Question

A gun fires a bullet of mass m with velocity v. The mass of the gun is M and it recoils with velocity V. If the kinetic energy of the bullet is ${E_b}$ and that of the gun is ${E_g}$, identify the relation between ${E_b}$ and ${E_g}$ :
A) ${E_b} = {E_g}$.
B) ${E_b} < {E_g}$.
C) ${E_b} > {E_g}$.
D) ${E_b} \geqslant {E_g}$.

Answer 1

Hint:Momentum is defined as the product of mass and velocity a momentum is a vector quantity. The linear momentum of a system is always constant until there is no interaction of any kind of external force on the system.

Formula used:The formula of the momentum is given by,
$p = m \cdot v$
Where momentum is p mass of the body is m and velocity is v.

Complete step by step answer:
It is given in the problem that a gun fires a bullet of mass m with velocity v the mass of the gun is M and it recoils with velocity V the kinetic energy of the bullet is ${E_b}$ and that of the gun is ${E_g}$ then we need to identify the relation between ${E_b}$ and ${E_g}$.
As there is no external force on the system therefore the linear momentum is conserved.
$ \Rightarrow m \cdot v = M \cdot V$
$ \Rightarrow V = \dfrac{{\left( {m \cdot v} \right)}}{M}$………eq. (1)
The kinetic energy of bullet is equal to,
$ \Rightarrow {E_b} = \dfrac{1}{2}m{v^2}$………eq. (2)
The kinetic energy of the gun is given by,
$ \Rightarrow {E_g} = \dfrac{1}{2}M{V^2}$
Replace the value of V from equation (1) in the above relation we get,
$ \Rightarrow {E_g} = \dfrac{1}{2}M{V^2}$
$ \Rightarrow {E_g} = \dfrac{1}{2}M{\left[ {\dfrac{{\left( {m \cdot v} \right)}}{M}} \right]^2}$
$ \Rightarrow {E_g} = \dfrac{1}{2} \cdot \dfrac{{{{\left( {m \cdot v} \right)}^2}}}{M}$………eq. (3)
Taking ratio of equation (3) and equation (2) we get,
$ \Rightarrow \dfrac{{{E_g}}}{{{E_b}}} = \dfrac{{\left( {\dfrac{1}{2}} \right)\dfrac{{{{\left( {m \cdot v} \right)}^2}}}{M}}}{{\left( {\dfrac{1}{2}} \right)m{v^2}}}$
$ \Rightarrow \dfrac{{{E_g}}}{{{E_b}}} = \dfrac{m}{M}$
Here the ratio $\dfrac{m}{M} < 1$ as the mass of the bullet is less than the mass of the gun. Therefore $\dfrac{{{E_g}}}{{{E_b}}} < 1$.
The relation between ${E_b}$ and ${E_g}$ is $\dfrac{{{E_g}}}{{{E_b}}} < 1$.

The correct answer for this problem is option C.

Note:The linear momentum is a vector quantity so if there is any change in the direction in the direction of movement of the body then we need to take momentum in different directions in which the linear momentum exits.