One end of a light elastic string of natural length 0.4 m and modulus of elasticity 8 N is attached to a fixed point O on a smooth horizontal plane. The other end of the string is attached to a particle P of mass 0.2 kg which moves on the plane in a circular path with centre O. The speed of P is v ms⁻¹ and the extension of the string is x m.
- (i)Given that v = 2.5, find x.[4]
- (ii)It is given instead that the kinetic energy of P is twice the elastic potential energy stored in the string. Form two simultaneous equations and hence find x and v.[5]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] Uses Newton's Second Law towards the centre of the circle.
- [B1] Uses T = λx/L.
- [M1] Eliminates T to find e.
- [A1] e = 0.12(0) m
- [B1] Uses KE = 2EE.
- [B1] Uses T = λx/L and T = m v² /r.
- [M1] Attempts to solve the 2 equations to find v or x.
- [A1A1] x = 0.4 and v = 5.66 or 4 √2
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