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This is an interesting situation.

You have said "low friction ice". We think what you're saying is that the ice is

doing nothing but holding them up, but nobody can get any traction with their

boots to pull against. The only way they can pull on the rope is with their arms.

So they might as well be two people, facing each other at opposite ends of a rope,

in outer space. We can't help visualizing it that way.

Everything starts with both people motionless and no slack in the rope, and then

the referee tosses the flag. (He can't just drop it because it won't go anywhere.)

If either person pulls on the rope, two things can happen: Either the other person

lets it go completely, or the other person holds onto it tight.

-- If the other person lets the rope go completely, then there's no way to get any tension

in the rope, and the person that's pulling winds up with a bunch of rope in his face.

-- The other person holds tight to the rope so that the rope develops tension in it:

It seems to us that whenever there's tension in the rope, whether both people are

pulling or one of them is just holding tight, the force of the end of the rope against

each person is going to wind up equal, and they'll accelerate equally toward each other.

Well, maybe not equally ... the skinny will accelerate toward the center faster than

the fatty does. F=M A, the force at each end is equal MfatAfat = MthinAthin, and the

accelerations are inversely proportional to their masses.

Athin = Afat (Mfat / Mthin) = 4 times the acceleration of the fatty.

We also think that since the whole thing was initially motionless, there's nothing

that either person can do to move their center of mass from where it was when

they began. The isolated system composed of two people and a rope began with

no momentum, so the whole isolated system never gets any, no matter what they do.

As long as both of them hold the ends of the rope, the point on the rope where their

center of mass is never moves anywhere.

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