A bullet of mass `0.01 kg` and travelling at a speed of `500 ms^(-1)` strikes a block of mass `2 kg` which is suspended by a string of length `5 m`. The centre of gravity of the block is found to raise a vertical distance of `0.2 m`. What is the speed of the bullet after it emerges from the block?

Options

A. `200 m//s`
B. `217 m//s`
C. `204 m//s`
D. `284 m//s`

Answer

B

Explanation

Let u and v be velocities of the bullet with which it enters and leaves the block and V is the velocity acquired by the block. Let m and M are masses of the bullet and the block and h is height raised by the block.
`implies mu = mv + mv and 1/2 MV^2 = Mgh implies V = sqrt(2gh) m//s`
`implies v = (mu - MV)/(m) = 500 - (2 sqrt2)/(0.01) = 217 m//s` .