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?

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200 N

300 N

100 N

400 N

answer is B.

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Detailed Solution

Change in momentum of any one ball

$\begin{array}{l} | Δ \vec{P} | = 2 \times 0 .05 \times 10 \\ | Δ \vec{P} | = 1 \\ |{\vec{F}}_{av}| = \frac{| Δ \vec{P} |}{Δt} \\ F_{av .} = 200 N \end{array}$

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