A 0.0300 kg bullet is accelerated from rest to 600 m/s in 2.00 ms inside a gun barrel. What is the average force exerted on the bullet?

1. The bullet's change in momentum is delta p equal to (0.0300 kg)(600 m/s minus 0 m/s), which is 18.0 kg m/s. 2. The time given is 2.00 ms, which is 2.00 times ten to the minus 3 seconds. 3. The average force is F_net equal to delta p divided by delta t, which is 18.0 kg m/s divided by 0.00200 s. 4. That division gives 9000 N. 5. 900 N would come from a tenfold error in the time, and 90,000 N from the opposite tenfold error. _Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.2 Impulse_