A gun fires $100bullet/\min $ with muzzle speed $600m/s$ . If each bullet is of mass $5g$, then the amount of force needed to hold the gun in position is
(A) $2.5N$
(B) $0.5N$
(C) $4N$
(D) $5N$

Hint We have the total number of bullets fired in $1\min $ and the speed of each bullet given. Using these values we can find out the total momentum of all the bullets fired in $1\min $. Then using Newton's second law of motion we can find the average force needed to hold the gun.

Complete Step by step solution
From the question we have the mass of each bullet, say $m$ and the speed of each bullet fired, say $v$ given.
Using the given values we can find out the value of momentum of each bullet, where
Momentum $(P) = m \times v$
Given, $m = 5g = 5 \times {10^{ - 3}}kg$
$v = 600m/s$
On putting the values we get,
$P = 5 \times {10^{ - 3}} \times 600$
$ \Rightarrow P = 3kgm/s$
We know that $100 bullets/\min $ is fired by the gun.
Now we calculate the total momentum, say ${P_T}$ carried by the bullets in $1\min $.
 ${P_T} = P \times 100$
$ \Rightarrow {P_T} = 300kgm/s$
From Newton’s Second Law of Motion, we know that,
$Average\,\,Force\,(F) = \dfrac{{Total\,\,Momentum}}{{Total\,\,Time\,\,taken}}$
We have the values for total momentum to be ${P_T}$ and total time is taken, say $T$ to be $1\min $ , or $60\sec $ .
Putting these values in the above equation we get,
$F = \dfrac{{{P_T}}}{T}$
$ \Rightarrow F = \dfrac{{300}}{{60}} = 5N$
Therefore the force needed to hold the gun in position is $5N$ .

Hence the correct answer is option D.

Note
Remember that the momentum of the gun is zero when the bullet is fired since there is no motion involved, hence the speed is zero. There is a forward momentum on the gun after the bullet is fired. The recoil of a gun is a result of the principle of conservation of momentum. This principle is responsible for balancing the forward momentum faced by the bullet by the backward recoil of the gun.