In the previous problem , the block of `mass M` is attached to a string of length `L`. Find the maximum angle made by the string with vertical . Take veloctiy of buller `u = 4 sqrt(g L)`. Take `M = 3m`.

Explanation

By the momentum conservation
`m u + M xx 0 = (m + M) V`
`V = ( m u)/(M + m) = (u)/(4) = (4 sqrt(g L))/(4) = sqrt(gL)`
Since `V lt sqrt(2 gL)`
Therefore , the combined mass will oscillate like a simple pendulum , hence , the maximum angle `theta lt 90^(@)`.
Between (1) and (2)
`0 = V^(2) - 2 gh`
`h = (V^(2))/( 2g)`
`L ( 1 - cos theta) = (g L)/( 2g) = (L)/(2) rArr cos theta = (1)/(2) rArr theta = 60^(@)`