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A-LevelMathematicsMechanics9709/52 · Feb/Mar 2018 · Q79 Marks

ABCD is a uniform square lamina with sides of length 0.6 m. A circular hole of radius r m is made in the lamina. The centre of the hole is 0.3 m from AB and 0.25 m from AD. The lamina is freely suspended at A and hangs with the axis of symmetry making an angle of 48° with the horizontal [Figure 7.1].

  1. (i)Show that r = 0.214, correct to 3 significant figures.[5]
  2. (ii)Find the weight of the original square lamina, before the hole was made.[4]

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Mark scheme

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

  1. [M1] tan48 = x/0.3 x is the distance of the centre of mass from AD
  2. [A1] x = 0.3332
  3. [M1] 0.6² × 0.3 = πr² × 0.25 + (0.6² – πr²)x OR 0.6² × 0.3 = πr² × 0.35 + (0.6² – πr²) × (0.6−x) Take moments about AD Take moments about BC
  4. [A1] πr² × (0.3332 – 0.25) = 0.6² × (0.3332 – 0.3) OR πr²(0.6 – 0.3332 – 0.35) = 0.6² (0.6 – 0.3332 – 0.3)
  5. [AG] r = 0.214 A1
  6. [M1] 0.3W=0.6 × 15cos60 Take moments about C.(W = weight of the lamina)
  7. [A1] W = 15
  8. [M1] Square = 15 × 0.6² / (0.6² – π × 0.214²) Recognise that the ratio of weights = ratio of areas
  9. [A1] Square = 25(.0) N

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About this question

Topic
Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source
A-Level Mathematics, Feb/Mar 2018 Paper 5 Variant 2, question 7.

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