Question

A gun of mass \[10\,kg\] fires four bullets per second. The mass of each bullet is \[20\,g\] and the velocity of the bullet when it leaves the gun is \[300\,m{s^{ - 1}}\] . The force required to hold the gun while firing is:
A. $6N$
B. $8N$
C. $24N$
D. $240N$

Answer 1

Hint:As the rate of firing as well as the mass of the gun and bullets are given in this question, we will first compute the total mass of the bullet being shot, then the change in momentum due to firing, and finally the force to suppress recoil by dividing it by the time of firing.

Complete step by step answer:
Let us consider mass of the gun, ${m_1} = 10\,kg$
Mass of the bullet ${m_2} = 20g = 20 \times {10^{ - 3}}\,kg$
The velocity of the bullet $v = 300\,m{s^{ - 1}}$
The rate at which bullets are discharged in a second is $n = 4$

When a bullet is shot, the shooter experiences a rearward force owing to recoil. This force may be determined by calculating the rate of change of momentum per second, which can be represented as,
$F = \dfrac{\text{Rate of change of momentum}}{\text{time}}$

Because four bullets are fired every second, the total mass of bullets fired per second is
$20 \times {10^{ - 3}}kg \times 4 = 80 \times {10^{ - 3}}\,kg/s $
As a result, the force needed to maintain the gun in place will be considerable.
$F = \dfrac{{mass \times velocity}}{{time}} \\
\Rightarrow F = \dfrac{{80 \times {{10}^{ - 3}} \times 300}}{1} \\
\therefore F = 24\,N $
The force required to hold the gun while firing is $24\,N$.

Hence, the correct option is C.

Note:It's worth mentioning that, recoil (also known as knockback, kickback, or simply kick) is the rearward force produced when a gun is fired. According to Newton's third law, the force required to accelerate something will elicit an equal but opposite reaction force, which means the forward momentum gained by the projectile and exhaust gases (ejectae) will be mathematically balanced out by an equal and opposite momentum exerted back upon the gun.