Physics > AS Mark Scheme > GCE Physics B H557/01: Fundamentals of physics Advanced GCE Mark Scheme for November 2020 (All)

GCE Physics B H557/01: Fundamentals of physics Advanced GCE Mark Scheme for November 2020

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GCE Physics B H557/01: Fundamentals of physics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Physics B H557/01: Fundamentals of physics Advanced GCE ... Mark Scheme for 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 2020H557/01 Mark Scheme November 2020 2 Annotations 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 equationH557/01 Mark Scheme November 2020 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 argumentH557/01 Mark Scheme November 2020 4 Section A: MCQs Question Answer Marks Guidance 1 C L 2 D L 3 D L 4 D L 5 B L 6 D M 7 C L 8 A L 9 B M 10 D M 11 D M 12 A M 13 D M 14 D L 15 D L 16 A M 17 C L 18 D H 19 B H 20 D H 21 A H 22 D M 23 C M 24 A H 25 C M 26 A M 27 C H 28 C H 29 B H 30 A M Total 30H557/01 Mark Scheme November 2020 5 Section B Question Answer Marks Guidance 31 (a) Xe �� 134  L 31 (b) ∆m = 0.20401 u  convert u to kg OR convert via 1 u ≡ 931 MeV  (∆E = ∆m c2 ) = 190(MeV)  LM M allow mark for evidence of correct approach, even if final (third) mark for evaluation is not awarded accept correct calculation of energy on each side of equation for the first mark leading to a correct evaluation for the second mark accept 189.9, 189.93, 189.933 or 189.9333 for full credit (as allows 2SF) allow 186.2 for MAX 2 as a result of using fewer SF for ∆m – annotate as SF error Total 4 Question Answer Marks Guidance 32 (a) method: n = PV / RT / = 450 x 103 x 4 / [8.3 x 288]  evaluation: = 75(3) moles  L L method in words / numbers / algebra accept 75(2) if using R = 8.31 allow calculations leading to values around 188 moles (as a result of dividing by 4 tyres) for MAX 1 32 (b) Any two from : molecules move faster / have more kinetic energy / collide more frequently / harder / momentum changes at collision are greater   P increases x 320 /288 = 1.1 OR P increases to 500 MM not reference to force increasingH557/01 Mark Scheme November 2020 6 Question Answer Marks Guidance kPa OR P increases by 50kPa  L MAX 2 for responses that are qualitative only allow one mark for just “pressure increases” within MAX 2 for qualitative only argument Total 5 Question Answer Marks Guidance 33 (a) curves path / slows velocity  L allow accelerates the α / changes direction / changes velocity / slows down 33 (b) (most) has been stored as / converted to electrical potential energy OR k Q1 Q2 / R  L allow (small) fraction converted to k.e. of recoiling nucleus (which carries original momentum of alpha at closest approach) 33 (c) (i) low Z and high k.e. i.e. bottom left of table  M 33 (c) (ii) method: R = k x 2 x 13 x e2 /[7.7.MeV]  OR = 9 x 109 x 2 x13 x 1.6 x 10-19 / [7.7 x 106] evaluation: = 4.9 x 10-15 m  M H allow 5.7 x 10-15 m as 15% alpha k.e. in Al nucleus at closest approach (due to momentum transfer) for 2 marks allow 4.85 x 10-15 as a result of using k=8.98 for 2 marks Total 5H557/01 Mark Scheme November 2020 7 Question Answer Marks Guidance 34 (a) method : calculation of initial gradient using values taken from 1.5 GPa and 10% strain  evaluation: = 1.5 x 1010 Pa  L M ignore POT errors on this graph for this marking point allow method for MAX 1 based on values taken around (4,0.4) not just a line or markings drawn on graph, must have values used to calculate a gradient from their values 34 (b) W = ρ AL g ∝ ρ for equal dimensions and gravity OR strength / weight ∝ σ B / ρ  silk / steel = [1.4 / 1.2 x 103 ] / [2.8 / 7.8 x 103] = 3.3 (≈3)  H H must have explanation of approach in words or symbols for first mark not just two calculations of σ B / ρ Total 4H557/01 Mark Scheme November 2020 8 Question Answer Marks Guidance 35 (a) as h increases the path difference (TM + MR – TR) increases  (two sets of waves superpose meaning) waves in phase at max and out of phase at min / whole number of wavelengths path difference gives constructive interference and (n+1/2)λ gives destructive interference  M M allow equivalent phasor description or in terms of wave amplitudes adding allow idea that waves are changing phase with respect to each other for MAX 1 35 (b) method: between two consecutive max ∆ p.d. = λ  substitution: 2{√[12+0.2132] - √[12+0.1232]}  evaluation: =2.98 x 10-2 m  H H H allow any valid method e.g. between adjacent max and min ∆ p.d. = λ / 2 answer of 1.49 x 10-2 scores 1 MAX as a result of omitting x2 Total Total section B 5 23H557/01 Mark Scheme November 2020 9 Section C Question Answer Marks Guidance 36 (a) (i) 44 x 103 x 16 x 2 = 1.4(1) x 106 bit s-1  L allow 1.3(4) Mbit s-1 using 1 k = 1024 36 (a) (ii) t = info / rate = 840 x 106 x 8 / [1.41 x 106 x 60]  = 79.