TMJC_2023_H2 Physics_P1_MS

1 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2023 JC2 H2 Physics Preliminary Examination Paper 1 Suggested Solution 1 A 11 A 21 B 2 C 12 C 22 D 3 C 13 C 23 A 4 B 14 A 24 A 5 B 15 D 25 C 6 D 16 C 26 B 7 D 17 A 27 B 8 B 18 A 28 D 9 D 19 A 29 A 10 C 20 B 30 C 1. Ans: A 2 21 [][] [ ][ ] kg m s[ ] k g s AA m F BIL FB IL B − −− = = = = 2. Ans: C 6 7 10 13 42.195 km 42195 m 4.2195 10 cm 4.2195 10 mm 4.2195 10 m 4.2195 10 nm µ = = × = × = × = ×

2 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 3. Ans: C ( ) ( ) 2 2 2 2 2 At time : 1 2 1 2 At time 2 : 1(2 ) 2 2 11 24 422 t s ut at x gt t s ut g t g t gt x = + = = + = = =  Hence distance fall in 2nd time interval t = 4x – x = 3x 4. Ans: B () W N ma N W ma W −= = −< By Newton’s third law, the magnitude of the force on boy by floor is the same as the magnitude of the force on the floor by the boy. weight of boy, W force on boy by floor, N

3 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 5. Ans: B At equilibrium, CW ACW PQ PQ Tx Tx TT = = = PQ PP QQ P X QY TT WUWU W gV W gVρρ = −=− −= − Since XYρρ > , PQUU> since both has the same volume. Therefore, PQWW> , PQmm> . TQ TP x x P Q WP TP UP WQ TQ UQ

4 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 6. Ans: D Taking right as positive 2 22 222 2 222 Using conservation of momentum: 0 ------ (1) U sing conservation of KE: 1 11 02 22 ------ (2) (1 ) (2) : () 0 P PQ P PQ P PQ P PQ PQ P Q mu mv mv uvv mu mv mv uvv vv v v += + = + += + = + − + −+= 20 0 and PQ P QPP P vv v v uv u = ⇒= ⇒=−= Hence, P becomes stationary and Q moves to the right with the same speed as P. When Q collides with R, the same occurs, Q becomes stationary while R moves to the right with the same speed. 7. Ans: D Force exerted by the motor, F = 40 – 25 = 15 N Output power of motor, P = F v , where v is speed of rotation (at circumference) of wheel. Hence P = 15 [ 2π(0.080) x 10 / 1] = 75 W 8. Ans: B Tension of cord provides for the centripetal force. ( ) ( )( ) 2 2 2 0.80 3.242 0.72 0.27 m mvT r mvkx r x x = = = = Unstretched length = 0.72 – 0.27 = 0.45 m

5 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 9. Ans: D 22 22 Net forces acting on satellite, S E S E Fnet ma Mm MmG G ma Rr M MaG G Rr = −= = − 10. Ans: C Escape speed, 2 14 2 0.2281 1 9 m mE E Em v GMv r Mv r v Mr v Mr = ∝ = × = ×== 11. Ans: A 221 2 PE mx ω= 12. Ans: C From Graph 1, 0.45 0.15 0.30 mλ =−= From Graph 2, 31.6 1.2 0.4 ms 0.4 10 sT −=−= =× 1 3 0.30 750 m s0.4 10vf T λλ − −= = = = × 13. Ans: C Average wavelength of light = 550 nm. Using Rayleigh criter