GCE Physics A H556/03: Unified physics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Physics A H556/03: Unified 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 2020H556/03 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. 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. 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. 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. A marks These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark to be scored. 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 Additional Guidance.H556/03 Mark Scheme November 2020 3 Annotations 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).H556/03 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 argumentH556/03 Mark Scheme November 2020 5 General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance Question Answer Marks Guidance 1 (a) (i) F = QE = QV/d or E = 5(.0) x 104 (Vm-1) F = 9.0 x 10-9 x 4000/ 8.0 x 10-2 (= 4.5 x 10-4 N) C1 A1 F = 5.0 x 104 x 9.0 x 10-9 (ii) weight; arrow vertically downwards tension; arrow upwards in direction of string electric (force); arrow horizontally to the right (not along dotted line) B1 x 2 All correct, 2 marks; 2 correct, 1 mark 1 mark maximum if more than 3 arrows are drawn Ignore position of arrows Allow W or 0.030(N) (not gravity or g) Allow T Allow F or E or 4.5 x 10-4(N) or electrostatic Ignore repulsion or attraction Not electric field / electric field strength / electromagnetic (iii) W x = F l 0.03 x = 4.5 x 10-4 x 120 or = 4.5 x 10-4 x 1.2 x = 1.8 cm or x = 0.018 m M1 M1 A0 Allow any valid alternative approach e.g. M1 deflection angle θ = 1o M1 x = 120sinθ 1 mark for each side of the equationH556/03 Mark Scheme November 2020 6 (b) Electric force/field (strength) increases Ball deflected further from vertical / moves to the right / touches negative plate Ball acquires the charge of the (negative) plate when it touches (Oscillates because) constantly repelled from (oppositely) charged plate B1 B1 B1 B1 Must be clear which force is increasing Must have the idea of a repeating cycle (c) I = Qf or Q = It f = 3.2 x 10-8/9.0 x10-9 = 3.6 (Hz) C1 A1 Total 12H556/03 Mark Scheme November 2020 7 Question Answer Marks Guidance 2 (a) (initial charge) Q = EC0 or (Q conserved so final) Q = V(C + C0) (as capacitors are in parallel) so EC0 = V(C + C0) (and hence V = C0 E / (C + C0)) M1 A1 At least one correct expression for Q for first mark The two correct expressions equated for the second mark (b) (i) 1/V = 1/E + C/EC0 (and compare to y = c + mx) B1 Mark is for rearrangement into linear equation (ii) 1/EC0 = 51 = 1/(9.1 C0) giving C0 = 1/(51 × 9.1) F C0 = 2.2 (mF) B1 B1 C0 = 2.1547 × 10-3 F Answer must be correct, rounded correctly and given in mF Candidate’s answer must be given to 2 SF (iii) (at least) one correct worst fit line drawn gradient calculated correctly using a large triangle uncertainty = C0 – 1/(wfl gradient x 9.1) uncertainty given is to the same number of decimal places as C0 B1 B1 B1 B1 Top and bottom points chosen must be from opposite extremes of uncertainty limits, accurate to within half a small square Δx ≥ 1.5 × 10-3; expect 59±1 or 44±1 (or 0.059 or 0.044); allow ECF from poorly drawn line; readings must be accurate to within half a small square ECF from b(ii); expect uncertainty of up to 0.4(mF) ECF from b(ii) If no value for C0 given in b(ii), allow any answer given to 1dp (c) Only effect is to slow the charging and/or discharging (of capacitor(s)) and so the final charges are unchanged / the values for V are unchanged / the graph is unchanged / the gradient is unchanged / there is no effect on the experiment (results) B1 Allow and so the experiment takes longer Total 10H556/03 Mark Scheme November 2020 8 Question Answer Marks Guidance 3 (a) (i) L (= 4πr2σT4) = 4 × π × (7.0 × 108)2 × 5.67 × 10-8 × 58004 L = 3.95 ×1026 (W) M1 A0 Mark is for substitution of values Allow σ for 5.67 × 10-8 (ii) By ratios: 25 = 1.72 