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A-LevelPhysicsDeformation of solids9702/23 · Oct/Nov 2020 · Q412 Marks

(a) State Hooke's law. (b) A spring is fixed at one end. A compressive force F is applied to the other end. The variation of the force F with the compression x of the spring is shown in Fig. 4.1. [Figure 4.1] Show that the elastic potential energy of the spring is 0.64 J when its compression is 16.0cm. (c) The spring in (b) is used to project a toy car along a track from point X to point Y, as illustrated in Fig. 4.2. [Figure 4.2] The spring is initially given a compression of 16.0 cm. The car of mass 0.076 kg is held against one end of the compressed spring. When the spring is released it projects the car forward. The car leaves the spring at point X with kinetic energy that is equal to the initial elastic potential energy of the compressed spring. The car follows the track around a vertical loop of radius 0.12m and then passes point Y. Assume that friction and air resistance are negligible. Calculate:

  1. (a)State Hooke's law. [1][1]
  2. (b)Show that the elastic potential energy of the spring is 0.64 J when its compression is 16.0cm. [2][2]
  3. (c)(i)the speed of the car at X speed = .............................................................. ms⁻¹ [2][2]
  4. (c)(ii)the kinetic energy of the car when it is at the top of the loop kinetic energy = ........................................................... J [3][3]
  5. (c)(iii)the speed of the car at Y. speed = ............................................................ ms⁻¹ [1][1]
  6. (d)In practice, a resistive force due to friction and air resistance acts on the car so that its kinetic energy at Y is 0.23 J less than its kinetic energy at X. Determine the average resistive force acting on the car for its movement from X to Y. average resistive force = ........................................................ N [3][3]

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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. [1] compression/extension is proportional to force (provided limit of proportionality is not exceeded)
  2. [1] (E) = ½Fx or ½kx² or area under graph
  3. [1] = ½ × 8 × 16 × 10⁻² = 0.64 (J) or = ½ × 50 × (16 × 10⁻²)² = 0.64 (J)
  4. [1] (E) = ½mv²
  5. [1] 0.64 = ½ × 0.076 x v²
  6. [1] v = 4.1 m s⁻¹
  7. [1] (Δ)(E) = mg(Δ)h
  8. [1] = 0.076 x 9.81 × 0.24
  9. [1] (= 0.18 (J)) kinetic energy = 0.64 – 0.18
  10. [1] 0.46 J
  11. [1] v = 4.1 m s⁻¹
  12. [1] W = Fs
  13. [1] d = 0.30 + (2π × 0.12) + 0.25 (= 1.3m)
  14. [1] F = 0.23 / 1.3
  15. [1] = 0.18 N

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About this question

Topic
Deformation of solids, A-Level Physics (syllabus 9702). Structured question, 12 marks.
Source
A-Level Physics, Oct/Nov 2020 Paper 2 Variant 3, question 4.

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