RI 9749 2022 Physics TYS Solutions

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2022 GCE A-Level Suggested Solutions Paper 1 1 A Since X and Y are vectors, they can be positive or negative. Given: XY< initially: XY xy− = −− finally: ( )XY x y xy− = −− − = −+ xy xy−− > −+ Note: General equation ( ) ( ) 22 22cos sin 2 cosX Y x y y x xy y θθ θ− = −− + = + + As θ increases from 0° to 180° , XY− decreases. 2 C ( )100 cos 2.0 cos 100 -----(1) xxs ut ut u θ θ = = = ( )sin sin 2.0 2 sin 2.0 -----(2) yyyv u at uug ug θθ θ = + −= − = ( ) ( ) ( ) ( )1 2 2.0 9.81: tan1 100 2.0 9.81 tan 11. 1 11100 θ θ − = = = °= ° 3 B Since collision is elastic, ( ) 1 2.0 0 0.50 2.0 0.50 1.5 m s PQQP Q Q uuvv v v − −=− − = −− = −= By conservation of momentum, ( ) ( ) ( )2.0 0 0.50 1.5 2.5 1.5 1.5 3 2 3 2.5 5 2 5 P PQ QP PQ Q P PQ PQ P Q mu mu mv mv m mm mm m m +=+ += − + = = = = X Y X Y X Y vQ P Q

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 4 B Consider vertical equilibrium, u mg T T u mg = + = − ( ) ( )( ) ( ) 3 3 3 14 23 2 3 2 0.500 1030 9.81 200 9.813 683.3 683 N ww w w T V g mg r g mg r g mg ρ ρπ πρ π = − = ×− = − = − = = 5 D Consider horizontal equilibrium, sin36 sin36 ----- (1) TF T ke °= °= Consider vertical equilibrium, cos36 ----- (2)TW °= ( ) ( ) ( ) 1 sin36 2 cos36 tan36 tan36 25 0.060 tan36 2.065 2.1 N ke W ke W keW ° =° °= = ° = ° = = 6 C 22 GM G Vg rr ρ= = ( )( ) 2 2 3 11 6 3 4 3 3 9. 81 4 6.67 10 6.37 10 5512 5510 kg m gr GV gr Gr ρ π π − − = =   = × ×× = = T mg u T W F

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 7 B Maximum height after 2 oscillations, ( )0.10 2 0.20 2 0.60 0.60he e − −= = At maximum height, pendulum is momentarily at rest i.e. K.E. = 0 From initial height to maximum height after 2 oscillations, by conservation of energy, work done against air resistance = decrease in G.P.E. ( ) ( )( ) ( ) 02 0.200.40 9.81 0.60 0.60 0.4268 0.43 J mg h h e− = − = − = = 8 C Angular velocity is the rate of change of angular displacement, which is the same for every point on the disc. Hence angular velocity is independent of the distance from the centre of the disc. 9 C Gravitational force on the satellite (mass m) by the Earth (mass M) provides the centripetal force. 2 2 2 2 132 2 2 4 GMm mrr GMm mr Tr GMTr ω π π = =  =   Option A is incorrect: From the equation above, r is independent of the mass m of the satellite. Option B is incorrect: The orbital period of all geostationary satellites is 24 hours. Option D is incorrect: 2v r r vr T πω = = ⇒∝ Option C is correct: The orbits of all geostationary satellites are in the same plane as the equator of the Earth and they orbit West to East, similar to the Earth’s rotation. 10 D Gravitational