A ball is dropped by the side of a vertical scale that is marked in centimetres. A video recording of the fall shows the position of the ball on the scale every 0.10s. The distance the ball falls is shown by the position on the scale. Table 1.1 shows the results. Table 1.1 time/s 0 0.10 0.20 0.30 0.40 0.50 distance/cm 0 5 20 44 78 123 [Figure 1.1]
- (a)(i)On Fig. 1.1, plot the distance-time graph for the ball. Draw a suitable line.[2]
- (a)(ii)Calculate the average speed of the ball between time t = 0 and time t = 0.40 s. speed = .................................................... cm/s[2]
- (a)(iii)Describe how to use the graph in Fig. 1.1 to determine the speed of the ball at time t = 0.40 s.[2]
- (b)(i)As the ball falls, the acceleration of the ball decreases and eventually becomes zero before it reaches the ground. Explain, in terms of the forces acting, why the acceleration decreases, and why it eventually becomes zero.[3]
- (b)(ii)Describe the appearance of a distance-time graph when the acceleration of the ball is zero.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [1] (a)(i) points plotted to within ½ a small square
- [1] (a)(i) smooth curve drawn
- [1] (a)(ii) (average speed =) d / t in any form
- [1] (a)(ii) 195 (cm/s)
- [1] (a)(iii) draw a tangent (at t = 0.40 s)
- [1] (a)(iii) find the gradient / slope (of the graph) or the gradient of the tangent
- [1] (b)(i) mention of air resistance / resistive force
- [1] (b)(i) resultant force decreases or air resistance / upwards force increases (as speed increases) / becomes high(er)
- [1] (b)(i) (eventually) air resistance = weight (and acceleration is zero) or air resistance balances / cancels weight
- [1] (b)(ii) straight line (of positive) or (line of) constant gradient / slope
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