GCE Physics A H156/02: Depth in physics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

GCE Physics A H156/02: Depth in physics Advanced Subsidiary GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H156/02: Depth in physics Advanced Subsidiary GCE Mark Scheme ...

GCE Physics A H156/02: Depth in physics Advanced Subsidiary GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H156/02: Depth in physics Advanced Subsidiary 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 2021H156/02 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. ^ Omission Used to indicate where more is needed for a mark to be awarded (what is written is not wrong but not enough).H156/02 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 argumentH156/02 Mark Scheme October 2021 4 Question Answer Marks Guidance 1 (a) Constant phase difference (between two or more waves) B1 Ignore in phase (b) (i) 0.08 (m) A1 (ii) π (rad) A1 (iii) path difference = λ / 2 or 2  0.08 0.16 (m) M1 A0 Allow ECF from (b)(i) Allow path difference shown at K (c) (i) period determined using timebase frequency = 1 / period B1 B1 Allow one mark for f = 1 / T without T being defined (ii) � = 2.1 × 103 × 0.16 340 (m s-1) C1 A1 Allow ECF from (b)(iii) 336 (3sf) Allow one mark for 0.336 or 0.34 Total 8H156/02 Mark Scheme October 2021 5 Question Answer Marks Guidance 2 (a) (i) (Area under graph =) energy / elastic potential energy B1 Allow work done on the elastic band (ii) (Area of) 1cm2 is 0.025 × 2.5 J or 0.0625 J (31 × 0.0625 =) 1.9 (J) C1 A1 Allow other alternatives. Do not accept 0.5Fx (iii) Energy transferred to the surroundings/ heating the rubber B1 (b) (i) 49 1.4 × 10−7 3.5 × 108 Pa C1 A1 (ii) 3.5 × 108 180 × 109 0.0019 C1 A1 Allow ECF from (b)(i) Ignore units (iii) 0.0019 ×2.2 0.0042 (m) C1 A1 Allow ECF from (b)(i) and (ii) Allow 0.0043 (m) (iv) 1 2 × 49 × 0.0043 0.10 (J) C1 A1 Allow ECF from (b)(i), (ii) and (iii) Do not accept 1sf (c) Area increases by a factor of four / extension decrease by factor of four Elastic strain energy decreases by a factor of four M1 A1 Total 14H156/02 Mark Scheme October 2021 6 Question Answer Marks Guidance 3 (a) Level 3 (5–6 marks) Clear explanation of terms and explanation of results correctly comparing momentum and kinetic energy. 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 explanation of terms and limited explanation of results comparing momentum or limited explanation of terms and some explanation of results or correct comparison of momentum and kinetic energy. 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) Has limited explanation of terms or limited comparison of momentum / kinetic energy. 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 x 6 Indicative scientific points may include: Explanation of terms • p = mv • �� = 1 2��2 • Total momentum conserved in all collisions • Total energy conserved in all collisions • Ek conserved in elastic collision • Ek NOT conserved in inelastic collision • Speed of approach = speed of separation in elastic collision Explanation of results • Initial pA = 15 kg cm s–1 or 0.15 kg ms–1 • Initial EkA = 0.015 J • Expt 1: o Speed of separation = 0.150 + 0.050 = 0.200 ms–1 o pA after collision = (-) 0.375 kg ms–1 o pB after collision = 0.1875 kg ms–1 o Total p after collision = 0.15 kg ms–1 o EkA after collision = 0.0009375 J o EkB after collision = 0.0140625 J o Total Ek after collision = 0.015 J o Collision is elastic since Ek conserved • Expt 2: o p after collision = 0.15 kg ms–1 o Ek after collision = 0.005625 J o Collison is inelastic since Ek not conserved • Momentum conserved in both collisions (b) Area under graph = 0.5 × 0.4 × 18 or 3.6 3.6 0.045 80 (m s–1) C1 C1 