In a simple Atwood machine, two unequal masses m₁ and m₂ are connected by a string going over a clamped light smooth pulley. In a typical arrangement shown in the figure, m₁ = 300 g and m₂ = 600 g . The system is released from rest. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.

Step-by-step solution

Given m₁ = 300 g and m₂ = 600 g . The initial acceleration of the system is: a = m₂ - m₁ m₂ + m₁ g = 600 - 300 600 + 300 g = 300 900 g = g 3 The velocity of the system just before the larger mass is stopped ( t = 2.0 s ) is: v = u + at = 0 + ( g 3 )(2) = 2g 3 When m₂ is momentarily stopped, its velocity becomes zero. However, m₁ is still moving upwards with velocity v = 2g 3 . At this point, the string becomes slack. Both masses now move freely under gravity. Let t be the time until the string becomes tight again.