How Does a Pellet Gun's Force Affect a Wall?

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The discussion focuses on calculating the average force exerted by a stream of pellets from a pellet gun on a wall. Each pellet weighs 2.14 grams and travels at a speed of 483 m/s, resulting in a specific momentum that contributes to the overall force. The average force exerted on the wall is derived from the impulse-momentum theorem, where impulse (J) equals the change in momentum. The confusion arises from integrating force over time to find the average force, particularly in relation to the contact time of 1.25 ms for each pellet.

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I am having trouble understanding another of my homework problems.

A Pellet gun fires ten 2.14 grams pellets per second with a speed of 483 m/s. The pellets are stopped by a rigid wall. a) find the momentum of each pellet. b) Calculate the average force exerted by the stream of pellets on the wall. c) if each pellet is in contact with the wall for 1.25ms, what is the average force exerted on th ewall by each pellet while in contact? Why is this so different from b?

I got the answers to a and c, but I don't understand b. My teacher mentioned something about integrating something, but I don't understand what. I know that J = Favg*deltaT and J = integral(F dt), but how do I use that to find Favg? He said you get an answer of something per second, because of the integral, but I'm just getting more and more confused about it.

Thanks!
 
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You realize that [tex]J = F_{ave} \Delta t[/tex], but what else does impulse equal? Answer/hint: The change in momentum.
 

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