Vertical Velocity Component Following Elastic Collision on Incline

To find the vertical component of the velocity immediately after the elastic collision at point $\text{B}$, we first determine the speed and direction of the block just before it strikes the second incline.

1. Motion along the first incline ($\text{AB}$):
From the given geometry:

• The angle of the first incline with the horizontal is $\theta_1 = 60^\circ$.
• The horizontal span of this section is $d_1 = \sqrt{3}\text{ m}$.

The vertical height through which the block falls from $\text{A}$ to $\text{B}$ is:
$h_1 = d_1 \tan 60^\circ = \sqrt{3} \times \sqrt{3} = 3\text{ m}$

Using conservation of mechanical energy, the speed of the block just before reaching point $\text{B}$ is:
$v = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 3} = \sqrt{60}\text{ m/s}$

The direction of this velocity vector is along the first incline, which is directed at an angle of $60^\circ$ below the horizontal:
$\vec{v} = v\cos(60^\circ)\hat{i} - v\sin(60^\circ)\hat{j}$

2. Elastic collision with the second incline:
The second incline makes an angle of $\theta_2 = 30^\circ$ below the horizontal.

Let us define a coordinate system with respect to the second incline:

• Parallel to the incline (downwards): $\hat{t} = \cos(30^\circ)\hat{i} - \sin(30^\circ)\hat{j}$
• Perpendicular to the incline (outwards/upwards): $\hat{n} = \sin(30^\circ)\hat{i} + \cos(30^\circ)\hat{j}$

The angle between the incoming velocity vector and the second incline's surface is:
$\phi = 60^\circ - 30^\circ = 30^\circ$

The velocity components just before the collision are:

• Parallel to the second incline:
  $v_\parallel = v \cos(30^\circ)$
• Perpendicular to the second incline:
  $v_\perp = -v \sin(30^\circ)$

For a completely elastic collision ($e = 1$):

• The component parallel to the surface remains unchanged:
  $v'_\parallel = v_\parallel = v \cos(30^\circ)$
• The component perpendicular to the surface reverses its direction:
  $v'_\perp = -e v_\perp = v \sin(30^\circ)$

3. Vertical component of the velocity immediately after impact:
The velocity vector immediately after the collision is given by:
$\vec{v}' = v'_\parallel \hat{t} + v'_\perp \hat{n}$
$\vec{v}' = \left(v \cos 30^\circ\right)\left(\cos 30^\circ\hat{i} - \sin 30^\circ\hat{j}\right) + \left(v \sin 30^\circ\right)\left(\sin 30^\circ\hat{i} + \cos 30^\circ\hat{j}\right)$

Taking the vertical (upward, $\hat{j}$) component:
$v'_y = -v \cos(30^\circ)\sin(30^\circ) + v \sin(30^\circ)\cos(30^\circ) = 0$

Alternatively, by the law of reflection, since the incoming velocity vector makes an angle of $30^\circ$ with the second incline (which is itself at an angle of $-30^\circ$ with the horizontal), the reflected velocity vector will make an angle of:
$-30^\circ + 30^\circ = 0^\circ$
with the horizontal. Thus, the velocity of the block immediately after the elastic collision is entirely horizontal, meaning its vertical component is zero.

Therefore, the correct option is C.