If soldiers had the options to use different weights but the bullets used are of a fixed weight. What would they prefer and why?
A) Light guns, because handing them is easy
B) Heavy guns, because they can be held firmly
C) Heavy guns, because they have less recoil
D) Light guns, because they can be carried easily

Hint: When a bullet is fired from a gun, it gets momentum in the forward direction. By the law of conservation of the momentum the gun also moves backward with the same momentum as the bullet. Hence there generates a recoil velocity of the gun.
So, use the law of conservation of momentum for the motions of both the gun and bullet and find the relation between the mass of the gun with its recoil velocity.

Formula used:
Before firing, we have to find out the total momentum of the bullet and the gun, $P = 0$
Let, \[{m_1}\] is the mass of the bullet and \[{m_2}\] is the mass of the gun.
After the firing,
The velocity of the bullet becomes ${v_1}$, hence the momentum $ = {m_1}{v_1}$
The velocity of the gun becomes ${v_2}$, hence the momentum $ = {m_2}{v_2}$
By the law of conservation of the momentum,
$P = {m_1}{v_1} + {m_2}{v_2}$
$ \Rightarrow {m_1}{v_1} + {m_2}{v_2} = 0$.

Complete step by step answer:
Before firing, both the gun and the bullet remain in the rest position. Hence The total momentum of the gun and the bullet before firing, $P = 0$.
When a bullet is fired from a gun, it gets momentum in the forward direction. By the law of conservation of the momentum the gun also moves backward with the same momentum as the bullet. Hence the gun holder feels a force in the backward direction – this is called the Recoil of a gun.
Let, \[{m_1}\] is the mass of the bullet and \[{m_2}\] is the mass of the gun.
After the firing,
The velocity of the bullet becomes ${v_1}$, hence the momentum $ = {m_1}{v_1}$
The velocity of the gun becomes ${v_2}$, hence the momentum $ = {m_2}{v_2}$
${v_2}$ is the Recoil velocity of the gun.
By the law of conservation of the momentum,
$P = {m_1}{v_1} + {m_2}{v_2}$
$ \Rightarrow {m_1}{v_1} + {m_2}{v_2} = 0$. [Since, $P = 0$]
$ \Rightarrow {m_1}{v_1} = - {m_2}{v_2}$
$ \Rightarrow {m_2} = - \dfrac{{{m_1}{v_1}}}{{{v_2}}}$
This relation shows that the mass of the gun is inversely proportional to the recoil velocity of its if the mass and velocity of a bullet is fixed. Hence for the lighter gun, the recoil velocity is very high and so the force on the gun holder is also very high which is not wanted.

So, for a soldier who is a gun holder, the preferable gun should be heavy because the heavier gun has less amount of recoil velocity.

Hence the right answer in option (C).

Note: In the equation, ${m_2} = - \dfrac{{{m_1}{v_1}}}{{{v_2}}}$, the negative sign defines that the velocity of the bullet ${v_1}$ and the velocity of the gun ${v_2}$ is in opposite direction to each other. That means the motion of the gun is opposite to the direction in which the bullet moves.