Physics Question

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... could not resist to answer some misconceptions ...

1. "zero gravity" - FALSE
Objects in space do affected by gravity. There are plenty of massive objects still around - planets, stars ... even if we are FAR away from everything, we still going to orbit at least the galaxy.
Now, there is no -weight-, but still -mass-. Those two quantities are DIFFERENT physics things, describing completely different things.
Mass is a measure of matter, it is a coefficient in a Newton's famous equation: F=ma. This quantity does not depent on where you are. The units of "m" is kilograms [kg].
Now -weight- is a FORCE which object produces on a support, measured in newtons [N]. This quantity does depend on how you move. When you are standing still, your weight = your mass X gravity constant of the place you are standing. NOTE: scales are measuring WEIGHT but shows it to you in kilos or pounds. Spring scales shows your mass correctly only when they are standing still. Now imagine you and the scales are jumping from a building ... you and scales are accelerating with the same g and flying the same way. You not going to exert any force on the scale, and they will show zero.
Now back to space ... spacecraft, astronaughts and ALL objects inside are orbiting the planet, affected by the same gravity acceleration and in never-ending free-fall, thus the -weight- of all objects are zero, but they still have mass.

2. "Does the muzzle flip?"
It depends on the design. On a typical handgun - yes, it will spin it.
The reason is that the line of linear momentum (not force as people before write) are not lined up with -center of mass- (not gravity)
The off-center linear momentum means we have angular momentum which will result in angular acceleration. Note that a shot has finite time of interaction with a gun, and thus resulting with finite final angular speed of the gun (it will be rotating). Every additional shot will increase gun's angular speed.

3. "Does the gun travel straight back?"
The gun will travel straight back while rotating. If you manage to fire second shot at the moment gun will point back compared to first shot (and assuming all ammo is perfectly identical) ... and NO YOU GUESSED WRONG ... the gun will not stop, because at the point of 2nd shot (compared to 1st) the gun is lighter by the mass of one cartridge - so it will recoil with slightly faster speed.
(of course, here I am not considering the momentum of the cartridge expelled to the side - we are talking only about gun and a bullet as they are the main contributors)

3. "do both the gun and the bullet travel equal distances?"
The gun and the bullet acquire equal MOMENTUM (magnitude same, opposite directions), but since they have different masses, they will have different speeds. Considering same amount of time they will travel different distances.
Is it infinite? - It depends. The gun and the bullet are on elliptical orbits, you would need to see if those orbits are intersecting the surface of the planet. If yes - they will perform a fiery reentry, if not - they will orbit the Earth like other cosmic junk.

4. Gravity or not - your slide/bullet still under 3rd Newton's law - it will operate same way anywhere. Now on the practical side, if your gun jams if you holding it light or "not holding it at all" on the planet - it will jam same way in space. But you your gun can fire and not jam while being "free" on the Earth, it will do same in space.

5. "no resistance" - FALSE, space have plenty of resistence. There is still molecules and ions, solar radiation etc. In fact "space vacuum" is considered pretty dirty and filled with stuff compared to laboratory and even microelectronic production vacuum chambers.



So ...
Gun will fire as usual and fly back while rotating.
Initially gun is on one elliptical orbit. The act of firing will send a bullet on an elliptical orbit and gun will change its orbit to a different one. Momentums are same and in opposite directions, M_gun > m_bullet --> gun travel back slower than a bullet. Bu conservation of momentum, the center of mass of gun+bullet is travelling on the original orbit (like if never fired).
 
... deleted my dup post ... and posting some add-ons

About hammer - Any internal part which stay inside (and not fly out) will NOT affect any motion of the gun due to conservation of linear and angular momentums. (This is the same thing as to why you can't pull yourself from a swamp by you own hair - all objects are within the system)

Rotation of the gun due to the rotation of the bullet (rifling) is around unstable inertial axis and thus will end up in rotations about two stable axises. Try spin and toss a book - there is three different ways you can spin it - you will quickly find that there is two axises (minimum and maximum ones) that are stable and the middle one will result in "wobble" - sum of rotations around two stable axises.
 
so, if the muzzle was pointed at earth when the gun fires, would the bullet "burn up" during re-entry? also, would it make a difference if it was plain lead, or a fmj?
 
