In the figure, the block `B` of mass `m` starts from rest at the top of a wedge `W` of mass `M`. All surfaces are without friction. `W` can slide on the ground. `B` slides down onto the ground, moves along it with a speed v, has an elastic collision with the wall, and climbs back on to `W`.

Options

A. `B` will reach the top of `W` again
B. from the beginning, till the collision of with the wall, the centre of mass `'B+W'` is stationary in horizontal direction
C. after the collision the centre of mass `'B+W'` moves with the velocity `(2 m v)/(m+M)`
D. when `B` reaches its highest position on `W,` the speed of `W` is `(2 mv)/(m+M)`

Answer

B, C, D

Explanation

From linear conservation
`mv=MV`
`V=(mV)/(M)`
After the elastic collision with wall speed of the block `B` remain same in the direction `V`
`V_(cm)=(m(v)+M((mv)/(M)))/(m+M)`
`=(2mV)/(m+M)`
When block `B` will reach at maximum height on wedge
From momentum conservation
`(mv)/(M)M+mv=(m+M)V_(c)`
`V_(C)=(2mv)/((M+m))`.