2021 A Level H2 Physics P2 Soln

1 (a) The block will accelerate down the slope (with a constant acceleration sin a g θ= directed along the slope) with increasing velocity. 1 (b) (i) Tangent must be drawn accurately at t = 20 s. 10.33 0 1.27 m s 0.36 0.1 v −−= = − 1 (b) (ii) Range of acceptable a values −− − = = = 21.27 0 6.35 m s 0.20 v u a t OR ( ) ( ) − = + = + = 2 2 2 2 2 1.27 0 2 0.13 6.20 m s v u as a a OR X (0.1,0) X (0.36,0.33)

( ) − = + = + = 2 2 2 2 0.5 0.13 0 0.5 0.20 6.50 m s s ut at a a 2 (a) Momentum just before contact with ground = 3.2 N s ( ) ( ) 22 K 3.2 8.26 J 2 2 0.62 pE m= = = 2 (b) Momentum is a vector and direction has to be taken into account. Based on the graph, downward is positive. Momentum changed from 3.2 N s to −1.8 N s. Time taken is from 0.53 s to 0.68 s. ( ) ( ) − − ∆= = = − ∆ − 1.8 3.2 N2L: Net force acting on the ball 33.3 N 0.68 0.53 p t (i.e. 33.3N upward) Magnitude of the average (normal contact) force the ground exerts on the ball = F net + mg = 33.3 + (0.62)(9.81) = 39.4 N (upward) 2 (c) ( )   = = =     < > 2 final 2 2 initial percentage of kinetic energy remaining after each b ounce 1.8 2 0.31641 3.2 2 Let number of bounces before energy drop s to less than 5% of initial be 0.31641 0.05 ln0.05 ln0.316 N p m p m N N = 2.60 41 Therefore, 3 bounces before energy drops below 5% of the initial energy. 3 (a) ( ) ( ) ( ) Taking moments about Q, 2 1 sin43 PQ cos58 3 2 3 cos58 3cos58 4 2.3 9.81 13.1 N sin43 4sin43 F mg PQ F mg     ° = °         ° °= = × = ° ° Comments: studnets have difficulty resolving F and W to obtain the component that was perpendicular to the line PQ. 3 (b) Force F has a horizontal component towards the right while W has no horiztontal component.

For the system to be in equilibrium, the contact force at Q should have a component directed towards the left to ensure that the sum of the forces in the horiziontal direction must be zero. Hence, contact force at Q needs to be at an angle to the left of the vertical. Comments: studnets have incomplete answres when they only state that F has a horizontal component. Answers were vague when it only talks about the need of forces to balance or cancel out W and F. 4 (a) 2Newton's law of gravitation: GMm F r= Since gravitational field strength is the gravitational force per unit mass placed at that point: 2 F GM g m r = = 4 (b) Orbit: involves circular motion. Gravitational force provides for centripetal force. ( ) 2 2 2 2 2 3 2 11 2 311 24 2 3 2 2 11 24 1 2 2 2 4 6.67 10 6 10 110 60 7614974 4 4 6.67 10 6 10 6.9 N kg 7614974 g c F ma GMm mr r GMm mr Tr GM r T GMT r m GM g r ω π π π π − − − = =   =     =   × × × × ×     = = =         × × × = = =

4 (c) Version 1: Radius of Earth: ( )( ) 2 11 24 6 E E 6.67 10 6 0 6.39 10 m 9.81 GM g r r −× × =  = = × Centripetal acceleration near surface of Earth: ( ) 2 2