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 rise'a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block?

Options

A. `580 ms^(-1)`
B. `220 ms^(-1)`
C. `1.4 ms^(-1)`
D. `7.8 ms^(-1)`

Answer

B

Explanation

The speed acquired by block, on account of collision of bullet with it, be `v_0 ms^(-1)`. Since the block rises by 0.1 in, hence
`0.1 = (v_0^2)/(2g) implies v_0^2 = 2 xx g xx 0.1 " or " v_0 = sqrt(2) ms^(-1)`
Now as per law of conservation of momentum for collision between bullet and block,
`m u = mv + Mv_0`
`implies v = u - M/m v_0 = 500 - (2 kg)/(0.01 kg) xx sqrt(2) ms^(-1) = (500 - 200 sqrt(2)) ms^(-1) = 220 ms^(-1)`