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