ASRJC 2022 Prelim P1 Guide

1 9749/01/ASRJC/2022PRELIM [Turn Over Anderson Serangoon Junior College 2022 JC2 H2 Physics Prelim Mark Scheme Paper 1 (30 marks) 1 2 3 4 5 6 7 8 9 10 B C A C D B B B A A 11 12 13 14 15 16 17 18 19 20 C B C B A D D B B A 21 22 23 24 25 26 27 28 29 30 B C D B B C B A C C 1 B Replace J with SI base unit of energy kg m2 s–2 in the original expression J kg–1 K–1 will give m2 s–2 K–1. 2 C In the calculation of fractional uncertainty of 3X, the coefficient 3 will cancel out, also, for calculation of uncertainty, we always add and not subtract because the uncertainty should increase with more terms involved rather than decrease, so the final answer is x + 2y. 3 A All responses show a flat constant v at the initial distance travelled. For the next part where a is constant, using v2 = u2 + 2as, we see that when we plot v vs s, we get a square root curve, so A is the answer. Alternatively, 1gradient of graph = =  = dv dv dt ads dt ds v As v approaches zero, 1 v approaches infinity, so the gradient approaches infinity. 4 C Since we are using the same scales, it is expected that the initial velocity is the same as the air resistance is negligible when the velocity is small. 5 D Consider horizontal direction, the change of momentum = p cos θ – p cos θ = 0 So average horizontal force = 0 Consider vertical direction, Change in momentum = p sin θ – (– p sin θ )= 2p sin θ Average vertical force = 2 sindp p dt t = 6 B Moment by weight about bottom left corner of book = Wx/2 clockwise

2 9749/01/ASRJC/2022PRELIM To avoid rotation (i.e. achieve rotational equilibrium), the resultant moment must be zero about bottom left corner of book. Hence, student must provide Wx/2 anticlockwise 7 B rate of increase of kinetic energy = Fnet × v = mav = (250000)(0.90)(5.0) = 1.1 MW Alternatively, output force of train’s engine, F, can be found using N2L: Fnet = ma F – 15000 = 250000(0.90) F = 240000 N output power of the train’s engine = Fv = 240000 × 5.0 = 1 200 000 W However, some of this output power is used against resistive force, while the rest accelerates the train and increases the kinetic energy. Hence, rate of increase of kinetic energy = 1 200 000 – 15000(5.0) = 1.1 MW 8 B T = 15.9 days = 15.9 × 24 × 3600 s r = 1.22 × 109 m centripetal acceleration = rω2 = (1.22 × 109)(2π/(15.9 × 24 × 3600))2 = 2.55 × 10−2 m s−2 9 A Since v = rω, and ω is constant and the same for every point on the record, v is then prop ortional to r. 10 A For the molecule to escape, total energy at Earth’s surface must at least be zero (min energy at infinity). Hence, − GMm r + 1 2 mv2 = 0 ➔ v = √ 2GM RE Therefore, vnew = √ 2GM 3RE = 1 √3 √ 2GM RE = 1 √3 × 1.1 × 104 = 6.4 × 103 m s−1 Alternatively, use ratio method  1v r , therefore = originalnew original new rv vr