Question

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 to keep the gun in position will be
(A) 30 N
(B) 40 N
(C) 72 N
(D) 400 N

Answer 1

In this question, the rate of firing and also the mass of gun and bullets are given, so we will first calculate the total mass of the bullet being fired, and then we will calculate the change in momentum due to firing, and by dividing it with the time of firing we will calculate the force to suppress recoil.

Complete step by step answer:
Mass of the machine gun, ${m_1} = 10kg$
Mass of the bullets, ${m_2} = 30g = 30 \times {10^{ - 3}}kg$
The velocity of bullets, $v = 400 {m}{s^{-1}}$
Number of bullets fired per second, $n = 6$
It is given that when a bullet is fired, then the shooter feels a back force due to recoil and this force can be calculated by finding the rate of change of momentum per second, and this can be written a $F = \dfrac{{{\text{ rate of change of momentum}}}}{{time}}$
Since six bullets are being fired in 1 second, hence the total mass of bullet being fired per second

$$= 30 \times {10^{ - 3}} \times 6 \\\ = 180 \times {10^{ - 3}}kg \\\$$

Hence the force required to keep the gun in position will be

$$F = \dfrac{{mass \times velocity}}{{time}} \\\ \Rightarrow F= \dfrac{{180 \times {{10}^{ - 3}} \times 400}}{1} \\\ \therefore F= 72N \\\$$

So the force required is $72N$.Hence,option C is correct.

Note: Momentum is the quantity of the motion that an object has. Mathematically, it is the product of the mass and the velocity of the object. It is denoted by the capital letter “M” and is expressed in the unit of kilograms meters per seconds.