2025 HCI C2 Prelim H2 Physics P2 Ans

1 HCi H2 Physics 9749 2025 C2 Preliminary Examination Paper 2 Suggested Solutions Q1 (a) ( ) forcePressure at depth in the fluid area of base of container weight of fluid column area mass of fluid column acceleration of free fall density volume Fph A W A mg A Vg A Ahg A gh    = = = = = = (b)(i) Let the volume of the object be V. Resultant force F = weight of object mg – upthrust U ( ) 1 1 F mg U Vd g V g Vdg d mg d    =− =− =−  =−  (b)(ii) At equilibrium, taking moments about the pivot, Clockwise moment due to sample = anticlockwise moment due to standard mass (Resultant force on sample) x = (Resultant force on standard mass) x ( ) 11 11 1.29 1.290.17851 1 1 8493 940.0 0.17873 kg (5 d.p.) s s s s mg m gdd mm dd   − = −    = − −     = − −        =

2 Q2 (a)(i) The mass passes through the equilibrium position P twice in a cycle, meaning that there are 100 complete cycles per minute. period, T = 1/f = 60/100 = 0.600 s (a)(ii) At equilibrium position x = 0, the kinetic energy of the mass Ek is at its maximum.  = 2 T  = 2 0.600  = 10.47 rad s-1 Ek = ( ) 2 2 2 211 (0.42) 10.4722 oom x x = Amplitude 2 2(0.500) 0.147 m = 14.7 cm = 15 cm (2 s.f.)0.42(10.47) ox == (a)(iii) Ek = 2 -121 2(0.500) =1.54 m s2 0.42 kEmv v m = = • Sinusoidal waveform • Maximum velocity of 1.54 m s-1 and time period of 0.60 s *No information is given whether mass starts oscillating from its equilibrium position or at its amplitude – sine or cosine graph (b)(i) No change (b)(ii) Frequency multiplied by 2 .

3 Q3 (a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point (without a change in kinetic energy) (b)(i) To just reach the neutral point from the Earth, gain in gravitational potential energy GPE = Loss in kinetic energy ( ) 8 7 -1 7 neutral point earth surface 7 neutral point earth surface 76 neutral point 6 -1 neutral point 10.0 6.10 10 J 6.10 10 J kg 6.10 10 6.10 10 6.10 10 62.3 10 1.3 10 J kg (2 s.f.) U m KE        =  =−  =   =  − =  =  + =  + −  =−  OR ( ) earth surface earth surface neutral point n eutral point 86 neutral point 6 -1 neutral point total energy at Earth surface = total energy at neutral point 6.10 10 10 62.3 10 0 10.0 1.30 10 J kg (3 s KE U KE U   + = +  + −  = −  =−  .f.) (b)(ii) The rock from the Moon must have enough energy to go past the neutral point, then resultant gravitational force of the Earth-mass system on the mass will accelerate it to the Earth. gravitational potential difference between the Moon’s surface and the neutral point ( ) neutral point moon surface 66 61 1.30 10 3.90 10 2.60 10 J kg    −  = − =−  − −  = Hence, the minimum kinetic energ