JEE Main202330 Jan 2023Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual
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 s - 2 )
Options
- A0
- B0 . 25   m   s - 1
- C2   m   s - 1
- D4   m   s - 1
Correct answer
C. 2   m   s - 1
Step-by-step solution
Let v p be the velocity of ball P just before collision. Therefore, applying conservation of energy for P (between the moment it is released and the moment just before collision), m g l + 0 = 1 2 m v p 2 ⇒ v p = 2 g l = 2 × 10 × 0 . 2 = 2   m   s - 1 Since both balls have equal masses and the collision is perfectly elastic, the velocities of both the balls will get interchanged. Therefore, the velocity of ball Q just after the collision is 2   m   s - 1 .