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The Atwood machine in figure has a third...

The Atwood machine in figure has a third mass attached to it by a limp string. After being released, the `2m` mass falls a distance x before the limp string becomes laut. Thereafter both the mass on the left rise at the same speed. What is the final speed ? Assume that pulley is ideal.

Text Solution

Verified by Experts

The correct Answer is:
`sqrt(3gx)/(8)`
`2 mg - T = 2ma, T - mg = ma, a = (g)/(3)`

Velocity of m & `2m` after falling through a distance `x = sqrt(2ax) = sqrt((2gx)/(3))`
Impulse equation
`T Deltat = 2m (v - sqrt((2gx)/(3)))`
`T Deltat - T' Deltat = m (v - sqrt((2gx)/(3)))`
`T' Deltat = m(v- 0), v = sqrt((3gx)/(8))`
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