A particle P of mass 0.1 kg is attached to one end of a light inextensible string of length 0.5 m. The other end of the string is attached to a fixed point A. The particle P moves in a circle which has its centre O on a smooth horizontal surface 0.3 m below A. The tension in the string has magnitude T N and the magnitude of the force exerted on P by the surface is R N.
- (i)Given that the speed of P is 1.5 m s¯¹, calculate T and R.[4]
- (ii)Given instead that T = R, calculate the angular speed of P.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [M1] 0.1×1.5²/0.4 = Tcos0 (Note r = 0.4, cos0 = 0.8, sin0 = 0.6. Use Newton's Second Law horizontally)
- [A1] T = 0.703
- [M1] R = 0.1g - Tsin0 (Resolve vertically for P)
- [A1] R = 0.578
- [M1] T + Tsin0 = 0.1g (Resolve vertically for P)
- [A1] T = 0.625
- [M1] 0.1ω²×0.4 = 0.625cos0 (Use Newton's Second Law horizontally)
- [A1] ω = 3.54 rad s⁻¹
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