GCE Physics B H557/02: Scientific literacy in physics Advanced GCE Mark Scheme for Autumn 2021

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GCE Physics B H557/02: Scientific literacy in physics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Physics B H557/02: Scientific literacy in physics Ad... vanced 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 2021H557/02 Mark Scheme October 2021 1. Annotations available in RM Assessor Annotation Meaning Benefit of doubt given Contradiction Incorrect response Error carried forward Level 1 Level 2 Level 3 Transcription error Benefit of doubt not given Power of 10 error Omission mark Error in number of significant figures Correct response Wrong physics or equationH557/02 Mark Scheme October 2021 2. 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 argumentH557/02 Mark Scheme October 2021 Question Answer Marks Guidance 1 a i Gravitational energy per kg (at a position in a gravitational field) . 1 Or work required per unit mass to bring a body from infinity to that point AW Or energy required per kg to travel to infinity. ii Gravitational potential is (set at) zero for bodies at infinite distance apart . It takes work to separate bodies therefore at distances less than infinity the potential must be negative. AW  2 Explanation in terms of potential well can gain both marks. (negative potential at mass going to zero at infinity) Second mark does not depend on the first. iii Electrical charges can attract each other (giving a negative potential) or repel each other (giving a positive potential)/ When two like charges interact energy is stored. AW  1 Accept that electrostatic force can be both positive and negative. Reference to charge only is not enough. b Value = (+)6.3 × 107 J . K.E. + G.P.E. = 0 for this value  AW Assuming no energy transfer through atmosphere/ No work done against frictional forces  3 When all KE has been transferred to PE, PE = zero so the body is out of the potential well. Allow KE greater than or equal to PE. Allow ‘energy losses’ if ‘in atmosphere’ specified c k.e. = 1.0 × 1011 J  g.p.e. = - 4.37 × 109 J  total energy = 1.0 × 1011 J  Total energy is positive – Voyager will continue to move further from Sun  4 1.04 × 1011 J to 3 s.f. Must have negative sign 9.97 × 1010 J to 3 s.f. Allow variations due to different numbers of sf used in calculations One mark for correct total energy from incorrect Ek and/ EP Accept ‘will leave Sun’s gravitational field’ Total 11H557/02 Mark Scheme October 2021 Question Answer Marks Guidance 2 a i V = 9.0 × e - 3.5 /( 4700 × 10^-6 × 1400)  = 5.29 V  2 Must give own value. a ii ∆E = ½ C (9.02 - 5.32) = 0.124 J  Power = 0.036 W  Current through/ p.d. across component not constant  3 3rd mark independent b p.d. across capacitor when E = 300 J, V 300 J = √(2E/C) = √(2×300 J/120 F) = √5 V = 2.24 V  p.d. across capacitor when E = 50 J, V 50 J = 0.91 V  time = -ln(0.91/2.24) × 30 × 10-3 × 120 = 3.2 s  minimum value because no external load  4 Other routes may be used. Bald correct answer gains all three marks for the calculation. Total 9 3 a Energy gained by block = 541 J  Power per m2 = 541/(600 × 0.0013) = 690 W m -2  2 Accept 540 J No s.f. penalty. Accept range of answers due to sf choice. Allow 700 W m-2 b Power output of Sun = 1.4 × 103 × 4 × π × (1.5 × 1011)2  = 3.96 × 1026 W  2 1 mark for correct calculation of area of sphere = 2.83 × 1023 m2 . Need own value c i Identification of positron as anti-lepton, neutrino as lepton  Lepton number on LHS = zero, lepton number on RHS = zero  2 c ii Mass loss from one three-stage reaction, ∆m = 0.0265 u  Energy released per reaction = ∆mc 2 =(0.0265× 1.661×10-27) kg × 9 × 1016 m2s-2= 3.96 × 10-12 J  Number of reactions s -1= 3.8 × 1026J/(3.96 × 10-12J × 0.98)  = 9.8 × 1037  4 Or 4.4 x 10-29 kg Ecf within question throughout 3 marks maximum for 9.6 or 9.4 × 1037 Correct bald answer gains four marks Total 10H557/02 Mark Scheme October 2021 Section B Question Answer Marks Guidance 4 a E = 3.43 N × 3.951 m / (5.9 × 10-8 m2 × 0.002 m)  = 1.15 × 1011 Pa  2 Or via ε = 5.062×10-4 & σ = 5.814×107 Pa Bald correct answer gains two marks. Accept two s.f. answer of 1.1 x 1011 Pa b i area occupied by one atom = (2.3 × 10-10 m)2 = 5.29 × 10-20 m2  Number of atoms in 5.9 × 10-8 m2 = 5.9 × 10-8 m2/5.29 × 10-20 m2 = 1.115 × 1012  Tension = 3.43 N/1.115 × 1012 = 3.1 × 10-12 N  3 Bald correct answer gains three marks. Allow use of π r2 giving area = 4.15 x 10-20 m2 and / 1.42 x 1012 atoms per layer. 