A bullet of mass m moving vertically upwards with a velocity 'u' hits the hanging block of mass 'm' and gets embedded in it. The height through which block rises after the collision, (assume sufficient space above block)

Options

A. `u^(2)//2g`
B. `u^(2)//2g`
C. `u^(2)//8g`
D. `u^(2)//4g`

Answer

C

Explanation

From momentum conservation
`m u =2mv rArr v=u/2`
from energy conservation
`1/2xx2mxx(u/2)^(2) =2 mgrArr h=(u^(2))/(8g)`