Originally Posted By: paul
the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.

Paul, here again you are making a simple mistake. Forces and momentums are vectorial - meaning they have both a magnitude and direction. Pressure is non-vectorial - it has a magnitude, but occurs equally in all directions.

It's not 100% clear what you're trying to claim, but I see two ways to interpret it.


The first interpretation is that you are claiming the pressure-mediated expantion of the air stream will dissipate its momentum, thus leaving only the momentum of the tank, and thus causing your pipe to move. This is wrong - think of the three spatial axis - x, y, z. Imagine your tank is expelling air in the -x direction - this would push the tank in the +x direction. Lets say 100kg*m/s of momentum is given to the tank. This means the tank has 100kg*m/s momentum in the x-axis, but 0 in the y/z axis. Likewise, the air expelled from the tank has -100kg*m/s momentum in the x-axis, but 0 in the y/x axis.

If I understand you correctly, you're saying the pressure of that air stream will counter the momentum of the air stream - as in the expansion of the air stream will dissipate the momentum. This is wrong. Keep in mind pressure is non-directional. So if there is 100PSI of pressure in the air stream, that means there is 100PSI pushing in the +x direction, -x directions, as well as the +/-y and +/-z directions.

So what happens? First, the obvious - the pressure expands the air stream in the +/-y and +/-z directions. In fact, if you were too look at any one air molecule it would now have more momentum, as it would have the initial -100kg*m/s of momentum along the x-axis, plus whatever momentum its now added on the y/z axis. Over all the stream of air has no net gain in momentum, as there will be equal expansion in the +/- directions on the y and z axis. Keep in mind, the expansion on the y/z axis has NO EFFECT on the momentum of the air stream, as this pressure-driven expansion along the y & z axis does not impact momentum on the x-axis.

The big Q is what happens on the x-axis, in regards to pressure. If you pick any one point along the x-axis of the air stream, and look immediately the the + and - of it, there will be essentially no pressure difference. Ergo, the pressure in the stream will equally push in the + and - x directions. Ergo, there is no net pressure-mediated force in the +/- x direction (i.e. expansion will be equal in the + and - x-directions), and thus no net change in momentum along the x-axis will occur.

Ergo, don't expect the pressure mediated expansion of the air stream to magically dissipate the momentum of the air stream - it won't. The net effect of the pressure of the air stream is it will expand it equally in all directions. What this means is any vectorial properties of the air stream (i.e its momentum or forces it applies) will not be affected by its pressure-mediated expansion, as that expansion has no vectorial value (i.e. its even in all directions). When that column of moving air hits the back of the pipe it'll still have a momentum of 100kg*m/s - the only difference is that momentum will be spread over a larger area.


The other way to interpret your claim is that the momentum of the air and tank will equal out (which they do), but the equalization of pressure will provide a net momentum. This interpretation too is wrong - because pressure is non-vectorial, it is incapable of inducing a vectorial force on your pipe. Equalization of pressure occurs equally in all directions; there is no net direction to the force (as air molecules move in a Brownian fashion). Since you have equal momentum of air in all directions the net force - and thus the net momentum - is zero.

Even taking into account simple mass-action, your net is still zero. You start by throwing a mass of air in the -x direction - this will give your pipe an equal momentum in the +x direction. However, the gas will want to reach equilibrium, meaning it'll expand towards the front of the pipe. The force of mass-action will equal (by definition) the amount of force needed to create the initial state. The movement of the air mass forward will be of equal momentum to the initial momentum of the air mass rear-wards. Net effect - pipe doesn't move.