GCE Physics A H556/02: Exploring physics Advanced GCE Mark Scheme for November 2020

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GCE Physics A H556/02: Exploring physics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Physics A H556/02: Exploring physics Advanced GCE Mark Scheme fo... r 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 2020H556/02 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. 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. A marks These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark to 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. 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. 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 Guidance.H556/02 Mark Scheme November 2020 3 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).H556/02 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 argumentH556/02 Mark Scheme November 2020 5 SECTION A Question Answer Marks Guidance 1 D 1 2 C 1 3 A 1 4 B 1 5 C 1 6 C 1 7 D 1 8 B 1 9 C 1 10 B 1 11 B 1 12 A 1 13 A 1 14 D 1 15 D 1 Total 15H556/02 Mark Scheme November 2020 6 SECTION B General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance Question Answer Marks Guidance 16 (a) (special coupling) gel is used that has the same /’matching’ (acoustic) impedance as skin / body Reduced / less / zero reflection (at the skin) B1 B1 Allow Z Allow gel and impedance is the same / matching for two materials / mediums Allow more transmission (b) (Pulses of) ultrasound sent into the eye Reflections from front and back of lens (and pulses displayed on oscilloscope) (Thickness of lens) determined from speed (of ultrasound) and time (difference) B1 B1 B1 Allow ultrasound reflected by any part of the eye Allow ‘sound’ / wave (since ultrasound is in the question) Ignore transducer placed close / next to eye Allow thickness = �� 2 with c = speed (of ultrasound) and t = time (difference) Allow this mark even when the reflections are from incorrect boundaries Total 5H556/02 Mark Scheme November 2020 7 Question Answer Marks Guidance 17 (a) (E =) 1.8 × 1.6 × 10-19 or 2.88 × 10-19 (J) 1.8 × 1.6 × 10-19 = 6.63 × 10−34 × 3.0 × 108 � λ = 6.9 × 10-7 (m) C1 C1 A1 (b) (VR =) 2.7 (V) or (current =) 0.018 (A) (ratio = 0.018 ×1.8 0.018 ×2.7 ) ratio = 0.67 C1 A1 Note the mark can be scored on circuit diagram Note values of powers are: 0.0324 W and 0.0486 W Allow 2/3; Not 0.66 (rounding error) (c) (i) In darkness LDR has more resistance / p.d. across LDR is large or In light LDR has less resistance / p.d. across LDR is small Clear idea that when the LED is on, this will force the p.d. across LED / LDR to decrease, forcing the LED to switch off (ORA) (The cycle of LED switching on and off is repeated) B1 B1 Note the explanation must be in terms of p.d. / potential divider. Ignore current (ii) A sensible suggestion, e.g. Point the LED away from the LDR / increase distance (between LED and LDR) / insert a card between (LED and LDR) B1 Total 8H556/02 Mark Scheme November 2020 8 Question Answer Marks Guidance 18* Level 3 (5–6 marks) Clear description and clear analysis of data 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 of data OR Clear description OR Clear analysis of data 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 and limited analysis OR Some description OR Some analysis of data 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 Indicative scientific points may include: Description • Circuit showing supply, ammeter, voltmeter and resistance wire /coil • Measure I (in coil) with ammeter • Measure V (across coil) with voltmeter • Power (for coil) calculated: P = VI • Resistance of thermistor either calculated using R = V/I or measured with ohmmeter • Change P / change V / use variable power supply / use variable resistor (to change I) • Keep the number of turns of coil constant throughout / no draughts / wait until the resistance stabilises Analysis • lgP = lgk +nlgR (or natural logs ln) • Plot a graph of lgP against lgR • If expression is correct, then a straight line with nonzero intercept • gradient = n • intercept = lgk • k = 10intercept (or k = eintercept for natural logs) Total 6H556/02 Mark Scheme November 2020 9 Question Answer Marks Guidance 19 (a) h → J s / h → N m s / J → kg m2 s-2 base unit = kg m2 s-1 C1 A1 (b) (i) Vq = ½ mv2 