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A-LevelPhysicsKinematics9702/22 · Oct/Nov 2018 · Q112 Marks

A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 ms⁻¹ at an angle θ to the horizontal, as illustrated in Fig. 1.1. [Figure 1.1] The magnitude of the initial vertical component v_y of the velocity is 4.8 ms⁻¹. Assume that air resistance is negligible.

  1. (a)Show that the magnitude of the initial horizontal component v_x of the velocity is 3.6 ms⁻¹.[1]
  2. (b)(i)On Fig. 1.2, sketch separate lines to show the variation with time t, until the ball returns to the ground, of the vertical component v_y of the velocity (label this line Y),[2]
  3. (b)(ii)the horizontal component v_x of the velocity (label this line X).[2]
  4. (c)Calculate the maximum height reached by the ball. maximum height = ..................................................... m[2]
  5. (d)For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height -------------------------------------------------------------- change in gravitational potential energy ratio = ....................................................................[4]
  6. (e)In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible.[1]

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Mark scheme

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

  1. [A1] vx = (6.0² – 4.8²)¹/² = 3.6 (ms⁻¹)
  2. or
  3. [A1] 6.0 sinθ = 4.8 (so θ = 53.1°) and vx = 6.0 cos 53.1° = 3.6 (ms⁻¹)
  4. [M1] straight line from (0, 4.8) to (0.49, 0)
  5. [A1] straight line continues with same slope to (0.98, −4.8) (labelled Y)
  6. [M1] a horizontal line
  7. [A1] from (0, 3.6) to (0.98, 3.6) (labelled X)
  8. [C1] s = ut + ½at²
  9. [C1] = (4.8 × 0.49) + (½ x −9.81 × 0.49²)
  10. or
  11. [C1] s = ½(u + v)t or area under graph
  12. [C1] = ½ × (4.8 + 0) × 0.49
  13. or
  14. [C1] s = vt - ½at²
  15. [C1] = ½ × 9.81 x 0.49²
  16. or
  17. [C1] v² = u² + 2as
  18. [C1] s = 4.8² / (2 × 9.81)
  19. [A1] s = 1.2 m
  20. [C1] (Δ)Ε = mg(Δ)h
  21. [C1] E = ½mv²
  22. [C1] ratio = (½ × m x 3.6²) / (m × 9.81 × 1.2)
  23. or
  24. [C1] ratio = [(½ × m × 6.0²) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2)
  25. or
  26. [C1] ratio = (½ × m × 3.6²) / (½ x m × 4.8²)
  27. [A1] ratio = 0.56
  28. [B1] (force due to) air resistance acts in opposite direction to the velocity
  29. or
  30. [B1] (with air resistance, average) resultant force is larger (than weight)

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About this question

Topic
Kinematics, A-Level Physics (syllabus 9702). Structured question, 12 marks.
Source
A-Level Physics, Oct/Nov 2018 Paper 2 Variant 2, question 1.

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