JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2022
- Medium
- Numerical
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 \(\dots N\)is the force exerted on the ball due to each other.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(200\)
Step-by-step Solution
Detailed explanation
Change in momentum of any one ball
\(|\Delta \overrightarrow{ P }|=2 \times 0.05 \times 10\)
\(|\Delta \overrightarrow{ P }|=1\)
\(\left|\overrightarrow{ F }_{x r}\right|=\frac{|\Delta \overrightarrow{ P }|}{\Delta t }\)
\(F _{ av }=200\,N\)
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