Question

A gun fires a small bullet with kinetic energy $K$. Then the kinetic energy of the gun while recoiling is.
$\begin{align}
  & \text{A) K} \\
 & \text{B) More than K} \\
 & \text{C) Less than K} \\
 & \text{D) }\sqrt{K} \\
\end{align}$

Answer 1

Hint: Students can apply the law of conservation of momentum. Then applying the formula for kinetic energy will lead to the required relation. Recoil is generally the thrust produced when a firearm is discharged.

Complete step by step solution:
Let the kinetic energy of the gun be ${{K}_{g}}$

Initially the system is at rest and so the momentum of the system is zero. By the law of conservation of momentum, the final momentum should also be zero.

Thus if ${{P}_{g}}$ is the recoil momentum of the gun and ${{P}_{b}}$ is the recoil momentum of the bullet, then by law of conservation of momentum,

${{P}_{g}}=-{{P}_{b}}$
We know that,

Kinetic energy of the bullet,

$K=\dfrac{{{(-{{P}_{b}})}^{2}}}{2{{m}_{b}}}$,
where ${{m}_{b}}$ is the mass of the bullet.

Kinetic energy of the gun,

${{K}_{g}}=\dfrac{{{({{P}_{g}})}^{2}}}{2{{m}_{g}}}$,
where ${{m}_{g}}$ is the mass of the gun.

Since,

$\begin{align}
  & {{m}_{g}} > {{m}_{b}},{{P}_{g}}=-{{P}_{b}} \\
 & \Rightarrow {{K}_{g}} < K \\
\end{align}$

Therefore the answer is option $\text{C) Less than K}$

Additional Information:
If the gun was placed in the hand of a person in motion and the person moves with a particular velocity, then the initial momentum of the system is not zero. In that case the velocity with which the person moves will also come into play.

Note: Students must ensure that the recoil momentum is equal, but opposite in direction to each other. In case, it is considered to be only equal and one misses the opposite in direction part, it will not produce an accurate result.