1.

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 ?

Answer»

`cos^(-1)[(m^(2)v^(2))/(2GL(M+m)^(2))]`
`tan^(-1)(m^(2)v^(2))/(2gL(M+m)^(2))`
`cos^(-1)[1-(m^(2)v^(2))/(2gL(M+m)^(2))]`
None of these

Solution :`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)]`


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