Firing a gun on a train doing 2000 mph

[Asphalt Alligator]

Imagine you are stood on the back of a train doing 2000mph. You have a gun which fires bullets at 2000 mph. You fire the gun in the opposite direction to the direction of the moving train. What happens to the bullet? Does it simply fall to Earth as the train speeds away at 2000 mph? Does it, at the moment of firing, behave like a normal bullet would fired from a stationary gun and speed away from the fired and momentarily (whilst the gun is fired) stationary gun? Or is there some other answer?

[steveupson]

It would fall straight down. This is the reason that the SR71 Blackbird wasn't armed. It really couldn't aim and fire a weapon effectively while moving at Mach-3.

 

As an amusing aside, North Korea fired a surface to air missile at one. That really made no sense unless they figured that the aircraft might turn around and go back in order to intercept it.

 

On edit >>> North Korea

Edited January 25, 2017 by steveupson

[imatfaal]

From the point of view of the shooter it speeds away from him of the back of the train at 2000mph - from the point of view of a person standing at the side of the track the train rushes away at 2000mph and the bullet falls vertically to the ground. Both are the same thing - but ou must take into account your frame of reference

[Strange]

From the point of view of someone on the train, the bullet will go past at 2000 mph (and gradually fall to the ground).

 

Form the point of view of someone standing beside the train as it passes, the bullet will just fall to the ground.

 

This is related to the interesting fact that bullet fired from a gun will fall to the ground at the same rate as one just dropped.

[Function]

It would fall straight down. This is the reason that the SR71 Blackbird wasn't armed. It really couldn't aim and fire a weapon effectively while moving at Mach-3.

 

Considering a projectile wouldn't go as fast, would the birdy then penetrate itself with its own bullet?

 

EDIT: on second thought, I think that'd be impossible too, given that the projectile will never go as fast as the birdy and upon firing (or rather, release), it will just stick to the bird, and when it (the bird) finally lands at a certain time, it will perhaps come roll out of its barrels?

Edited January 25, 2017 by Function

[MigL]

The SR-71 ( since its been brought up ) was originally developed to also do hi-speed/hi-altitude interception as the YF-12A, so yes, missile launches are usually done from a hi-energy ( hi-speed ) state to maximize missile range/persistence. The only caveat is that the missile launch has to be cleared for various speeds as aerodynamic interference between missile and airplane at separation could cause it to strike the plane ( most are only cleared for low-supersonic launches, not M3, the speed of an SR-71 ).

 

Also SAMs are launched on intercept courses not a chase, so the problem with shooting down an SR-71 is not catching it, but after you've used up all the rocket's fuel to reach the 70000 cruising altitude ( of the SR-71 ) there's nothing left for final maneuvering.

[Bender]

While the original question is adequately answered, I like to add an interesting point (at least, I think it is interesting):

the gun used the same amount of energy to fire the bullet, there is no difference in the combustion process, but to someone standing besides the rail, the bullet now has no kinetic energy. Where did the energy go?

Edited January 26, 2017 by Bender

[DrP]

The bullet has a starting v of -2000mph to the observer on the ground.. The energy went into accelerating the bullet to zero mph.

[Bender]

Of course, Mythbusters did it.

The bullet has a starting v of -2000mph to the observer on the ground.. The energy went into accelerating the bullet to zero mph.

Sorry, you have violated conservation of energy . To slow the bullet down, you need to absorb energy (so on top of the energy from firing the gun, the kinetic energy of the bullet is now missing too).

[DrP]

No - think about it.

 

Stationary gun. Bullet starts at 0mph. Bang! it is accelerated up to 2000mph. It gains KE from the explosion.

 

In reverse it is exactly the same, except this time the bullet is already doing 2000mph in the opposite direction. -2000mph. Bang! it is accelerated to 2000mph ion the opposite direction of travel - just as before... but the net speed is now 0mph. The KE it had was absorbed/cancelled out by the KE it gained from the explosion.

 

Any clearer?

[Asphalt Alligator]

What about turning round the other way and firing the gun in the same direction as the train? Will the bullet ever emerge from the barrel? Or will the back of the gun barrel be constantly catching up with the bullet and so the bullet remains in the barrel?

Edited January 26, 2017 by Asphalt Alligator

[imatfaal]

What about turning round the other way and firing the gun in the same direction as the train? Will the bullet ever emerge from the barrel?

 

Yes of course. You need to think of your frames of reference. You can throw a ball on a train can't you? From the perspective of the shooter the bullet will fly forward through the train at 2000mph. From the perspective of the person on the platform watching the train go past the bullet will have a speed of 2000+2000=4000mph.

