Physics Question

The muzzle flips because the force is off of the center of gravity. This is true everywhere. Then when the slide chambers the new round the flip should be slightly reduced, but I imagine that it will keep going end over end until it hits something.

Although it has no 'weight' it still has mass. The gun would do everything it does on Earth except fall to the ground. Due to the mass differential between the slide area and the grip, there would be a slight amount of rotation imparted to it, but not much. What causes muzzle flip is the slide trying to move rearward while the grip's movement is resisted by the hand of the shooter. The return of the slide will have little effect on the movement of the gun, the net result will be:

m(bullet) x a(bullet)=m(pistol) x a(pistol)

Because the pistol weighs more than the bullet, its rearward acceleration will be a lot less than that of the bullet.

Since there is no resistance, do both the gun and the bullet travel equal distances?

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If infinity = infinity, then yes.

If not, then I don't know.

It seems likely that the slide wouldn't cycle fully. Without a supporting hand, arm, and body, the frame and loaded magazine alone probably wouldn't provide enough inertia to resist the recoil spring.

It would be like the worst possible case of limp wristing.

stovepipe.

The pistol would spin backwards like a top, with a sideways motion imparted by the rifling torquing from the bullet mass, and whether it fully cycled would depend on the strength of the ammo vs. the timing of the gun and recoil spring strength.

with a sideways motion imparted by the rifling torquing from the bullet mass

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Ah yes, I forgot to take that into consideration.

I think you are in the realm of calculus because of so many different factors involved. The muzzle flips because the center of mass will be below the barrel. It will spin because even though the pistol has more mass it will still spin the opposite of the bullet spin. With each shot the pistol will gain speed as each shot accelerates the gun a little more. You have the gun spinning on a vertical axis and a lateral axis and depending what the attitude of the gun is when the next shot is fired you won't know the direction of travel, we need to know how often the gun is fired, exact time intervals. Each shot will affect the gun differently because it will have less mass with the ejection of the bullet, powder and brass. We need to know where the nearest gravity wells are and distance traveled by the projectile and the pistol will be dependent on that. I expect you are going to need a pretty good computer system to handle the trajectory of the pistol with pitch and yaw. The bullets will be slightly easier if you know the exact direction the gun was pointed at when fired. I almost forgot the mass of the falling hammer will affect the motion of the pistol too. Interesting question.

Even though the gun is now spinning around its Z-axis (from muzzle flip) and its X-axis (from rifling), the three points of gun, point of ignition, and bullet will always form a straight line (Euclidian straight if there are no other gravitational bodies, or Einsteinian time-space curved if there are), as long as the brass stove-pipes (I think it would, although with a light or no recoil spring, it obviously wouldn't).

If the brass is ejected from the gun, the gun would now get a different trajectory (no longer on the same line as the bullet and point of ignition), as well as slightly change its spin. All objects would now move in straight lines, and the gun, brass, and point of ejection would always form a straight line. Therefore, the bullet, gun, brass, point of ejection and point of ignition would all remain in the same Euclidian plane (discounting gravity).

Since space is not a perfect vacuum (solar wind, etc.), it is likely the objects would eventually fall off their origninal lines and plane, and the brass would eventually stop, then the gun, then the bullet. However, there is too little gravitational attraction between the gun, brass and bullet to ever cause them to be attracted back to each other.

And they'd most likely eventually crash onto another body, or burn up approaching it.

See what happens when you ask a question?

As the bullet was launched during muzzle flip, its axis of rotation (from the rifling) will be angeled away from the trajectory, with bullet nose "down" (toward the 6 o'clock position on the muzzle, at the time of launch).

If there is not perfect bullet symmetry and perfect rifling, the bullet will also precess (wobble) as it spins.

Put a gun on it's side on a very smooth table in a bullet proof chamber.

Same effect.

The table will not allow x- or y- axis rotation from rifling and brass ejection.

Clarification: muzzle flip is not cause by slide action, but by the bore being above the center of gravity of the pistol.

It would behave pretty much the same as if you threw it out a window here on earth and fired it remotely as it was falling. The round in the chamber would fire normally, the round would exit the barrel in the direction the pistol was pointing, the pistol would be propelled backward in the opposite direction (not terribly fast, maybe 10 mph or so?), it would be spinning, and the next round probably wouldn't chamber. It would still be falling, of course, but all else would be as it would be in space until it started falling fast enough for aerodynamic forces to come into play.

No way would the next round chamber, the force required to operate the slide would be transferred into backward movement of the gun, the exaggerated version of "weak" gripping a semi-auto and causing FTF. In a perfect vaccuum, the gun would be propelled backwards with the same energy(E)as was expended by the bullet mass( M) times the square of the velocity. With the mass of the gun differing significantly from the bullet, it would not react with the same velocity as the bullet.

"the three points of gun, point of ignition, and bullet will always form a straight line (Euclidian straight if there are no other gravitational bodies, or Einsteinian time-space curved if there are)"

All the bodies are going to move about their respective centers of gravity (since it is also the center of mass).
The CGs will describe a straight line at short distances until other masses (and their gravity if they are large) come into play.

This is fun!

Any more hypothetical questions?

Does the muzzle flip?

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Yes. It will flip end over end backward, and continue to do so until colliding with some matter.

Does the gun travel straight back?

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Not sure. In addition to the spinning movement, it will also travel backward - whether it's perfectly horizontal - parallel to the bullet line of travel, I'm not sure - it may go mostly back but slightly down, or mostly back but slightly up. IOW, it will go back but probably at a small angle one way or another. Also, due to the rifling twist, in addition to the end over end movement and backward movement, it will also take on a right-to-left spinning movement.

Since there is no resistance, do both the gun and the bullet travel equal distances?

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Yes, which is to say, an infinite distance, or until they collide with matter.

Does the returning slide push the gun forward?

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It will, but probably does not have nearly enough magnitude to overcome the recoil forces which have first acted upon it - it may stop or greatly decrease the backward spin due to muzzle flip.

Those are my somewhat-educated guesses.

I don't know, But I'm going to wack the guy that own's the 1911 with my light saber, before he figures it out!

OK, Next question, When the gun in space goes bang, and no one is there, Does it really go BANG?

Claymore1500 said:

OK, Next question, When the gun in space goes bang, and no one is there, Does it really go BANG?

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Zen riddles aside:
Although there would obviously be no atmosphere to convey sound, if you were in a spacesuit close enough to the muzzle when it discharged, you would probably hear some of the propellant gas impacting your suit.

Dont fire your guns into space gave me headaches..

The muzzle flips because the force is off of the center of gravity.

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The center of exactly what gravity in a weightless environment?

I'm an idiot, with a degree in physics. What I originally said was incorrect upon further review.