Question

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 , has an elastic collision with the wall and climbs back onto W. Then,

• (A) B will reach the top of W again
• (B) from the beginning, till the collision with the wall, the centre of mass of 'B plus W' does not move horizontally
• (C) after the collision, the centre of mass of 'B plus W' moves with the velocity 2mv/m+M
• (D) when B reaches its highest position on W, the speed of W will be 2mv/m+M

Answer 1

The Correct Option is D

Explanation:
As there is no force acting on the system in horizontal direction, hence momentum of the system will remain conserved. Hence, centre of the block and wedge system will remain at rest along the horizontalline.Assume the right direction as positive. Let $V$ be the velocity of the wedge, when the block reaches the bottom and starts moving horizontally on the surface with velocity $v$ Hence, $\mathrm{MV}+\mathrm{mv}=0$ ( lnitially both are at rest)$MV=-mv\Rightarrow V=-\frac{mv}{M}$As the collision of the block with wall is perfectly elastic, block will rebound with same speed in opposite direction.So, the impulse imparted on the block by the wall $=-2\mathrm{mv}$Velocity of the centre of mass in horizontal direction after collision$V=\frac{-MV-mv}{M+m}\Rightarrow -\frac{2mv}{M+m}$Negative direction indicates that centre of mass is on left side.As there is no external force acting on the system in the horizontal direction, there will be no change in the speed of the $\mathrm{CM}$ in horizontal direction. When the block reaches at the highest point on wedge, the speed of the system will remain the same.$V=\frac{2mv}{M+m}$