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.
- A \(100\)
- B \(200\)
- C \(300\)
- D \(400\)
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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