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. From the beginning, till the collision with the wall, the centre of mass of `'B` plus `W'` does not move horizontally.
B. After te collision, the centre of mass of `B` plus `W` moves with the velocity `(2mv)/(m+M)`
C. When `B` reaches its highest position of `W`, the speed of `W` is `(2mv)/(m+M)`
D. When `B` reaches its highest position of `W`, the speed of `W` is `(mv)/(m+M)`

Answer

A, B, C, D

Explanation

i. force acting on the horizontal direction is zero. Since block and wedge both are initially at rest. So velocity of `COM` (`B` plus `W`) does not move horizontally.
`ii. v_(cm)=(mv-(-mv))/(M+m) [:'v_(cm)=(mv_(1)+m_(2)v_(2))/(m_(1)+m_(2))]`
`v_(cm)=(2mv)/(M+m)`
ii When `B` reaches its highest position of `W`, the speed `W` is equal to velocity of `COM` i.e. `v=(2mv)/(M+m)`