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. `15ms^(-1)`
B. `20ms^(-1)`
C. `100ms^(-1)`
D. `50ms^(-1)`

Answer

C

Explanation

Conservastion of momentum just before and after the colision yields
`mu=mv_(1)+Mv_(2)`……..i

Conservation of energy of the block between the point 1 and 2 after the bullet pieces
It yields `/_\KE+/_\PE=0`
`-1/2Mv_(2)^(2)+Mgh=0impliesv_(2)=sqrt(2gh)`........ii
By using i and ii we obtain
`u=v_(1)+M/m sqrt(2gh)impliesv_(1)u-M/m sqrt(2gh)`
`v_(1)=500-2/0.01 sqrt(2xx10x0.2)=100ms^(-1)`