Two particles A and B of masses 2 kg and 3 kg respectively are connected by a light inextensible string. Particle B is on a smooth fixed plane which is at an angle of 18° to horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle A hangs vertically below the pulley and is 0.45 m above the ground (see diagram). The system is released from rest with the string taut. When A reaches the ground, the string breaks. Find the total distance travelled by B before coming to instantaneous rest. You may assume that B does not reach the pulley.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. [M1] Apply Newton's 2nd law to either particle A, or to particle B or the system. Correct number of terms.
2. [A1] A and B correct or system correct. Particle A: 2g - T = 2a. Particle B: T - 3g sin 18 = 3a.
3. [M1] Attempt to find a using equations with correct number of terms.
4. [A1] a = 2.145898034
5. [M1] Use of constant acceleration equations with their a≠±g_to find v² when A reaches the ground. v² = 2xa×0.45.
6. [A1] ν² = 1.931308... [v=1.389715162]
7. [M1] Attempt to find a for the motion of B when string becomes slack. Allow sin/cos error, no extra terms. T = 0, ±3g sin 18 = 3a.
8. [M1] Use constant acceleration equations, using a new a≠±g, to find the further distance, s, travelled by B before coming to rest. [0=1.93-2×3.09×s]
9. [A1] Total distance moved by B = 0.45 + 0.312 = 0.762m