(4) mins  MM method allow 4760 s for first mark evaluation 36 (b) (i) there is a high f wave whose amplitude varies regularly at a lower f  M allow AW that convincingly explains there are two distinct frequencies with associated amplitude variation present 36 (b) (ii) noise is present with the signal (and should be ignored)  L 36 (b) (iii) 11 bits (211 ≈ 2048 (> 1600 Hz) )  M 36 (b) (iv) evaluation: 4 x 24 x 100 = 9.6 k bit s-1  show that fraction: 9.6 k / 1.4 M = 0.0069 ≈ 1/146  H H allow 9.6 k / 1.0 M = 0.0096 ≈ 1/104 allow ecf from a(i) allow 0.0068 if using 1.41M Total 8H557/01 Mark Scheme November 2020 10 Question Answer Marks Guidance 37 (a) (i) should cut 9th large square exactly  L 37 (a) (ii) (100 √2 ) = 14(1) OR (√{1002 + 1002 ] ) = 14(1) m s-1  M allow by discussion of equal x and y velocity components of 100 m s-1 accept 141.4 37 (a) (iii) 1 each large square represents a displacement of 100 m s-1 x 2.0 s = 200 m  2 so range is 200 x 9 = 1800 m  LM allow correct evaluation of any incorrect answer from 1 multiplied by 9 37 (a) (iv) R = 1412 sin 90° / 10 = 19(90) (m)  M allow 2000 (m) allow use of 1002 + 1002 instead of 1412 (from part a(ii) leading to either 2000 (from g=10) or 2038 (from g=9.81) allow use of g=9.81 leading to 2026 (m) allow use of 140 (from “show that” in a(ii) ) for acceptable values of g 37 (b) horizontally: (R = 1402 sin 30° / 10 = 980 m ) times of flight t75°= 980 / 140cos75° = 27.0 s  and t15° = 980 / 140cos15° = 7.2 s  all 3 marks for ∆t = 27 – 7.2 = 19.8 s  OR vertically: t75° = 2 x 140 sin75° /10 = 27.0 s  and t15° = 2 x 140 sin15° /10 = 7.2 s  all 3 marks for ∆t = 27 – 7.2 = 19.8 s  S & C S & C S & C for both horizontal and vertical approaches allow use of g = 9.8 m s-2 leading to 27.6 - 7.4 = 20.2H557/01 Mark Scheme November 2020 11 37 (c)* Level 3 (5–6 marks) Marshals argument in a clear manner and includes clear explanation of all strands including : • origin of air resistance • x and y components of v • trajectory 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) covers all strands at a superficial level and does not include enough depth for level 3. 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) Makes at least two independent points (possibly from only one strand), that are relevant to the argument but does not link them together and shows only superficial engagement with the argument. 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 LL MM HH accept labelled diagrams or graphs with “exaggeration for clarity” Indicative scientific points may include: origin of air resistance • projectile collides with air molecules / knocks them out of the path, transfers momentum / which exert a backwards force opposing the velocity • Fres = ∆(mv)/∆t ∝ ρAv x v ∝ ρAv2 • surface friction / viscous drag concepts x and y components of v horizontally • component of v is no longer constant / but decreases more quickly at start when v is larger / rate of acceleration less noticeable as v slower vertically • acceleration no longer constant / g but starts larger due to extra downwards force of drag / equals g when v = 0 because no vertical drag at max height of trajectory / becomes less than g on way down because drag force is now upwards opposing gravity v sin75° / v cos75° = 3.7 / vertical component is affected more because > horizontal component trajectory shape not parabolic / not symmetric about maximum height / descent covers shorter horizontal distance than ascent height travels less far / less than v2 sin150° / g = 980m less area under v(t) graph to max height / less than ½ g t 2 = ½ x 10 x 13.52 = 910 m / reaches max sooner / t descent > t ascent for equal area under v(t) graph range is smallerH557/01 Mark Scheme November 2020 12 credit diagrams / sketch graphs indicative allow exaggeration for clarity Total 14 Question Answer Marks Guidance 38 (a) pattern of lines is moved towards the red end of spectrum / blue line becomes blue-green, nearer the red end of spectrum  L remember red wavelengths and longer (i.r., μ , radio) are shifted away from red end of spectrum during red-shift! 38 (b) method: ∆λ / λ = constant  eval: 22/434 = 0.051 24/486 = 0.049 33/656 = 0.050 so sensibly constant  M M allow any two correct checks 38 (c) v = 0.050 x 3 x 108 = 15 x 106 (m s-1)  L Total 4H557/01 Mark Scheme November 2020 13 Question Answer Marks Guidance 39 Level 3 (5–6 marks) Marshals argument in a clear manner and includes clear explanation of all strands including : • why ratio 235 U / 238 U changes in time • estimation of age of atoms • assumptions 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) Covers all strands at a superficial level and does not include enough indicative points for level 3. 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) Makes at least two independent points (possibly from only one strand), that are relevant to the argument but does not link them together and shows only superficial engagement with the argument. 