Freshman-level justification for the use of $(x,p)$ in thermodynamic phase space I'm looking for an elementary explanation, at the freshman level, of why we use position and momentum for phase space rather than, say, position and velocity.
At this level, it's not going to work to appeal to generalized coordinates and such. I think there are probably arguments that can be constructed based on Liouville's theorem, which basically tells us that $dx dp$ is the natural measure for probability, but again, that doesn't fly at this level, nor can I appeal to symplectic-blah-blah.
Reif is a book at this level that generally pays careful attention to such foundational issues, but all he seems to have is the following footnote on p. 226:

If q denotes an ordinary cartesian coordinate and if no magnetic field is present, the momentum p is simply related to the velocity v of the particle of mass m by the proportion­ality p = mv. The description in terms of the momentum p rather than the velocity v is, however, valid in more general cases and is thus the one commonly used.

This seems to be both obscure and not a real justification.
The other undergrad book that I'm familiar with is Kittel, which is an upper-division text. Kittel's philosophy is to aggressively use quantum mechanics. The publisher's blurb says that the book "offers a modern approach to thermal physics that is based on the idea that all physical systems can be described in terms of their discrete quantum states, rather than drawing on 19th-century classical mechanics concepts." Perhaps for this reason, there does not seem to be any discussion of this in Kittel. There is not even an entry in the index for "phase space."
The best hand-wavy justification I've been able to come up with is the following. It's not hard to see what goes wrong if we try to use velocities instead of momenta. When object of different mass collide, they cannot transfer energy in an arbitrary way while still obeying conservation of momentum. For example, if you throw a golf ball with an energy of 10 joules, and the golf ball hits a bowling ball, it is not possible for the bowling ball to absorb all 10 joules of energy from the golf ball in the form of kinetic energy. The result is that statistically, in such collisions, there is a tendency for the less massive object to undergo bigger accelerations and have larger velocities. This means that it is not reasonable to assign the same probability per unit velocity to the golf ball as to the bowling ball.
Can anyone supply something that's more of a real justification than this, while not appealing to lots of knowledge beyond the freshman physics level?
 A: Here is an elementary intuitive justification of the Liouville theorem. You may want to add pictures. It is not a proof.
1 - The Liouville theorem applies to complex functions that have continuous derivatives of all orders. 
The derivative of a complex function is a much stronger condition than of a real function. It is the slope of a tangent plane instead of tangent line. You can draw a tangent line through a point from any direction. All such lines must have the same complex slope. This must hold true on both sides of the point. All lines must be coplanar. 
2 - Even though the magnitude of a complex function can be concave up or down at a point, the magnitude cannot be concave down at the maximum. That is, the magnitude cannot have a hilltop or ridgetop at its maximum. 
Suppose it did have such a maximum. The tangent plane must be horizontal. The derivative of the magnitude must be $0$.
We said nothing yet about the phase of the slope. It must be the same in all directions through the point. But if the magnitude of the slope is $0$, the phase doesn't matter. 
The second derivative is also a continuous, differential function. For any line through the point, the magnitude cannot be positive on one side and negative on the other. The magnitude of this must also have a horizontal tangent plane at the point. The derivative must be $0$ at the point. 
A real function can be concave down at the max, and still have first and second derivatives be $0$. E.G. $y = -x^4$. Some higher order derivative in a neighborhood of the point must be non-zero. For a complex function, We can make an argument like above that derivatives of all orders must be $0$ at the point. 
A power series expansion implies the function must be constant for some neighborhood of the point. 
3 - The neighborhood where the function is constant cannot be bounded. 
If it was, consider a point on the boundary. Any order derivative would be $0$ on one side. It would therefor have to be $0$ on the other. If all order derivatives are $0$, there is a neighborhood of this boundary point where the function is constant. This isn't a boundary point.  
Therefor the region where the function is constant must cover the entire complex plane. 
