GCE Physics B H557/02: Scientific literacy in physics Advanced GCE Mark Scheme for November 2020

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GCE Physics B H557/02: Scientific literacy in physics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Physics B H557/02: Scientific literacy in 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 2020H557/02 Mark Scheme November 2020 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 equationH557/02 Mark Scheme November 2020 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 November 2020 Question Answer Marks Guidance 1 a i Data logger allow higher frequency of p.d. readings/reading a (changing) voltmeter would introduce greater uncertainty/datalogger timing will be precise 1 Don’t accept bald ‘more accurate’ or ‘less uncertainty’ ii Either: 0.63V0 = 2.8 V (1), time taken to reach this value (� ), 2.3 s <τ < 2.4 s (1) giving C between 450 and 480 µF (1) Or: Use of � = �0(1 − �− � �� ) Correct pair of V and t (1) Correct rearrangement to make C the subject(1) Value = 470 µF (1) 3 Don’t accept answers from Q = It when I is held constant. Range at SSU SSU to consider 1/e approximated to 1/3, i.e. (2/3) V0 = 2.8 V (Haven’t seen this in first 20% of scripts) Range at SSU e.g. V = 2.5 V at t = 2 s (1) C = - 2/(5000 x ln(1 – 2.5/4.4) (1) = 0.00047 F (1) iii E = 0.5 x 5 x 10-4 x 4.22 (1) = 0.0044 J (1) 2 Use either own value from a ii or 0.0005 F b i Line of decreasing negative gradient starting at 4.2 V (1) Gradient -0.5 x gradient of charging line at any time, by eye (1) Explanation: RC value in the discharge circuit is twice that of the charging circuit. AW(1) 3 Accept any clearly lower magnitude of gradient. ii E ∝ V 2 ⇒ p.d. when energy stored has halved = 4.2 V × √0.5 = 3.0 V (2.97 V) (1) � = �0 �− � �� where R= 10 kΩ & C= 470 µF ⇒ RC = 4.7 s (1) -t/4.7 s = ln(3.0 V/4.2 V) -t = 4.7 s × -0.336 ⇒ t = 1.58 s = 1.6 s (1) 3 Range at SSU Ecf from a iii if used Or RC = double τ from (a)(ii) If 500 x 10-6 F used, RC = 5 s leading to value 1.68/1.7 s. Bald in-range answer gains all marksH557/02 Mark Scheme November 2020 Question Answer Marks Guidance 2 a 0.8/3 x 108 = 2.7 x 10-9 light seconds (1) 1 No s.f. penalty b i Light (is observed to) travel a greater distance (1) at the same velocity (1) 2 e.g. 2ct > 2cτ ii �2� 2 = � 2� 2 + �2�2 (1) � 2(1 − �2 �2) = �2 (1) (leading to � = � �1− � �22 ) 2 Expect intermediate stages. Any correct routes gain both marks. c � = 611 �1− ( (5 3..4 0 × ×1 10 07 8) )2 2 (1) = 621 s (1) 2 Calculating γ to 1.017 gains first mark. Accept answer rounded to 620 s. d For photon v = c so denominator (or �1− ��2 2 ) is zero/ gamma factor is infinite (1) Therefore �/� ( = �) is zero (for any value of t) (1) 2 AW but both steps needed for first mark. AWH557/02 Mark Scheme November 2020 Question Answer Marks Guidance 3 a wavelength of emitted light = (6.63 x 10-34 x 3.00 x 108)/ (1.88 x 1.60 x 10-19) (1) = 6.61 x 10-7 m (1) 2 Accept 2 or 3 s.f. No ecf from wrong value used in calculation. b i Linear graph with x-axis intercept of 1.25 ± 0.03 x 106 m-1 1 Examiner to extrapolate line if it doesn’t extend to axis. ii ℎ = �� � gradient = Δ�/Δ�−1 = Δ� × ∆� (1) � = �� (1) (therefore, required equation) 2 Can get the second mark as a lone mark. Accept lack of delta notation if algebra correct, but working from Δ�/Δ�−1 needed0. iii gradient working using x-interval of at least 0.4 x 106 m-1 (1) gradient in range 1.0 x 10-6 to 1.2 x 10-6 (V m) (1) h in range 5.3 x 10-34 to 6.4 x 10-34 (J s)(1) 3 Ecf from b i clear values from graph required for first point expected value = 1.1 x 10-6 V m expected value = 5.9 x 10-34 J s iv Steepest-possible value = 9.3 x 10-34 J s. (1) Uncertainty = ± (9.3 – 5.9) x 10-34 = 3.4 x 10-34J s (1) Answer given as 5.9 x 10-34 J s ± 3.4 x 10-34 J s (1) 3 ecf from (iii) ecf from (iii) Allow 3 sf in final answer and uncertainty. Sf of value and uncertainty must match.Accept 1 s.f. uncertainty and valueH557/02 Mark Scheme November 2020 Question Answer Marks Guidance Section B 4 a 1/3.85 x 10-3 = (+) 260 D (1) 1 b Waves will strike lens with curvature more negative than – 260D (1) Curvature still negative (or zero) on leaving the lens (1) 2 accept attempt to calculate v using lens equation plus comment on (non) result. c for u = 1.2 m, v = 0.00386 m (1) for u = 0.050 m, v = 0.0042 m(1) In the first case the image is formed very close to the sensitive surface, this is not so when u = 0.050 m (1) 3 Can’t use argument that power of the lens is insufficient Accept correct answers derived from u and v confusion d length on surface = (0.00385/0.35) x 0.090 (1) = 9.9 x 10-4 m (1) 2 Or calculates