A-LevelPhysicsKinematics9702/22 · May/June 2020 · Q210 Marks

(a) Fig. 2.1 shows the velocity-time graph for an object moving in a straight line. [Figure 2.1] (b) A ball is kicked with a velocity of 15ms⁻¹ at an angle of 60° to horizontal ground. The ball then strikes a vertical wall at the instant when the path of the ball becomes horizontal, as shown in Fig. 2.2. [Figure 2.2] Assume that air resistance is negligible.

(a)(i)Determine an expression, in terms of u, v and t, for the area under the graph.[1]
(a)(ii)State the name of the quantity represented by the area under the graph.[1]
(b)(i)By considering the vertical motion of the ball, calculate the time it takes to reach the wall.[3]
(b)(ii)Explain why the horizontal component of the velocity of the ball remains constant as it moves to the wall.[1]
(b)(iii)Show that the ball strikes the wall with a horizontal velocity of 7.5ms⁻¹.[1]
(c)(i)Use the information from (b)(iii) to calculate the change in momentum of the ball due to the collision.[2]
(c)(ii)Calculate the magnitude of the average force exerted on the ball by the wall.[1]

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Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[A1] area = ut + 1/2(v – u)t or area = vt - 1/2(v – u)t or area = 1/2(u + v)t
[A1] displacement
[C1] u = 15 sin 60° (= 13 m s-1)
[C1] t = 15 sin 60°/9.81
[A1] = 1.3 s
[B1] the force in the horizontal direction is zero
[A1] (velocity =) 15 cos 60° = 7.5 (m s−1) or (velocity =) 15 sin 30° = 7.5 (m s−1)
[C1] Δp = mv or 0.40 × 7.5 or 0.40 × 4.3
[A1] Δp = 0.40 (7.5 + 4.3) = 4.7 kg m s-1
[A1] force = 4.7/0.12 or 0.40 × [(7.5 + 4.3) / 0.12] = 39 N

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