Physics (AS & A Level) · Oct / Nov 2016 · Paper 22 · Question 2
9702
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A ball of mass 0.030 kg moves along a curved track, as shown in Fig. 2.1. ball mass 0.030 kg 0.31 m wall B A speed 1.3 m s–1 Fig. 2.1 The speed of the ball is 1.3 m s–1 when it is at point A at a height of 0.31 m. The ball moves down the track and collides with a vertical wall at point B. The ball then rebounds back up the track. It may be assumed that frictional forces are negligible.
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2(a)[2]
∆E = mg∆h = 0.030 × 9.81 × (–) 0.31 = (–) 0.091 J [2]
2(b)[3]
E = ½mv 2 (initial) E = ½ × 0.030 × 1.32 (= 0.0254) 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1 or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 [3]
2(c)(i)[2]
0.096 = 0.030 (v + 2.8) v = 0.40 m s–1 [2]
2(c)(ii)[2]
F = ∆p / (∆)t or F = ma = 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 = 4.8 N [2]
2(d)[1]
kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or (relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. [1]
2(e)[2]
force = work done / distance moved = (0.091 – 0.076) / 0.60 = 0.025 N [2]
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