2025 HCI C2 Prelim H2 Physics P3 Ans

HCI H2 Physics 9749 2025 C2 Preliminary Examination Paper 3 Suggested Solutions Q1 (a) (i) Vertical component of the ball’s initial velocity, uy = u sin θ = 25 sin 30° = 12.5 m s-1 = 13 m s-1 (ii) Considering the vertical component motion, At maximum height, the vertical component of the ball’s velocity will decrease to zero. Using vy2 = uy2 + 2aysy and taking upwards direction as positive, 0 = (12.5)2 + 2(-9.81)sy Vertical displacement (i.e. max height reached), sy = 7.96 = 8.0 m (iii) Initial total energy = 21 2K mu= At maximum height of 8.0 m, ball’s vertical component velocity, vy = 0. Its velocity will be its horizontal component velocity, vx = u cos 30 = 3 2 u kinetic energy at 8.0 m = 2 21 3 3 1 0.752 2 4 2m u mu K ==   Gain in potential energy of system = Loss in kinetic energy of ball Potential energy at 8.0 m = 3 0.254K K K−= OR: Initial total energy 2 2 12 2 KK mu m u=  = at 8.0 m, ( )( ) ( ) ( ) ( ) 22 2 9.81 8.02 0.25 2 s.f. 25 0.25 0.75 2 d.p. KPE mgh gh K Ku KE K PE K K K = = = = = − = − =

2 (b) (i) Correct graph Ep parabolic curve starts from zero, peaks at K/4 at the middle, ends at zero. Ek parabolic curve starts from K, decreases to 3K/4 at the middle, ends at K. Correct labels, clear markings, symmetry ` 0 x energy 0.5x Fig. 1.2 Ek Ep K parabolic horizontal distance

3 Q2 (a) ( )( )( ) 2 2 1 108 10001.20 0.30 2.502 60 60 405 N 1 2 ddF C Av   = = =  ( ) ( ) 2 220.05 0.02 0.05 405 1.20 0.30 2.50 108 70 N 1s.f. dd dd d d FC Av F C A v F F      = + + +  = + + + = Fd   Fd = 410  70 N (b) velocity of car A relative to car B, AB A Bv v v=− By Cosine Rule, ( )( ) 22 -1 40.0 50.0 2 40.0 50.0 cos135 83.2 km h ABv = + −  = By Sine Rule, sin sin135 50.0 83.2  = , hence 25.1 = direction: bearing = 90.0° + 25.1° = 115.1° 135° 40.0 km h-1 50.0 km h-1 θ

4 Q3 (a) Let mass of Planet Z be M, mass of argon molecules be m, Consider an argon molecule escapes from the planet’s surface to infinity. By conservation of energy, Loss in KE = Gain in GPE 2 3 2 1 002 2 42 3 8 ()3 surface surfaceKE KE U U GMmmv r GMv r Gr r G r shown   − = − − = − − =  = = (b) 2 -11 3 2 -1 8 3 8 (6.67 10 ) (413 10 ) (5500)3 724 ms v G r   = =   = (c) Assume that escape velocity v is equal to the root-mean-square speed crms of the argon molecules. By Kinetic Theory, average KE of one argon molecule = 3/2 kT Average KE of one mole of monatomic gas is therefore 3/2 RT. ( ) 2 max 2 max 2-3 13 where is th