Question 78 of 130

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

A$100N$B$200N$C$300N$D$400N$

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Answer: (B) $200N$

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Change in momentum of any one ball
$\left|\Delta \vec{p}\right|=2\times 0.05\times 10$
$\Rightarrow \left|\Delta \vec{p}\right|=1kgm{s}^{-1}$
The force exerted on the ball is $\left|{\vec{F}}_{av}\right|=\dfrac{\left|\Delta \vec{p}\right|}{\Delta t}=\dfrac{1}{0.005}$
${F}_{av}=200N$