2024 JPJC H2 Prelim P2_solutions_Printed

2024/JPJC/Prelim/9749/02 Answers to 2024 JC2 H2 Preliminary Examinations Paper 2 Suggested Solutions: No. Solution Remarks 1(a) 22 22 f A vk k f A v = = I I I 2 2 3 2 3 2 2 1 3 3 S base units of W m kg m s m kg m s m m s s m k − − − − − − −= = = [1] for correct units of frequency and amplitude [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 11 33 11(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)(i) ( )( ) ( )( ) 33 4 Applying Newton's second law, 4 10 9.81 4 10 0.32 4.05 10 N T T −  =  = [1] for correct substitution [1] for answer 2(a)(ii)1. ( ) ( ) ( ) ( ) 2 mass per unit time vol per unit time 1.3 10 408 410 Av v v v    = = =  = = [1] for correct substitution [1] for correct intermediate value 2(a)(ii)2. Applying Newton’s second law, ( ) 4 1 9.1 410 14.9 10 ms net dp dmFv dt dt vv v −  ==  = = [1] for correct substitution [1] for correct answer

2 2024/JPJC/Prelim/9749/02 2(b) For vertical equilirium, ( )( ) ( ) 3cos65 4 10 9.81 --- 1T mg = =  Horizontally, ( ) ( ) ( ) 33 33 sin65 6 10 4 10 sin65 4 10 6 10 --- 2 T ma a Ta −  = =  =  +  ( ) ( ) 3 22 2 6 10: tan651 19.5 m s 20 m s ma mg a −− += == [1] for correct substitution [1] for correct answer 3(a)(i) The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force. [1] 3(a)(ii) Let N be the number of 5.0 g mass needed for equilibrium. Taking moments about the edge of the table, anti-clockwise moments = clockwise moments ( ) ( ) ( ) ( ) ( ) ( )1.2 0.05 0.11 0.40 0.005 0.75 4.3 g g N g N =+ = Since N is a whole number, the maximum number is 4. [1] correct substitution [1] correct answer 3(b)(i) [1] for T 1.0 kg 1.2 kg 0.75 m string pulley 0.10 m table T 65° mg T air resistance

3 2024/JPJC/Prelim/9749/02 [Turn over 3(b)(ii) Taking moments about the edge of the table, anti-clockwise moments = clockwise moments ( ) ( ) ( ) ( ) ( ) ( ) ( )1.2 0.05 sin60 0.40 0.11 0.40 1.0 0.75 0.346 7.2 20.8 N 21 N g T g g T T +  = + = == [1] correct substitution [1] correct answer 4(a) Since the direction of motion of the body is always changing, its velocity is not constant and thus a resultant force acts on the body. Since the speed of the body is constant, the resultant force acts perpendicular to its velocity and thus, direction of motion, which is towards the centre of the circle. [1] [1] 4(b)(i) The net force towards centre of the