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Applying Newton's 3rd Law after an object breaks

  1. Jul 31, 2014 #1
    In a martial arts breaking demo (boards, concrete blocks, or whatever) the amount of pain one's hand or other body part feels correlates with the amount of force applied, up to the point where the force is enough to break the object. If you hit it lightly, there is only a little pain, or if you hit it hard but not hard enough to break it there is a lot of pain. But if you hit it hard enough to break it, there is hardly any pain either--it feels like your hand goes right through it.

    I assume that the equal and opposite reaction to the force applied by your hand goes into your hand up to the point that you hand breaks through the object, but when the object breaks then most of the reaction gets absorbed by the object breaking. I'm sure this must be a thing in physics, but I don't know what it is called, which makes it hard to find more information on how to quantify the phenomenon. Also, I assume that probably impulse or maybe power are the key things I'm working with rather than force.

    I'd like to be able to calculate the behavior of one object as it encounters another object and breaks it, hopefully using an equation that I can look up values for different materials to plug into the equation.
     
  2. jcsd
  3. Jul 31, 2014 #2

    A.T.

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    That doesn't make any sense. If you call the force applied to the hand "reaction", why would it go into the object? When the object breaks the equal and opposite forces simply both go down.

    If the object doesn't break, all the kinetic energy of the arm has to be dissipated quickly. And that hurts.
     
  4. Jul 31, 2014 #3

    CWatters

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    Your hand has KE at the moment of impact.

    Think about where the energy goes if it doesn't break? Perhaps the KE is absorbed by the hand?

    Where does the energy go if it does break? Perhaps the energy is dissipated in breaking bonds in the object rather than in your hand?
     
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