Two masses A & B each of `5 kg` are suspended by a light inextensible string passing over a smooth massless pulley such that mass A rest on smooth table & B is held at the position shown. Mass B is now gently lifted up to the pulley and allowed to fall from rest. Determine up to what height will A rise for the ensuing motion.

Answer

`1.25 m`

Explanation

Velocity of B when string is in natural length
`= u_(B) = sqrt(2gh) = sqrt(2g) (h = 1)`
Impulse equation
`rArr - TDelta t = m [v - sqrt(2g)]` ….(i)
`T. Deltat = m[v - 0]` ….(ii)

On solving eq. (i) & (ii) `: v = (sqrt2g)/(2) = (g)/(sqrt2)`
`:.` Distance travelled by A begore coming to rest,
`s = 1 + (v^(2))/(2g) = 1.25m`