Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms^-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?

Correct Option - 2


Detailed Solution:

$\Delta p$ = Change in momentum of each ball

$\Delta p=2mv=2\times 0.05\times 10=1Ns$

$\Rightarrow F=\frac{\Delta p}{\Delta t}=\frac{2\times 0.05\times 10}{0.005}=200N$

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