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Old 04-16-2007 | 12:59 PM
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bow_hunter44
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From: Idaho
Default RE: Who said that physics wasn't fun?

ORIGINAL: doublelunginem

Bowhunter44, Thanks for the terminology...it's all coming back to me...lol
So the KE would have to be a parabola since it is the integral of the draw force curve or the area under the curve.

Actually the graph of KE is parabolic because it is a quadratic function (the velocity squared). The integral of the draw force curve (yes the area under the curve) represents the total potential energy stored in the bow at full draw.

Just trying to refresh here, so how would the momentum graph reach a vertex (which we both agree that it must) since it is a linear function. I guess my thoughts are leading me to the fact that in this situation of dealing with bow performance, momentum would have to be graphed in relation to KE. As I stated before, when we get to the vertex of the KE graph, we are dealing with the maximum efficiency of the bow in the test.

Where the maximum effeciency of the bow is realtive to the vertex of the parabola of the KE graph, I honestly don't know. I don't have an intuative feel foreither. Great questions! Arthur, can you help out?

Once this is fixed, velocity decreases exponentially as weight goes up. So would this mean that the graph of the momentum be a " / " with half of an upside down parabola following it and connecting at the highest point?

I doubt it - unfortuantely. I suspect if that were to happen it would be serendipitous and not out of necessity.

The 2nd half would resemble y= 1/x ? Or would it be y = -x? Dude, you ask great questions!!

What I am attepting to prove with out the math proplems or actual test results is that momentum and KE are so closely related that they would maximize at the same point. If infact, velocity is decreasing "exponetially" once the bow has maximized efficiency and KE is maximized, then momentum would have to decrease exponentially at that point as well.

Again, I doubt that is necessairly true. Instead of the decrease of momentum being a graph of P = 1/mv it could be P = -mv, simply a negative slope of the linear momentum graph (P = mv). But again, you ask good (tough to answer) questions.

P.S...I had a physics and chemistry teacher simular to you in HS....he's the one that got me so interested in this stuff. We had a lab every week and it was all real life scenerios. We would spend Monday learning the physics and the rest of the week performing the physics that we learned...Lab report was due by end of class on Friday for a grade. He never taught from a book and never helped with the Labs...you had to listen on Monday very closely and figure it out on your own. He was the BEST DAMN TEACHER I ever had!!!!

Well, I doubt I'm the best damn teacher around. But it sure is nice to hear that some of us do make a difference. Obviously your physics teacher did (make a difference) as your recall of physics is great and you analytical skills are great as well!
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