Question

A gun of mass $M$ fires a bullet of mass $m$ with maximum speed $v$. Given that m is less than M,The kinetic energy of the gun will be:

$\begin{aligned} & \text{A}.\text{ }\dfrac{1}{2}M{{v}^{2}} \\\ & \text{B}\text{. }\dfrac{1}{2}m{{v}^{2}} \\\ & \text{C}\text{. More than}\dfrac{1}{2}M{{v}^{2}} \\\ & \text{D}\text{. Less than}\dfrac{1}{2}M{{v}^{2}} \\\ \end{aligned}$

Answer 1

This type of question is solved by using the law of conservation of momentum. The total linear momentum after firing is given by the law of conservation of momentum. Using this law, calculate the velocity of gum. Use kinetic energy expression which gives the relation between velocity and mass. This way you will get the answer.

Formula used:
Formula of kinetic energy is given by,
Kinetic energy $\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\times \text{ mass }\times \text{ }{{\left( \text{velocity} \right)}^{\text{2}}}$
Using conservation of linear momentum
${{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}=0$
${{m}_{1}}$ = mass of the bullet
${{\vec{v}}_{1}}$ = velocity of bullet
${{m}_{2}}$ = mass of the gun
${{\vec{v}}_{1}}$ = velocity of gun

Complete answer:
We have given the mass of the gun and the mass of the bullet $M$ and $m$ respectively.
Before firing, the gun and bullet both are at rest, hence total linear momentum is zero. The total linear momentum after firing is given by
${{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}$
Using conservation of linear momentum
${{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}=0$
${{m}_{1}}$ = mass of the bullet
${{\vec{v}}_{1}}$ = velocity of bullet
${{m}_{2}}$ = mass of the gun
${{\vec{v}}_{1}}$ = velocity of gun
$\therefore -{{m}_{1}}{{\vec{v}}_{1}}={{m}_{2}}{{\vec{v}}_{2}}$
$\therefore {{\vec{v}}_{2}}=\left( \dfrac{{{m}_{1}}}{{{m}_{2}}} \right)\left( -{{{\vec{v}}}_{1}} \right)$
The negative sign in the above equation shows that the direction ${{\vec{v}}_{2}}$ is exactly opposite to ${{\vec{v}}_{1}}$.
So apply the law of conservation over here,
$MV=mv$
m = mass of the bullet
v=velocity of bullet
M = mass of the gun
V = velocity of gun
$V=\dfrac{m}{M}v$
We know that kinetic energy is given by,
Kinetic energy $\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\times \text{ mass }\times \text{ }{{\left( \text{velocity} \right)}^{\text{2}}}$
The kinetic energy of mass $M$ is given by.
Kinetic energy = $\dfrac{1}{2}M{{V}^{2}}$
Put the value of $V$ in the above equation, we get
Kinetic energy $=\dfrac{1}{2}M{{\left( \dfrac{mv}{M} \right)}^{2}}={{\left( \dfrac{m}{M} \right)}^{2}}\times {{\left( \dfrac{1}{2}M{{v}^{2}} \right)}^{2}}$
It is given that the mass of the bullet is small than the mass of the gun

So, the correct answer is “Option D”.

Note:
When a bullet is fired from a gun, the gun recoils. The bullet and gun move in the exact opposite direction. In this case, the linear momentum of the bullet is equal to the linear momentum of the gun. Mass of the gun is very large compared to the mass of the bullet, the recoil velocity of the gun is very small as compared to the velocity of the bullet in the forward direction, since mass has an inversely proportional relationship with velocity.