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 LL MM S&C S&C Indicative scientific points may include: ratio 235 U / 238 U changes in time • 235 U has shorter half-life, decays more quickly • 235 U / 238 U drops < 1 after atoms have formed • accept sketch decay graphs differing t ½ • 235 U / 238 U = N0 e- λ1 t / N0 e- λ2 t estimation of age of atoms • 235 U / 238 U = N0 e- λ1 t / N0 e- λ2 t • = e ln2(λ2 – λ1) t = 0.00725 • t = ln{0.00725} / ln2 {1/λ2 – 1/λ1} t = - 4.926 / {0.693 x [1/4.5 – 1/0.71]} Gyears = 5.99 x 109 years assumptions • early universe had only H and He atoms • U only formed later after star formation and supernova explosions • probability of forming U nuclei is small but equal for both isotopes of similar complexity so 235 U / 238 U ≈ 1 at start • aging is from formation of U atoms rather than the formation of the rocks • the Earth’s U is from one short epoch / no new material added from later supernovae which would re-enrich the ratio • all the U nuclei remaining today are only produced by this process and have not been added to since by other processesH557/01 Mark Scheme November 2020 14 Total 6 Question Answer Marks Guidance 40 (a) (i) method: E = 40000 / 6 x 1023  evaluation: 1023 = 6.66 x 10-20 (J)  L L Must be ‘show that’ 40 (a) (ii) kT = 1.38 x 10-23 x [273 +70] = 4.7(3) x 10-21 (J)  L 40 (a) (iii) E = {6.66 x 10-20 / 4.7(3) x 10-21} kT ≈ 14.(1) kT  f = e-E/kT = e-14 = 8.3 x 10-7  H H OR E / kT ≈ 14.(1) allow ecf from a(i) and (ii) 40 (a) (iv) molecules making many collisions per second (≈ 1010) so lots of opportunities to break hydrogen bonds OR energetic molecules are replaced by new ones by those molecules that “get lucky” in random collisions and keep gaining energy up to the bond breaking level  MM 40 (b) same BF at x10 T so bond energy is x 10 = 6.7 x 10-19 (J)  H OR may involve more complex calculations using BF e.g. ln10-7 = - E / [k x 3000] → E = 6.7 x 10-19 J Total 8H557/01 Mark Scheme November 2020 15 Question Answer Marks Guidance 41 (a) (i) γ only penetrating radiation getting deep inside the food α absorbed by few cms in air / in surface layer of solids β absorbed in surface layer mm of food / would not irradiate whole sample  LL not has best penetration not α stopped more easily not β stopped more easily any two points fromγ, α, β (mention of two points about same radiation type is MAX 1). To score 2 marks , the response must mention γ 41 (a) (ii) a 1 s dose received would be [500 x 1]/300  = 1.7 [Gy]  MM accept 1.67 41 (b) (i) exponential dilution due to absorption of γ-rays by water OR fixed small probability / fraction removed from each equal thickness layer → exponential decay with distance  γ- rays spread in all spatial directions diluting over the surface of sphere of surface area 4 π R 2 gives inverse square law dilution due to geometry  H H allow linear absorption coefficient μ for water as probability of absorption per track length OR half-thickness = ln2/μ = 0.11 m / 11 cm allow diagram explanation OR doubling R quadruples area exposed arguments expect high level reasoning including 4 π R 2 not descriptions of exponential relationships for either marking point since the question requires an explanation of terms in the equation and/or the context. 41 (b) (ii) method: 3/4 x flux I x A x time x E photon-1 / mass worker   evaluation: 1.3 x 10-3 [Sv]  S & C S & C S & C OR [3/4 x1.2 x 1016 x e-6.3 x 2 x0.5x1200x1.3x106x1.6x10-19] Sv [4π22 x 60] Credit part calculations for 1 mark e.g. e-6.3 x 2 = 3.4 x 10-6H557/01 Mark Scheme November 2020 16 Question Answer Marks Guidance OR 1/[4π22] = 2 x 10-2 OR calculating I from formula given in b(i) Total 9H557/01 Mark Scheme November 2020 17 Question Answer Marks Guidance 42 (a) (i) straight horizontal line at 1/3 AB from A  L 42 (a) (ii) linear decrease from 12 to 0 V from A to B  L 42 (a) (iii) 2400 (V m-1)  L 42 (b) (i) method: Vc = 49/10 = 4.9 V  v = √[2 e Vc / m] evaluation = 1.3 x 106 m s-1  M M accept values for anode potential of 48< V < 50 (b) (ii) Method : 4x3.7x105 m2 /[3.70001 x 105 m]2 = 1.08 x 10-5  H must show full evaluation, not just 10-5 (b) (iii) must be inelastic collisions removing electrons k.e. so they can no longer climb the potential hill of Vback off means mercury atom must have an internal energy level at 4.9 eV above ground state / evidence of a quantized electrical potential energy level inside mercury atom  S&C S&C any two points allow 4.9 + 0.1 V OR electron from ground state can be promoted by sufficient energy to a higher energy state, but cannot exist in between states etc… Total 8 Total section C Total sections B & C 57 80OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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