Question

A machine gun fires a bullet of mass $65g$ with a velocity of $1300m{{s}^{-1}}$. The man holding it can exert a maximum force of $169N$ on the gun. The number of bullets he can fire per second will be
$\begin{align}
  & A)1 \\
 & B)2 \\
 & C)3 \\
 & D)4 \\
\end{align}$

Answer 1

Hint: When a person fires a gun, an impulse is generated on the gun. Impulse refers to the product of force and the time interval, during which the force acts on an object. Impulse is also equal to the change in linear momentum of an object. Linear momentum is defined as the product of mass of an object and velocity of the object.
Formula used:
$1)J=Fdt$
$2)J=m{{v}_{2}}-m{{v}_{1}}$

Complete answer:
When a bullet is fired by a person using a gun, an impulse is generated on the gun. Impulse is defined as the product of force acting on an object and the time interval, during which the force acts on the object. Mathematically, impulse is given by
$J=Fdt$
where
$J$ is the impulse generated on an object
$F$ is the force acting on the object over a time interval $dt$
Let this be equation 1.
Impulse on an object is also equal to the change in linear momentum of the object. Linear momentum refers to the product of mass of an object and velocity of the object. Mathematically, impulse can also be represented as
$J=m{{v}_{2}}-m{{v}_{1}}$
where
$J$ is the impulse acting on an object
$m{{v}_{2}}-m{{v}_{1}}$ is the change in linear momentum of the object
Let this be equation 2.
Combining equation 1 and equation 2, we have
$Fdt=m{{v}_{2}}-m{{v}_{1}}$
Let this be equation 3.
Coming to our question, we are provided that a machine gun is used to fire a bullet of mass $65g$ with a velocity of $1300m{{s}^{-1}}$. It is also said that the man holding it can exert a maximum force of $169N$ on the gun. We are required to find the number of bullets he can fire per second.
Let us assume that the person can fire $n$ bullets per second. If $m$ represents the mass of a bullet, the total mass of bullets, the person can fire per second is equal to $nm$. Using equation 1, let us calculate the impulse generated per second. Clearly, using equation 1, we have
$J=Fdt=165N\times 1s=165Ns$
where
$J$ is the impulse generated in $dt=1s$
$F=165N$ is the maximum force, the person exerts on the gun
Let this be equation 4.
Now, let us also use equation 2 to determine the impulse generated per second on $n$ bullets. Clearly, using equation 2, we have
$J=n(m{{v}_{2}}-m{{v}_{1}})=nm({{v}_{2}}-{{v}_{1}})=nm{{v}_{2}}=n\times 65g\times 1300m{{s}^{-1}}$
where
$J$ is the impulse generated per second on $n$ bullets
$nm=n\times 65g$ is the total mass of the bullets fired in one second
${{v}_{1}}=0m{{s}^{-1}}$ is the initial velocity of the bullets
${{v}_{2}}=1300m{{s}^{-1}}$ is the final velocity of the bullets
Let this be equation 5.
Substituting equation 4 and equation 5 in equation 3, we have
$Fdt=n(m{{v}_{2}}-m{{v}_{1}})\Rightarrow 165Ns=n\times 65g\times 1300m{{s}^{-1}}$
On further simplification, we have
$165Ns=n\times 0.065kg\times 1300m{{s}^{-1}}\Rightarrow n=\dfrac{165Ns}{0.065\times 1300kgm{{s}^{-1}}}\approx 2$
Therefore, the man can fire $2$ bullets per second.

Hence, the correct answer is option $B$.

Note:
Students need to be thorough with conversion formulas. Conversion formulas used in the above solution are given below.
$\begin{align}
  & 1kg=1000g \\
 & 1Ns=1kgm{{s}^{-2}}\times 1s=1kgm{{s}^{-1}} \\
\end{align}$
Students need not get worried if they do not arrive at the exact number of bullets given in the options. But it should be made sure that the value closest to the calculated answer is selected without any confusion.