1. Speed of ball $P$ just before collision

Since ball $P$ slides बिना friction down the quadrant, mechanical energy is conserved.

From the figure, the vertical drop of $P$ is the radius of the quadrant. Let it be $R$.

So, $m g R = \frac12 m v_P^2$

Hence, $v_P = \sqrt{2gR}$

From the figure, $R = 0.2\,\text{m}$ and $g=10\,\text{m/s}^2$.

Therefore, $v_P = \sqrt{2\cdot 10\cdot 0.2} = \sqrt{4} = 2\,\text{m/s}$

2. Elastic collision between equal masses

Ball $Q$ is initially at rest, and the collision is elastic.

In a head-on elastic collision between two equal masses, the velocities are exchanged.

So after collision:

• ball $P$ comes to rest,
• ball $Q$ moves with the initial speed of ball $P$.

Thus, $v_Q = 2\,\text{m/s}$

3. Check options

• A: $0.25\,\text{m/s}$ ❌
• B: $4\,\text{m/s}$ ❌
• C: $0$ ❌
• D: $2\,\text{m/s}$ ✅

Therefore, the correct answer is Option D.