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Wraaaaaaaaaastling w/physics PART II



I talked about in a previous blog post about the physics that goes into getting an effective and violent take down, but I realized I could apply even more of my physics knowledge to this specific case. Because I know I can find the momentum of a system of two wrestlers during a takedown!!!!!! How awesome. Lets start.

The equation for momentum is of course p = mv, p being momentum, m being mass and v being velocity. I know mass pretty easily since we wrestle in weight classes, I am 170 pounds (77 kg) and so is my opponent! Now velocity is a little bit tricky, we can assume that my opponent is at 0 m/s since I have caught him completely off guard. For me I know that my shot takes about 6 feet (1.829 meters) and it takes about .79 seconds based on my official testing. Plugging in to some kinematic equations se know x = (1/2) (v final + v initial) t so solving for v final (v initial is 0) we get my final velocity as 4.63 m/s. So lets CALCULATE!!!!

We start with our equation, m1v1 + m2v2 = v (m1+m2)

We find my initial momentum to be 356.5 kg/(m/s), then solving for v final using conservation of momentum we find our final velocity with both bodies moving as 2.31 m/s. That is pretty fast for two big guys on a mat I think!

I thought it was atleast pretty cool to use my physics skillz to figure out just exactly how fast and how much momentum is in a system of wrestlers! Hope you did too!


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