Physics question about pulleys.

Question

Two particles P and Q have masses m kg and (1 - m) kg respectively. The particles are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. P is held at rest with the string taut and both straight parts of the string vertical. P and Q are each at a height of h meters above horizontal ground. P is released and Q moves downwards. Subsequently Q hits the ground and comes to rest 0.5 seconds later. The max velocity reached by Q is 2m/s.

I've calculated h to be 0.5m. I can't figure out what the masses of the two particles are though. Any help would be greatly appreciated.

Answer 1

Consider free body diagrams and net forces on the two particles: For P, F_net = T – m_P g = m_P a. For Q, F_net = T – m_Q g = - m_Q a (negative because its acceleration is downward). Because T has to be the same for both, m_P a + m_P g = m_Q g – m_Q a. Solve for a = (m_Q – m_P) g / (m_P + m_Q) and substitute their values: a = (1 – 2m) g.

Meanwhile from kinematics v_f = v_i + a*t, 2 = 0 + a*0.5 and a = 4 m/s². Δy = -h = v_i t + ½ a*t² = 0 – ½(4)(0.5)² = -1/2, so height h = 0.5 m.

With a = 4 = (1 – 2m)(9.8), 19.6m = 9.8 – 4 = 5.8 and m = 0.296 kg.