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A-LevelPhysicsDeformation of solids9702/21 · Oct/Nov 2019 · Q49 Marks

The variation with extension x of the force F applied to a spring is shown in Fig. 4.1. [Figure 4.1] The spring has an unstretched length of 0.080 m and is suspended vertically from a fixed point, as shown in Fig. 4.2. [Figure 4.2] A block is attached to the lower end of the spring. The block hangs in equilibrium at position X when the length of the spring is 0.095 m, as shown in Fig. 4.3. [Figure 4.3] The block is then pulled vertically downwards and held at position Y so that the length of the spring is 0.120m, as shown in Fig. 4.4. The block is then released and moves vertically upwards from position Y back towards position X. [Figure 4.4]

  1. (a)Use Fig. 4.1 to determine the spring constant of the spring. spring constant = ............................................... Nm⁻¹ [2][2]
  2. (b)Use Fig. 4.1 to show that the decrease in elastic potential energy of the spring is 0.055 J when the block moves from position Y to position X. [2][2]
  3. (c)The block has a mass of 0.122kg. Calculate the increase in gravitational potential energy of the block for its movement from position Y to position X. increase in gravitational potential energy = ................................................. J [2][2]
  4. (d)Use the decrease in elastic potential energy stated in (b) and your answer in (c) to determine, for the block, as it moves through position X:
  5. (d)(i)its kinetic energy kinetic energy = ................................................. J [1][1]
  6. (d)(ii)its speed. speed = .................................................... ms⁻¹ [2][2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

  1. [C1] k = F/x or k = gradient; e.g. k = 4.0/0.050
  2. [A1] k = 80 N m⁻¹
  3. [C1] E = ½Fx or E = ½kx² or E = area under graph
  4. [A1] (Δ)Ε = (½ × 3.2 × 0.040) – (½ x 1.2 x 0.015) = 0.055 J; or (Δ)Ε = (½ × 80 × 0.040²) – (½ × 80 × 0.015²) = 0.055 J; or (Δ)Ε = ½ × (1.2 + 3.2) × 0.025 = 0.055 J
  5. [C1] (Δ)Ε = mg(Δ)h; = 0.122 x 9.81 × (0.120 – 0.095)
  6. [A1] = 0.030 J
  7. [(C1)] (Δ)Ε = W × (Δ)h; = 1.2 × 0.025
  8. [(A1)] = 0.030 J
  9. [A1] E = 0.055 -0.030; = 0.025 J
  10. [C1] E = ½mv²; V = [(2 × 0.025) / 0.122]⁰.⁵
  11. [A1] = 0.64 m s⁻¹

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About this question

Topic
Deformation of solids, A-Level Physics (syllabus 9702). Structured question, 9 marks.
Source
A-Level Physics, Oct/Nov 2019 Paper 2 Variant 1, question 4.

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