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A-LevelPhysicsKinematics9702/23 · May/June 2018 · Q313 Marks

A ball is thrown vertically upwards towards a ceiling and then rebounds, as illustrated in Fig. 3.1. [Figure 3.1] The ball is thrown with speed 9.6 m s⁻¹ and takes a time of 0.37 s to reach the ceiling. The ball is then in contact with the ceiling for a further time of 0.085 s until leaving it with a speed of 3.8 m s⁻¹. The mass of the ball is 0.056 kg. Assume that air resistance is negligible.

  1. (a)Show that the ball reaches the ceiling with a speed of 6.0 m s⁻¹.[1]
  2. (b)Calculate the height of the ceiling above the point from which the ball was thrown. height = ........................................................................................... m[2]
  3. (c)Calculate
  4. (c)(i)the increase in gravitational potential energy of the ball for its movement from its initial position to the ceiling,[2]
  5. (c)(ii)the decrease in kinetic energy of the ball while it is in contact with the ceiling.[2]
  6. (d)State how Newton's third law applies to the collision between the ball and the ceiling.[2]
  7. (e)Calculate the change in momentum of the ball during the collision. change in momentum = ......................................................................... N s[2]
  8. (f)Determine the magnitude of the average force exerted by the ceiling on the ball during the collision. average force = ................................................................................... N[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

  1. [A1] v = u + at v = 9.6 – (9.81 × 0.37) = 6.0 ms⁻¹
  2. [C1] s = ½ × (9.6 + 6.0) × 0.37 or 6.0² = 9.6² – (2 × 9.81 x s) or s = (9.6 × 0.37) – (½ x 9.81 × 0.37²) or s = (6.0 × 0.37) + (½ x 9.81 × 0.37²)
  3. [A1] s = 2.9 m
  4. [C1] (∆)Ε = mg(∆)h
  5. [A1] ∆Ε = 0.056 × 9.81 x 2.9 = 1.6 J
  6. [C1] E = ½mv²
  7. [A1] ∆Ε = ½ × 0.056 × (6.0² – 3.8²) = 0.60 J
  8. [M1] force on ball (by ceiling) equal to force on ceiling (by ball)
  9. [A1] and opposite (in direction)
  10. [C1] (p =) mv or 0.056 × 6.0 or 0.056 × 3.8
  11. [A1] change in momentum = 0.056 x (6.0 + 3.8) = 0.55 N s
  12. [C1] resultant force = 0.55/0.085 (= 6.47N)
  13. [A1] force by ceiling = 6.47 – (0.056 × 9.81) = 5.9 N

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About this question

Topic
Kinematics, A-Level Physics (syllabus 9702). Structured question, 13 marks.
Source
A-Level Physics, May/June 2018 Paper 2 Variant 3, question 3.

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