A machine gun of mass $10 k g$ fires $30 g$ bullets at the rate of $6$ bullet/s with a speed of $400 m / s$ . The force required to keep the gun in position will be

30 N

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40 N

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72 N

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400 N

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Solution

The correct option is C $72 N$
Given, $m_{2} = 10 k g; m_{1} = 30 g, v_{1} = 400 m / s$
Number of bullets fired/second $= 6$
$F = ?$
We know, $F =$ rate of change in linear momentum
$∴ F = \frac{change in linear momentum}{t i m e}$
In $t = 1 s$ , total mass of bullets fired
$= 30 \times 6 = 180 g$
$= 180 \times 10^{- 3} k g$
So, $F = \frac{180 \times 10^{- 3} \times 400}{1}$
$= 72 N$ .

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A machine gun of mass 10 kg fires 30 g bullets at the rate of 6 bullet/s with a speed of 400 m/s. The force required to keep the gun in position will be

The correct option is C 72 NGiven, m2=10 kg;m1=30 g,v1=400 m/sNumber of bullets fired/second =6F=?We know, F= rate of change in linear momentum∴F=change in linear momentumtimeIn t=1s, total mass of bullets fired=30×6=180 g=180×10−3kgSo, F=180×10−3×4001=72 N.

A machine gun of mass $10 k g$ fires $30 g$ bullets at the rate of $6$ bullet/s with a speed of $400 m / s$ . The force required to keep the gun in position will be