As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball Q after collision will be

JEE Mains - Elastic Collision after Sliding Down a Quadrant

Question: As per the given figure, a small ball \(P\) slides down a smooth, frictionless quadrant of a circle of radius \(20\,\text{cm}\) (i.e. \(0.2\,\text{m}\)) and collides elastically with an identical ball \(Q\) that is initially at rest. Assuming the collision is head‑on, determine the velocity of ball \(Q\) after the collision. (Take \(g = 10\,\text{m/s}^2\))

[Shift 2, 30-Jan-2023]

Select the correct answer:

(A) \(1\,\text{m/s}\)

(B) \(2\,\text{m/s}\)

(C) \(3\,\text{m/s}\)

(D) \(4\,\text{m/s}\)

Solution

Step 1: Find the speed of ball \(P\) at the end of the slide

Ball \(P\) starts from rest and slides down a frictionless quadrant. The vertical drop is equal to the radius of the circle, i.e.: \[ h = 0.2\,\text{m}. \] Using energy conservation: \[ mgh = \frac{1}{2} m v^2 \quad \Longrightarrow \quad v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2\,\text{m/s}. \]

Step 2: Analyze the collision

In an elastic head‑on collision between two identical masses, the moving ball \(P\) comes to rest and the stationary ball \(Q\) acquires the velocity of \(P\).

Thus, the velocity of ball \(Q\) after collision is: \[ v_Q = 2\,\text{m/s}. \]