Question

A machine gun fires ‘ $x$ ’ bullets per second into a target. Each bullet weighs ‘ $m$ ’ ${\text{kg}}$ and has a speed of ‘ $y$ ’ ${\text{m}}{{\text{s}}^{ - 1}}$. Find the force necessary to hold the gun in position.
A. $\left( {2mxy} \right)\,{\text{N}}$
B. $\left( {mxy} \right)\,{\text{N}}$
C. $\left( {mx3y} \right)\,{\text{N}}$
D. $\left( {m2xy} \right)\,{\text{N}}$

Answer 1

First of all, we will find the time taken by one bullet to come out of the barrel. Then we find the acceleration. After that we will put the required values in the formula given by Newton’s second law of motion and manipulate accordingly to find the force.

Complete step by step answer:
In the given question, we are supplied with the following data:
The number of bullets that a machine gun fires in one second is $x$ .
The mass of each bullet is ‘ $m$ ’ ${\text{kg}}$ .
Each bullet has a speed of ‘ $y$ ’ ${\text{m}}{{\text{s}}^{ - 1}}$ .
We are asked to find out the force necessary to hold the gun in the desired firing position.
To begin with, we will first find out the time taken by one bullet to come out of the barrel.
Since, the time taken by ‘ $x$ ’ bullets to come out of the barrel is $1$ second.
So, the time taken by one bullet to come out of the barrel is $\dfrac{1}{x}$ seconds.
Since, we the velocity of the bullet is given as ‘ $y$ ’ ${\text{m}}{{\text{s}}^{ - 1}}$, we can literally say that the bullet travels a distance of ‘ $y$ ’ ${\text{m}}$ in a duration of time which equals to $\dfrac{1}{x}$ seconds.
Here, the initial velocity of the bullet is $0$ ${\text{m}}{{\text{s}}^{ - 1}}$ .
The final velocity of the bullet is ‘ $y$ ’ ${\text{m}}{{\text{s}}^{ - 1}}$ .
Now, we will find the acceleration of the bullet, which is given by the formula:
$a = \dfrac{{v - u}}{t}$ …… (1)
Where,
$a$ indicates acceleration of the bullet.
$v$ indicates the final velocity of the bullet.
$u$ indicates the initial velocity of the bullet.
$t$ indicates the time taken by the bullet.
Substituting the required values in the equation (1), we get:
a = \dfrac{{y - 0}}{{\left( {\dfrac{1}{x}} \right)}} \\\
\implies a = \dfrac{y}{{\left( {\dfrac{1}{x}} \right)}} \\\
\implies a = xy\,{\text{m}}{{\text{s}}^{ - 2}} \\\
Now, to find the force required to give this acceleration to the bullet, we use the formula:
$F = ma$ …… (2)
Where,
$F$ indicates force.
$m$ indicates the mass of the bullet.
$a$ indicates the acceleration.
Substituting the required values in the equation (2), we get:
F = ma \\\
\implies F = m \times xy \\\
\therefore F = mxy\,{\text{N}} \\\
Therefore, the force required to give this acceleration to the bullet is $\left( {mxy} \right)\,{\text{N}}$ .
Again, we know,
Force necessary to hold the gun is equal to the recoil force of the gun.
Hence, the force necessary to hold the gun in position is $\left( {mxy} \right)\,{\text{N}}$ .

So, the correct answer is “Option B”.

Note:
It is important to remember that to solve a question becomes really easier when you consider one bullet out of the rest. Many students seem to have confusion regarding time taken by one bullet. Suppose $20$ bullets are fired in one second time, then the fraction of second taken by one bullet to come out of the barrel is $\dfrac{1}{{20}}\,{\text{th}}$ of a second.