Question

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gunpowder. What is the average force exerted on a 0.0100 kg bullet to accelerate it to a speed of 450 m/s in a time of 1.80 milliseconds?

Answer 1

The average force exerted on the bullet can be calculated by using Second Newton Law:

`F_{\text{avg}}=ma`

where m is the mass of the bullet and a its acceleration. Consider that acceleration is determined by the following expression (from kinematics):

`a=(v-v_o)/(t)`

where v is the final speed of the bullet, vo the initial speed and t the time of the change of the velocity.

Then, by replacing the previous expression for a into the formula for Favg, you have:

`F_{\text{avg}}=m((v-v_o)/(t))`

In this case,

m = 0.0100 kg

v = 450m/s

vo = 0m/s (the bullet starts from rest)

t = 1.80ms = 1.80*10^-3 s

Replace the previous values of the parameters into the formula for Favg and simplify:

`\begin{gathered} F_{\text{avg}}=(0.0100kg)((450(m)/(s)-0(m)/(s))/(1.80\cdot10^(-3)s)) \\ F_{\text{avg}}=2500N \end{gathered}`

Hence, the average force exerted on the given bullet is 2500N