A-LevelMathematicsMechanics9709/51 · May/June 2017 · Q26 Marks
One end of a light inextensible string is attached to a fixed point A. The other end of the string is attached to a particle P of mass m kg which hangs vertically below A. The particle is also attached to one end of a light elastic string of natural length 0.25 m. The other end of this string is attached to a point B which is 0.6 m from P and on the same horizontal level as P. Equilibrium is maintained by a horizontal force of magnitude 7 N applied to P (see Fig. 1).
- (i)Calculate the modulus of elasticity of the elastic string.[2]
- (ii)P is released from rest by removing the 7N force. In its subsequent motion P first comes to instantaneous rest at a point where BP = 0.3 m and the elastic string makes an angle of 30° with the horizontal (see Fig. 2). Find the value of m.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [M1] 7 = 0.35λ / 0.25
- [A1] λ = 5
- [B1] EE =0.35² × 5 / (2 × 0.25) or 0.05² × 5 / (2 × 0.05)
- [B1] PE = mg × 0.3sin30
- [M1] mg × 0.3sin30 = 0.35² × 5 / (2 × 0.25) – 0.05² × 5 / (2 × 0.25)
- [A1] m = 0.8
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