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A-LevelPhysicsForces, density and pressure9702/23 · May/June 2021 · Q310 Marks

Define the moment of a force about a point.

  1. (a)Define the moment of a force about a point.[2]
  2. (b)Fig. 3.1 shows a type of balance that is used for measuring mass. [Figure 3.1] A rigid rod is pivoted about a point 6.2cm from the centre of a pan which is attached to one end. The object being measured is placed on the centre of this pan. A spring, attached to the rod 1.8cm from the pivot, is attached at its other end to a fixed point P. The spring obeys Hooke's law over the full range of operation of the balance. A pointer, on the other side of the pivot, is set against a millimetre scale which is a distance 52.6 cm from the pivot. When the system is in equilibrium with no mass on the pan, the rod is horizontal and the pointer indicates a reading on the scale of 86 mm. An object of mass 0.472kg is now placed on the pan. As a result, the pointer moves to indicate a reading of 123mm on the scale when the system is again in equilibrium.
  3. (b)(i)Show that the increase in the length of the spring is approximately 1.3mm.[2]
  4. (b)(ii)Calculate the magnitude of the moment about the pivot of the weight of the object. moment = ............................................................ Nm[2]
  5. (b)(iii)Use your answer in (b)(ii) to determine the increase in the tension in the spring due to the 0.472kg mass. increase in tension = ............................................................ N[2]
  6. (b)(iv)Use the information in (b)(i) and your answer in (b)(iii) to determine the spring constant k of the spring. Give a unit with your answer. k = ............................................................ unit ............[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

  1. [M1] force × distance
  2. [A1] perpendicular distance of (line of action of) force from the point
  3. [C1] distance moved by pointer = 123 – 86 (= 37 mm)
  4. [A1] (extension =) 37 × (1.8 / 52.6) = 1.3 (mm) OR sin or tan θ = 37/526 (so θ = 4.0° so extension =) sin or tan θ × 18 = 1.3 (mm)
  5. [C1] moment = 0.472 × 9.81 × 6.2 × 10⁻²
  6. [A1] = 0.29 N m
  7. [C1] (Δ)F × 1.8 × 10⁻² = 0.29
  8. [A1] ΔF = 16 N
  9. [C1] k = F/x
  10. [A1] = 16 / (1.3 × 10⁻³) = 1.2 × 10⁴ N m⁻¹

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About this question

Topic
Forces, density and pressure, A-Level Physics (syllabus 9702). Structured question, 10 marks.
Source
A-Level Physics, May/June 2021 Paper 2 Variant 3, question 3.

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