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. `sqrt((5gh)/(4))`
B. `sqrt(gh)`
C. `sqrt((gh)/(4))`
D. `(sqrt(10gh))/(3)`

Answer

D

Explanation

`v_(2)=sqrt(2g((h)/(2)))=sqrt(gh)`
`t=sqrt((2((h)/(2)))/(g))=sqrt((h)/(g))`
Now. using relative motion
`t=(h)/(v-0)rArrsqrt((h)/(g))=(h)/(v)rArrv=sqrt(gh)`
`v_(1)^(2)=v^(2)-2g((h)/(2))rArrv_(1)^(2)=gh-gh=0rArrv_(1)=0`
using conservation of linear momentum
`3"m v"_(3)="m v"_(2)-2"m v"_(1)rArr3"m v"_(3)=msqrt(gh)-2m(0)" "rArrv_(3)=(1)/(3)sqrt(gh)`
`v_(4)^(2)=v_(3)^(2)+2g((h)/(2))" "rArrv_(4)^(2)=(gh)/(9)+gh=(10gh)/(9)" "thereforev_(4)=(sqrt(10gh))/(3)`