and λ = ℎ �� Clear algebra leading to �2 = ℎ2 2�� × 1 � M1 A1 Allow p for mv Allow e for q in (b)(i) – this is to be treated as a ‘slip’ (ii)1 (% uncertainty in λ2 =) 10% (% uncertainty in λ =) 5% C1 A1 Note 10 (%) on answer line will score the C1 mark (ii)2 Straight line of best fit passes through all error bars B1 (ii)3 gradient = 1.0 (× 10-22) ℎ2 2�� = gradient (6.63×10−34)2 2 × � × 3.2 × 10−19 = gradient m = 6.9 × 10 -27 (kg) (hence about 10-26 kg) C1 C1 C1 A1 Ignore POT for this mark; Allow ± 0.20 (× 10-22) Possible ECF for incorrect value of gradient Note check for AE (condone rounding error here) and answer must be about 10-26 (kg) for any incorrect gradient value for this A1 mark Special case: 1.37 × 10-26 kg scores 3 marks for q = 1.6 × 10-19 C because answer is about 10-26 kg Total 11H556/02 Mark Scheme November 2020 10 Question Answer Marks Guidance 20 (a) (i) sensible diameter, e.g. 7 (mm) (power = 4.8 × 10-7 × π × (0.0035)2 ) power = 1.8 × 10-11 (W) C1 A1 Allow 2 – 16 (mm) Not πd2; this is XP Note check for AE (condone rounding error here) Possible ECF for diameter outside the range 2 – 16 (mm) Allow 1 SF answer here (ii) (I ∝ A2; intensity doubles) A = √2 × 7.8 (or equivalent) A = 11 (nm) C1 A1 Allow the C1 mark for 4.8 (× 10-7) = k × [7.8 × (10-9)]2 (b) (When two or more waves meet at a point) the resultant displacement is (equal to) the sum of the (individual) displacements (of the waves) B1 Allow sum / total / net for resultant Ignore vector sum (c) (i) phase difference = n × 360(°) for bright (fringes) / constructive (interference) phase difference = (n + ½) × 360(°) for dark (fringes) / destructive (interference) B1 B1 Allow zero or n × 2π(rad) or even number of π (rad) or even number of 180(°) Allow 180(°) or (n + ½) × 2π(rad) or odd number of π (rad) or odd number of 180(°) Special case: 1 mark for ‘completely in phase for bright fringes/constructive (interference) and in anti-phase / completely out of phase for dark fringes /destructive (interference)’ (ii) λ = 3.0 × 108 4.75 × 1014 or λ = 6.316 × 10-7 (m) x = 6.316 × 10−7 × 8.2 0.20 × 10−3 or x = 0.0259 (m) t = 0.14 (s) C1 C1 A1 Note the answer must be given to 2 SF for this mark Special case: allow 1 mark for 8.6 × 10-11 s on the answer line; incorrect physics using 0.18 = 4.75 × 1014 λ Total 10H556/02 Mark Scheme November 2020 11 Question Answer Marks Guidance 21 (a) Electrons and quarks identified as fundamental particles There are 6 electrons, 6 protons and 8 neutrons Composition of proton → u ud Composition of neutron → u d d B1 B1 B1 B1 Allow e for electron, p for proton, and n for neutron throughout Allow 6 electrons, 20 u and 22 d Do not award this mark if electron has quark-composition Allow ‘2 up and 1 down’ Allow ‘2 down and 1 up’ (b) (i) (decay constant =) ln2 5700 decay constant = 1.2(2) × 10-4 (y-1) C1 A0 (ii) 0.78 = e− � � ln0.78 = (-) 1.2 × 10-4 × t age = 2100 (y) C1 C1 A1 Note 1 = 0.78e− � � is XP; answer is negative (- 2100 y) There is no ECF from (b)(i) Note 1.22 × 10-4 gives an answer of 2040 y or 2000 y (iii) The ratio (of carbon-14 to carbon-12) has remained constant B1H556/02 Mark Scheme November 2020 12 (c)* Level 3 (5–6 marks) Some description and clear analysis for r ∝ A1/3 and correct calculation of mean density 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 for r ∝ A1/3 or some calculation of mean density OR Some description and clear analysis for r ∝ A1/3 OR Some description and correct calculation of mean density OR Clear analysis for r ∝ A1/3 and correct calculation of mean density 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) Some description OR Limited analysis for r ∝ A1/3 OR Limited calculation of mean density 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 Indicative scientific points may include: Description • The density is independent of A • The density is constant for most of d • Nucleus with bigger A is larger (d / volume / mass) Analysis for r ∝ A1/3 • r ≈ 3.6 (× 10-15 m) for Al-27 / r ≈ 5.5 (× 10-15 m) for Mo- 96 / r ≈ 7.0 (× 10-15 m) for Hg-200 • r/A1/3 = constant (or equivalent) • Evidence for r ∝ A1/3 with at least 2 nuclei (Note: 3.6 (×10-15)/271/3 ≈ 5.5 (×10-15)/961/3 ≈ 7.0 (×10-15)/2001/3 ≈ 1.2 (×10-15) or • r3/A = constant (or equivalent) • Evidence for r3 ∝ A with at least 2 nuclei (Note. 3.63 (×10-45)/27 ≈ 5.53 (×10-45)/96 ≈ 7.03 (×10-45)/200 ≈ 1.7 (×10-45) Calculation for density • ρ = M/V • ρ = Amn ÷ 4 3 π r3 or ρ ≈ Amn ÷ diameter3 • mn≈ 1.7 × 10-27 (kg); ρ = 2.3 × 1017 (kg m-3) for at least one of the nuclei given in the figure or table Total 15H556/02 Mark Scheme November 2020 13 Question Answer Marks Guidance 22 (a) Magnetic field (around current-carrying wire) (Fleming’s) left-hand rule mentioned (Magnetic) field into page, (current is up the page) and force is to the left / towards X B1 B1 B1 Not magnetic force Allow ‘field into page and wires attract’ Note the field direction and force direction can be shown on the figure (b) (i) (induced) e.m.f. is (directly) proportional / equal to the rate of change of (magnetic) flux linkage B1 Not current Allow ‘rate of cutting’ for ‘rate of change’ (ii) Connect the primary (coil) to an alternating voltage / current Oscilloscope connected across secondary coil / to measure E A graph of E against N will be a straight line through the origin. B1 B1 B1 Allow AC (can be on the figure) Not changing / variable for alternating Allow voltmeter (can be on the figure) Allow p.d. / voltage for e.m.f. / E throughout Ignore any component (e.g. lamp or resistor) connected across the secondary coil Allow (E ÷ N) = constant Total 7H556/02 Mark Scheme November 2020 14 Question Answer Marks Guidance 23 (a) (i) (force =) (1.6 × 10−19)2 4π�0×(1.0 × 10−15)2 (F =) 230 (N) F2 = 2302 + 2302 – 2 × 230 × 230 × cos120° or F = 2 × 230cos30° F = 400 (N) C1 C1 C1 A1 Special case: � = �� 4π�0�2 = 2 × 1.6 × 10−19 4π�0×(1.0 × 10−15)2 loses this C1 mark, then ECF for the rest of the marks Not the first two C1 marks for incorrect charge, then allow ECF for the final C1A1 marks Note force to 4 SF is 230.2 N Allow sine rule / scale drawing Allow this mark for 230cos30° or 200 (N) Allow ± 10 (N) if scale drawing used (ii) F / arrow vertical up the page B1 Allow correct arrow direction anywhere on the figure (iii) Strong (nuclear) force (acts on the protons) The strong (nuclear) force is attractive B1 B1 Ignore gravitational force Allow pulls / holds (the protons) / binds (the protons) for ‘attractive’ (b) (i) 12 000 = � 4π�0� 12 000 = � 4π�0 × 0.19 Q = 2.5(4) × 10-7 (C) C1 C1 A0 Allow E = (V/d =) 6.316 × 104 C1 and E = 6.316 × 104 = � 4π�0 × 0.192 C1 (ii)1 t = 78 × 3600 (I =) 2.5 × 10−7 78 ×3600 I = 8.9 × 10-13 (A) C1 C1 A0 There is no ECF from (b)(i) Note 2.54 × 10-7 gives an answer 9.0 × 10-13 A (ii)2 (R =) 6000 9.0 × 10−13 or 6.7 × 1015 (Ω) or V = IR and � = �� � 6000 9.0 × 10−13 = � × 0.38 1.1 × 10−4 ρ = 1.9 × 1012 (Ω m) C1 C1 A1 There is no ECF from (b)(ii)1 Take 12000 V as TE for this C1 mark, then ECF Note 8.9 × 10-13 (A) gives an answer 2.0 × 1012 (Ω m) Total 14H556/02 Mark Scheme November 2020 15 Question Answer Marks Guidance 24 (a) Emits gamma (photons / radiation / waves / rays) Any one from: (Diagnosing the) function of organ Detecting tumour Small half-life (Gamma rays) can be detected outside body / passes through patient / least ionising Position of tracer located B1 B1 Not injected into a patient / non-invasive Allow for half-life is a few hours (b) Collimator: Allows gamma (photons) parallel to the axis of the tubes to pass through Scintillator: gamma (photons) produces (many) light (photons) Photomultiplier (tubes): light (photons) produces electrons / current / electrical pulse / p.d. / signal B1 B1 B1 Ignore any other components named / described Allow photon / waves / rays Allow idea of tubes allowing the gamma (photons) to travel in the same direction Allow crystal (or named crystal) for scintillator Allow high-energy photons produce (many) low-energy photons Total 5H556/02 Mark Scheme November 2020 16 Question Answer Marks Guidance 25 (a) All except pair production / PP B1 Allow PE, S and C (b) (energy =) 9.11 × 10-31 × (3.0 × 108)2 (energy =) 2 × 9.11 × 10-31 × (3.0 × 108)2 / 1.60 × 10-19 lg1.0(2) × 106 = 6 (as on graph) OR (energy =) 1.0 × 106 (eV) or lg1.0 × 106 = 6 (from graph) (energy =) 1.6 × 10-13 J and evidence of mc2 2 × 9.11 × 10-31 × (3.0 × 108)2 ≈ 1.6 × 10-13 B1 B1 B1 B1 B1 B1 Note this is 8.2 × 10-14 (J) Note this is 1.0(2) × 106 eV Note this can be shown in a variety of ways Total 4OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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