Question

A bullet is fired from a rifle. If the rifle recoils freely, then the kinetic energy of the rifle is

• A. Less than that of the bullet
• B. More than that of the bullet
• C. Same as that of the bullet
• D. Equal or less than that of the bullet

Answer 1

The correct option is A

Less than that of the bullet

Step 1. Given data

A bullet is fired from a rifle and the rifle recoils freely.

Step 2. Explanation for the correct option

According to the law of conservation of momentum, the momentum of the bullet, $p_b$ is equal to the momentum of the rifle, $p_r$

By using the formula of momentum, momentum is equal to the product of the mass and the velocity.

$p_b=p_r$

$mv=MV$ [Where, $m$ is mass of the bullet, $v$ is velocity of the bullet, $M$ is the mass of the rifle, $V$ is the velocity of the rifle.]

$V=\frac{mv}{M}$ [$eq\left(1\right)$]

Now, the Kinetic energy is equal to the half of the product of the mass and the square of the velocity.

Kinetic energy of bullet,

${\left(K.E.\right)}_b=\frac{1}{2}m{v}^2$

Kinetic energy of rifle,

${\left(K.E.\right)}_r=\frac{1}{2}M{V}^2$

From $eq\left(1\right)$ putting the value we get,

${\left(K.E.\right)}_r=\frac{1}{2}M{\left(\frac{mv}{M}\right)}^2$

${\left(K.E.\right)}_r=\frac{m}{M}\times \frac{1}{2}m{v}^2$

${\left(K.E.\right)}_r=\frac{m}{M}\times {\left(K.E.\right)}_b$

Since, the mass of the bullet is less than the rifle

$m<M$

$\frac{m}{M}<1$

${\left(K.E.\right)}_r<{\left(K.E.\right)}_b$

Therefore, the kinetic energy of the rifle is less than the kinetic energy of the bullet.

Hence option $\left(A\right)$ is correct.