MI 23M1PrelimAS (H2 Physics P3)

1 2023 MI PU3 H2 Physics Prelim Paper 3 Suggested Solution 1a(i) No external forces act on the system of two objects. OR The two objects form a closed system. B1 (ii) Straight line from (1.0, -3.0u) to (2.0, 4.2u) Horizontal line at p = 4.2u from t = 2.0 to 3.0 s B1 B1 (iii) Since the total momentum before the collision is 6.0u, the total momentum of the system remains constant at 6.0u at all times. So final p of object B = 6.0u – 1.8u =4.2u OR Since p of object A decrease by 7.2u after the collision, p of particle B will increase by 7.2mu. So final p of particle B = -3.0u + 7.2u = 4.2u B1 1b vA = 1.8u / 9.0 = 0.2u vB = 4.2u / 1.0 = 4.2u 9.0u 0 3.0 2.0 –3.0u 1.0 t p 4.2u 1.8u

2 relative speed of approach= 𝑢𝐴 − 𝑢𝐵 = 𝑢 − (−3𝑢) = 4𝑢 relative speed of separation= 𝑣𝐵 − 𝑣𝐴 = 4.2𝑢 − 0.2𝑢 = 4𝑢 Since the relative speed of approach and the relative speed of separation are equal, the collision is elastic. OR kinetic energy, 𝐸𝑘 = 1 2 𝑚𝑣2 = 1 2 (𝑚𝑣)2 𝑚 = 𝑝2 2𝑚 before collision: 𝐸𝑘,𝑏𝑒𝑓𝑜𝑟𝑒 = 𝑝𝐴,𝑏𝑒𝑓𝑜𝑟𝑒2 2𝑚𝐴 + 𝑝𝐵,𝑏𝑒𝑓𝑜𝑟𝑒2 2𝑚𝐵 = (9.0𝑢)2 2(9.0) + (−3.0𝑢)2 2(1.0) = 9.0𝑢2 after collision: 𝐸𝑘,𝑎𝑓𝑡𝑒𝑟 = 𝑝𝐴,𝑎𝑓𝑡𝑒𝑟 2 2𝑚𝐴 + 𝑝𝐵,𝑎𝑓𝑡𝑒𝑟 2 2𝑚𝐵 = (1.8𝑢)2 2(9.0) + (4.2𝑢)2 2(1.0) = 9.0𝑢2 Since the kinetic energy of the system before and after the collision remains the same , the collision is elastic. M1 M1 A1 (M1) (M1) (A1) 1c(i) 𝐹 = ∆𝑝 ∆𝑡 = (1.8𝑢 − 9.0𝑢) (2.0 − 1.0) = ( - ) 7.2u (accept + or – answer) Therefore, magnitude of F is 7.2u M1 A0 (ii) By Newton’s second law, gradient of p vs t graph represents resultant force. During collision, gradient of graph of object A and object B have equal magnitude, showing that forces acting on A and B have equal magnitude. The gradients have opposite signs indicate that the two forces act in opposite directions. Hence the graphs are consistent with Newton’s third law B1 B1 Total: 10

3 2a Total volume of molecules is negligible compared with volume occupied by the gas B1 2b pV = NkT 2.10 × 105 × 950 × 10–6 = N × 1.38 × 10–23 × (280) N = 5.16 × 1022 volume of one molecule = (4 / 3)r3 (= 1.41 × 10–29 m3) volume of all molecules = 5.16 × 1022 × 1.41 × 10–29 = 7.3 × 10–7 m3 = 7 × 10–7 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) OR volume of one molecule = d3 (= 2.7 × 10–29 m3) volume of all molecules = 2.7 × 10–29 × 5.16 × 1022 = 7 × 10–7 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) C1 C1 A1 (C1) (C1) (A1) 2c Since volume of all atoms (1 × 10–6 m3) is 3 orders of magnitude less than volume occupied by the gas (950 × 10 –6 m3  1 x 10-3 m3) OR 0.1% of volume occupied by the gas so assumption in (a) is justified. B1 2d(i) Internal energy depends on temperature. Since final temperature equal initial temperature so no change in total internal energy B1 B1 2d(ii) For P → Q: work done on gas = 0 J and increase in internal energy, U = Q + W = +97.0 + 0 = 97.0 J For Q → R: increase in internal energy, U = 0 -42.5 = –42.5 J For R→P: work done on gas, W = p ∆V = 2.10 × 105 × (1125 – 950) × 10–6 = 36.8 J since total change in internal energy is zero, +97 + (–42.5) + ∆URP = 0 increase in internal energy, ∆URP = –54.5 J thermal energy supplied, Q = ∆U – W = −54.5 – 36.8 = –91.3 J A1 A1 A1 A1 A1 Total: 12

