2021 GCE A Level 9814 H3 P1_SS [HCI]

1 2021 H3 Physics A Level Paper Suggested Solutions 1a) In vacuum, the object will fall at a constant acceleration of g. Taking direction downward as positive, and applying the kinematics equation s = ut + 1 2at2. 100 = (0)t + 1 2 (9.81)t2 t = 4.5152 = 4.5 s (2 s.f.) [1] 1bi) time-of-flight t’ = 2 x 4.5152 = 9.0304 s Taking direction downward as positive, and apply the kinematics equation s = ut + 1 2at2. s = (0)(9.0304) + 1 2(9.81)(9.0304)2 = 399.96 m = 400 m (2 s.f.) [1] 1bii) Instead of dropping the object from the top of the tower, the same double time- of-flight could be achieved by throwing the object up from the ground of the tower with the same speed as the speed of the object just before it reaches the ground from its fall from rest as indicated in (b)(i). [1] 1cii) Let h be the initial height of the safety net and vi be the final velocity of the ‘weightless’ (free-fall) phase and initial velocity at the safety net phase. Final speed at the safety net phase is zero. Taking direction downward as positive, and apply the kinematics equation v2 = u2 + 2as For the ‘weightless’ phase, vi2 = 02 + 2 (g)(100 – h) …… (1) For the safety net phase, 0 = vi2 + 2(-4g)(h) ⇒ vi2 = 8gh……(2) Subst (2) in eqn (1) 8gh = 2(g)(100 – h) ∴ h = 20 m (to 2 s.f.) [1] [2] [1] top of tower bottom of tower safety net h 100-h

2 1ci) Taking direction downward as positive, and apply the kinematics equation s = ut + 1 2at2. For the ‘weightless’ phase, (100 – 20 ) = (0) + 1 2(9.81)t2 t = 4.0386 s Hence the ‘weightless’ time of flight is reduced by 4.5152 - 4.0386 = 0.4766 = 0.5 s (1 s.f.) [1] 2a) By PCOLM, 4 4 44 44 mm mm mv mv mv v v v =+ =− By RSOA = RSOS, 44m m m mv v v v v v= − → = + 4 4 4 4 4 4 3 5 3 / 5 8 / 5 m m m m m v v v v v v v v vv + = − → = → = = % of kinetic energy transferred to m, 2 2 18 6425 1 4(25)42 0.64 64% vm mv   = =→ Method II: Using zero-momentum frame (center-of-mass frame) 4 0 4 55 cm mv m vv m +== The velocities of particle in the zero-momentum frame before the collision, 4 , 4 , , , , , 4 55 440 55 m cm m Earth Earth cm m cm m Earth Earth cm vvv v v v vvv v v = + = − = = + = − =− After the collision, the sign of velocities will change, hence, 4 4 and 55 mm vvvv =− = Rewriting the velocities after the collision in the Earth frame, 4 , 4 , , , , , 43 5 5 5 4 4 8 5 5 5 m Earth m cm cm Earth m Earth m cm cm Earth v v vv v v v v vv v v = + =− + = = + = + = % of kinetic energy transferred to m, [1]