x (T/5800)4 T4 = 9.8 × 1015 T = 9950 (K) C1 C1 A1 or T4 = 25L/4πσ(1.7r)2 = 25 x 3.95 x1026 4π x 5.67x 10−8 x (1.7 x 7 x 108)2 ECF for L in a(i) but only if L = 4 x 1026 to 1s.f. ECF for incorrect σ in a(i) Allow 9.9 × 1015 (using L = 4 x 1026) Allow 10,000 K (b) see next page B1 x 6 Total 10H556/03 Mark Scheme November 2020 9 Question Answer Marks Guidance 3 (b) * Level 3 (5 - 6 marks) Clear expansion of three statements There is a well-developed line of reasoning which is clear and logically structured. The information presented is clear, relevant and substantiated. Level 2 (3 – 4 marks) Clear expansion of two statements or Limited attempt at all three 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 attempt at one or two statements 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 x 6 Use level of response annotations in RM Assessor, e.g. L2 for 4 marks, L2^ for 3 marks, etc. Indicative scientific points may include: statement 1 • fusion reactions are occurring • which change H into He • and mass is lost which releases energy • energy released = c2Δm • Δm per second = luminosity / c2 statement 2 • average k.e. of each proton is 3 2 kT • high T means protons are travelling at high speed • so fast enough to overcome repulsive forces • and get close enough to fuse • p.e. = e2/4πε0r so T must be high enough for 3 2 kT> e2/4πε0r • r is approximately 3fm statement 3 • k.e. ∝ T so average energy at 107 K is only one thousandth of the average energy at 1010 K when protons might fuse • but M-B distribution applies so at the high energy end there will be a few p with enough energy • quantum tunnelling across potential barrier is possible • small probability of many favourable collisions to boost energy of p • 4 p must fuse to produce He; it is complicated process making probability of fusion much less • number of p in Sun is so huge that, even with such a small probability, 4 x 109 kg of p still interact s-1 • a larger probability means lifetime of the Sun would be shorterH556/03 Mark Scheme November 2020 10 Question Answer Marks Guidance 4 (a) There is no contact force between the astronaut and the (floor of the) space station (so no method of measuring / experiencing weight) B1 Allow astronaut and the space station have same acceleration (towards Earth) / floor is falling (beneath astronaut) (b) (i) M = 5.97 x 1024(kg) or ISS orbital radius R = 6.78 x 106(m) or g ∝ 1/ r2 (gr2= constant so) g x (6.78 x 106)2 = 9.81 x (6.37 x 106)2 g = 8.66 (N kg-1) C1 C1 A1 or g (= GM/R2) = 6.67 x 10-11 x 5.97 x 1024 / (6.78 x 106)2 Allow rounding of final answer to 2 SF i.e. 8.7 (N kg-1) (ii) 2πr /T = v or T = 2 × 3.14 × 6.78 × 106 / 7.7 × 103 T = 5.5 × 103 s (= 92 min) M1 A1 ECF incorrect value of R from b(i) (c) ½Mc2 (½NAmc2) = 3 2 RT c2 = 3 × 8.31 × 293 / 2.9 × 10-2 = 2.52 × 105 √c2 = 500 (m s-1) (= 7.7 × 103 / 15) C1 C1 A1 A0 or ½mc2 = 3 2 kT or c2 = 3kT/m or c2 = 3 ×1.38 x 10-23 × 6.02 x 1023 x 293/2.9 × 10-2 = 2.52 × 105 not (7.7 × 103 / 15) = 510 (m s-1) (d) power reaching cells (= IA) = 1.4 × 103 × 2500 = 3.5 × 106 W power absorbed = 0.07 × 3.5 × 106 = 2.45 × 105 W cells in Sun for (92 - 35 =) 57 minutes average power = 57/92 × 2.45 × 105 = 1.5 × 105 (W) C1 C1 C1 A1 mark given for multiplication by 0.07 at any stage of calculation (90 – 35 =) 55 minutes using T = 90 minutes ECF value of T from b(ii) 55/90 × 2.45 × 105 = 1.5 × 105 (W) using T = 90 minutes Total 13H556/03 Mark Scheme November 2020 11 Question Answer Marks Guidance 5 (a) c = fλ or f = 3.0 × 108 / 0.60 f = 5.0 × 108 (Hz) C1 A1 Allow v = fλ Allow 5 × 108 (b) amax = - ω2A or amax = - (2πf)2A ω = 2 × π × 5.0 × 108 = 3.1(4) × 109 (rad s-1) or 109π amax (= π2 × 1018 × 4.0 × 10-6) = 3.9 × 1013 (m s-2) C1 C1 A1 Allow without the minus sign ECF from (a) Not 4.0 × 1012π2 (c) the current (induced in the aerial) is alternating (5 × 108 times per second) (so the meter would register zero) / AW or the diode (half-)rectifies the current / changes