Question

A machine gun has a mass of $20\,Kg$ it fires $35\,g$ bullets at a speed of $400$ bullets per minute with a speed of $400\,m{s^{ - 1}}$ ,what average force must be applied to the gun to keep it in position ?

Answer 1

Hint:In order to solve this question, we will use the concept of Newton’s second law of motion which states that, force on a body is equal to rate of change in momentum of the body, where momentum is defined as the product of mass and velocity of the body.

Formula used:
Mathematically Newton's second law is written as $F = \dfrac{p}{t}$ where $p$ denotes momentum of the body and defined as $p = m \times v$ .

Complete step by step answer:
According to the question we have,
Mass of the bullet is $m = 35g = 0.035\,Kg$.
Velocity of the bullet firing is $v = 400\,m{s^{ - 1}}$.
Total number of bullets fired in one minute is $n = 400$.
Time taken to fire these bullets is one minute $t = 60\,s$.
Now, according to the definition of force, we need to find the ratio of momentum and time taken. So, momentum of the single bullet is $p = mv$
Momentum of total bullets is $p' = nmv$
$p' = 400 \times 400 \times 0.035$
$\Rightarrow p' = 5600\,Kgm{s^{ - 1}}$
Now, using formula of force we have,
$F = \dfrac{{p'}}{t}$
\[\Rightarrow F = \dfrac{{5600}}{{60}}\]
$\therefore F = 93.3N$

Hence, the average force needed for fire gun to stay in its position is $F = 93.3\,N$.

Note: It should be remembered that, basic units of conversions are as $1\,g = 0.001\,Kg$ and $1\min = 60\sec $ and the SI unit for force is newton and $1\,N$ is defined as the product of $1\,kg$ mass of a body moving with an acceleration of $1\,m{\sec ^{ - 2}}$, force and momentum are both a vector quantity such that they have magnitude a well as direction and in given question the direction of momentum of bullet and momentum on fire gun is opposite which makes fire gun stable in its respective position.