2024 RI Prelim H2 Phy Paper 2 Answers

Raffles Institution Year 5-6 Physics Department 1 2024 H2 Physics Preliminary Examination Solution and Comments Paper 2 1 (a) (i) Take direction up the slope as positive. ( ) 22 0 2 0 0 2 0 7.0 2 9.81sin30 4.9949 4.99 m v u as s s =+ = + −  == OR By the principle of conservation of energy, increase in G.P.E. = decrease in K.E. ( ) 2 2 2 0 2 1 0, where is the max. vertical height fr om ground2 2 7.0 2(9.81) sin30 7.0 sin302(9.81) 4.9949 4.99 m mg h mv h vh g hs  = −  = = =  =   == (ii) *Both graphs aligned at the same total energy and 0s and clearly labelled. 1.  ( ) ( ) 22 2 11 2 sin22 1 sin2 K K E mv m u g s E mu mg s = = + −  = −  Graph of EK-s is a straight line with negative gradient and vertical intercept 21 2 mu . 2. ( ) ( ) ( ) sin sin P P E mg h mg s E mg s =  =  = Graph of EP-s is a straight line through the origin with positive gradient. energy EP EK s / m 0 0

Raffles Institution Year 5-6 Physics Department 2 (b) (i) ( ) 22 22 1 2 4.014.0 2 9.81sin30 sin30 10.841 10.8 m s (shown) v u as v v − =+ = + −    == OR decrease in K.E. = increase in G.P.E. ( ) ( ) ( )( ) 22 2 2 2 1 11 22 2 14.0 2 9.81 4.0 10.841 10.8 m s (shown) mu mv mg h v u g h v − − =  = −  = − == (ii) Take directions to the right and upwards as positive. Time of flight after the ball leaves the top of the slope to the ground: ( ) ( ) 21 2 214.0 10.8sin30 9.81 2 1.6080 s or 0.50713 s (NA) y y ys u t a t tt t =+ − =  + − =− Horizontal distance travelled: ( )( ) 21 2 10.8cos30 1.6080 0 15.040 15.0 m x x xs u t a t=+ =  + == *State both values of t and reject the negative one.

Raffles Institution Year 5-6 Physics Department 3 2 (a) The initial total momentum of both balls is not zero. Since there is no net external force acting on the balls as a system, by the principle of conservation of momentum, the total momentum of both balls must remain unchanged and cannot be zero. Hence, the balls could not be stationary at the same time. (b) By the principle of conservation of momentum, 4.0 ( 1.0) 3.0 ----- (1) A A B B A A B B A B A B AB AB m u m u m v m v u u v v vv vv + = + + = + + − = + += Since collision is elastic, ( 1.0) 4.0 5.0 ----- (2) B A A B AB AB u u v v vv vv − = − − − = − − =− 1 (1) (2) 2 3.0 ( 5.0) 4.0 m s (shown) B B v v − − = − − = (c) By Newton’s second law, the average force on ball B by ball A is ( )( ) , 0.50 4.0 1.0 0.25 10 N B net B pF t =  −− = = By Newton’s third law, the average force on ball A by ball B has the same magnitude of 10 N. OR From equation (1) or (2) in part (b), vA = 3.0 − vB = 3.0 − 4.0 = −1.0 m s−1. By Newton’s second law, the average force on ball A by ball B is ( )( ) , 0.50 1.0 4.0 0.25 10 N A net A pF t =  −− = =

Raffles Institution Year 5-6 Physics Department 4 (d) Balls A and B will exchange velocities and momenta as both balls have the same mass. From (b), ( )( ), 0.50 4.0 2.0 N sBfp == . Since duration of collision is 0.25 s, constant final momenta to start from 0.75 s to 1.5 s, with lines joining 0.50 s to 0.75 s during the collision. -3 -2 -1 0 1 2 3 momentum / N s t / s pA pB 0 1.0 0.5 1.5

