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 :
(g = 10 m/s²)

Solution

Physics sir · 4 steps

Find P's speed at the bottom from energy, then let the equal-mass elastic collision hand it to Q.

1. Height of the drop. P starts level with the centre of the quadrant, so it falls through one radius: h = 20 cm = 0.2 m.

   mgh = 1/2mv² ⇒ v = √2gh = √(2 × 10 × 0.2) = 2 m s⁻¹

2. Speed at the bottom. There is no friction, so all the potential energy becomes kinetic energy. P reaches Q at 2 m s⁻¹.
3. Equal masses, elastic. In a head-on elastic collision between equal masses, the two balls exchange velocities. P stops and Q moves off with P's speed.
4. So. Q moves off at 2 m/s, which is option (d).

The idea to keep. Equal masses in a head-on elastic collision swap velocities. Get the speed just before impact from energy conservation, and that is the target's speed after impact.

Answer from the NTA final answer key. Solution by PrepWiser.