A bullet of mass `20 g` and moving with `600 m/s` collides with a block of mass `4 kg` hanging with the string. What is the velocity of bullet when it comes out of block, if block rises to height `0.2 m` after collision?

Options

A. `200 ms^(-1)`
B. `150 ms^(-1)`
C. `400 ms^(-1)`
D. `300 ms^(-1)`

Answer

A

Explanation

Velocity of block just after collisioin = sqrt(2gh)`
`=sqrt(2 xx 10xx 0.2)`
`2ms^(-1)`
Now applying conservation of linear momentum just before and just after collision.
`0.02 xx 600 = 4 xx2+0.02xxv`
`v=200 ms^(-1)`