Question

Find the power of a machine gun that can fire $240$ bullets per minute, each weighing $50\,g$, with a velocity of $100\,m{s^{ - 1}}$
A) $10\,kW$
B) $1\,kW$
C) $5\,kW$
D) $7\,kW$

Answer 1

You can solve this question by considering the fact that power is just the measure of ability to do a given work, or ability to convert energy from one form to another, here we can see that the machine gun is giving energy to the bullets, so you can easily find the power of the machine gun by measuring the total energy given by it to the bullets, per unit time.

Complete step by step answer:
We will be proceeding with the solution exactly as we told in the hint section of the solution to the question. We will first measure the energy of a single bullet when it just exits the machine gun, after doing that, we will find the energy of all the bullets and thus, the total energy provided to bullets in each second by the gun, which will be the final required answer.
Weight of each bullet, $\left( m \right) = 50g = 0.05\,kg$
Velocity of each bullet, $\left( v \right) = 100m{s^{ - 1}}$
Using this, we can find the kinetic energy of bullet when it just exits the machine gun as:
$K = \dfrac{1}{2}m{v^2}$
We already know the values of mass and velocity of the bullets, substituting them in the formula, we get:
$K = \dfrac{1}{2}\left( {0.05} \right){\left( {100} \right)^2} \\\ K = 250\,J \\\$
Since, in a minute, $240$ bullets are being shot, we can find the change in total kinetic energy given by the machine gun to the bullets as:
$\Delta {K_{net}} = 240 \times K \\\ \Delta {K_{net}} = 240 \times 250\,J \\\ \Delta {K_{net}} = 60000\,J = 60kJ \\\$
We will be assuming that during this short instance of time when the bullet just exits the machine gun, there is no change in the height of the bullet or machine gun from the earth’s surface, hence, change in potential energy can be neglected, so, we can write as the work done by the machine gun in a minute as:
$W = \Delta {K_{net}} + \Delta {U_{net}}$
After substituting the values that we found out:
$W = 60\,kJ$
Since, power is given as the ratio of work done and the time taken to do the work, we can mathematically see that:
$P = \dfrac{W}{t}$
Time taken is given to us as a minute, which means $60\,s$ .
Substituting the values, we get:
$P = \dfrac{{60\,kJ}}{{60\,s}} \\\ P = 1\,kW \\\$

Hence, the correct answer is option (B).

Note: Another way of reaching the answer could have been to first find the number of bullets being shot per second and then multiplying the kinetic energy change of one bullet with only that much number of bullets. Many times, students take time in minutes instead of seconds and lose marks.