A-LevelMathematicsMechanics9709/43 · Oct/Nov 2021 · Q612 Marks

The diagram shows a particle of mass 5 kg on a rough horizontal table, and two light inextensible strings attached to it passing over smooth pulleys fixed at the edges of the table. Particles of masses 4kg and 6kg hang freely at the ends of the strings. The particle of mass 6kg is 0.5m above the ground. The system is in limiting equilibrium.

(a)Show that the coefficient of friction between the 5 kg particle and the table is 0.4.[2]
(b)The 6 kg particle is now replaced by a particle of mass 8 kg and the system is released from rest. Find the acceleration of the 4 kg particle and the tensions in the strings.[5]
(c)In the subsequent motion the 8 kg particle hits the ground and does not rebound. Find the time that elapses after the 8 kg particle hits the ground before the other two particles come to instantaneous rest. (You may assume this occurs before either particle reaches a pulley.)[5]

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Mark scheme

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

[M1] For resolving forces to find F and R. (R = 5g, F = 6g - 4g)
[A1] AG (μ = 2g/5g = 0.4)
[M1] For applying Newton's 2nd law to the 4 kg particle or the 8 kg particle. (T₁ - 4g = 4a or 8g - T₂ = 8a)
[A1] Both equations correct. (T₁ - 4g = 4a and 8g - T₂ = 8a)
[B1] T₂ - T₁ - F = 5a and F = 0.4 × 5g
[M1] Attempt to solve for a, T₁ or T₂. (Adding gives 8g - 4g - 2g = 17a leading to a = ...)
[A1] a = 1.18 m s⁻², T₁ = 44.7 N, T₂ = 70.6 N
[M1] Applying Newton's 2nd law to both active particles. (T - 4g = 4a, -T - F = 5a, F = 2g or 4g - 2g = 9a, a = 60/9)
[A1] a = 60/9
[M1] Use of v² = u² + 2as or equivalent to find v or v² when the 8 kg particle reaches the ground. (v² = 2 × (20/17) × 0.5 = 20/17 leading to v = ... [v = 1.0846...])
[M1] Use of v = u + at or equivalent to find t. (0 = √20/17 - (60/9) * t)
[A1] From t = 0.1626978... (t = 0.163 s)

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