Two particles A and B, of masses 0.8 kg and 1.6 kg respectively, are connected by a light inextensible string. Particle A is placed on a smooth plane inclined at an angle θ to the horizontal, where sin θ = 3/5. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely (see diagram). The section AP of the string is parallel to a line of greatest slope of the plane. The particles are released from rest with both sections of the string taut. Use an energy method to find the speed of the particles after each particle has moved a distance of 0.5 m, assuming that A has not yet reached the pulley. [Figure 4.1]
- Two particles A and B, of masses 0.8 kg and 1.6 kg respectively, are connected by a light inextensible string. Particle A is placed on a smooth plane inclined at an angle θ to the horizontal, where sin θ = 3/5. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely (see diagram). The section AP of the string is parallel to a line of greatest slope of the plane. The particles are released from rest with both sections of the string taut. Use an energy method to find the speed of the particles after each particle has moved a distance of 0.5 m, assuming that A has not yet reached the pulley.[6]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [M1] [½ × 0.8 x v²] or [½ × 1.6 x v²] For KE of either particle
- [A1] Gain in KE = ½×0.8×v²+½×1.6×v² Total KE
- [M1] [Gain in PE_A= 0.8 g × 0.5 × sinθ] or [Loss in PE_B= 1.6 g × 0.5] For PE change of either particle (irrespective of sign)
- [A1] Loss in PE = 1.6 g × 0.5 – 0.8 g × 0.5 × 0.6 Change of PE
- [M1] [1.2v² = 8-2.4] Energy equation originating from 4 terms
- [A1] Speed is 2.16 (m s⁻¹)
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