A1 Allow alternative methods Total 9H156/02 Mark Scheme October 2021 7 Question Answer Marks Guidance 4 (a) (i) � × (2.9×10−2)2 4 or � × (1.45 × 10−2)2 6.605  10-4 m2  6.6  10-4 M1 A0 (ii) � = 6.6 × 10−4 × 12.0 or 7.92 × 10−5 (m3) � = 400 × 7.92 × 10−5 or 0.03168 kg W = 0.31 (N) C1 C1 A1 Ignore POT (b) � = 0.31 1000× 9.81 or 3.16 × 10−5 y = 3.16×10−5 6 .6×10−4 y = 0.048 (m) C1 C1 A1 Mass of water displaced = 0.31 9.81 = 0.316 y = 0 .316 1000×6.6×10−4 (c) y = 0.053 m Same weight/mass displaced of oil Smaller density implies larger volume of oil displaced y is larger OR y α 1/ρ B1 B1 B1 B1 Total 11H156/02 Mark Scheme October 2021 8 Question Answer Marks Guidance 5 (a) ΣE = ΣV or ΣE = ΣIr E = V + Ir  V = E − Ir C1 A1 (b) E = y–intercept r = – gradient B1 B1 E must be the subject R must be the subject Do not accept gradient = − r (c) (i) (� = 05.025 .68 =) 230 Ω A1 Allow 227 (ii) ( 5 (c)(i) .682 or 0.0252 × (c)(i) or 0.025 × 5.68 =) 0.14 0.14 x 300 = 42 (J) C1 A1 Allow ECF from (c) (i) 0.140 or 0.142 or 0.144 Allow 43 (J) (for 0.142 or 0.144) (iii) (� = (5c) .68 (ii) or 0.025 × 300 = ) 7.4 or 7.5 C B1 B1 Allow ECF from (c) (ii) Total 9 1H156/02 Mark Scheme October 2021 Question Answer Marks Guidance 6 Level 3 (5–6 marks) Correct circuit diagram and explanation including detailed calculations and explanation of circuit for different light intensities. 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 diagram, some calculations / explanation. 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 diagram with incorrect position of voltmeter and limited calculations / explanation OR correct diagram with correct symbols. 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: Circuit Diagram • Potential divider circuit • Correct symbols • Battery/power supply of at least 6.0 V • Voltmeter • Voltmeter correctly positioned across fixed resistor. Explanation and calculations • Potential divider equation • Appropriate value of fixed/variable resistor • Vout calculated when LDR is in very bright light / resistance value calculated • Vout calculated when LDR does not receive light / or resistance value calculated. Explanation for different light intensities • Use of variable resistor • Effect of increasing/decreasing the resistance of the fixed resistor. Total 6H156/02 Mark Scheme October 2021 10 Question Answer Marks Guidance 7 (a) (i) points from the graph read to the nearest half square size of triangle is greater than half the length of the drawn line and Δy / Δx with correct power of ten shown B1 B1 Allow Δy and Δx to less than half a small square Ignore POT Note triangle may be determined from read-offs Δx must be greater than 0.3  106 (ii) ℎ = 1.2 × 10−6 × 1.60 × 10−19 3.00 × 108 or (a)(i) × 5.333 × 10−28 6.4  10-34 (J s) given to 2 significant figures C1 A1 (iii) steepest or shallowest line that passes through all the error bars B1 Note steepest line passes inside the lowest error bar (1.76) since it just cuts the bottom of second error bar. (iv) gradient determined: 1.0  10-6 Vm or 1.4  10-6 V m ∆gradient=0.2× 10−6 V m ∆gradient gradient × 100 = 15% M1 C1 A1 Allow ecf from (iii) Allow h = 1.06 or 1.067 Allow 16.7% (b) (i) 4.1 eV = 6.56  10-19 (J) 6.63 × 10−34 × 3.00 × 108 98 × 10−9 �� 2.03 × 10−18 1.4  10-18 (J) C1 C1 A1 Allow 1.96 × 10−18 for use of 6.4  10-34 (J s) Allow 1.3  10-18 (J) for use of 6.4  10-34 (J s) (ii) KE is independent of intensity (for constant wavelength) / intensity only affects the number of photons No change in KEmax M1 A1 Allow decreasing intensity, decreases the number of photons / energy of a photon only depends on frequency or wavelength Total 13OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA d