2024 JPJC H1 Prelim P1 Solutions

Answers to 2024 JC2 Preliminary Examination H1 Paper 1 1 C 6 C 11 B 16 D 21 B 26 D 2 B 7 C 12 C 17 A 22 D 27 D 3 B 8 A 13 C 18 B 23 A 28 D 4 A 9 C 14 A 19 B 24 D 29 B 5 A 10 A 15 A 20 C 25 A 30 C 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 nano should be 10−9 Answer: B 3 Answer: B 4    2 2Using 2 0 2(9.81)(2.5)v u as   1 7.0 m sv 21 2 21 2 Using 0.12 7.0 (9.81) 0.0167 s s ut at t t t        Answer: A change in velocity

2 2024/JPJC/PHYSICS/8867 5   21 2Horizontally, using s ut at 900 (450 cos31.6 ) 0 2.35 s t t               21 2 21 2 Vertically, using (450 sin31.6 )(2.35) ( 9.81)(2.35) 527 m s ut at s Answer: A 6 In throwing vertically upward, the ball has the same horizontal velocity as the thrower. So the ball would be caught if the thrower maintain a constant velocity. Answer: C 7 At equilibrium, force by air on four propellers = weight of drone        air drone4 0 dmv m g dt          air4 0 0.40 1.2 9.81v  1 air 7.4 m sv Answer: C 8 Let T be the tension in the spring. At equilibrium,    0.30 9.81T Immediately after the thread is cut,             2 0.30 9.81 0.20 9.81 0.20 4.9 m s T mg ma a a Answer: A 9 Momentum has to be conserved since there is no external net force acting on the system (comprising the two asteroids). As the collision is inelastic, some of the kinetic energy of the system could be converted to other forms of energy such as thermal energy which raises their temperature. Answer: C 10 Torque of couple  5.0 (0.60sin40 ) 1.9 N m   Answer: A

3 2024/JPJC/PHYSICS/8867 11 Area under the F-x graph =   2 1 2 1 1 2 T T x x  Answer: B 12 Resolving the forces,        2 1 2 1 1 sin30 (1) cos30 (2) (1): tan30(2) tan30 15tan30 8.66 8.7 N (2s.f.) N N N W N W N W    1 (Alternative: Using vector triangle) tan30 15tan30 8.66 8.7 N (2s.f.)N W Answer: C 13 By conservation of energy, loss in K.E. = work done against frictional force.      2 22 1 02 800 30 7200 N2 2 50 mu fd muf d      Answer: C 14 v u at gt     222 21 1Kinetic energy 2 2 2 mgmv m gt t         . This is given by Option A or C. Based on conservation of energy, Potential Energy = Total Energy – Kinetic energy Answer: A N1 W N2 N1 W N2 30° 30°

4 2024/JPJC/PHYSICS/8867 15 Usin