Question

As shown in figure, the block B of mass m starts from the 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 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 center of mass of ${}^{\prime }B$ plus $W^{\prime }$ is stationary
• C. After the collision, the center of mass of ${}^{\prime }B$ plus $W^{\prime }$ moves with a 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: (C), (D)

When $B$ reaches its highest position on $W$, the speed of $W$ is $\frac{2mv}{m+M}$

From beginning to the collision no net external force is applied so CM of (W + B) remains stationary conserving momentum on x-direction.

0 = $m\times v-M{v}_1$

$v_1$ = $\frac{mv}{M}$ i.e., $\frac{mv}{M}$ toward left

$B$ has an elastic collision with wall

$\Rightarrow$ it's velocity direction will be reversed

$v_{CM}$ = $\frac{mv+M{v}_1}{M+m}$ = $\frac{2mv}{M+m}$ toward left

When $B$ reaches the highest point on W

$\therefore$ conserving momentum

$(M+m)\frac{2mv}{M+m}$ = $M\times v^{\prime }+m{v}^{\prime }$

$v^{\prime }$ = $\frac{2mv}{M+m}$