2023 HCI H2 PH Prelim P2 SS

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 1 2023 H2 Physics Preliminary Examination Paper 2 Suggested Solutions Q1 Suggested Solutions Mark 1 (a) In the direction normal to the slope, the block is in equilibrium. net force on the block = 0 N thus, Normal contact force by the scale on the block, N = weight component of the block normal to the slope = 1.6 g cos 30 The scale measures the normal contact force exerted by the block on the scale, that is equal and opposite to the normal contact force by the scale on the block. (Newton’s 3rd Law of Motion) Thus, the reading on the scale = normal contact force by the block on the scale/ g = 1.6 g cos 30 / g = 1.39 or 1.4 kg. B1 B1 B1 (b) (i) Consider the block moving from the bottom to maximum height: Let the velocity of the block immediately after collision be v. By conservation of energy, Loss in KE = Gain in GPE ( )( ) 2 -1 1 02 2 2 9.81 1.3 1.3cos37 2.3 m s mv mgh v gh −= = = −  = B1 B1 (ii) applying N2L on the block: ( ) ( )1.6 2.3 0 0.2 18.4 N m v uF t −−==  = B1 A1 30 wooden block weighing scale normal contact force by scale on block, N weight of block, W = 1.6 g W⊥ = 1.6 g cos 30 Normal contact force by block on scale, N’ 37 1.3 m h

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 2 OR variations: ( )J F t m v u=  = − ; ( )vuF ma m t −==  (iii) By conservation of linear momentum, mballuball = mblockvblock + mballvball uball = (mblockvblock / mball ) + vball = ( ) -11.6 2.3 19 44.4 m s0.058 −= Or ( ) ( ) on ball -1 -1 0.058 1918.4 0.2 44 m s i.e. v = 44.4 m s opposite to the final velocity. m v uF t u u −=  −= =− B1 A1 (c) Relative speed of approach = 44 m s-1 -1 h R 1elat ive s speed o c f separ 2.3 19.1 2 .4 m rel p ati e on ative s e d of approa = + =  Thus, the collision is not elastic. OR Initial KE = ½ (0.058)(44)2 = 56 J Final KE = ½ (0.058)(19)2 + ½ (1.6)(2.3)2 = 14.7 J Since total KE s not conserved/ initial KE is greater than final KE, the collision is not elastic. B1 A1 1.6 kg before after 58 g u 2.3 m s-1 1.6 kg 58 g 19 m s-1 0.058 kg u 19 m s -1 F

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 3 Q2 Suggested Solutions Mark 2 (a) Newton’s Law of Gravitation states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. B1 (b) (i) The gravitational force on the satellite is always directed towards the centre of the Earth and so any circular orbit must have its centre at the centre of the Earth. The orbit must be in the equatorial plane, otherwise the satellite will sometimes be over the northern hemisphere and sometimes over the southern