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Question: A bullet of mass \(15{\text{ }}g\) leaves the barrel of a rifle with a speed of \(800{\text{ }}\dfra...

A bullet of mass 15 g15{\text{ }}g leaves the barrel of a rifle with a speed of 800 ms800{\text{ }}\dfrac{m}{s}. If the length of the barrel is 75 cm75{\text{ }}cm, what is the average force that accelerates the bullet?

Explanation

Solution

For this question, we have to use motion’s equation to find the acceleration of the bullet. Then by using Newton’s Formula of force we will find the average force associated with the bullet when it accelerates.

Complete step by step answer:
It is given in the question it is stated that a bullet of mass 15 g15{\text{ }}g leaves the barrel of a rifle which is of length 75 cm75{\text{ }}cm with a speed of 800 ms800{\text{ }}\dfrac{m}{s}. We have to find the average force with which the bullet accelerates.
Let the acceleration of the bullet be aa.
We have to use the motion’s equation to find the acceleration. The motion’s equation is,
v2−u2=2as−−−−−−(1){v^2} - {u^2} = 2as - - - - - - \left( 1 \right)
The variables are defined as,
v=v = final velocity of the object
u=u = initial velocity of the object
a=a = acceleration of the object
s=s = distance covered by the object
For the given question,
As the bullet is fired from rest position then the initial velocity, u=0 msu = 0{\text{ }}\dfrac{m}{s}
The final velocity of the bullet is the velocity achieved by it when it was fired, v=800 msv = 800{\text{ }}\dfrac{m}{s}
The distance covered is considered to be the length of the barrel of the gun, s=75 cm=0.75 ms = 75{\text{ }}cm = 0.75{\text{ }}m
Substituting all the values in equation (1)\left( 1 \right) we get,
(800)2=2a×0.75{\left( {800} \right)^2} = 2a \times 0.75
⇒a=64×1041.5=426666.66\Rightarrow a = \dfrac{{64 \times {{10}^4}}}{{1.5}} = 426666.66
The acceleration of the bullet is 426666.66 ms2426666.66{\text{ }}\dfrac{m}{{{s^2}}}
From the Newton’s Law we get,
F=ma−−−−−−(2)F = ma - - - - - - \left( 2 \right) where F=F = force, m=m = mass and a=a = acceleration
It is given mass of the bullet m=15 g=0.015 kgm = 15{\text{ }}g = 0.015{\text{ }}kg and we got a=426666.66 ms2a = 426666.66{\text{ }}\dfrac{m}{{{s^2}}}
Substituting the values in equation (2)\left( 2 \right) we get,
F=0.015×426666.66=6399.99F = 0.015 \times 426666.66 = 6399.99
Hence, force F≈6400 NF \approx 6400{\text{ }}N
The average force that accelerates the bullet is 6400 N6400{\text{ }}N.

Note: It must be noted that we have to convert every value in SI unit for the ease of calculation, or there appear various mistakes. The final velocity is considered to be velocity given as, it is asked to find the force just when the bullet is fired.