3rd marking point available as ecf from number of atoms. (3.075 × 10-12 N to 4 s.f.) b ii x = FL/AE = 3.43 N × 2.3 x 10-10 m/(5.9 × 10-8 m2 ×1.15 × 1011 Pa)  = 1.163 × 10-13 m = 1.2 × 10-13 m  2 Ecf from (a) if this method used. (precise value of E gives 1.16 × 10-13 m) Alternative methods possible, e.g. simple ratio: x = 0.002 m × 2.3 × 10-10 m/3.591 m  = 1.16 × 10-13 m  Or x = εL= 5.062×10-4 × 2.3 x 10-10 m  = 1.16 × 10-13 m  b iii Force constant = 3.1 × 10-12 N/1.16 × 10-13 m  = 26.7 N m-1 = 27 N m-1 2 Ecf from b(i) and b(ii). No credit if 3.43 N used. Unrounded answers give acceptable 26 N m-1 to two s.f.H557/02 Mark Scheme October 2021 Question Answer Marks Guidance c Level 3 (5–6 marks) Clear explanation of yield stress; clear explanation of the limitations of the model; clear explanation of the effect of mobile dislocations in materials and its effect on yield stress. (can be helped by a correct diagram) Clear explanation of the effects of adding alloying atoms to the metal. 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 yield stress. Gives a clear explanation of either the limitations of the model or the effect of mobile dislocations and considers the effect of alloying atoms. The explanations are correct but lack either breadth or depth. There is a line of reasoning presented with some structure. The information presented is relevant and supported by some evidence. Level 1 (1–2 marks) Description/explanation of yield stress; Gives a superficial description of either the limitations of the model or the effect of mobile dislocations. Mentions the role of alloying atoms. 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 6 Indicative scientific points may include: • Description/explanation of term ‘yield stress’ • Model suggests perfect (crystal) structure • Better model includes missing atoms/dislocations in layers of atoms. • Dislocations in metals are mobile • (Dislocations allow) bonds to be broken one at a time rather than the whole plane failing at once • (dislocations) reduce stress required for material to yield • Dislocations can be ‘pinned’ by alloying atoms • Dislocations can be pinned (tangled together) by work hardening • Alloying/work hardening (usually) leads to a harder material • Alloying/work hardening (usually) leads to a more brittle material • Diagrams of dislocations • Effect of grain size/rate of cooling Total 15H557/02 Mark Scheme October 2021 Question Answer Marks Guidance 5 a k = 4π2 × 0.84 kg/(1.3 s)2  = 19.62 N m-1 = 20 (N m-1)  2 Bald answer of 19.62/19.6/20 gains two marks. b k.e.max = ½ kx 2 = 0.5 × 19.6 N m-1 × (0.24 m)2 = 0.565 J vmax = (2 × [0.5 × 19.6 N m-1 × {0.24 m}2] J /0.84 kg)1/2  = 1.16 m s-1  3 One mark for correct k.e. (0.56 J) Ecf from (a) OR vmax = 2 π f A  = 2 × π × (1.3 s) -1 × 0.24  = 1.16 m s-1  1.2 m s-1 c p.e. is proportional to x2 so p.e. will fall to half maximum value when x2 = (xmax)2/2 so p.e. is half maximum value when x = x/(2)1/2 AW Total energy = k.e. + p.e. so when p.e. = ½ p.e.max so is kinetic energy AW  2 Or via p.e. = 0.565 J/2 = 0.2865 J 0.2865 J = ½ kx 2 x = √(0.565 J/19.62 N m-1) = 0.170 m and 0.24 m / √2 = 0.170 m  d 1/(21/2) = cos (2 π t/1.3 s)  t = 0.163 s  2 Or via 0.170 m = 0.24 m cos (2 π t/1.3 s) e Changing flux/magnetic field in metal sheet  Emf generated  Current driven by emf  (I2r) heating effect of current  4 ‘Changing flux generates emf’ gains 2 marks. Or work done against Lenz force AW – need work done/energy transferred rather than just ‘against force’ 4th mark can be independent. f Pendulum will resonate with (wind-induced) oscillations of the tower  Energy of movement of tower is transferred to pendulum  Energy of pendulum transferred to internal energy of block  3 AW throughout Total 16H557/02 Mark Scheme October 2021 Question Answer Marks Guidance 6 a Any two from: • Heating effect of current • Changes resistance of wire • Higher current will lead to higher equilibrium temperature 2 Accept ‘maintain constant temperature of wire’ b i Non-zero p.d. intercept  Contact resistance  2 Accept low resistance voltmeter/ systematic error in length measurement/zero error on meter b ii Method of establishing gradient clearly shown, ∆L ≥ 0.5 m  Gradient 1.8 (V m -1) to 2 s.f.  2 b iii σ = GL/A = I/V × L/A = I/A ÷ [V/L]  = 0.30 A/{π × (2.8 × 10-4m)2} ÷ 1.8 V m -1  = 6.8 × 105 (S m-1)  3 Bald answer of 6.8/6.77/6.767 × 105 gains two marks. e.c.f. own gradient from (b)(ii) If using specific I and V values (instead of gradient) two marks max. c Any two from: • same current will require greater p.d. across