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

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)`