A-LevelMathematicsMechanics9709/51 · May/June 2017 · Q69 Marks
A particle P of mass 0.15 kg is attached to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The particle P moves in a horizontal circle which has its centre vertically below A, with the string inclined at θ° to the vertical and AP = 0.5 m.
- (i)Find the angular speed of P and the value of θ.[5]
- (ii)Calculate the difference between the elastic potential energy stored in the string and the kinetic energy of P.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] T = 12 × 0.1 / 0.4 ( = 3 N)
- [M1] 3sinθ = 0.15 ω² (0.5sinθ)
- [A1] ω = 6.32 rad s⁻¹
- [M1] Tcosθ = 0.15g (cosθ=0.5)
- [A1] θ = 60
- [B1 FT] v = 6.32 × 0.5sin60
- [B1] KE = 0.15(6.32 × 0.5sin60)² / 2 (=0.5625J)
- [M1] Difference = 0.5625 – 12 × 0.1² / (2 × 0.4)
- [A1] Difference = 0.4125 J
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