Yes, the bullet would most likely burn up upon re-entry into the atmosphere. It would most likely melt into a blob of whatever metal it is made of. At the speed it will be traveling, it will most likely vaporize into very minute particles.
 
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If the bullet was HP, and travelled an infinite distance without impacting on anything, would it expand based solely on its speed an friction with all those cosmic particles?
 
Every single movement would have an effect.

The trigger being moved the hammer falling

the discharge

bullet rotation slide movement

everything comes into play no matter how small it is.
 
While space is not a perfect vacuum, it is still pretty darn good.

"Cosmic particles" are mostly electrons, protons, and individual atomic nuclei.

The nuclei are mostly in the lower mass part of the periodic chart.

Heavier elements are present, but at very decreased rates compared to lighter elements.

Think many orders of magnitude like ~10^-8 (or more, tables are at work) from hydrogen to even iron.

We still worry about them hitting satellites though.

There are charts with spans of 16 orders of magnitude covering atomic distribution in cosmic rays.
 
Do we have any empiricists here who would like to suspend a pistol by a thread and touch it off with a cable release? Probably have to do it several times hanging in different orientations to average out surface gravity. Or with some little accelerometers stuck to it.

Otherwise, Axtal has the physics down best.

And just remember, if you have a space vessel equipped with a Bergenholm, your velocity will immediately assume the value at which the thrust of the conventional drive is equaled by the frictional resistance of the medium being transited.
 
so it will recoil with slightly faster speed.
Click to expand...

If you assume that all the cartridges are identical, then it will recoil at the same speed, the speed of cycling is independent of how many rounds are in the magazine. If is base on the spring constant of the recoil spring, and the force of the shot (to determine how much time it will take for the spring to oscillate one cycle (compression and decompression).....unless you are counting the little change in friction of the round beind stripped since there is a little less compression in the magazine spring

About hammer - Any internal part which stay inside (and not fly out) will NOT affect any motion of the gun due to conservation of linear and angular momentums. (This is the same thing as to why you can't pull yourself from a swamp by you own hair - all objects are within the system)
Click to expand...

Sorry but this is not true, the hammer CAN give the gun an angular acceleration, although it will be extremely small. Try this, if you are in a computer chair that swivels, hold your arm out, then quickly move it to the side, you will spin, because you give yourself an angular acceleration. Not to mention the movement of any part changes the center of mass of the gun.
 
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Brings all sorts of questions to mind.

What effect would the vacuum and extreme temperatures of space have on a steel 1911?
Would being in a vacuum increase the stress on the barrel?




With all that high pressure gas INSIDE and negative pressure OUTSIDE, with the 1911 either brittle cold or blazing hot, I think you'd possibly have a ka-boom and ruin your gun. I could be wrong, somebody should try it.
 
eyebrows said:
With all that high pressure gas INSIDE and negative pressure OUTSIDE, with the 1911 either brittle cold or blazing hot, I think you'd possibly have a ka-boom and ruin your gun. I could be wrong, somebody should try it.
Click to expand...

There's no such thing as negative pressure. The vacuum of space is roughly 0 psi. Compared with the roughly 14 psi of sea level on Earth and the 21,000 psi chamber pressure, you'd see about a 0.06% increase in relative pressure when firing in space. Probably significantly less than the typical shot to shot variation.
 
Star Wars!!!!!:D (sorry i had to do that) I would say teh question has been awnsered and that this post is utterly useless except for this. MAKE IT A STICKY PLEASE!!!

:):):)
 
Negative pressure means less than atmospheric pressure. Therefore, one could say that a perfect vacuum is -14.7psi.
 
the hammer CAN give the gun an angular acceleration, although it will be extremely small.
Click to expand...

No. Think of the hammer as a separate body, with its own center of mass, and the gun with a larger mass, and the two connected by a spring under tension.

The spring is released: the smaller mass speeds toward the larger mass--but the larger mass also moves toward the smaller, more slowly, but with equal momentum. The two collide (inelastically--like two balls of stickum with a stretched rubber band between them), and the momentum of one cancels out the other, so the stuck-together balls no longer move in either direction. And the center of gravity of the two-objects-considered-together never moved throughout the event.

