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 smooth. $W$ 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 onto $W$.

• A. $B$ will reach the top of W again.
• B. From the beginning, till the collision with the wall the centre of mass of ${}^{\prime }B$ plus $W^{\prime }$ does not move horizontally.
• C. After the collision, the centre of mass of ${}^{\prime }B$ plus $W^{\prime }$ moves with the velocity $\frac{2mv}{m+M}$.
• D. When B reaches its highest position on $W$, the speed of $W$ is $\frac{2mv}{m+M}$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B From the beginning, till the collision with the wall the centre of mass of ${}^{\prime }B$ plus $W^{\prime }$ does not move horizontally.
C After the collision, the centre of mass of ${}^{\prime }B$ plus $W^{\prime }$ moves with the velocity $\frac{2mv}{m+M}$.
D When B reaches its highest position on $W$, the speed of $W$ is $\frac{2mv}{m+M}$.

Since there is no force acting on the system till the collision, the COM doesn't move till the collision. during collision the wall imparts an impulse of 2mv to the block and thus the velocity of the COM becomes $\frac{2mv}{(m+M)}$. When m reaches the top of the wedge both of the will be moving with same velocity.