A-LevelMathematicsMechanics9709/52 · Feb/Mar 2019 · Q58 Marks
A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.4 m vertically below O.
- (i)Find the greatest speed of P.[5]
- (ii)Calculate the greatest distance of P below O.[3]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.
- [M1] Note greatest speed occurs at the equilibrium position. Use T=λx/L
- 0.3g = 24e/0.6
- [A1] Fall 0.275 m
- e = 0.075 m
- [B1]
- PE Change = 0.3g × 0.275
- [M1] Set up a 3 term energy equation
- 0.3 v² /2 = 0.3g × 0.275 – 24 × 0.075²/(2 × 0.6)
- [A1]
- v = 2.18 m s-1
- [M1] Set up an energy equation. Note v = 0 at the greatest distance
- 0.3g(0.2 + E) = 24 E² /(2 × 0.6)
- [M1] Attempt to solve a 3 term quadratic equation
- 20 E² – 3E – 0.6 = 0
- [A1]
- E = 0.264 and so greatest distance is 0.864 m
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