Raffles Institution Year 5-6 Physics Department 5 3 (a) When the disc is just about to rotate, the contact force by the ground just becomes zero. Perpendicular distance from corner of box to line-of-action of F is 2 R . Perpendicular distance from corner of box to line-of-action of W is 2 2 3 22 RRR −= . Applying the principle of moments about the corner of box, 3 22 3 1.7321 1.73 RF W R F W  =  = = = (b) F acting at O needs to be inclined upwards such that it is at an angle above the horizontal to produce a clockwise moment about the corner to overcome the anticlockwise moment due to the weight. OR F acting at O needs to be inclined upwards so that there is an upward vertical component to produce a clockwise moment about the corner to overcome the anticlockwise moment due the weight. OR F needs to be shifted upwards above O so that there is a moment arm from the corner of the box to the line -of-action of F, to produce a clockwise moment about the corner to overcome the anticlockwise moment due to the weight. Note: Increasing the magnitude of the horizontal force F acting at O will not cause any rotation as F has no moment about the corner of the box because the perpendicular distance from the corner to the line-of-action of F is zero. F O R box ground disc force by ground force by box W

Raffles Institution Year 5-6 Physics Department 6 4 (a) (i) At the top of the circle, 2 T T mvT mg L+= For the ball to just complete the vertical circle, the tension TT at the top of the circle is zero. 2 2 T T T mvmg L v gL v gL = = = (ii) By the principle of conservation of energy, as the ball moves from the top to the bottom of the circle, it s gravitational potential energy decreases and its kinetic energy increases. This means the speed at the bottom is greater than the speed at the top. Hence 1B T v v  . (iii) 3 3 B T BT v v vv = = As the ball moves from the top to the bottom, increase in K.E. = decrease in G.P.E. ( ) ( ) ( ) 22 2 2 22 2 11 222 11 3222 91 222 42 1 2 BT TT TT T T mv mv mg L m v mv mg L mv mv mgL mv mgL v gL −= −= −= = = As 1 2 gL gL , where gL is the value of vT at which the string just goes slack, the ball will not be able to complete a full circle if 3B T v v = . Hence, 3B T v v = is not possible to achieve. (b) Considering forces along the radial direction, 2sinT mr= ( ) 2 2 Since sin , sin sin rL T m L T mL      = = = Since m and L are constants, 2T  . Hence when the angular velocity is doubled, the tension in the string is 4T.

Raffles Institution Year 5-6 Physics Department 7 5 (a) (i) Gravitational potential at a point in a gravitational field is the work done per unit mass by an external force in bringing a small test mass from infinity to that point. (ii) Gravitational potential at infinity is zero. Since gravitational force is attractive in nature, to bring a mass from infinity to a point in the gravitational field, the direction of the external force is opposite to the direction of displacement of the mass. This results in negative work done per unit mass by the external force. Hence, based on its definition, gravitational potential is a negative value. (b) (i) ( )( )( ) ( ) ( ) 11 24 3 37 10 10 11 116.67 10 6.0 10 1600 6400 10 6400 10 2.1 10 7.6681 10 7.67 10 J EE P E E E GM m GM mE xR GM m Rx −  = − − −   =−    =   −    +  =  =  (ii) 1. Gravitational force provides the centripetal force on the satellite. 2 2 where is the mass of the satelliteE E GM m mv mrr GMv r = = 2. By the principle of conservation of energy, if the satellite has just enough energy to escape to infinity, its total energy is zero. Let 1KE be the kinetic energy of satellite just after the boost. 1 1 1 0 0 PK E K E K EE GM m Er GM mE r += − + = = Just before the boost, kinetic energy of satellite in orbit, 2 2 1 2 1 2 2 K E E E mv GMm r GM m r =  =  = 1ratio 2 2 K EE K E GM m GM m rrE= = =

Raffles Institution Year 5-6 Physics Department 8 6 (a) (i) Effective resistance of Q and LDR, 1 1 11 11 6.0 8.0 3.4286 k eff Q LDR R RR − − =+ =+ = ( ) 1 3 3 3 9.0 3.4286 4.0 10 1.2115 10 1.21 10 A A T EI R −− = = + =  =  (ii) Potential difference across Q and LDR, 3.4286 9.03.4286 4.0 4.1539 V eff eff eff P RVE RR= + = + = 2 3 4 4 4.1539 8.0 10 5.1924 10 5.19 10 A LDR A LDR VI R −− = =