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A-LevelPhysicsDynamics9702/21 · Oct/Nov 2020 · Q38 Marks

A ball falls vertically downwards towards a horizontal floor and then rebounds along its original path, as illustrated in Fig. 3.1. [Figure 3.1] The ball reaches the floor with speed 3.8 m s⁻¹. The ball is then in contact with the floor for a time of 0.081 s before leaving it with speed 1.7 m s⁻¹. The mass of the ball is 0.062 kg.

  1. (a)Define force.[1]
  2. (b)(i)Calculate the loss of kinetic energy of the ball during the collision.[2]
  3. (b)(ii)Determine the magnitude of the change in momentum of the ball during the collision.[2]
  4. (b)(iii)Show that the magnitude of the average resultant force acting on the ball during the collision is 4.2 N.[1]
  5. (b)(iv)1.Use the information in (iii) to calculate the magnitude of: the average force of the floor on the ball during the collision[1]
  6. (b)(iv)2.the average force of the ball on the floor during the collision.[1]

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Mark scheme

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

  1. [B1] (force =) rate of change of momentum
  2. [C1] E = ½mv² or ½ x 0.062 x 3.8² or ½ x 0.062 × 1.7²
  3. [A1] loss of KE = ½ × 0.062 × (3.8² – 1.7²) = 0.36 J
  4. [C1] p = mv or 0.062 × 3.8 or 0.062 × 1.7
  5. [A1] change in momentum = 0.062 × (1.7 + 3.8) = 0.34 N s
  6. [A1] (average resultant force =) 0.34 / 0.081 = 4.2 (N) or (average resultant force =) 0.062 × (1.7 + 3.8)/0.081 = 4.2 (N)
  7. [A1] average force = 4.2 + (0.062 × 9.81) = 4.8 N
  8. [A1] average force = 4.8 N

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About this question

Topic
Dynamics, A-Level Physics (syllabus 9702). Structured question, 8 marks.
Source
A-Level Physics, Oct/Nov 2020 Paper 2 Variant 1, question 3.

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