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?
Solution
<p>Change in momentum of one ball</p>
<p>= 2 $\times$ (0.05)(10) kg m/s</p>
<p>= 1 kg m/s</p>
<p>$\Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}$ N</p>
<p>= 200 N</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.