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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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.

  1. A \(100\)
  2. B \(200\)
  3. C \(300\)
  4. D \(400\)
Verified Solution

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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