Question

Typically a bullet leaves a standard 45 caliber pistol at 262 m/s. If it takes 1 millisecond to transverse the barrel, determine the average acceleration experienced by the 16.2-gram bullet within the gun's barrel and then compute the average force exerted on it.

Answer 1

By definition, the acceleration is

$$a = \dfrac{\Delta v}{t}$$

where $\Delta v = 262m/s-0m/s = 262m/s$ is the change in speed and $t = 1ms = 10^{-6}s$ is the time. Thus

$$a = \dfrac{\Delta v}{t} = \dfrac{262m/s}{10^{-6}s} = 2.62\times 10^8m/s^2$$

According to the Newton's second law, the force is

$$F = ma = 16.2\times 10^{-3}kg\cdot 2.62\times 10^8m/s^2 = 4.2444\times 10^6N$$

Answer. $a = 2.62\times 10^8m/s^2$ and $F = 4.2444\times 10^6N$.