A ball is thrown vertically downwards to the ground, as illustrated in Fig. 2.1. ball speed u path of ball 1.5m speed 8.7 ms⁻¹ ground Fig. 2.1 The ball is thrown with speed u from a height of 1.5m. The ball then hits the ground with speed 8.7ms⁻¹. Assume that air resistance is negligible.
- (a)Calculate speed u.[2]
- (b)State how Newton's third law applies to the collision between the ball and the ground.[2]
- (c)(i)The ball is in contact with the ground for a time of 0.091 s. The ball rebounds vertically and leaves the ground with speed 5.4ms⁻¹. The mass of the ball is 0.059 kg. Calculate the magnitude of the change in momentum of the ball during the collision.[2]
- (c)(ii)Determine the magnitude of the average resultant force that acts on the ball during the collision.[1]
- (c)(iii)Use your answer in (c)(ii) to calculate the magnitude of the average force exerted by the ground on the ball during the collision.[2]
- (d)The ball was thrown downwards at time t = 0 and hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from t = 0 to t = T. Numerical values are not required. speed 0 0 T t Fig. 2.2[1]
- (e)In practice, air resistance is not negligible. State and explain the variation, if any, with time t of the gradient of the graph in (d) when air resistance is not negligible.[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [C1] v² = u² + 2as; u² = 8.7² – (2 × 9.81 × 1.5)
- [A1] u = 6.8 m s⁻¹
- [B1] (magnitude of) force on ball (by ground) equal to force on ground (by ball)
- [B1] (direction of) force on ball (by ground) opposite to force on ground (by ball)
- [C1] (p = ) 0.059 × 8.7 or 0.059 × 5.4; change in momentum = 0.059 (8.7 + 5.4)
- [A1] = 0.83 N s
- [A1] resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4)/0.091]; = 9.1 N
- [C1] (W=) 0.059 × 9.81; (W=) 0.58 (N)
- [A1] force = 9.1 + 0.58; = 9.7 N
- [B1] straight line with a positive gradient and starting from a non-zero value of speed at t = 0 and ending when t = T
- [B1] air resistance increases
- [B1] resultant force/acceleration decreases so gradient (of curve) decreases
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