Question

A bullet of mass $50\,g$ is fired from a $5\,kg$ gun with a velocity of $1\,km{s^{ - 1}}$. The speed of recoil of the gun is ?
A. $5\,m{s^{ - 1}}$
B. $1\,m{s^{ - 1}}$
C. $0.5\,m{s^{ - 1}}$
D. $10\,m{s^{ - 1}}$

Answer 1

Hint: The rule of conservation of linear momentum will be used to solve this problem, and after that, equations based on this theory and others will be used to calculate the speed of the gun's recoil.

Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where $P = mv$ denotes the momentum of a body.

Complete step by step solution:
According to the question, we have given that mass of the bullet is $m = 50g = 0.05kg$ and velocity of the bullet when it’s forced from the gun having mass $M = 5kg$ is $v = 1km{s^{ - 1}} = {10^3}m{s^{ - 1}}$.

So, momentum of the bullet is $mv = 0.05 \times {10^3}\,kgm{s^{ - 1}}$ and momentum of the gun is $MV = - 5V$ where V is the velocity of the recoil of the gun and gun will recoil in opposite direct as compared to direction of bullet, so equating momentum of bullet and the gun as we get,
$- 5V = 0.05 \times {10^3} \\
\therefore V = - 10\,m{s^{ - 1}}$
So, the magnitude of the velocity of the recoil of the gun is $10\,m{s^{ - 1}}$.

Hence, the correct answer is option D.

Note: It should be remembered that, backward momentum of the recoil gun is equal to the forward momentum of the bullet because initially the gun and bullet were at rest so final momentum must be equal to the final momentum of the system as per law of conservation of linear momentum.