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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?
AAbout 300 N
BAbout 900 N
CAbout 90,000 N
DAbout 9000 N
Answer & Solution
Correct answer: D. About 9000 N
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_
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