A bullet of mass m moving with a horizontal velocity u strikes a stationary block of mass M suspended by a string of length L. If the bullet gets embedded, to what maximum angle, with vertical, the block would risc ?

Options

A. `cos^(-1)[(m^(2)v^(2))/(2gL(M+m)^(2))]`
B. `tan^(-1)(m^(2)v^(2))/(2gL(M+m)^(2))`
C. `cos^(-1)[1-(m^(2)v^(2))/(2gL(M+m)^(2))]`
D. None of these

Answer

c

Explanation

`h=L-x=L-L cos theta=L(1-cos theta)`
By law of conservation of momentum mv=(M+m) V where V is the velocity of combination.By conservation of energy
`1/2(m+M)V^2=(M+m)gh` or `V^2=2gh`

Squaring the first eqn.
`V^2=(m^2v^2)/((m+m)^2)`.Thus putting for `V^2` and .h.
`2gL(1-cos theta)=(m^2v^2)/((M+m)^2)`
or `theta=cos^(-1)[1-(m^2v^2)/(2gL(M+m)^2)]`