ice) through same temperature / gradient of graph is greater for ice / gradient of graph is smaller for water/AW Water has greater specific heat capacity M1 A1 Allow calculation of gradients (b) (i) Molecules in X vibrate about fixed positions /AW Molecules in Z are free to move/random/AW B1 B1 Allow references to ice for X and water / liquid for Z Allow one correct for B1 from: • Molecules in X have lower KE/speed/velocity • Speed/velocity of molecules increases with temp/time • Amplitude or frequency increases with temp/time in XH556/01 Mark Scheme October 2021 9 (ii) Region Physical quantity, or quantities, that increase as time increases Physical quantity, or quantities, that remain constant as time increases X K P Y P K Z K P B1×3 Note that each B1 mark is for a correct row Allow KP/- for both X and Z (iii) Absolute zero / 0 K / - 273 °C B1 Total 13H556/01 Mark Scheme October 2021 10 Question Answer Marks Guidance 19 (a) For a body in (rotational) equilibrium the sum of the clockwise moments (about any point) is equal to the sum of the anticlockwise moments (about the same point) B1 Note Accept Σ / total (AW) for sum (b) (i) (horizontal component of F =) F × cos20° F cos20° × 1.30 = 0.30 × 40 × 9.81 F = 96.4 (N) M1 M1 A1 Allow ECF for incorrect trig i.e. use of sine (gives F =265) or cos(20 radians) which gives F = 222 for 2 marks. Allow ECF for incorrect units for angle and incorrect trig sin(20 radians) which gives F = 99(.2) for 1 mark (ii) R = F cos20° or 96(.4) × cos 20° (R =) 91 (N) C1 A1 Allow ECF from (b)(i) Answer is 90.6 (N) to 3sf if 96.4 used. Answer is 90(.2) (N) to 3sf if 96 used (c) When (line of force of the) weight falls to the right of the (bottom of the) wheel/AW B1 Total 7H556/01 Mark Scheme October 2021 11 Question Answer Marks Guidance 20 (a) (i) Straight symmetrical radial field lines and correct direction of field B1 Ignore field lines inside the Earth (ii) X and Y labelled which should be an equal distance away from the centre of the Earth B1 Note Judge by eye Allow X and Y both on the surface of the EarthH556/01 Mark Scheme October 2021 12 (b)* Level 3 (5–6 marks) Clear description and correct calculations leading to value of total energy (must include the negative sign) 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 correct calculations or Correct calculations (including the negative sign) 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 The information is basic and communicated in an unstructured way. The information is supported by limited evidence and the relationship to the evidence may not be clear. 0 marks No response or no response worthy of credit. B1×6 Indicative scientific points may include: Description • Orbit above the equator / equatorial orbit • Orbit from west to east/same direction of orbit as Earth’s rotation • Orbital period is 24 hours / 1 (sidereal) day /23hrs 56mins (4 s) • Orbit is circular / above the same point on the Earth Calculation • � = (−) ��� � • � = 6.67×10−11× 5.97 × 1024 × 2500 4.22 × 107 = (-) 2.4 × 1010 J • � = 2�� � = �� • � = 2� × 4.22 × 107 24 × 3600 = 3.07 × 103 m s-1 • � = 1 2 ��2 • � = 1 2 × 2500 × [3.07× 103]2 = 1.2 × 1010 J • Total energy = - 2.4 × 1010 + 1.2 × 1010 = - 1.2 × 1010 J • Allow full credit for algebraic proof using ��� �2 = ��2 � , � = (−)��� � , � = 1 2 ��2 and total energy = KE + PE Allow higher order answers in terms of Lagrange's Identity Total 8H556/01 Mark Scheme October 2021 13 Question Answer Marks Guidance 21 (a) ω2 = k/m or 60/0.080 or ω2 = 750 T = 2π/27.39 or T = 0.2295 (s) t = ¼ × 0.2295 t = 0.057 (s) C1 C1 C1 A1 Allow correct algebraic expression for T Allow incorrect value for omega Allow incorrect value of T (b) (i) � = 1 2 ��2 or � = ��ℎ or 0.080 × 9.81 × 0.20 or 1 2 × 60 × �2 0.080 × 9.81 × 0.20 = 1 2 × 60 × �2 x = 0.072 (m) C1 C1 A1 (ii)1 Time of flight is independent of speed/AW Because distance of fall is the same and initial velocity vertically is zero / velocity is horizontal at X B1 B1 Allow algebraic answers that assume initial vertical velocity is zero/velocity is horizontal at X. (ii)2 D increases as speed at X increases because the time of flight is constant/AW D is directly proportional to speed at X M1 A1 Allow d = vt idea “D is directly proportional to speed at X because the time of flight is constant” scores 2. Total 11H556/01 Mark Scheme October 2021 14 Question Answer Marks Guidance 22 (a) (i) {v = ωr and ω = 2πf} or v = 2 π fr Comparison with y = mx leading to gradient = 2 π r or ∆v/∆f = 2πr B1 B1 Allow v/f = 2πr (ii) Line of best fit drawn Gradient = 62.5 (m) 2πr = 62.5 r = 9.9 (m) B1 M1 M1 A0 Allow ± 3 Allow ECF on gradient (iii) � = ��2 � or � = �� and � = �2 � � = 1.7 × 10−27 × [2.0 × 107]2 9.9 F = 6.8 × 10-14 (N) C1 C1 A1 Allow use of candidate’s answer for (ii) or use of ‘10’ Expect answers of 6.8 or 6.9 × 10-14 (N) (b) r ∝ v2 / speed increases by a factor of √2 maximum speed = 2.8 × 107 (m s-1) C1 A1 Allow substitution into correct equation with r doubled Allow recalculation from previous value of force in (a)(iii) Total 10H556/01 Mark Scheme October 2021 15 Question Answer Marks Guidance 23 (a) � = � × 6.31 × 1030 (1.90 × 109)2 g = 117 (N kg-1) 2 × 0.14 1.90 + 0.42 6.31 or 0.21 or 21% (absolute uncertainty =) 25 (N kg-1) C1 C1 C1 A1 Reject 0.42/(0.142) = 21.4 Note: final answer of g = 117 ± 25 (N kg-1) with no working scores all 4 marks. Allow correct identification of %(r) and %(m) for 1 mark max if no other marks scored. Allow alternate method using max/min values of m and r that give the correct absolute uncertainty (28) (b) (i) The (total radiant) power (of a star) /AW B1 (ii) (L = 4πr2σT4) ratio = �(6 10.92 .0 × × 7500 450044) ratio = 2.3(1) C1 A1 Allow 1 mark for 5.3; square root omitted Allow 1 mark for 1:2.3 or 0.43 (iii) � (max) ∝ 1 � Lower temperature star will have the longest wavelength, so it is Aa2 B1 B1 Allow word equation Note Must mention wavelength and temperature (iv) From their (different) colours B1H556/01 Mark Scheme October 2021 16 (v) Any three from: • Continuous spectrum • Light / radiation / photons passes through cooler gas/star’s atmosphere • Photon(s) absorbed by electron(s) • Electron(s) excited / jump / make transition to higher energy level(s) • Electron only promoted if energy of photon matches energy gap between two given levels • Photons remitted in different directions • (so) idea of contrast with non-absorbed wavelengths B1 × 3 (c) Any two from: • Black hole has smaller mass / radius / size • Black hole has higher density/gravitational field strength/ stronger gravitational field • black hole absorbs light / does not emit visible light • Has an escape velocity => c • No fusion in a black hole (ORA) B1 B1 Allow black hole emits Hawking radiation Total 15OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsim