JEE Main 2022 Physics question
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other ?
Answer: (2)
Solution
The change in momentum of one ball is kg m/s. So = = 1/0.005 = 200 N.