GCE Physics A H556/03: Unified physics Advanced GCE Mark Scheme for Autumn 2021

GCE Physics A H556/03: Unified physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H556/03: Unified physics Advanced GCE Mark Scheme for Autumn 2021Oxford Camb ...

GCE Physics A H556/03: Unified physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H556/03: Unified 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 2021H556/03 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.H556/03 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 argument .H556/03 Mark Scheme October 2021 4 General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance Question Answer Marks Guidance 1 (a) (i) (F = ma =) 190 × 103 = 2.1 × 105 a a = 0.90 (m s-2) M1 A0 a = 0.905 to 3 SF (ii) (v2 = u2 + 2as gives) 36 = 2 × 0.90 × s s = 20 (m) C1 A1 Allow any valid suvat approach; allow ECF from (a)(i) Note using a = 1 gives s = 18(m) (iii)1 P = Fv One correct calculation e.g. F = 100 × 103 and v = 42 gives P = 4.2 ×106 (W) Fv = constant B1 B1 B1 Equation must be seen (not inferred from working) Allow any corresponding values of F and v; working must be shown. No credit for finding area below curve Allow F is proportional to 1/v or graph is hyperbolic or correct calculation of Fv at two points (or more) (iii)2 (P = VI = 4.2MW so) 4.2 × 106 = 25 × 103 × I I = 170 (A) C1 A1 Allow P = 4MW or ECF from (iii)1 Expect answers between 160 - 170 (A) (b) (i) R (= ρL/A) = 1.8 × 10-8 × 1500/1.1 × 10-4 R = 0.25 (Ω) C1 A1 (ii) E = σ/ε = T/Aε (so T = EAε) T = 1.2 x 1010 x 1.1 x 10-4 x 0.013 T = 1.7 x 104 (N) or 17 (kN) C1 C1 A1 or calculation of σ =1.56 x 108 (Nm-2) or T = 1.56 x 108 x 1.1 x 10-4 Total 13H556/03 Mark Scheme October 2021 5 Question Answer Marks Guidance 2 (a) (i) R = 3000 + 1500 V = 12 × 1500/4500 = 4(.0) (V) C1 A1 R = 4500 (Ω) or I = V/R = 12 /4500 = 2.67 mA V1500 = 2.67 mA x 1.5 kΩ = 4.0 (V) (ii) V (= 12 × 1500/1600) = 11.25 (V) ΔV = 11.25 – 4.0 = 7.25 (V) C1 A0 (b) see next pageH556/03 Mark Scheme October 2021 6 Question Answer Marks Guidance (b) * Level 3 (5–6 marks) Clear description of a valid experiment which would lead to accurate results, sensible suggestions for table, graph and accuracy 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) Reasonable description of experiment and sensible suggestion for table or graph or accuracy, or 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) Attempt at experiment and attempt at table or graph or accuracy 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 Indicative scientific points may include: Experiment • Diagram(s) to show electrical circuit and/or set-up of apparatus • Use ammeter in series to measure current I through LED • Use micrometer to measure thickness of (one sheet of) tracing paper • or use calliper with m sheets of thickness t; measure t and use x = mt • Measure I for various x • Calculate ln I Results table • Need columns for total thickness x (or number of sheets m and x = mt), current I, ln (I) • Units if included should be appropriate and presented in an appropriate format e.g. ln(I/mA) Graph • plot ln I against x • expect straight line graph with negative gradient and non-zero intercept • gradient = - n and y–intercept c = ln k • k = ec (alternatively, k is the current when no sheets of paper are used) Accuracy • work in darkened room/constant low light conditions • keep distance between light-source and LDR constant • use same power light source and same LDR throughout • position yourself so as not to cast shadow on LDR • clamp equipment to bench to ensure distances do not change Total 9H556/03 Mark Scheme October 2021 7 Question Answer Marks Guidance 3 (a) GPE is the work done in bringing an object from infinity (to that point) B1 Ignore any equations (b) (i) GPE = (-) GMm/r GPE = (-) 6.67 × 10-11 × 2 × 1030 × 810/1.5 × 1011 GPE = (-) 7.2 × 1011 (J) C1 C1 A0 Mark is for full substitution, including 6.67 × 10-11 for G (ii) v = 2πr/T = 2π × 1.5 × 1011 / 3.16 × 107 (= 29.8 km s-1) KE = ½mv2 = 0.5 × 810 × (29.8 × 103)2 KE = 3.6 × 1011 (J) C1 M1 A1 Allow proof by algebraic method for full marks e.g. mv2/r = GMm/r2 so mv2= GMm/r Therefore KE/GPE = ½mv2/(GMm/r) = ½ (iii) total energy = (-) (7.2 × 1011 - 3.6 × 1011) total energy = (-) 3.6 × 1011 (J) M1 A0 working must be shown; ECF (i) and (ii) (c) (i) A = 470/8.8 × 10-13 = 5.3 × 1014 (Bq) λ = ln 2/(88 × 3.16 × 107) (= 2.5 x 10-10 s-1) (A = λN); N (= 5.3 x1014 / 2.5 x 10-10) = 2.1 x 1024 C1 C1 A1 Mark is for correct calculation of A (in Bq or decays per s) Mark is for correct working to give λ in s-1 (ii) P = Po exp (- λt) P = 470 exp (- ln 2 x 100 / 88) P = 210 (W) C1 C1 A1 Allow formula in terms of N or A Allow calculation in terms of N or A; allow ECF for N or A Total 13H556/03 Mark Scheme October 2021 8 Question Answer Marks Guidance 4 (a) * Level 3 (5 - 6 marks) Clear explanation using kinetic theory ideas and either a clear proof