Mike,
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Mike Sigman wrote:
The equation M1V1 = m2v2 can be looked at as the basic explanation of why a billiard ball can roll into another billiard ball and the first ball stops while the second ball begins rolling away. If we make the mass (M1) of the first ball to be twice the mass of the second ball, then the second ball (in a perfect, inelastic collision, neglecting friction, etc.) will wind up rolling away at twice the velocity the first ball arrived with.

I *think* you mean 'elastic collision' rather than inelastic. An elastic collision preserves the most kinetic energy by conserving all of it. I was bored a while ago, so I did the math on this. It doesn't really change your point, but it's a detail that's worth considering if one starts really thinking about what can happen:
If the first ball is twice the mass of the second ball, then actually (in an elastic collision) the second ball will only roll away at 1 1/3 the speed the first ball started with. This is because in order for kinetic energy to be conserved, the first ball has to continue rolling at 1/3 of its original speed. Once the first ball's mass exceeds the second ball's, it is not possible for it to transfer all of its momentum. The reason this is possible with pool balls is because they each have the same mass.
The best you can do with an elastic collision is for the second ball to leave with double the speed the first ball arrived at  but this is the theoretical limit as the ratio of the masses approaches infinity. For example, if the first ball is 100 times heavier than the second, the second ball leaves at 1.98 times the original speed.
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In a lot of ways, you can look at the torso/middle of our body as the larger first ball and the arm/fist as the smaller second ball. The same transfer of momentum can happen if you connect things up correctly.

And the above analysis points out that the proportionally bigger the first ball, the closer you can come to getting double the velocity out of the second ball. However, that's the best you can do. So once you've gotten those proportions worked out, the only way (within this formula) to add more kinetic energy to the second ball is to increase the velocity of the first ball. You can only go so far by increasing mass (to the first ball; increasing mass of the whole system will always help, but any increase in mass of the second ball needs to be proportionally matched in the first ball in order not to slip back in terms of velocity transfer). All else being the same, the faster you can get the first ball going, the faster the second ball will go.
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Better yet, make it 3 balls. The earth is the first ball, the torso/tanden is the second ball, the arm/fist would be the third ball.

But of course you cannot actually get the earth moving. What you *can* do is take advantage of its mass to accelerate the second ball by knocking it off the first ball. In an elastic collision, this won't actually buy you anything. However, if you are adding power to the ball over time, it gives you a longer path over which to accelerate the middle ball (more work = more energy). Better still, if you can make the collision with the ground be 'superelastic' then the impact itself will effectively add energy. The only way to accomplish that is for some form of stored energy to be brought into play. If you imagine colliding with the sprung part of a mousetrap, you can see that you might be able to leave the collision with more energy than you entered with. Of course this only works if someone else put the energy there by setting the mousetrap in the first place.
I know *you* know this Mike, more clearly I am sure than I do. I just threw it out there for the number geeks to chew on. If my math's wrong, please point it out  but I think I got the equations right.
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