2024 JPJC H1 Prelim P2 solutions

2024/JPJC/Prelim/8867/02 Answers to 2024 JC2 H1 Preliminary Examinations Paper 2 Suggested Solutions: No. Solution Remarks 1(a)   4 4 P kAT Pk AT I      2 3 3 4 2 4 2 4 S base units of kg m sW kg s K m K m K k [1] for correct base units of power [1] for correct answer 1(b) 3 3 133 84 10 53.133 m s0.56 P v b Pv b      3 3 1 1 1 1 3 3 1 1(0.05) (0.07)53.133 3 3 2.1253 m s 2 m s P v b Pv b v P b v P b v v v                [1] for correct value of v [1] for correct substitution [1] for correct answer to 1 s.f. 2(a) Using 21 2s ut at  for vertical direction 21 2 11 2 0 (1.5) ( 10)(1.5) (10)(1.5) 7.5 m s u u       OR Using v u at  for reaching maximum height, 1 2 11 2 0 ( 10)( 1.5) (10)( 1.5) 7.5 m s u u         [1] substitution [1] answer 2(b) [1] line starts at 0 m [1] ends at 15 m w.r.t. label at y-axis 1.5 t / s Sh / m 10 20

2 2024/JPJC/Prelim/8867/02 2(b) [1] horizontal line at 10 m s1 w.r.t. label at y-axis 2(b) [1] line starts at 7.5 m s1 and cuts x-axis at 0.75 s [1] ends at 7.5 m s1 w.r.t. label at y-axis 3(a) The rate of change of momentum of a body (system) is proportional to the resultant force that acts on it and the momentum change takes place in the direction of the force. [1] 3(b) Impulse is defined as the product of force and time of impact. [1] 3(c)(i) [1] for drawing and labelling   2 sin35M m g Or in words [1] for drawing and labelling T 3(c)(ii) Trolley A:       2 sin35 2M m g T M m a ------ (1) Trolley B:   sin35T Mg Ma ------(2) (1)+(2):       2 sin35 sin35 2 2M m g Mg M m a   2 sin35 2 2mg M m a [1] for correct equation [1] for correct equation [1] for showing 1.5 t / s 1.5 t / s vh / m s1 vv / m s1 10 10 10 35° T (M + 2m) g sin 35°

3 2024/JPJC/Prelim/8867/02 [Turn over   sin35mga M m      2 sin35 2 2 mg M m a 3(c)(iii)          225 30 9.81 sin35 0.338 m s100 95 30a    219.0 0.3382 7.3 s t t [1] for correct acceleration [1] for correct substitution [1] for correct answer 3(c)(iv) The flexible buffer stops the trolley over a prolonged period of time. Applying Newton’s second law, this means that the rate of change of momentum of the trolley is reduced . The resultant force acting on the trolley is therefore reduced. [1] [1] 4(a) [2] all correct Or [1] any 3 correct (ignore length of arrows) 4(b)(i) Taking moments about point A, sum of anti-clockwise moments = sum of clockwise moments 4.0 4.0 9.81 1.0 25.0 9.81 2.0 60.0 9.81 3.6 662 N B B T T            [1] sub [1] ans 4(b)(ii) Sum of forces acting on scaffold = 0 N Upward forces = downward forces 4.0 25.0