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Nov 2020 p41 q5
3956
Two particles of masses 0.8 kg and 0.2 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The system is released from rest with both particles 0.5 m above a horizontal floor (see diagram). In the subsequent motion the 0.2 kg particle does not reach the pulley.
(a) Show that the magnitude of the acceleration of the particles is 6 m s-2 and find the tension in the string.
(b) When the 0.8 kg particle reaches the floor it comes to rest. Find the greatest height of the 0.2 kg particle above the floor.
Solution
(a) Apply Newton's second law to each particle:
For the 0.8 kg particle:
\(0.8g - T = 0.8a\)
For the 0.2 kg particle:
\(T - 0.2g = 0.2a\)
For the system:
\(0.8g - 0.2g = (0.8 + 0.2)a\)
Solve for \(a\):
\(a = 6 \text{ m s}^{-2}\)
Solve for \(T\):
\(T = 3.2 \text{ N}\)
(b) Use the equation \(v^2 = u^2 + 2as\) to find the velocity \(v\) of the 0.8 kg particle as it reaches the ground:
\(v^2 = 2 \times 6 \times 0.5\)
Find the extra height reached by the 0.2 kg particle using \(v^2\):