JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 3)
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms$$-$$1 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?
100 N
200 N
300 N
400 N
Explanation
Change in momentum of one ball
= 2 $$\times$$ (0.05)(10) kg m/s
= 1 kg m/s
$$ \Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}$$ N
= 200 N
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