using formulae or a correct calculation 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) A partial explanation using kinetic theory ideas and either a partial proof using formulae or a partial calculation 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) An attempt at either explanation or proof or calculation 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 Indicative scientific points may include: Explanation using kinetic theory • pressure = force/area • force is caused by air molecules colliding with oven walls • Newton’s 2nd Law states force = rate of momentum change • increased temperature means each molecule has greater KE • hence greater velocity and hence greater momentum • and more collisions with walls per second • hence greater rate of momentum change on hitting walls. • This would lead to greater pressure if N remained constant • so number of molecules in oven must decrease (air escapes) • so fewer but 'harder' collisions at higher temperatures giving constant pressure. • Rms velocity c increases with temperature but number N decreases and so effects balance out to keep total KE (½Nmc2) constant Proof using formulae • equate pV = NkT and E = 32NkT to show E = 3 2 pV • in an ideal gas, all internal energy E is kinetic energy • so E is independent of temperature Calculation • Internal energy = 32pV = 1.5 x 0.065 x 1.0 x 105 = 9.8 kJ • At T = 293K, N = pV/kT = 1.6 x 1024 and n = 2.7 moles • At T = 473K, N = 1.0 x 1024 and n = 1.7 moles • so we can show that NT (and/or nT) remain constantH556/03 Mark Scheme October 2021 9 Question Answer Marks Guidance 4 (b) (i) B1 B1 One correct line (or dot and cross) drawn Line must go through centre of coil Allow an incomplete line or a complete circle round the coil Ignore direction of arrow More than one line drawn All lines drawn must go through centre of coil and follow correct shape and direction of field Ignore spacing of lines Ignore any lines to the right of the coil (ii) • (the magnetic) flux (of the coil) links the base / saucepan • (the size/direction of) the flux linkage (constantly) changes/alternates (causing an alternating induced e.m.f.) • (induced) current is large because metal/base/ saucepan has low resistance B1 x 2 2 out of 3 possible marking points Allow (the magnetic) field lines cut the (base of the) saucepan Allow the (magnetic) field constantly changes/alternates Allow a bald statement of Faraday’s Law (iii) The resistance of glass-ceramic/the (cook’s) hand is (very) large So (induced) current (or heating effect of current) is zero/negligible M1 A1 Allow glass-ceramic/hand is an insulator/not a (good) conductor Do not allow the induced e.m.f. is (very) small Total 12H556/03 Mark Scheme October 2021 10 Question Answer Marks Guidance 5 (a) the (sound) wave reflects at the water (surface) reflected wave interferes/superposes with the incident wave to produce a (resultant) wave with a node at the water surface and an antinode at the top of the tube l = λ/4 B1 B1 B1 B1 Allow the (two) waves interfere/superpose Do not allow interact/combine (b) (i) l = (v/4)(1/f) – k Correct comparison with y = mx + c M1 A1 Correct manipulation of equation must be shown (ii) large triangle used to determine gradient gradient calculated correctly v = 320 (m s-1) B1 B1 B1 Δx > 0.6 x 10-3s Expect between 80 and 82 (m s-1) Allow 320 ± 20; allow ECF from an incorrect gradient (c) (i) Value of 1/F determined correctly from graph F = 350 (Hz) C1 A1 Allow values between 2.83 x 10-3s and 2.84 x 10-3s Allow only alternative methods which use values from line of best fit (ii) (100 (ΔF/F) =) 100 Δv/v + 100 (Δl + Δk) (l + k) B1 B1 Total 13H556/03 Mark Scheme October 2021 11 Question Answer Marks Guidance 6 (a) At t = 0 (and t = 15, 30) the (magnitude of the) centripetal force equals R – W (as only vertical forces act on the tourist) B1 Allow at t = 0 (or the bottom of the circle) the centripetal force is provided by the resultant/ upwards/vertical force (b) (i) (For circular motion) there must (always) be a resultant force towards the centre The resultant force is not always vertical/sometimes has a horizontal component This can only be provided by friction/cannot be provided by R and W / R and W are always vertical/only F is horizontal B1 x 2 any 2 from 3 marking points Allow F provides the horizontal (component of the) centripetal force (ii) Sine wave with period 30 min and amplitude 0.050 (N) Correct phase, i.e. negative sine wave B1 B1 Must start at the origin (iii) F = 0.050 cos 40o F = 0.038 (N) C1 A1 Allow alternative methods e.g. triangle of forces Allow ECF from graph if used (c) m = 650/g or m = 650/9.81 (= 66.3 kg) (F = mrω2 gives) d = 0.050 / mω2 = 0.050 / 66.3 x (3.5 × 10-3 )2 d = 62 (m) C1 C1 A1 Not m = 650 kg or m = 65 kg or ( F = mv2/r and v = 2πr/T gives) d = 0.050 x (30 x 60)2 / (4π2 x 66.3) Total 10OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA