Center of Mass and Linear Momentum of pellet gun

[michaelmayhem]

1. A pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall. What are (a) the magnitude of the momentum of each pellet, (b) the kinetic energy of each pellet, and (c) the magnitude of the average force on the wall from the stream of pellets? (d) If each pellet is in contact with the wall for 0.60 ms, what is the magnitude of the average force on the wall from each pellet during contact? (e) Why is this average force so different from the average force calculated in (c)?



2. Can someone please explain to me step by step how I might go about solving this problem? I am stuck. Thanks. FYI this isn't really homework due tomorrow but I though I would try and get ahead.



3. I was going to post my work here but I'm sure it's completely wrong.

 

[CompuChip]

Welcome to PF Michael. It's a good thing that you're trying to get ahead; now that you got stuck you have plenty of time to really understand the problem without the pressure of time.

a) and b) are rather easy, basically it's just plugging given numbers into the correct formula. Can you write down the formula for momentum and kinetic energy (you should know it by heart)?

Actually for c) I think you cannot really calculate the number, unless I am missing something. For d) you can again use an appropriate formula, which relates the average force to the change in momentum... can you find that formula?

 

[michaelmayhem]

well K=1/2mv^2
and P=mv
I'm not really sure how to incorporate these into the equation.
I'm not really sure about the formula that relates average force to the change in momentum.

 

[Doc Al]

Actually for c) I think you cannot really calculate the number, unless I am missing something.

The only difference between finding the average force in c) versus the average force in d) is the time interval over which you are averaging. For c) consider a time interval of seconds.

For michaelmayhem: Look up the impulse-momentum theorem.

 

[michaelmayhem]

Oh yes the impulse is equal to the change in momentum. I can find the impulse by multiplying the force times the change in time. The momentum is simply P=mv.

 

[CompuChip]

The only difference between finding the average force in c) versus the average force in d) is the time interval over which you are averaging. For c) consider a time interval of seconds.

Yes, I thought something like that too, but to really calculate anything quantitatively you would need the number (e.g. the distance from the wall or the time-of-flight). Assuming that t ~ seconds will only give you an estimate.

Oh yes the impulse is equal to the change in momentum. I can find the impulse by multiplying the force times the change in time. The momentum is simply P=mv.

Very well, that's the results I was hinting at.

 

[Doc Al]

Yes, I thought something like that too, but to really calculate anything quantitatively you would need the number (e.g. the distance from the wall or the time-of-flight). Assuming that t ~ seconds will only give you an estimate.

You are given that 10 bullets, of known momentum, are fired every second--thus 10 bullets hit the wall every second. That's all you need to calculate the average force.

 

[CompuChip]

You are given that 10 bullets, of known momentum, are fired every second--thus 10 bullets hit the wall every second. That's all you need to calculate the average force.

OK that was kind of stupid... completely missed that piece of info.

 

[michaelmayhem]

Ok thank you. I'll give it another go.