Wedge problem

Suppose the wedge ends up moving to the left with speed $s_x$ .

By conservation of momentum in the horizontal direction, the block will be moving to the right with speed $2 s_x$ .

Then just before the block slides off the wedge, it will be going downward at $s_y \equiv (s_x + 2s_x)\tan(30)=\sqrt 3 s_x$ .

The total speed of the block will be $s_{Block} \equiv \sqrt{ (\sqrt 3 s_x)^2 + (2 s_x)^2 } = \sqrt 7 s_x$ .

The kinetic energy of the wedge as the block slides off is $\frac{1}{2} (2 m) s_x^2=m s_x^2$ .

The kinetic energy of the block as it slides off is $\frac{1}{2} m (7 s_x^2)=\frac{7}{2}m s_x^2$ .

By conservation of energy, the total kinetic energy must equal the initial potential energy, which is entirely due to the block: $4.5 m s_x^2 = mgh$ .

Solving $s_x = \sqrt{\frac{g h}{4.5}} \approx 6.67$

Use momentum conservation along x-axis to find relation between velocities of the block and the wedge .Then finally apply energy conservation for the wedge-block system to get individual velocities. Easy question :p!!!