2024 JPJC H2 Prelim P1 Solutions - Printed

[Turn over Answers to 2024 JC2 Prelim Exam Paper 1 (H2 Physics) 1 C 6 B 11 B 16 C 21 D 26 A 2 A 7 C 12 C 17 D 22 B 27 A 3 C 8 A 13 A 18 B 23 A 28 B 4 B 9 D 14 D 19 C 24 D 29 C 5 A 10 A 15 B 20 D 25 B 30 A Suggested Solutions: 1 Assume that the diameter of the cross-section of the wire in a paper clip is 1 mm. Cross-sectional area of the wire in a paper clip  −−= =    2 7 2 7 20.001 7.85 10 m 8 10 m2 Answer: C 2 22 1 Using 2 0 2(9.81)(2.5) 7.0 m s v u as v − = + = + →= =+ → = + →= 21 2 21 2 Using 0.12 7.0 (9.81) 0.017 s s ut at tt t Answer: A 3 Applying conservation of momentum, ( )1 1 2 2 1 1 2 2 12 21 0 m v m v m v m v vm vm = + − = = Answer: C 4 By resolving each forces into its vertical and horizontal components, only option B is most likely to be in equilibrium. Answer: B

2 2024/JPJC/PHYSICS/9749 5 Using conservation of energy, loss in G.P.E. = gain in K.E. + work done against frictional force ( ) ( ) ( ) ( )( ) ( ) =+ =+ = 2 2 1 2002 160 50 60 20 2002 87 N mgh mv f gf f Answer: A 6 For circular motion at the highest point, 2 2 2 2 3 3 2 3 2 3 mvmg N r W mvW r mg mvmg r mg mv r grv −= −= −= = = Answer: B 7 −=gcF N F −=  2 2 2GMm N mR TR Since N = 0, =  2 2 2GMm mR TR = 3 2 RT GM Answer: C

3 2024/JPJC/PHYSICS/9749 [Turn over 8 acceleration, = 0 sina a t 0 0 0 0 displacement, sin 2 sin 3 2 sin 3 0.87 x x t Tx T x x    =−   =−      =−  =− Answer: A 9 A and B are correct as they are valid assumptions in the simple kinetic theory of gases, i.e. how ideal gases particles should behave. C is correct as the total energy of the system must remain constant due to the system being isolated. D is not correct – given that the initial momentum of a molecule before collision is p, there will be a change in momentum of 2p because the final momentum of the molecule will be –p (same magnitude, opposite in direction). Answer: D 10 The pressure of a gas is derived from the forces of collision between the molecules and the container walls, and not intermolecular collisions nor the energy transfer to the walls. C will lead to a decrease in pressure instead, as the time interval between collisions will increase. Answer: A 11 Using the equation Q Tmc=  2 2 1 2 3 3 8 4 mcmv T vT c  = = Answer: B 12 Phase difference   =  = 5.0 sin30 210 1.6 rad Answer: C

4 2024/JPJC/PHYSICS/9749 13 By Malu’s law, cosfAA = Since 2AI , 2 2 2)( cos cosf k A kA ==I Since power is proportional to intensity, power is proportional to 22cosA  . Answer: A 14 In order to get destructive interference at Q, the path