wire • greater heating effect • higher resistance therefore lower conductance (therefore lower conductivity 2 Or same I & larger R gives larger I2R  d • metals have free electrons/electron cloud/sea of free electrons/metallic bonding allows electrons to migrate through lattice under potential difference • insulators have far fewer free electrons/electrons are bound to individual atoms in insulators • more electrons (therefore more charge) flow under given potential difference in metals so conductivity far higher 3 Total 14H557/02 Mark Scheme October 2021 Section C Question Answer Marks Guidance 7 a 0.0119 m  1 no s.f. penalty (so 0.011935 is OK, as is 0.012)) b Number of waves in pulse = 1 × 10-6 × 3.5 × 106 = 3.5  Wavelength = 4.4 × 10-4 m  Resolution = 4.4 × 10-4 × 3.5/2 = 7.7 × 10-4  3 Ecf from number of waves in a cycle Or pulse duration x velocity/2. 1.54 x 10-3 credited two marks Total 4 8 a Width of Fig 8 = 53 mm Number of pixels along length = (25/53) × 920 = 434  Resolution = 39/434 = 0.090 mm  2 Expect to see one stage calculation. Allow horizontal length or length along arrow ECF from length of image. Range: rounds to 0.09 mm to 1 s.f. b 1 bit per pixel so only choice of 2 possibilities (0 & 1)  Density of white pixels in image  2 ‘one bit per pixel’ on its own is not enough for mark Total 4H557/02 Mark Scheme October 2021 Question Answer Marks Guidance 9 a i The value of the variable concerned falls by a factor of the square of the distance between the source and detector  Intensity will have fallen to 1/R2 at the object (this now acts as the source) this reflected intensity falls by a factor of 1/R2 again; intensity of reflection signal at source= 1/R2 ×1/R2 = 1/R4  2 AW – clear explanations gain the mark A complete and clear statement required. a ii Calculation of intensity ratio = 1/(2.4)4  = 0.03014 = 0.030  Power difference in dB = 10 log10 0.030 = ( -) 15 dB  3 10 log10 0.03014 = ( -) 15.2 dB b i ln 0.93 = -0.07257 = -� × 0.1  � = 0.07257/0.1 = 0.726  2 Evaluation needed for second mark b ii � = �0 �−�(2�) �4 ⇒�4 × ��0 = �−�(2�)  ln �3.04 × ��0 � = ln 3.04 + ln��0 = −2 � � = −6.0�  Using α = 0.7 m-1 ⇒ P/P0 = 0.000186  10 log (0.000186) = -37.3 dB  4 Bald correct answer gains all the marks. Ecf own value of α. If the factor of two left out in the attenuation expression, leading to an answer of – 28.2 dB (α = 0.7 m-1) or –28.5 dB (α = 0.726 m-1), three marks. Using α = 0.726 m-1 ⇒ P/P0 = 0.000158 & ∆P = -38.0 dB Ecf from third to fourth mark. Total 11H557/02 Mark Scheme October 2021 Question Answer Marks Guidance 10 Level 3 (5–6 marks) Superposition: Clear explanation of the principle of superposition including the concept of in phase superposition producing highest amplitude resultant and the link between amplitude and power. Bats: Clear explanation of superposition from the two sources and explanation/description of energy distribution of the sound in front of the bat. Medical ultrasound: Clear explanation of delaying pulses so that they meet in front of the transducer in phase at a (chosen) depth. Link between amplitude and intensity/power of beam at given depth and why this is important. 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) Gives a clear explanation principle of superposition and its relevance to the beam from bats or medical ultrasound. Or a superficial explanation of all three sections attempted. There is a line of reasoning presented with some structure. The information presented is relevant and supported by some evidence. Level 1 (1–2 marks) Gives a superficial description of any two of the three areas of interest. 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 6 Indicative scientific points may include: • Credit clear diagram showing waves from two sources meeting in phase. • Principle of superposition clearly stated • Waves from two sources will always meet in phase along the line at right angles to the sources at the midpoint of the sources. • Energy ‘focused’ /redistributed along line where waves meet in phase • Relationship between amplitude and power • More intense beam will have same proportion of energy at ‘target’ but greater value of energy. • Medical ultrasound concentrates energy at a depth • Concentrating energy in this fashion means greater energy reflected • Greater return energy delivers more detail/information • Concentrating energy in this manner allows greater depths to be imaged Total 6OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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