RI 2021 Y5 H2 Physics CT Soln

Raffles Institution Year 5-6 Physics Department 1 2021 Year 5 H2 Physics Term 3 Common Test Solution Section A Qn Ans Solution 1 A change in velocity, fivv v∆= − Since ifvvv−= = , vector triangle is an isosceles triangle. 180 45 67.52β °− °= = ° ( ) 2 22 22 2 22 2 2 cos 45 2 cos 45 222 2 22 i f ifv v v vv vv v vv v ∆ = − + −− ° =+− ° = −  = − 0.765 0.77v vv∆= = 45 67.5 112.5θ = °+ °= ° 2 C mass of a donut 50 g≈ outer diameter 8 cm≈ inner diameter 2 cm≈ height 3 cm≈ ( ) 3 23 22 mass 0.050density 354 kg m 3.54 10 kg mvolume 0.08 0.02 0.0322π −−= = = = ×    −     3 D Both X and Y experience the same acceleration of free fall g regardless of their masses. vertical displacement, 2 221 11 02 22s ut at gt gt= += += Using the graph of s against t, where t = 0 is the time X is dropped, compare the displacements of X and Y at t1, t2 and t3. The distance between X and Y increases with time and this is independent of their masses. 45° vi vf 45° ∆v θ s t 0 X Y t1 t2 t3

Raffles Institution Year 5-6 Physics Department 2 4 A Option A: At the mid-point of the graph, the gradient is not zero, which indicates that the speed of the object is non-zero at the highest point, which is incorrect. Options B, C and D: Each graph shows that velocity is zero momentarily at the highest point of the motion and the motion is symmetrical about the highest point. It also shows that the object reverses its motion at the highest point and returns to its original position. 5 B ( ) − = + −= = 22 22 2 2 20 10 1.5 m s2 100 v u as a 6 C At 8.0 st = , 28.0 0 2.0 m s6.0 10.0a −−= = −− t = 8.0 s ( )( ) 62 2.0 9.81 484 N N mg ma N ma g −= = += − + = where N is the force on the passenger by the weighing scale. By Newton’s third law, the force on the weighing scale by the passenger is also 484 N in magnitude. Hence the weighing scale will read 484 N. Answer can be obtained by analysing the upwards motion of the elevator: Between 0 s to 2.0 s, the elevator is accelerating upwards. Hence the resultant force acting on the passenger is upwards. This implies N > mg. Option A is incorrect. Between 2.0 s and 6.0 s, the elevator is moving at constant velocity upwards. Hence the resultant force on the passenger is zero. This implies N = mg. Option B is incorrect. Between 6.0 s and 10.0 s, the elevator is decelerating upwards. Hence the resultant force acting on the passenger is downwards. This implies N < mg. Option D is incorrect. 7 C Consider forces parallel to plane on Q: Consider forces parallel to plane on P: sin ... (1)QQmg T f maθ − −= sin ... (2)PPT mg f ma θ− −= (1) + (2): ( ) ( ) ( )( ) sin 2 0.900 0.100 9.81 sin 2(0.10) (0.900 0.100)(3.0 ) 24 QP QPm mg f m ma θ θ θ − −= + − −= + = ° T f mQgsinθ Q T mPgsinθ P