Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

12 of 130

Question 12 of 130

A machine gun of mass 10kg10kg fires 20g20g bullets at the rate of 180180 bullets per minute with a speed of 100ms–1100m{s}^{–1} each. The recoil velocity of the gun is :

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Answer: (D) 0.6ms−10.6m{s}^{-1}

Number of bullets fired per second 18060=3\dfrac{180}{60}=3.
Applying momentum conservation, we get
3[(20×10−3)×100]=10v3\left[\left(20\times {10}^{-3}\right)\times 100\right]=10v
⇒v=0.6ms−1\Rightarrow v=0.6m{s}^{-1}

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