Question

A bullet is fired from a rifle. If rifle recoils freely, then K.E of the rifle is:
A. \[Less\text{ }than\text{ }that\text{ }of\text{ }bullet\]
B. \[More\text{ }than\text{ }that\text{ }of\text{ }the\text{ }bullet\]
C. \[Same\text{ }as\text{ }that\text{ }of\text{ }the\text{ }bullet\]
D. \[Equal\text{ }or\text{ }less\text{ }than\text{ }that\text{ }of\text{ }the\text{ }bullet\]

Answer 1

Hint: Consider the law of conservation of momentum to solve this question. The law states that, unless an external force is applied, even if two or more bodies in an isolated system act upon each other, their total momentum will be a constant. But while comparing the mass and velocity of a bullet and gun we can understand that kinetic energy is different for both.
Formula used:
\[mv=MV\]
\[K.E=\dfrac{1}{2}m{{v}^{2}}\]

Complete answer:
According to law of conservation of momentum,
\[\text{Momentum of bullet = momentum of gun}\]
\[mv=MV\]
\[\text{m- mass of bullet}\]
\[M-mass\text{ }of\text{ }gun\]
\[V-\text{ }velocity\text{ }of\text{ }gun\]
\[v-velocity\text{ }of\text{ }bullet\]
Then,
\[V=\dfrac{mv}{M}\]
Considering the equation for Kinetic energy,
\[K.E=\dfrac{1}{2}m{{v}^{2}}\]
\[\text{Kinetic energy of gun, K}\text{.}{{\text{E}}_{g}}=\dfrac{1}{2}M{{V}^{2}}\]
Substitute, \[V=\dfrac{mv}{M}\]
\[K.{{E}_{g}}=\dfrac{1}{2}M{{\left( \dfrac{mv}{M} \right)}^{2}}=\dfrac{m}{M}\times \dfrac{1}{2}m{{v}^{2}}=\dfrac{m}{M}\times K.{{E}_{b}}\]
Where,
\[\text{ }\!\!~\!\!\text{ K}\text{.}{{\text{E}}_{\text{b}}}\text{- Kinetic energy of bullet}\]
Since the mass of bullet is less than that of gun,
\[\dfrac{m}{M} < 1\]
Hence,
Kinetic energy of bullet > Kinetic energy of gun

So, the correct answer is “Option A”.

Additional Information:
Recoil of a gun means sudden backward movement or jerk of gun when a bullet is fired from it. When a bullet is fired, the force sending the bullet forward is the same as the force sending the gun backward. But, due to the high mass of the gun, it moves only a small distance backward. This force gives a backward jerk or kick to the shoulder of the gunman. The gun is said to have recoiled. Initially before firing, both the bullet and the gun are at rest. Hence, before firing, the initial momentum of the bullet and gun is zero (Because their velocities are zero). When a bullet is fired, then it has the momentum given by,
\[Mass\text{ }of\text{ }bullet\times ~velocity\text{ }of\text{ }bullet\]
The bullet imposes an equal and opposite momentum to the gun due to which the gun jerks. The gun is said to recoil. The backward velocity of the gun is known as recoil velocity. The momentum of gun is,
\[Mass\text{ }of\text{ }gun~\times recoil\text{ }velocity\text{ }of\text{ }gun\]

Note:
When a gun exerts force on a bullet while firing it forward, the bullet will also exert an equal force in the opposite direction on the gun, hence the gun moves backward. Even though the force of action and reaction has the same magnitude, the effect on the gun and the bullet are not the same, since the mass of the gun is greater than that of the bullet. The acceleration of the bullet would be much greater than the acceleration of the gun.