the current (from a.c.) to d.c. / AW B1 Allow ‘a diode only lets current pass through in one direction’ AW (d) see next page B1 x 6 Total 12H556/03 Mark Scheme November 2020 12 Question Answer Marks Guidance 5 (d) * Level 3 (5 - 6 marks) Response shows clear distinction between investigations; clear and correct reasoning is given for the situations which give maximum/minimum readings in both cases, including correct numerical values 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) Response refers to both investigations; some reasoning is given for the situations which give maximum/minimum readings in both investigations, including some numerical values 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 reasons are given for the situations which give maximum/minimum readings in either investigation 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 x 6 Use level of response annotations in RM Assessor, e.g. L2 for 4 marks, L2^ for 3 marks, etc. Indicative scientific points may include: explanation 1 • receiver aerial vertical – electrons are driven (maximum distance) up and down along the length of the aerial because the oscillations (of the electric field) are vertical, causing maximum (a.c.) current • receiver aerial horizontal – electrons are driven (minimum distance) across the aerial because the oscillations (of the electric field) are only in the vertical plane (no oscillation along the aerial to cause current), so zero / minimum current • rotation of receiver aerial by ± 90o (or 90o and 270o) from vertical leads to zero current explanation 2 • reflected wave superposes with incident wave at receiver aerial • coherent waves as from same source • constructive interference/waves in phase gives max current • reflected wave has travelled nλ further, n = 0,1, etc • so max current when plate is at λ/2, 2λ/2, etc from receiver aerial, i.e. 30, 60 cm • destructive interference/waves 180o (π rad) out of phase gives zero current • reflected wave has travelled (2n +1)λ/2 further, n = 0,1, etc • so zero current when plate is at λ/4, 3λ/4, etc from receiver aerial, i.e. 15, 45 cm • reflected signal will be weaker the further it has to travel so no longer complete cancellation (ammeter reads close to zero) Note: Give full credit to candidates who take the 180o (π rad) phase change on reflection into account, which gives max current at 15, 45 cm etc and zero current at 30, 60 cm etc.H556/03 Mark Scheme November 2020 13 Question Answer Marks Guidance 6 (a) B1 B1 Path is initially horizontal and further up the page than original Path ends at 300 to horizontal (angle must be labelled) in the direction shown (b) Q = 79e and q = 2e F = (1/4πε0)Qq/r2 = 79 × 2 × (1.6 × 10-19)2/ [4π × 8.85 × 10-12 ×(6.8 × 10-14)2] = 7.9 (N) C1 C1 C1 A1 Apply ECF for wrong charge(s), e.g. Q and/or q = e, or Q = 79 and/or q = 2, etc 6 (c) (k.e. =) E = 5.0 × 106 × 1.6 × 10-19 v = √(2E/m) or = √(2 x 8.0 x 10-13 / 6.6 × 10-27) = 1.6 x 107 (ms-1) p (= mv) = 6.6 × 10-27 x 1.6 x 107 giving p = 1.1 × 10-19 (kg m s-1) C1 C1 A1 E = 8(.0) × 10-13 J or (E = p2/2m so) p = √(2mE) Note: A value of v = 1.6 x 107 (ms-1) automatically scores both C1 marks even if the calculation for E is not shown or p (= √(2mE) = √(2 × 6.6 × 10-27 × 8.0 × 10-13) giving p = 1.0 × 10-19 (kg m s-1) Full substitution of values must be shown and answer (if calculated) must be correctH556/03 Mark Scheme November 2020 14 6 (d) (i) Example (not to scale): p(A) B1 B1 horizontal arrow (judge by eye), in the direction shown arrow drawn at an angle of 60o to the horizontal (angle must be shown), in the direction shown (ii) Example (not to scale): (Can apply principle of) conservation of momentum (since no external forces are acting) B1 B1 arrow drawn at an angle of 60o to the horizontal (angle must be shown), in the direction shown Total 13 p(B) 60o 60oOCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]
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