This is easier to envision linearly, but the same effect holds with angular moments. The reason the swivel chair thing "works" is because the effect of friction allows your chair to not move at all when you start your arm moving (so that your body does NOT begin to move in the opposite direction) but when it decelerates abruptly (as your forearm hits your opposite shoulder) now your chair budges a bit in the direction of arm movement (and stops) since there is no counter-movement of your body to counter it.

In a frictionless chair, the chair would move (in the opposite direction) when the arm moved, and stop when the arm stopped. Anyone who has attempted to walk on wet, smooth ice knows it's very difficult in a frictionless environment to acclerate your center of mass in the desired direction, no matter your arm (or leg) movements--until gravity kindly solves the problem for you!

(The position of the gun's center of mass within the gun silhouette WILL change, slightly, as the postition of the hammer changes, and to this extent the position of the gun silhouette in reference to space will change ever so slightly during hammer fall. But the position of the gun's center of mass in reference to space will not change as a result of hammer fall.)
 
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Loomis said:
Negative pressure means less than atmospheric pressure. Therefore, one could say that a perfect vacuum is -14.7psi.
Click to expand...

If one said that, they would be wrong. A perfect vacuum is 0 psi and you can't get less than 0 psi.

You can talk about relative pressures if you like, but then you really should be specifying what they are relative to.

It would be awfully silly to say that the air I'm breathing here at 5200ft altitude is -2.47 psi, wouldn't it?
 
The spring is released: the smaller mass speeds toward the larger mass--but the larger mass also moves toward the smaller, more slowly, but with equal momentum and energy. The two collide (inelastically--like two balls of stickum with a stretched rubber band between them), and the energy of one cancels out the other, so the stuck-together balls no longer move in either direction. And the center of gravity of the two-objects-considered-together never moved throughout the event.
Click to expand...
Close, but no cigar.

Momentum and energy are almost never equal.

Momentum = mass * velocity
Energy = 1/2 * mass * velocity^2

For the two to be equal, mass * velocity = 1/2 * mass * velocity * velocity

This can be satisfied, but is very rare. In most (nearly all) cases, it won't be. This is an inelastic collision. Since inelastic collisions are governed by conservation of momentum, the action-reaction pair will result in a net motion (equations for final velocity are available at the link provided; I'd include them here, but mathematical markup isn't supported in vBulletin).



Incidentally, thank you very much for making me remember my long-repressed high school physics. I'm going to go have a drink. :neener:
 
Loosedhorse - Yes (about friction{less}) ! In fact if you keep thinking in that direction, you will realize that the only force that propell you forward is a friction force! Thinking more you will realize that this frinction is a simple static one. So if we have a friction coefficient [mu] between two surfaces, and the mass of the object (think your car) is M, than maximum allowed friction force (before we will go into slipping mode) will be [mu]Mg. Sticking this into Newton's F=Ma will give you your maximum acceleration:
a_max = [mu]g
( here we assumed flat road )
So if you want to build a cool race car and thus maximize your acceleration, you need to either increase planet's gravity (realistically impossible) or maximize friction coefficient - get wider tires with better rubber.
Note that if you go into slipping, you will be under mercy of kinetic friction coefficients which is typically smaller than the static one.
Another note: Can your engine exert enough torque to match maximum-optimal force? Well ... depends on engine.
Another note (practical): you are having max acceleration if you are just on the brink of going into "peeling" but not entering it.

Flyboy - You just got big fat "F" for equating apples and oranges - energy and momentum

Guys, once again:
ANY moving part which STAYS with a gun (slide/hammer/whatever) will NOT affect the gun's final state of motion in any way. Yes, DURING the time of motion of said parts they have effect, but after the motion is done the gun moves like if that action had never happen.
This comes from conservation of linear and angular momentums.

So considering that I dont really care how slide moves unless my task is to describe the motion DURING the shot. If my task is to find final gun's motion AFTER the shot, I do not need any information on moving parts which stays with the gun.

