1. Motion from A to B on the first incline

The surface is frictionless, so from $A$ to $B$, mechanical energy is conserved.

If the vertical drop from $A$ to $B$ is $h_1$, then $\frac12 M v_B^2 = M g h_1$ so $v_B^2 = 2 g h_1.$

From the figure, the block moves a vertical height of $6\,\text{m}$ from $A$ to $B$ (equivalently, the first incline is at $60^\circ$ and gives this drop), hence $v_B^2 = 2 \cdot 10 \cdot 6 = 120.$ Thus, $v_B = \sqrt{120}\,\text{m/s}.$

2. Collision at the kink B

At $B$, the incline angle suddenly changes from $60^\circ$ to $30^\circ$. The collision is totally inelastic with the new incline.

This means the component of velocity perpendicular to the new incline is lost, while the component parallel to the new incline remains.

Just before collision, the velocity is along the first incline, i.e. at $60^\circ$ to horizontal. After collision, the velocity must be along the second incline, i.e. at $30^\circ$ to horizontal.

So the angle between the incoming velocity and the new incline is $60^\circ - 30^\circ = 30^\circ.$

Hence the speed just after collision is $v_{B'} = v_B \cos 30^\circ.$ Therefore, $v_{B'}^2 = v_B^2 \cos^2 30^\circ = 120 \cdot \frac34 = 90.$

3. Motion from B to C on the second incline

Again the plane is frictionless, so after the collision, energy is conserved from $B$ to $C$.

From the figure, the vertical drop from $B$ to $C$ is $\Delta h = \frac34\,\text{m}$. So the gain in speed satisfies $v_C^2 = v_{B'}^2 + 2g\Delta h.$ Substitute values: $v_C^2 = 90 + 2\cdot 10 \cdot \frac34 = 90 + 15 = 105.$ Thus, $v_C = \sqrt{105}\,\text{m/s}.$

4. Checking options

• A: $\sqrt{120}$
• B: $\sqrt{105}$
• C: $\sqrt{90}$
• D: $\sqrt{75}$

So the correct option is $\boxed{\text{B } (\sqrt{105}\,\text{m/s})}.$