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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.

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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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