2023 JPJC Prelim H2 Phy P1 Solutions

1 2023/JPJC/Prelim/9749/01 Answers to 2023 JC2 Preliminary Examination Paper 1 (H2 Physics) Answer: 1 B 2 B 3 C 4 C 5 A 6 A 7 D 8 B 9 C 10 B 11 C 12 A 13 D 14 C 15 D 16 B 17 A 18 D 19 A 20 A 21 C 22 A 23 C 24 D 25 B 26 D 27 B 28 B 29 D 30 A Suggested Solutions: Qn Ans Solution 1 B R = σT4  4 R Tσ = Unit of 12 4 J s m Kσ −− = = (kg m2 s−2) s−1 m−2 K−4 = kg s−3 K−4 2 B Z = X − Y  X = Z + Y 3 C By conservation of energy, 22 fi 11 22mv mv mg h= +∆ Since, the magnitude of the loss in gravitational potential energy is the same for stones P and Q, with the same initial kinetic energy, the stones will end up with the same magnitude of kinetic energy, implying they will hit the ground with the same speed. 4 C Using 21 2s ut at= + , distance travelled in first 10 s is 24 m, ( ) 2 2 124 102 0.48 m s a a − = ∴= Distance travelled in the first 25 s, ( )( ) 21 0.48 252 150 m s = = Distance travelled from 10 st = to 25 st = 150 24 126 m = − = 5 A ( ) ( ) 2 For A: 2.0 ........... (1) For B: (5.0)(9.81) 5.0 ............ (2) Sub (1) into (2): 49.05 2 5 7.0 m s Ta Ta aa a − = −= −= =

2 2023/JPJC/Prelim/9749/01 Qn Ans Solution 6 A By principle of conservation of linear momentum, ( )2 3( ) ( 3)m v m v m mV+ −= + 1 4Vv=− The speed of mass m is 1 4 v after the collision. 7 D Weight of boat = upthrust = weight of fresh water displaced = 35.6 kN Weight of boat = weight of salt water displaced = mg = ρVg = 35.6 kN 33 3 35.6 10 35.6 10 (1024)(9.81) 3.54 m V gρ ××= = = 8 B Extension for spring X, x1 = 1 1.5 0.30 m5.0 F k = = Extension for spring Y, x2 = 2 1.5 0.15 m10.0 F k = = Elastic P.E. stored = 21 2 kx Total elastic P.E. stored = 1 2 (5)(0.3)2 + 1 2 (10)(0.15)2 J = 0.34 J 9 C Loss in G.P.E. by Y = gain in G.P.E. by X + gain in K.E. by X and Y ( ) 213.0 9.81 3.0sin40 2.0 9.81 3.0sin30 2.0 3.0 2 v×× ° = ×× ° + + 3.3v ≈ ms−1 10 B For mass m1: 2 1 2 1 () mv keLe ke L ev m =+ += For mass m2: 2 2 2 2 2 2 (2 2 )2( ) ( 2)2( ) 2 ( 2 )( ) mv kL eLLe mv kL eLe kL e L em v = +−+ = ++ ++= 1 1 2 ( 2 )( ) k( ) 2 ( 2) kL e L em eL e mL e e ++= + +=

3 2023/JPJC/Prelim/9749/01 Qn Ans Solution 11 C total energy at top of ramp = total energy at lowest point 2 2 1 mvmgh =  ghv 22 = By Newton’s second law of motion, at the lowest point, r mvmgR 2 = − ( ) mgr ghmR + =2 Since r = h, mgR 3= 12 A Gravitational potential at infinity is zero. Moving from −30 MJ kg−1 to −70 MJ kg-1 means moving closer to Earth. Potential energy change = 50 [−70 − (−30)] = −2000 MJ 13 D A is true: From pV = nRT, at constant pressure, when volume is doubled, temperature is doubled. B is true: work done W = p∆V = 4Po(2Vo − Vo) C is true: W = 0  ∆U = Q. From pV