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human
The physics behind breaking boards in karate

I was just curious and decided to work up a simple physical model of breaking boards, so I was hoping you guys could verify this idea:

**Motivation:** As with many other martial artists, I've broken my share of boards in my day.  I've found that if you hit a board and it doesn't break, it hurts like an SOB.  But if you do break it, it doesn't hurt at all.  You'd think hitting the board harder would hurt more (Newton's 3rd Law).

**Model:** We have a fist flying at the board with initial velocity v0. I modeled the board as a simple spring, F = k x.  If the board is stretched past a certain point, x=x_b, it will break and exert no more force on the fist. 

**Analysis:** *For the analysis, I dropped the spring constant, the mass, and various factors of 1/2. They aren't really relevant to the analysis, and they'd sort of muddle the point a bit. So I'll be using kinetic energy E=v^2 , potential energy U=x^2 , and momentum p=v.*  The initial energy of the fist goes as E_0 = v_0^2 , the initial momentum goes as p_0 = v_0. Once the board reaches x_b, it has absorbed the most energy it can absorb, U_b = x_b^2 (The corresponding momentum is p_b=x_b). If anymore energy is introduced it will snap; any less energy and it will return to its starting position.  The final energy of the fist then will be either E_f= 0 or E_f = E_0-U_b=v_0^2 - x_b^2 .  The final velocity then comes from: v_f^2 =v_0^2 -x_b^2 . That makes our final momentum v_f=√(v_0^2 - x_b^2 )

**Conclusions:** The energy taken away from the fist will always be U_b or less.  If E_0>U_b, the energy loss will always be U_b.  The momentum loss will be dp=v_0-v_f=v0-√(v_0^2 -x_b^2 ). The energy/momentum statistics of breaking a board will be the same if you just barely manage to break it vs if you just barely don't break it, ie. you've still lost U_b energy and p_b momentum over the same amount of time. Since x_b is constant, as v_0 gets higher and higher, dp actually decreases. That means the impulse from the board actually diminishes with harder punches.  This is the reason breaking the board doesn't hurt.

**Summary (TL;DR):** The reason breaking a board doesn't hurt, while not breaking it does, is that you break the board so quickly the board cannot deliver an impulse to your hand.  The amount of force the board resists with will always be the same, no matter how hard you hit it, but breaking through the board quickly makes it so the force can't do much damage.  Consequently, if you just barely manage to break the board, it will hurt almost as much as if you didn't break it, but hitting it even a little harder will drop the impulse considerably.
gpt
The conclusion is qualitatively correct, but you should analyze collisions from a momentum point of view instead of an energy point of view. The reason is that energy needs to take into account the sound, heat, deformation, etc. that are generated from the collision. Momentum is just straight up conserved.


Your fist has momentum p0, if it breaks the board, it has a lower momentum p1, but not by much. If it doesn't break the board, your fist essentially stops, and the board has to provide all of the momentum change, -p0, which hurts like a sonovabitch.