Two particles, P and Q, of masses 0.3 kg and 0.6kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley fixed at a point B where the inclined planes AB and BC meet. P lies on the smooth plane AB which is inclined at an angle θ° to the horizontal where sin θ° = 0.4. Q lies on the plane BC which is inclined at 30° to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram). The particles are released from rest.
- (a)It is given that the plane BC is smooth. Find the tension in the string and the acceleration of Q.[5]
- (b)It is given instead that the plane BC is rough. The work done against the frictional force when Q moves 2m down the plane is 1.8 J. You should assume that P does not reach the pulley and that Q does not reach C. Use an energy method to find the speed of Q when it has moved 2m down the plane.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [*M1] Use of Newton's second law for P or Q or system. Allow g missing. Correct number of terms. Allow sin/cos mix. Allow sign errors. Allow using θ=24° or better.
- [A2] A1 For any one correct equation. A2 For any two correct equations. (0.6g sin 30 - T = 0.6a; T - 0.3g × 0.4 = 0.3a; 0.6g sin 30 - 0.3g × 0.4 = (0.3+0.6)a)
- [DM1] Must get to ‘T =' or ‘a ='. From equations with the correct number of relevant terms. For attempt to solve for T or a
- [A1] Allow ±2. (a = 2 ms⁻²; T = 1.8N)
- [B1] PE change for P = ±0.3g × 2 × 0.4 [=±2.4] For either.
- [B1] PE change for Q = ±0.6g × 2sin30 [=±6]
- [M1] Attempt at work-energy equation. 4 terms; dimensionally correct. Allow sign errors. Do not allow missing g. (½ × (0.3+0.6)v² = 0.6g × 2sin30 - 0.3g × 2 × 0.4 - 1.8)
- [A1] Speed = 2ms⁻¹
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