A block of mass `m` slides down a wedge of mass `M` as shown . The whole system is at rest , when the height of the block is `h` above the ground. The wedge surface is smooth and gradually flattens. There is no friction between wedge and ground.

If there is no friction any where, the speed of the wedge, as the block leaves the wedge is `:`

Options

A. `msqrt((2gh)/((M+m)M))`
B. `Msqrt((2gh)/((M+m)m))`
C. `(sqrt(2gh))(m)/(M+m)`
D. `(sqrt(2gh))(M)/(M+m)`

Answer

A

Explanation

Linear momentum is conserved only in horizontal direction.
`mv_(1)=Mv_(2) …..(i)`

`(1)/(2) mv_(1)^(2)+(1)/(2)Mv_(2)^(2)=mgh ....(ii)`
From `(i) & (ii), v_(2)=msqrt((2gh)/((M+m)M))`