Jump to content
  • entries
    20
  • comment
    1
  • views
    3,676

Wraaaaaaaaaastling w/physics PART II


csoup88

518 views

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!

0 Comments


Recommended Comments

There are no comments to display.

Guest
Add a comment...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...