# Punch Physics

The total energy available to transfer to the opponent will be the difference in the total kinetic energy before and after the collision. This difference is:

DE = 1/2 M1V12 - 1/2 (M1 + M2)V'2

Plugging in the above expression for V' yields:

DE = 1/2 M1V12 - 1/2 M12V12/(M1 + M2)DE = 1/2 M1V12(1 - M1/(M1 + M2))DE = 1/2 M1V12(M2/(M1 + M2)DE = 1/2 M1M2V12/(M1 + M2)

In general, the mass of a person's arm is about 10% of the total mass of his or her body. So if we assume that the puncher and the opponent have about the same mass we have M1 ~ 0.1M2, and the above expression reduces to:DE = 1/22 M2V12If we assume a standard 70-kilogram person, and that the punch makes contact at its maximum velocity (~7 meters per second for a black belt) then we have:DE = 156 JAt first, this looks more than sufficient to break an arm bone. But the real-life situation is more complicated; in most cases an arm is more likely to just move aside when hit, rather than deform and break. Ribs, however, may move very little. It is advantageous to break the bones of your opponent, but you would prefer to avoid breaking the bones in your own handâ€”these bones are even smaller (and thus more vulnerable) than arm or rib bones. So how do Taekwondo punchers break bones, or boards, without damaging their hands?