A body of mass `m` is dropped from a height of `h`. Simultaneously another body of mass `2m` is thrown up vertically with such a velocity `v` that they collide at height `h/2`. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be

Options

A. (a) `sqrt((5gh)/(4))`
B. (b) `sqrt(gh)`
C. (c) `sqrt((gh)/(4))`
D. (d) None of these

Answer

D

Explanation

`m` mass falls a distance `h/2` in time, say t.

`h/2=1/2g t^2implies:. t=sqrt(h/g)`
For mass `2m`,
`h/2=vt-1/2g t^2`
`=vsqrt(h/g)-h/2`
`:. v=sqrt(gh)`
Now, let u is the velocity with which combined mass collides with ground, then
`u=sqrt((2/3v)^2+2g(h/3))`
`=sqrt(4/9gh+2/3gh)`
`=sqrt10/3sqrt(gh)`