Interesting Collision

Velocity of the ball perpendicular to incline = $\frac{4}{\sqrt{2}} m/s$

Perpendicular vel of ball w.r.t inclined plane = $\frac{4}{\sqrt{2}} + \frac{4}{\sqrt{2}} = 4\sqrt2 m/s$

If after collison perpendicular vel of ball $= V$ then perpendicular velocity w.r.t inclined plane $= V - \frac{4}{\sqrt{2}}$ , this should be equal to $4\sqrt2$

so $V = \frac{12}{\sqrt2} m/s$ and along inclined plane velocity will be same as $\frac{4}{\sqrt{2}}$ , so net speed $= \sqrt{(\frac{12}{\sqrt2})^2+(\frac{4}{\sqrt2})^2} = \sqrt80 m/s = 4\sqrt5 m/s$

As we know, when small object with momentum $p$ strikes elastically a massive one, like a wall, then it rebounds with the same value of momentum but in opposite direction (velocity vector perpendicular to the wall). So change in momentum is $2p$ . Relative velocity of the ball with cart is $-V$ along x-axis, which is directed to the right and $-V$ along y-axis, which is directed upward. So the ball rebounds with the velocity $+V$ along x-axis, and $+V$ along y-axis in cart's frame of reference. Now we go back to our lab frame of reference where ball moves with velocity $+2$ V along x-axis (cart's velocity + ball's velocity relative to the cart) and $+V$ along y-axis. Then velocity of rebounded ball is $\sqrt{(2V)^2+V^2}=V \sqrt{5}=4\sqrt{5}$ .

so the relative velocity ball with cart is 4sqrt(2). because it strikes elistacally so,it will bounce with 4sqrt(2) too... but remember,we must make it back to relative with earth so,the velocity is 4sqrt(5) equals to,8.9