Question

A bullet of mass 10g is fired from a gun. The bullet takes 0.003s to move through its barrel and leaves it with a velocity of 300m/s. Find the force exerted on the bullet by the gun? [ Ans: 1000N]​

Answer 1

Answer:

1,000 N

Explanation:

$v=300 \text{ms}^{-1} ;u=0,t=000.3\text{s},m=0.01\text{kg}$

Force exerted by the bullet on the rifle = $\frac{m(v-u)}{\text{time}}$

F = 1000N

Contextual Way:

Newton's second law of motion.

F=m a

Now to solve based on the current info, we shall assume that:-

The force exerted on the bullet was uniform across the entire duration of bullet leaving the barrel, i.e., 0.003 seconds {Not necessarily true for real life applications as the force will not be uniform from the point of hammer impact till the point of leaving the barrel. In reality you will get a Force profile across that entire duration}

We are not distinguishing between bullet and cartridge. {What you shall hold in your hand and load in a revolver is a cartridge containing the gunpowder, bullet etc. The bullet is the projectile at the mouth of the cartridge that actually leaves the barrel and hit the target. So when you are weighing in real life, you are not weighing the bullet, rather the cartridge as a whole}

Getting back to the question

Impulse equation for the bullet

∫F∗dt=∫m∗dv

Average impulse delivered= Change in momentum of the bullet

Assuming average force delivery

Favg∗∫dt=m∗∫dv

Favg∗0.003=0.010∗300

Favg=300∗10/3

Favg=1000N