RVHS H2 Physics P2 Soln

River Valley High School Page 1 of 9 J2 H2 Physics 9749 Preliminary Examinations RVHS JC2 H2 Physics Preliminary Examinations Paper 2 Mark Scheme Questions Answers Marks 1 (a) Impulse is defined as product of (average) force exerted on wall and time of impact B1 (b) Δ𝑝𝑝 = 0.058 × (−34 − 34) M1 = −3.944 kg m s−1 A1 (c) impulse = change in momentum = area under F – t graph M1 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 × 4 1000 = 3.944 giving 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 = 986 = 990 N (2 s.f.) A1 (d) From t = 0 to t = 2 ms: decrease at increasing rate. From t = 4 ms to t = 6 ms: decrease at decreasing rate. B1 From t = 2 ms to t = 4 ms: decrease at constant rate. B1 after t = 6 ms, velocity at −34 m s-1 B1 Example Questions Answers Marks 2 (a) molecules has component of velocity in 3 directions, hence 𝑐𝑐2 = 𝑐𝑐𝑚𝑚2 + 𝑐𝑐𝑦𝑦2 + 𝑐𝑐𝑧𝑧2 ------- (1) M1 their motions are random, so averaging gives 〈𝑐𝑐𝑚𝑚2〉 = 〈𝑐𝑐𝑦𝑦2〉 = 〈𝑐𝑐𝑧𝑧2〉 --------- (2) M1 Hence substitute the result of (2) into (1), then 〈𝑐𝑐2〉 = 3〈𝑐𝑐𝑚𝑚2〉 A1 velocity / m s-1 0 2 4 6 t / ms 34 −34

River Valley High School Page 2 of 9 J2 H2 Physics 9749 Preliminary Examinations Hence for N molecules, we have 𝑝𝑝𝑝𝑝 = 1 3 𝑁𝑁𝑁𝑁〈𝑐𝑐2〉 A0 (b) (i) pressure 𝑝𝑝𝑝𝑝 = 𝑛𝑛𝑛𝑛𝑛𝑛 or 𝑝𝑝 = 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿3 = 3×8.31×(20.0+273.15) 0.2003 M1 so force on each sides 𝐹𝐹 = 𝑝𝑝𝑝𝑝 = 3×8.31×(20.0+273.15) 0.2003 × 0.2002 M1 𝐹𝐹 = 36 500 N A1 (ii) particles would attract one another B1 so average force on wall will decrease B1 3 (a) (i) 𝜔𝜔 = 2𝜋𝜋𝜋𝜋 [M1] → 𝜋𝜋 = 1 2𝜋𝜋 � 10.5 0.450 = 0.76879 ~ 0.77 Hz [A1] (ii) From 50 cm to 80 cm position, a change of 0.300 m: 𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥 = 𝑀𝑀𝑀𝑀(𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑛𝑛𝑒𝑒 𝑒𝑒𝑒𝑒 𝑛𝑛) = 0 .450 × 9.81 × 0.300 = 1.32435 J [M1] W hen mass was initially held at 50 cm position, spring was already experiencing EPE, as unstretched length of spring is 32 (or 32.5 cm), therefore initial EPE is due to e = 18 cm. When mass ends up at 80 cm position, spring would have stretched by e = 48 cm. 𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥 = 1 2 𝑘𝑘(𝑒𝑒22 − 𝑒𝑒12) = 1 2 (10.5)(0.4802 − 0.1802) = 1.0395 J [M1] So, at the position of 80 cm, the KE present in the mass would be: 𝛥𝛥𝛥𝛥𝛥𝛥 = 1.32435 − 1.0395 = 0.28485 J [C1] 0.28485 = 1 2 𝑁𝑁𝑣𝑣2 → 𝑣𝑣 = 1.12517 ~ 1.13 m s−1 [A1]

River Valley High School Page 3 of 9 J2 H2 Physics 9749 Preliminary Examinations When considering energy conversion for a vertical spring-mass system, there are GPE, EPE and KE. This is at a position close to the maximum displacement. (b) (i), (ii) Fig. 3.3 Labels – Displacement of +/- 24.0 cm, 2 cycles of period 1.3 s (for Line W) [B1] When spring-mass is at equilibrium, reading off the graph, the length of the spring should be 74.0 cm. Given the start length of the spring is 50.0 cm, therefore the displacement of free oscillations will be 24.0 cm. Line W –