magnification as 0.011 for first mark Accept rounding to 1(.0) x 10-3 m as long as method clear If u = 1.2 is used, one mark for correct answer, 2.9 x 10-4 m. e length of pixel = (4.89 x 10-3 x 3.65 x 10-3/12 x 106)0.5 (1) = 1.2 x 10-6 m (1) Number of pixels along 9.9 x 10-4 m = 825 (1) Resolution = 0.090/825 =1.1 x 10-4 m 4 First mark can be awarded for area of a pixel calculated as 1.49 x 10-12/1.5 x 10-12 m. If area of pixel is used to calculate resolution, no marks. f Increase brightness (1) changing each pixel by a fixed value (until brightest pixel is coded at 255) (1) OR Increase contrast (1) stretching pixel values to cover full range (1) Benefit: improving images, removing noise, aesthetic changes etc Problems: pictures can be manipulated AW 4 Total 16H557/02 Mark Scheme November 2020 Question Answer Marks Guidance 5 a i k = (0.059 x 9.81)/(1.31 – 1.15)= 3.62 (~4 ) (1) 1 ii E = 0.5 x 3.62 x (1.31 – 1.15)2 (1) = 0.046 J (1) 2 Iii Level 3 (5–6 marks) Marshals argument in a clear manner. Clearly links acceleration of the ball with the forces acting upon it throughout the fall. Makes a clear statement of the position of greatest velocity and calculates the maximum upwards acceleration. 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 of the motion of the ball but does not consider forces or gives incomplete description including forces. Makes some accurate quantitative statements. 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 incomplete description which includes some correct physics, for instance initial acceleration, position of maximum velocity and position of maximum acceleration. 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: • Acceleration = 9.81 m s-2 from release until length of cord = 1.15 m • as only force acting on ball is mg. • Describes velocity increasing at a decreasing rate until 1.31 m • From 1.15 to 1.31 m acceleration is decreasing • as net force on ball = mg – kx • Greatest velocity at 1.31 m • From 1.31 m to 1.78 m the ball is decelerating at an increasing rate • as kx > mg and the upward force increases as x increases • maximum upwards acceleration occurs at lowest point • upwards acceleration = (kx – mg)/m; • upwards acceleration = ((3.62 x 0.63) - (0.059 x 9.81))/0.059 = 28.8 m s-2H557/02 Mark Scheme November 2020 Question Answer Marks Guidance 5 a iv Sensible, practicable suggestion (1) Description of method (1) 2 E.g. video against a ruled background (1) Analyse frame by frame (1) NOT just ‘use a video camera’ OR Place board on floor and move up until the ball just touches the board at lowest point (1) Judge by sight or sound Measure distance (1) b 4 points from: • (Such a polymer is an example of) a long chain molecule • which can rotate about its bonds/untangle • Bonds are strong so polymer has high breaking stress • Elastic region has relatively low k, reducing force on bungee jumper • Metal structure is an array of positive ions in a sea of delocalised electrons • Mobile dislocations lead to plastic behaviour • Metals are only elastic for small strains as the deforming force pulls bonds apart • Plastic deformation occurs in metal for small strains • Decelerating force on jumper would be very great as metals have higher k 4 AW throughout. First and fourth bullet points can be given diagrammatically. total 15H557/02 Mark Scheme November 2020 Question Answer Marks Guidance 6 a i Starting height = 6.40 x 106 m & ending height = 7.225 x 106 m (1) Change in Vgrav. is the area between the line and the axis (bounded by the starting and ending height) (1) Change in energy = change in Vgrav x 2300 kg. Value between 1.55 x 1010 J and 1.75 x 1010 J (1) 3 Must show own value; from graph should be ~ 7.2 × 10 6 J kg -1. Accept calculating to values of GM/r by multiplying g values by r: First mark for identifying heights correctly, Second mark for method Third mark for evaluation. Candidate must show how the result is achieved, even if a full explanation is not present. Accept ½ base x height calculations(plus rectangle) giving answers of around 1.7 x 1010 J Use of E = mgh is neutral in this case. Calculated value = 1.64 x 1010 J a ii 1.6 x 1010 = GMm�6.4 ×1106 − 7.2251×106� (1) = 1.78 x 10-8 GMm M = 1.6 x 1010/(6.67 x 10-11 x 2300 x 1.78 x 10-8 ) (1) = 5.8 x 1024 kg (≈ 6 × 10 24 kg) (1) 3 e.c.f. own answer to ai. Must show working but steps in the working may be conflated. b i F = (−) 6.67 ×10−11×2300 ×6 ×1024 (7.225 ×106)2 (1) = (-) 1.763 × 10 4 N = 1.8 x 104 N (1)0 2 Bald correct answer gains two marks. Allow ecf from POT error in method – if clearly shown. ii v = √1.8 ×104×7.225 ×106 2300 = 7520 m s-1 (1) T = 2� ×7.225 ×106 7520 (1) = 6037 s = 6040 s (1) OR a = - ω2r 1.8 x 1044/2300 = 4πf2 x 7.225 x 106 (1) 3 Allow for rounding differences. Bald correct answer gains three marks. Ecf from b(i). If 1.763 x 104 used, T = 6100 sH557/02 Mark Scheme November 2020 f2 = 2.74 x 10-8 (1) f = 1.66 x 10-4 s-1 T = 6037 s (1) OR v = (GM/r)0.5 = 7520 m s-1 (1) for first mark c Any three from: Advantages to low polar orbit: • High(er) resolution imaging AW • Image more of the planet as the Earth spins underneath the satellite Geostationary: • Remain at the same position in the sky so dishes can keep locked on to signal AW • Higher orbit means greater coverage 3 Not just ‘image more of planet’ or ‘see more clearly’ Note stem of question informs candidates that geostationary satellites are always above the same point on the Earth’s surface. total 14H557/02 Mark Scheme November 2020 Section C Question Answer Marks Guidance 7 a 400 Hz (1) 1 b Highest frequency component = 800 Hz (1) Need twice this to avoid aliasing (AW) (1) Minimum sampling frequency = 1600 Hz (1) 3 The time period of the ‘wobble’ is ~ 0.0006 s. This may lead students to a high frequency component of 1670 Hz and therefore a minimum frequency of around 3300 Hz. Two marks max for this. (SSU) Can get the second two marks from the wrong highest frequency component. Two marks max for doubling the frequency in (a) c i Velocity increases whilst wavelength constant (1) 1 Need wavelength constant ii 445 440 = ��2 √285 (1) T2 = 291.5 Hz so temperature rise = 6.5 K (1) 2 Bald answer gains both marks Accept 7 K total 7H557/02 Mark Scheme November 2020 Question Answer Marks Guidance 8 a particle identified as anti-neutrino (1) (No leptons on LHS of equation), anti-lepton (anti-neutrino) balances lepton (electron) on RHS (1) Electrons released in beta decay have range of energies, (1) (the energy of the neutrino makes up the difference between energy released and the energy of the electron) 3 AW Mark for recognising the spectrum of energies of electrons emitted. b Using � = − � ln ���t 0�: 9000 = −5700 ln2 ln(1.4 ×�1t0−12) (1) Ratio = 4.7 x 10-13 (1) OR λ = ln 2/5700 years = 1.216 × 10-4 year-1 (1) Rt= R0 e - λ t = 1.4 × 10 -12 exp(-1.216×10-4 year -1 × 9000 year) = 4.7 × 10 -13 (1) 2 Other methods possible. Correct bald answer gains both marks. Accept 3.8 x 10-12 (s-1) for λH557/02 Mark Scheme November 2020 Question Answer Marks Guidance 8 c Level 3 (5–6 marks) Marshals argument in a clear manner. Explains the idea of ‘old carbon’ released from volcanoes having lower 14C/12C ratio than living material. Clearly explains the use of tree ring data (which can also have its 14C/12C ratio measured) and links use of tree ring data back to measurement of plant age. Includes a clear and correct calculation of the effect of ‘old-carbon’ contamination. 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 and complete qualitative explanation of the effect of old carbon and the use of tree ring data but the calculation is incomplete or incorrect in some aspects. 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 description of the effect of old carbon and the use of tree rings but the explanation is superficial or incomplete. Calculation, if attempted, is limited or incorrect. 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: • Carbon-14 originally present in volcanic material has decayed • Carbon from volcanoes combines with oxygen to give carbon-dioxide which plants take up (through photosynthesis) • Take up of ancient carbon reduces 14C/12C ratio • Reduced 14C/12C ratio suggests a greater age • Trees absorb CO2 during growth cycle • Age of trees can be directly measured by counting tree rings • 14C/12C ratio in each ring can be measured and compared with organic material in the vicinity. • Calculation: • 10% ancient carbon will reduce 14C/12C ratio to 1.26 x 10-12 • Apparent age = −5700 ln2 ln(11.26 .4 ) • Apparent age = 866 years (2 s.f.) total 11H557/02 Mark Scheme November 2020 Question Answer Marks Guidance 9 a � = �2��� = �2��� (= 2.50 × 10 -19 N s) (1) r = �2 ×2.33 ×10−26 ×4.2 ×106/3.2 × 10−19 0.72 (1) = 1.1 m (2 s.f.) (1) 3 Other routes possible. Correct bald answer gains three marks. Correct calculation following incorrect velocity value gains two marks ecf. Double error in charge (giving v = 7.59 x 106 and r = 1.54 gains two marks) 0 1.55 or 1.6 m gains two marks (use of 1.6 x 10-19 rather than 3.2 x 10-19) b These have the same mass as 14C (1) and those with the same charge will be deflected the same amount in the magnetic field, (1) producing spurious 14C results. 2OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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