A-LevelPhysicsDynamics9702/23 · May/June 2021 · Q211 Marks

Define acceleration.

(a)Define acceleration.[1]
(b)A stone falls vertically from the top of a cliff. Fig. 2.1 shows the variation with time t of the velocity v of the stone. [Figure 2.1]
(b)(i)Explain, with reference to forces acting on the stone, the shape of the curve in Fig. 2.1.[3]
(b)(ii)Use Fig. 2.1 to determine the speed of the stone when the resultant force on it is zero. speed = ............................................................ ms⁻¹[1]
(b)(iii)Use Fig. 2.1 to calculate the approximate height through which the stone falls between t = 0 and t = 30 s. height = ............................................................ m[3]
(b)(iv)On Fig. 2.2, sketch the variation with t of the acceleration a of the stone between t = 0 and t = 30s. [Figure 2.2][3]

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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.

[B1] change in velocity / time (taken)
[B1] air resistance increases (with speed/with time)
[B1] resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time)
[B1] when air resistance equals the weight the speed/velocity/v becomes constant
[A1] speed = 36 m s⁻¹
[C1] height given by area under the curve
[A2] height = 950 m (Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range 900-910 m or 990-1000 m.)
[B1] line starting at (0, 9.8)
[B1] curve with negative gradient between t = 0 and t = 20 s
[B1] line showing zero acceleration between t = 20 s and t = 30 s

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