So, during the 1st shot, my gun's mass is (M) and I have n bullets each of mass m. Let's say bullet exits with speed v. Then by conservation of linear momentum (in center-of-mass coordinates):

0 = [ M + (n-1)m ]V1 + mv, so the speed of the gun after 1st shot will be
V1 = -{ m / [ M + (n-1)m ] }v
(Negative sign shows that they are in opposite directions)

Note that i purposfully here writtem the answer as V1 = {something unitless (kg/kg) } times v

Let's say second shot is directed against the gun's motion (as to "stop" it).
The gun is moving with above-found speed V. Applying same principle (but remember now we have one less bullet). Here we are taking just the value of the V1 and the signs are as follows: V1 is to the right (+), bullet exits also to the right (+), final gun's V2 is unknown, so keep it with (+) - if it comes out as (+) then gun flies right, if negative, then gun flies left.

[ M + (n-1)m ]V1 = [ M + (n-2)m ]V2 + m( v+V1 ), so
V2 = { [ M + (n-1)m ]V1 - m( v+V1 ) } / { [ M + (n-2)m ] }; substituting V1 from above:

V2 = { [ M + (n-1)m ]{ m / [ M + (n-1)m ] }v - m( v + { m / [ M + (n-1)m ] }v) } / { [ M + (n-2)m ] }

After simple algebra we will find the velocity of the gun after 2nd shot:

V2 = - { { m^2 } / { [ M + (n-1)m ][ M + (n-2)m ] } } v

WHICH IS ALWAYS NEGATIVE no matter what the (physycally real*) values of those variables are.

(Physically real means all your masses are positive, n>=2 to be able to produce two shots, v>0 )

So if 1st shot was pointing left and sent the gun flying to the right, the second shot aimed to the right will not stop the gun but send it (quite slower) to the left.


Uff ... formulas on paper (compared to text) is so easier so see ... hope you got through that.
 
Flyboy--

Inelastic collisions (in effect, all real collisions) result in a loss of kinetic energy AFTER impact: the two moving bodies had more energy (summed) before the collision than after it.

What I was speaking of was a spring simultaneously acting (with the same force) on two bodies, one heavier than the other, before collision. The force will act for the same amount of time on each body (starts when hammer is released, ends when it impacts).

So, agreed, the two bodies will head toward each other with equal momentums, but not with equal energies.

Previous post amended.
 
Hot diggety dog, this is a fun thread. I have lain awake a couple of nights just thinking about all the factors that changes the equations after each shot, like the reduced mass after each shot, the effect of lubrication on the moving parts after it warmed up, The effect of the nearest or strongest gravity well in relationship to the gun and on and on.
 
A few tidbits, which may already have been mentioned since I didn't read every reply yet.

1) "Space" is not free from gravity. When astronauts are orbiting in a Shuttle, both the shuttle and everything in it are free falling fully within the effects of Earth's gravity. If you think of a typical schoolroom globe, a shuttle orbits at perhaps 225 miles or so above the surface, which would be only about maybe a half inch or so above the globe, keeping the same scale.

2) There is no need to go into space to perform this experiment unless you had to reduce the minor effects of air resistance to a bare minimum. Merely drop the gun from something at altitude, like a weather balloon, high bridge, etc. Track the gun with a camera with a very high power lens if you want to examine the motion in detail.

3) Upon remote firing, the reaction forces on the gun will cause it to change it's motion relative to itself in several ways. Assuming the axis of the forces are above the gun's center of gravity, the gun will rotate around the center of gravity and there will be a slight sideways rotation imparted by the rifling of the barrel acting against the inertia of the bullet as it travels through the bore. The *exact* motion of the gun is based on the complex interactions of all the factors in play, barrel length, gun design, spring weights, time of the bullet in the bore, slide mass, all kinds of things. A revolver's motion would likely be easier for a really good physicist to calculate than a semi-auto pistol due to the fewer moving parts in play.

4) If you did do this experiment in space by having an astronaut simply place the gun in orbit along side the Shuttle, the basic results would be the same, with a little more velocity due to the lack of air resistance. The gun and bullet, however, would, in this case, not travel in straight lines, being captured in Earth orbit, and would eventually fall to Earth as the orbits decayed.

K
 
There's no such thing as negative pressure.
Click to expand...

While I can not truthfully say that space is "Negative air pressure" ('cause I've never been there), Negative air pressure is for real, Think about flight, What gives an airplane it's lift, (or at least what do they call it) also, If you talk about carburetors on an internal combustion engine, The fuel is drawn out of the float bowl by,.......wait for it........Negative air pressure.

Like I said, While I can't say for sure that space has neg. air pressure, you cannot say it doesn't exist.
 
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