Question

A cubical block of mass $m$ is released from rest at a height $h$ on a frictionless surface of a movable wedge of mass $M$, which is, in turn is placed on a horizontal frictionless surface as shown in the figure. Find the velocity of the triangular block when the smaller block reaches the bottom.

Answer 1

Let finel velocity of masses $M$ and $m$, be $v_1$ and $v_2$ here, two things will be conserved.

$\left(i\right)$Momentum in horizontal direction

$\left(ii\right)$Energy

$m{v}_1=M{v}_2.........................\left(1\right)$

$\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_1^2=mgh...............\left(2\right)$

$\frac{1}{2}m\left({\frac{M}{m}}^2\right)v_2^2+\frac{1}{2}m{v}_1^2=mgh$

$v_2^2[m+\frac{M^2}{m}]=2mgh$

$v_2^2=\frac{2mgh\times m}{m^2+M^2}$

$v_2=\sqrt{\frac{2m^{2}gh}{m^2+M^2}}$