GCE Physics A H556/01: Modelling physics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Physics A H556/01: Modelling 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 2020H556/01 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 Mmark, 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 Cmark 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.H556/01 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.H556/01 Mark Scheme November 2020 4 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.H556/01 Mark Scheme November 2020 5 ^ 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 argumentH556/01 Mark Scheme November 2020 7 SECTION A Question Answer Marks Guidance 1 C 1 2 C 1 3 A 1 4 C 1 5 B 1 6 C 1 7 B 1 8 B 1 9 D 1 10 D 1 11 B 1 12 B 1 13 D 1 14 C 1 15 A 1 Total 15H556/01 Mark Scheme November 2020 8 SECTION B General rule: For substitution into an equation, allow any subject – unless stated otherwise in the guidance Question Answer Marks Guidance 16 (a) Arrow vertical down and an arrow opposite to the frictional force. Both arrows labelled correctly. M1 A1 Allow weight / mg / W for the downward arrow and tension / T / ‘force in rod’ / ‘force in tow bar’ /’driving force’ for the ‘upward’ arrow (b) (Ws =) 1100 × 9.81 × sin 10° or 1100 × 9.81 × cos 80° (Ws = 1874 N or 1900 N) C1 A0 Allow g instead of value (c) force = 1900 + 300 force = 2200 (N) A1 Allow 1870 + 300 = 2170 (N) (d) (distance =) 120 / sin 10° or 691 (m) (work done =) 2200 × 691 work done = 1.5 × 106 (J) C1 C1 A1 Allow ECF from (c) Allow ECF from an incorrect attempt at first mark. (e) (A =) π × 0.0062 or 1.1 × 10-4 (m2) (stress =) 2200 � × 0.0062 and 2.0 × 1011 = stress strain x = 4.8 × 10-5 (m) C1 C1 A1 Allow ECF from (c) Allow x (=FL/EA) = 2174×0.5 2.0×1011×1.1 × 10−4 Allow 2 marks for 1.2 × 10-5; 1.2 × 10-2 m used as radius Allow answer between 4.7 and 5.1 × 10-5 (m) Total 10H556/01 Mark Scheme November 2020 9 Question Answer Marks Guidance 17 (a) (i) Any THREE from: Atoms of metal vibrate (about fixed points) Water molecules have translational KE The motion of the water molecules is random Metal atoms and water molecules have the same KE B1x3 Allow particles for atoms / molecules throughout Allow idea that water particles move past each other Not idea that the water molecules have more KE than metal atoms (ii) (Eheater =) 200 × 10 × 60 or 120000 (J) (Ewater =) 0.5 × 4200 × 40 or 84000 (J) (energy transferred = 120000 – 84000) energy transferred = 3.6 ×104 (J) C1 C1 A1H556/01 Mark Scheme November 2020 10 (b)* Level 3 (5–6 marks) Clear description and explanation and correct calculations leading to value of Lf 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 description and explanation or Correct calculations leading to value of Lf or Some description or explanation and some correct calculations 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 or explanation or Limited calculations 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. B1×6 Indicative scientific points may include: Description and explanation • m ∝ t (for both) • Greater gradient for funnel with heater / greater rate of water from funnel with heater • Energy supplied to the ice is at a constant rate (for both beakers) • Idea that arrangement in Fig 17.2 is a control • Beaker in 17.2 heated just by surroundings / air / room • Arrangement in Fig. 17.1 gains energy from heater and surroundings / air / room Calculation • Gradient(s) calculated • Δm = 45 × 10-3 kg • ΔE = mL(f) • ΔE = 5 × 12 × 240 = 14400 J • L(f) = 14400 / 45 × 10-3 = 3.2 × 105 • Units: J kg-1 Note: L(f) can be calculated using L(f) = VI ÷ |∆gradient| Total 12H556/01 Mark Scheme November 2020 11 Question Answer Marks Guidance 18 (a) (Kinetic energy) reduces (with height) At maximum height, KE is minimum / non-zero B1 B1 Allow idea that KE is transferred to GPE/KE store reduces and GPE store increases Not references to KE being a vector/having components for second mark (b) (u =) 68 sin 11° or 13.0 (m s-1) t = 13.0 / 9.81 and t correctly evaluated t = 1.3(2) (s) C1 C1 A0 Not t=90/(68cos(11)) =1.35 for zero marks. Allow any subject (c) (t =) 2 × 1.3 or 2.6 (s) (x =) 68 cos11° × 2.6 or 174 (m) horizontal distance = 174 – 90 horizontal distance = 84 (m) C1 C1 A1 Note answer is 86 (m) if 1.32 s is used Note answer is 87 (m) if 1.3226... s is used Allow 1.3 x 68 cos11° for 1 mark Allow 3 or –3 m for 2 marks (d) (i) A collision in which kinetic energy is lost B1 Allow KE is not conserved (ii) Conservation of momentum Idea that velocity is to the right and velocity is very small / much smaller than 68 (m s-1) B1 B1 Not ‘goes backwards’ Total 10H556/01 Mark Scheme November 2020 12 Question Answer Marks Guidance 19* Level 3 (5–6 marks) Clear description of experiment and measurements and clear analysis. 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 of experiment and some measurements and some analysis. 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 of experiment or Limited measurements or Limited analysis 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× 6 Indicative scientific points may include: Description • Release method • Ensure bob is not pushed • Repeat experiment for same H • Repeat for different H • Centre of mass of single bob and joined bob considered • Keep bob string taught Measurements • Measure heights h and H with ruler • Use centre of mass of bob or another suitable method • Use video camera to record motion • Use of datalogger and appropriate sensor to measure H and h • Measure mass with (top pan) balance Analysis • Construct a table of h and H • Plot graph of h against H • LoBF should pass through origin. • Determine gradient or calculate h/H repeatedly • gradient = ���+��2 (gradient must be consistent with the plot) • Masses substituted into above expression and checked against experimental gradient Total 6H556/01 Mark Scheme November 2020 13 Question Answer Marks Guidance 20 (a) F = mω2r and ω = 2πf kmg = mω2r Clear algebra leading to �2 = �4�� �2� × 1 � M1 M1 A1 Allow F = mv2/r and v = 2π fr Allow this mark for kmg = mv2/r (b) y-intercept = - 0.45 1 2 lg �4�� π2� = −0.45 �4�� �2� = 10−0.9 � = 0.126× 4 × �2 9.81 k = 0.51 C1 C1 C1 A1 Allow ± 0.05 Allow attempt at calculating y-intercept using gradient and a point on the line. Not e-0.9 wrong physics Allow k in range 0.48 to 0.63 Note Answer must be to 2 SF Total 7H556/01 Mark Scheme November 2020 14 Question Answer Marks Guidance 21 (a) (i) (pV = nRT) 100 × 103 × (0.46)3 = n × 8.31 × (273 + 20) n = 4.0 C1 A1 Note T = 20 is XP Not 1 SF answer of 4 Note answer is 4.00 to 3SF (ii) 100 293 = � 1573 or p × (0.46)3 = n × 8.31 × 1573 pressure = 540 (kPa) C1 A1 Note T = 1300 is XP Allow use of correct, unrounded n (b) (i) (p =) 6.6 × 10-26 × 990 or 6.5(3) × 10-23 (kg m s-1) (Δp =) 2 × 6.6 × 10-26 × 990 Δp = 1.3 × 10-22 (kg m s-1) C1 A1 Ignore sign of answer (ii)1 990/[2 × 0.46] (= 1080) B1 (ii)2 (F = Δp/∆t) (F =) 1.3 × 10-22 × 1000 F = 1.3 × 10-19 N C1 A1 Possible