4 3a upthrust and weight B1 3b upthrust greater than weight so resultant force is upwards B1 3c(i) A, g and ρ all constant so F is proportional to x minus sign means F and x are in opposite directions B1 B1 3c(ii) 𝑎 = 𝐹 𝑚 = (−) 𝐴𝑔 𝑚 𝑥 compare with SHM defining equation 𝑎 = −𝜔2𝑥 𝜔2 = 𝐴𝜌𝑔 𝑚 𝜔 = √𝐴𝑔 𝑚 M1 M1 A0 3d(i) Damping due to viscous forces B1 3d(ii) − 𝜔2 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 2⋅30 0.020 = 115 (2𝜋𝑓)2 = 115 f = 1.71 = 1.7 Hz C1 A1 3d(iii) 𝐸 = 1 2 𝑘𝑥2 = 1 2 𝑚𝑤2𝑥 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐸 = 1 2 (𝑂. 57)(115)(0.020)2 = 0.01311 𝐽 𝑓𝑖𝑛𝑎𝑙 𝐸 = 1 2 (0.57)(115)(0.016)2 = 0.0083904 J 𝛥𝐸 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐸 − 𝑓𝑖𝑛𝑎𝑙 𝐸 = 4.7 𝑥 10−3𝐽 OR accept if students use area under graph to final initial and final E read off first point (-0.02,2.3) gives initial E = 0.01311 J accept either: read off second point as (0.016, -1.8) gives final E = 0.008208, hence 𝛥𝐸 = 4.9 𝑥 10−3𝐽 read off second point as (0.016, -1.84) gives final E = 0.0083904 hence 𝛥𝐸 = 4.9 𝑥 10−3𝐽 C1 C1 A1 Total: 12

5 4a Zero electric field strengths in sphere A(between x = 0 and x = 1.4 cm) and in sphere B (between x = 11.4 and x = 12.0 cm) B1 4b Since the field strength is zero at a point between the spheres OR the electric fields are in opposite directions, the charges on the spheres are of the same sign. M1 A1 4c At x = 0.08 m, the electric field strength due to sphere A cancels out the electric field strength due to sphere B. EA = EB 𝑄𝐴 4𝜋𝜀𝑜 (0.08)2 = 𝑄𝐵 4𝜋𝜀𝑜 (0.04)2 𝑄𝐴 𝑄𝐵 = (0.08 0.04) 2 = 4 (Allow estimation from graph, 7.8 cm < x < 8.2 cm) B1 C1 A1 4d ● Correct field line direction (either all inwards or all outwards) and shape ● neutral point nearer to sphere B and any of the following: ● field lines more closely spaced for sphere B ● field lines perpendicularly from the spheres B1 B1 B1 4e change in electric potential = area under graph between x = 1.4 cm and x = 3.0 cm Estimated area between x = 1.4 cm and x = 3.0 cm = ½ (0.8 x 10-2)(11.0 + 5.0)107 + ½ (0.8 x 10-2)( 5.0 + 2.5)107 = 940 kV Or by counting squares under graph. Acceptable range for either method is between 850 kV and 1050 kV. Energy gained by proton = (940 x 103)(1.60 x 10-19) = 1.5 x 10-13 J Accept (1.4 to 1.6) x 10-13 J C1 C1 A1 Total: 12

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7 5a(i) (𝐸 = 𝐵𝐸𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐵𝐸𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑎𝑛 𝑡 𝑠) E = (BE per nucleon of C) x 12 – 3 x (BE per nuceon of He x 4) 𝐸 = [12 × (7.680) − 3 × (4 × 7.074)] =7.272 MeV C1 A1 5a(ii) 𝑛 𝑡 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑡 × 3 = (2.3635 × 1038)(3) 𝑛 𝑡 = 7.09 × 1038 ℎ𝑒𝑙𝑖𝑢𝑚 𝑛𝑢𝑐𝑙𝑒𝑖 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 C1 C1 A1 5b(i) 714𝑋 −10𝑒 B1 B1 5b(ii)1. Observation 1: electrons/-particles (emitted from the nucleus) have a (continuous) range of/ different (kinetic) energies OR Observation 2: electrons/-particles and daughter nuclei X are not in opposite directions B1 5b(iii)2. Explanation for observation 1: Since the energy released in each decay is same/fixed by the principle of conservation of energy there must be neutrinos (emitted) to take varying amounts of the (same total) energy (released in the decay) so that electrons can take different/range of kinetic energy OR Explanation for observation 1: By conservation of momentum there must be neutrinos (emitted) such that the sum of momentum of neutrino and electron is in the opposite direction of the daught er nuclei X direction of neutrinos varies so electrons can be emitted in varying directions B1 B1, (B1) (B1) 5c(i) tangent drawn and gradient calculation attempted activity = 1.3 × 106 Bq (accept answer within ±0.2 × 106 Bq) B1 A1 5c(ii) A = λN λ = (1.3 × 106 )/(3.05 × 1010) = 4.3 × 10–5 s –1 (≈ 4 × 10–5 s –1 ) B1 M1 A0 Total: 14

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9 6a distance moved by wavefront/energy during one cycle/oscillation/period (of source) B1 6b T = 2.0 × 2.5  (= 5.0 ms) f  = 1 / (5.0 × 10–3) = 200 Hz A1 6c(i) (incident) wave reflects at end/top of tube (in