GCE Physics A H156/01: Breadth in physics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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GCE Physics A H156/01: Breadth in physics Advanced Subsidiary GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics A H156/01: Breadth in physics Advanced Subsidia... ry 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 2021H156/01 Mark Scheme October 2021 2 RM ASSESSOR 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.H156/01 Mark Scheme October 2021 3 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). 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 argumentH156/01 Mark Scheme October 2021 4 SECTION A Question Answer Marks Guidance 1 D 1 2 B 1 3 D 1 4 C 1 5 B 1 6 C 1 7 C 1 8 D 1 9 A 1 10 D 1 11 A 1 12 A 1 13 B 1 14 B 1 15 D 1 16 C 1 17 D 1 18 A 1 19 D 1 20 C 1 Total 20H156/01 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 21 (a) Separation between droplets increases (further down) B1 (b) ‘No motion’ explained either in terms of the first law or second law There is no / negligible resultant force B1 B1 Allow F = ma, since F = 0, a is zero (hence at rest) Allow an object continues in a state of rest or uniform motion unless acted upon by a (resultant) force. ALLOW no frictional/extra/new force (c) (i) Tangent drawn at t = 4.0 s Attempt at calculating the gradient v calculated from gradient and between 9.50 - 10.50 (m s-1) OR s = 20 (m) and s = ½ at2 20 = ½ a × 4.02 or a = 2.5 (m s-2) v = 2.5 × 4.0 or v2 = 2 × 2.5 × 20 v = 10 (m s-1) C1 C1 A1 C1 C1 C1 A0 Allow other correct methods Note working required for this mark (c) (ii) change in momentum = 1200 × 10 or 12000 (kg m s-1) rate of change of momentum = 3000 unit: kg m s-2 or N OR F = 1200 × 2.5 rate of change of momentum = 3000 unit: kg m s-2 or N C1 A1 B1 C1 A1 B1 Allow ECF from (c)(i) Allow 2850 - 3150 Allow newton Allow ECF from (c)(i) Allow newton Total 9H156/01 Mark Scheme October 2021 6 Question Answer Marks Guidance 22 (a) Ruler (with mm scale) B1 Allow any instruments with a mm scale e.g. tape measure NOT calipers and micrometre (b) 0.002 0.1000 (× 100) or 0.1 1.4 (× 100) or g = 1.42 2 ×0.100 (2 × 0.071… + 0.02) or 0.1628 …. or 16.3 % absolute uncertainty = 1.6 (m s-2) OR gmax = 1.52 2 × 0.098 (= 11.48) or gmin = 1.32 2 ×0.102 (= 8.28) range = 3.2 (m s-2) absolute uncertainty = 1.6 (m s-2) C1 C1 A1 C1 C1 A1 Allow 1SF answers here for uncertainties Not g = 9.8 for this C1 mark; must see working Allow 0.16 or 16% Note: The answer must be given to 2 SF Ignore value of g given on the answer line, e.g. 9.8 ± 1.6 Note: The answer must be given to 2 SF Total 4H156/01 Mark Scheme October 2021 7 Question Answer Marks Guidance 23 (a) Circuit diagram with LED, voltmeter and ammeter connected to a supply with correct polarity and some means of adjusting the p.d. Increase the p.d. until the LED is just lit / current is shown in the circuit ORA To observe the light from LED use a tube / turn lights off B1 B1 B1 Allow variable (power) supply / variable resistor in circuit Allow this on a labelled diagram e.g. tube drawn around the diode (b) (i) 0.045 1.6 × 10−19 number of electrons = 2.8 × 1017 C1 A1 (ii) A = π × (0.12 × 10-3)2 or 4.5(2) × 10-8 (m2) 0.045 = π × (0.12 × 10-3)2 × 6.3 × 1028 × 1.6 × 10-19 × v v = 9.9 × 10-5 (m s-1) C1 C1 A1 Allow 2 marks for 2.5 × 10-5 (m s-1); 0.24 mm and POT error (c) (i) (Current causes) increase in temperature of thermistor Resistance of thermistor decreases (and hence V decreases) or Current in the circuit increases, p.d. across resistor increases (and hence V decreases) B1 B1 Allow warms up/heat ups Ignore increase temperature of the circuit (ii) V = 2.4 (V) or VR = 3.6 (V) I = 0.30 (A) resistance = 8.0 (Ω) OR V = 2.4 (V) and a potential divider equation / idea 2.4 = � � +12 × 6.0 or � 2.4 = 12 3.6 resistance = 8.0 (Ω) C1 C1 A1 C1 C1 A1 Not V = 2.2 (V); misreading Allow ECF if V = 2.2 (V) is used Allow 8 (1 SF answer) Not V = 2.2 (V); misreading Allow ECF if V = 2.2 (V) is used Allow 8 (1 SF answer) Total 13H156/01 Mark Scheme October 2021 8 Question Answer Marks Guidance 24 (a) Both travel at the same speed / speed of light/ 3.0 × 108 m s-1 B1 Ignore travels at ‘c’ (b) 3.0 × 108 = f × 2.5 × 10-11 f = 1.2 × 1019 (Hz) C1 A1 (c) Use of E = hf or E = hc/λ EITHER (λ) number = (500 × 10-9)/(2.5 × 10-11) number = 2.0 × 104 OR (E) number = (7.96 × 10-15)/ (3.98 × 10-19) number = 2.0 × 104 OR (f) number = (1.2 x 1019) / (6 x 1014) number = 2.0 × 104 C1 C1 A1 C1 A1 C1 A1 Allow ECF from (b) Allow ECF from (b) Total 6H156/01 Mark Scheme October 2021 9 Question Answer Marks Guidance 25 (a) (i) It is longitudinal B1 (ii) Loudspeaker, microphone/ear and slit Sound spreads from the slit AW Size of slit comparable to the wavelength (of sound) B1 B1 B1 Allow doorway for a slit/gap – receiver for microphone Not diffraction (since it is in the stem of the question) (b) (i) (refraction index) = speed of light in vacuum ÷ speed of light in material B1 Note light must be mentioned at least once Allow n = c/v if all terms defined Allow ration of speed of light in vacuum to speed of light in material NOT speed of light in air for c (ii)1 Frequency (of light) is the same (in A and B) (Light travels) slower in B or vB = 0.77vA ORA v = fλ and λB < λA B1 B1 B1 Allow f for frequency Allow v directly proportional to λ (ii)2 sin60° = 1.3 × sinθ θ = 42 (°) C1 A1 (ii)3 (No total internal reflection) Internal reflection / critical angle can only occur for light travelling from B to A AW B1 Allow TIR can only occur for light entering an optically less dense material/lower refractive index ORA Not θ < φ Total 11H156/01 Mark Scheme October 2021 10 Question Answer Marks Guidance 26 (a) weight × y = Fx (ALρg) × y = Fx � = ����� � �� M1 M1 A0 Allow W or mg Wy = Fx or mgy = Fx (b) (i) Straight line of best fit drawn through the data points Gradient = 1.5 B1 B1 Allow gradient in the range 1.40 - 1.60 (ii) ����� � � = 1.5 6.8 6.4 × 10−5 × 0.90 × � × 9.81 = 1.5 ρ = 8.0 × 103 (kg m-3) C1 C1 A1 Allow ECF from (b)(i) Allow 8 × 103 (1 SF answer) Note must be consistent with gradient value from (b)(i) Total 8OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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