The system consists of the block ($m$) and the cart ($M$). Since there are no external horizontal forces, the horizontal momentum of the system is conserved. As the track is smooth, the total mechanical energy is also conserved. Let $v_m$ and $v_M$ be the final speeds of the block and the cart relative to the ground.

Conservation of horizontal momentum (taking the direction of the block's final velocity as positive):

$$m v_m - M v_M = 0$$

Conservation of mechanical energy (initial potential energy equals final kinetic energy):

$$mgR = \frac{1}{2}mv_m^2 + \frac{1}{2}MV_M^2$$

From the momentum equation, $V_M = \frac{m}{M}v_m$. Substituting this into the energy equation:

$$mgR = \frac{1}{2}mv_m^2 + \frac{1}{2}M\left(\frac{m}{M}v_m\right)^2 = \frac{1}{2}mv_m^2\left(1 + \frac{m}{M}\right)$$

Solving for $v_m$:

$$v_m^2 = \frac{2mgR}{m(1 + m/M)} = \frac{2MgR}{M+m} \implies v_m = \sqrt{\frac{2MgR}{M+m}}$$

Then, solving for $V_M$:

$$V_M = \frac{m}{M}v_m = \frac{m}{M}\sqrt{\frac{2MgR}{M+m}} = \sqrt{\frac{2m^2gR}{M(M+m)}}$$