JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2023
- Medium
- Numerical
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 \(............\,m/s\) \(:\left( g =10\,m / s ^2\right)\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
The velocities will be interchanged after collision.
Speed of \(P\) just before collision \(=\sqrt{2 gh }\)
\(=\sqrt{2 \times 10 \times 0.2}=2\,m / s\)
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