ECF from (b)(i) Note 1080 would give 1.4 × 10-19 (N) (iii) Use of p = F/A or pressure = (total) force/area Idea of multiplying by total number of atoms B1 B1 Allow particles or molecules for atoms Total 11H556/01 Mark Scheme November 2020 15 Question Answer Marks Guidance 22 (a) (i) The upthrust (on an object in a fluid) is equal to the weight of fluid (it displaces) B1 Note ‘fluid’ or ‘liquid’ must be mentioned at least once. Allow a named fluid, e.g. water (ii) (p = hρg) 1.9 × 103 = 0.15 × ρ × 9.81 ρ = 1.3 × 103 (kg m-3) C1 A1 (b) (i) x = A cos (ωt) or x = A cos (2πft) x = 2.0 cos (2π × 1.4 × 0.60) displacement = 1.1 (cm) C1 C1 A1 Note: Treat use of sine as TE Note answer is 1.07 (cm) to 3SF Note answer if calculator left in degrees of 1.99cm scores 2 marks. (ii) (vmax =) 2π × 1.4 × 0.02 maximum speed = 0.18 (m s-1) C1 A1 (iii) 1 Larger (amplitude) 2 Same (period) B1 B1 Total 10H556/01 Mark Scheme November 2020 16 Question Answer Marks Guidance 23 (a) (i) X at closest point on orbit to the Sun B1 Allow X on the orbit to the left of the Sun (ii) (When the asteroid orbits the sun a) line segment joining the asteroid to the Sun sweeps out equal areas in equal time (intervals) Longer distance (in orbit for the same time) B1 B1 Allow this mark on diagram (no labelling required) Allow ‘equal area swept in same time’ (b) (i) Work done per unit mass to move an object from infinity (to that point) B1 Not ‘work done on 1 kg’ (ii) Manipulation of V(g) = (-) GM/r B1 (iii) gradient = (-)30.4 or equivalent working candidate's gradient or expression = 6.67 × 10-11 × M and M calculated correctly from that gradient M = 4.6 × 1011 (kg) C1 C1 A0 Allow ± 2 Possible ECF from incorrect gradient Allow any subject (c) Method 1: Evidence of 2.3 × 10-3 and 600-1 or (2.3 × 10-3)-1 and 600 ½ v2 = 6.67 × 10-11 × 4.6 × 1011 × (2.3 × 10-3 – 600-1) v = 0.20 (m s-1) Method 2: Evidence of 7.0 × 10-2 and 5.1 × 10-2 from graph ½ v2 (= ∆V(g) ) = 7.0 × 10-2 – 5.1 × 10-2 v = 0.19 (m s-1) C1 C1 A1 (C1) (C1) (A1) Possible ECF from (b)(iii) for either value of GM or M Allow ½ v2 = 30 × (2.3 × 10-3 – 600-1) Note answer can be 0.19 or 0.20 or 0.2 m s-1 Note answer can be 0.19 or 0.20 or 0.2 m s-1 Allow correct use of one piece of data arriving at a value for v for 1 mark max Total 10H556/01 Mark Scheme November 2020 17 Question Answer Marks Guidance 24 (a) (i) Any four from • reduction in energy released by fusion • gravitational force is greater than that from radiation and gas pressure • core collapses • fusion no longer takes place in the core • fusion continues in the shell around the core • outer layers of star expand and cool • outer layers are released • reference to planetary nebula • reference to white dwarf (left as remnant hot core) B1 x 4 Ignore current or previous stages of the Sun’s evolution (b) (i) λT = constant however expressed 500 × 5.8 × 103 = 240 × T and T correctly evaluated T = 12000 (K) C1 C1 A0 Note answer is 12080 (K) to 4 SF Allow any subject (ii) (L = 4πr2σT4) 4.62 × 1031 = 4π × 5.67 × 10-8 × r2 × 120004 radius = 5.6 × 1010 (m) C1 A1 Note 12080 K gives 5.5 × 1010 (m) (c) (A white dwarf has mass equal to or) less than 1.4(4) solar masses / Chandrasekhar limit (ORA) B1 Allow M⨀ for solar mass Allow reference to neutron star (over Chandrasekhar limit) Allow correct reference to electron degeneracy pressure/Pauli Exclusion Principle Total 9OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]
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