ACJC 2025 J2 H2 Physics Prelim P1 Guide

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 1 of 6 Qn Ans Discussion 1 C The object travels at the same speed. Hence, fivv = . ( ) fi fi v v v vv  = − = + − 2 A For the upward motion, air resistance and gravitational force are both acting downwards, acceleration is greater than the acceleration of free fall and hence it decelerates faster to reach the maximum height. As for the downward motion, air resistance and gravitational force are in opposite direction s and the resultant force equals to gravitational force minus the air resistance. Acceleration is smaller than the acceleration of free fall and hence it accelerates slower downwards than it decelerates upwards. 3 D Consider cart 3 and cart 4 as single object of mass 800 kg. Only T is acting on this object. F = ma T = 800 (2) = 1600 N 4 A Assuming the mass of each ball is m. Since total initial momentum = mv By Conservation of linear momentum, total final momentum should be mv A) Total final momentum = 0 + mv = mv B) Total final momentum = 22 mv mv mv+= C) Total final momentum = 31 4 4 2 mv mv mv−= D) Total final momentum = mv – mv = 0 Only A and B’s momentum are conserved. Elastic means relative velocity of approach = relative velocity of separation By relative velocity of approach and separation, KE not conserved for option B. 5 A By conservation of energy, elastic potential energy is converted to kinetic energy and work done against friction. As height is the same at initial and final positions, gravitational potential energy is unchanged. ( ) ( ) friction 2 2 1 2 1 since 02 loss gain final initial final initial EPE KE WD kx KE KE fd KE kx fd KE =+ = − + = − =

Anglo-Chinese Junior College 2025 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2025 Page 2 of 6 6 D 2 2 29.30.52 9.811.3 39.7 40 N mvN mg r mvN mg r −= =+ =+  =  7 A 11 2 2 2 11 1 (6.67 10 )(30)Resultant 2 cos 45 ( 10 10 ) 1.4 10 N kg upwards g − −− =   + = 11 22 10 1 (6.67 10 )(30)Resultant 2 ( 10 10 ) 2.8 10 J kg  − −− = − + =−  8 C k p p at 3.2 MJ2 at 6.4 MJ at 2 3.2 MJ2 GMmEr R GMmEr R GMmEr R == = − = − = − = − 9 B Since they have the same volume and pressure initially, they must have the same T. Therefore when the partition is removed, the P and T remains the same. 10 B Since temperature is the same, the average kinetic energy is the same. 21 2 kE m c= 2 , , 1 1 2 0.71 rms X Y rms Y X c m c m cm  = = = 11 B U Q W = + For path 1, (