Speed of Block Before Leaving Second Inclined Plane

To find the speed of the block at point $\text{C}$, we analyze the motion of the block in three stages:

1. Motion along the first incline from $\text{A}$ to $\text{B}$:
The first incline is at an angle of $\theta_1 = 60^\circ$ to the horizontal.
The horizontal distance between points $\text{A}$ and $\text{B}$ is given as $x_1 = \sqrt{3}\text{ m}$.
The vertical height fallen by the block from $\text{A}$ to $\text{B}$ is:
$h_1 = x_1 \tan 60^\circ = \sqrt{3} \times \sqrt{3} = 3\text{ m}$

Since the block is released from rest at $\text{A}$ on a frictionless surface, its speed $v_1$ just before reaching $\text{B}$ is obtained using the work-energy theorem:
$v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 3} = \sqrt{60}\text{ m/s}$

The velocity vector $\vec{v}_1$ is directed along the first incline at an angle of $60^\circ$ below the horizontal.

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2. Inelastic collision at point $\text{B}$:
The second incline is inclined at an angle of $\theta_2 = 30^\circ$ to the horizontal.
The angle between the two inclines is:
$\Delta\theta = 60^\circ - 30^\circ = 30^\circ$

During the totally inelastic collision at point $\text{B}$:

• The normal component of velocity relative to the second incline becomes zero.
• Since the incline is frictionless, there is no impulsive force parallel to the second incline, so the component of velocity along the second incline is conserved.

Thus, the speed of the block $v_B$ immediately after striking the second incline is:
$v_B = v_1 \cos(\Delta\theta) = \sqrt{60} \cos 30^\circ = \sqrt{60} \times \frac{\sqrt{3}}{2} = \sqrt{60 \times \frac{3}{4}} = \sqrt{45}\text{ m/s}$

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3. Motion along the second incline from $\text{B}$ to $\text{C}$:
The horizontal distance between points $\text{B}$ and $\text{C}$ is $x_2 = 3\sqrt{3}\text{ m}$.
The vertical height fallen by the block from $\text{B}$ to $\text{C}$ is:
$h_2 = x_2 \tan 30^\circ = 3\sqrt{3} \times \frac{1}{\sqrt{3}} = 3\text{ m}$

Since the surface from $\text{B}$ to $\text{C}$ is frictionless, mechanical energy is conserved:
$\frac{1}{2} M v_C^2 = \frac{1}{2} M v_B^2 + M g h_2$

Dividing both sides by $\frac{1}{2} M$:
$v_C^2 = v_B^2 + 2gh_2$
$v_C^2 = 45 + 2 \times 10 \times 3 = 45 + 60 = 105\text{ m}^2/\text{s}^2$
$v_C = \sqrt{105}\text{ m/s}$

Therefore, the correct option is B.