 

If you fired out the window - but in the same direction as the train is travelling then the air resistance would quickly slow the bullet / cause it to lose spin and tumble; but it would still leave the muzzle at 2000mph according to shooter and be travelling at 4000mph according to platform observer

 

FYG - It is this form of adding / substracting speeds that stops working accurately when you get close to light speed, and doesn't work at all for light

[Bender]

No - think about it.

 

Stationary gun. Bullet starts at 0mph. Bang! it is accelerated up to 2000mph. It gains KE from the explosion.

 

In reverse it is exactly the same, except this time the bullet is already doing 2000mph in the opposite direction. -2000mph. Bang! it is accelerated to 2000mph ion the opposite direction of travel - just as before... but the net speed is now 0mph. The KE it had was absorbed/cancelled out by the KE it gained from the explosion.

 

Any clearer?

Perfectly. Of course, both energies are positive and thus cannot cancel each other out.

 

On the subject of firing in the same direction as the train: now the bullet has 4 times the kinetic energy it would have when fired from a stationary position. It already had the KE from the train, and you add the energy from the gun, but where do the other 2KE come from?

[Janus]

Perfectly. Of course, both energies are positive and thus cannot cancel each other out.

 

On the subject of firing in the same direction as the train: now the bullet has 4 times the kinetic energy it would have when fired from a stationary position. It already had the KE from the train, and you add the energy from the gun, but where do the other 2KE come from?

You're neglecting recoil. The firing of the gun doesn't just propel the bullet to speed, it also pushes back on the gun, and through it the person holding it, and through them the train.

The easiest way of considering it is if we consider the train as coasting on friction-less bearings. Then when the gun fires backwards, the recoil adds velocity to the train, and that's where your "missing" KE went. (I should point out here that if the final velocity of the bullet is going to be 2000 mph relative to the train, it will end up with just slighty more than a zero velocity with respect to the ground.)

 

Now you might be tempted to argue that the train is not coasting without friction, but that it is being driven by an engine with a good solid contact to the rails such that it s velocity relative to the rails does not change upon the bullet firing, and thus it does not change with respect to you.

 

But here's the thing. Then the recoil is transferred to the Earth, and through the Earth to you. This means that you are in a different inertial frame after the bullet was fired than you were before the bullet was fired.

 

And while KE is conserved within any inertial frame, it is not conserved across inertial frames.

[RiceAWay]

Of course, Mythbusters did it.

Sorry, you have violated conservation of energy . To slow the bullet down, you need to absorb energy (so on top of the energy from firing the gun, the kinetic energy of the bullet is now missing too).

Look at DrP's answer just above your comments.

You're neglecting recoil. The firing of the gun doesn't just propel the bullet to speed, it also pushes back on the gun, and through it the person holding it, and through them the train.

The easiest way of considering it is if we consider the train as coasting on friction-less bearings. Then when the gun fires backwards, the recoil adds velocity to the train, and that's where your "missing" KE went. (I should point out here that if the final velocity of the bullet is going to be 2000 mph relative to the train, it will end up with just slighty more than a zero velocity with respect to the ground.)

 

Now you might be tempted to argue that the train is not coasting without friction, but that it is being driven by an engine with a good solid contact to the rails such that it s velocity relative to the rails does not change upon the bullet firing, and thus it does not change with respect to you.

 

But here's the thing. Then the recoil is transferred to the Earth, and through the Earth to you. This means that you are in a different inertial frame after the bullet was fired than you were before the bullet was fired.

 

And while KE is conserved within any inertial frame, it is not conserved across inertial frames.

The added energy is immeasurably small or the man could not hold the weapon's recoil. How strong do you believe the human spine to be?

[Asphalt Alligator]

Thank you all for your inputs, most informative.

[Bender]

You're neglecting recoil. The firing of the gun doesn't just propel the bullet to speed, it also pushes back on the gun, and through it the person holding it, and through them the train.

The easiest way of considering it is if we consider the train as coasting on friction-less bearings. Then when the gun fires backwards, the recoil adds velocity to the train, and that's where your "missing" KE went. (I should point out here that if the final velocity of the bullet is going to be 2000 mph relative to the train, it will end up with just slighty more than a zero velocity with respect to the ground.)

 

Now you might be tempted to argue that the train is not coasting without friction, but that it is being driven by an engine with a good solid contact to the rails such that it s velocity relative to the rails does not change upon the bullet firing, and thus it does not change with respect to you.

 

But here's the thing. Then the recoil is transferred to the Earth, and through the Earth to you. This means that you are in a different inertial frame after the bullet was fired than you were before the bullet was fired.

 

And while KE is conserved within any inertial frame, it is not conserved across inertial frames.

correct

Look at DrP's answer just above your comments.

I did. It was not correct.

 

The added energy is immeasurably small or the man could not hold the weapon's recoil. How strong do you believe the human spine to be?

I don't see how the strength of a human spine is relevant in a thought experiment. A train also doesn't go 2000 mph. This is not the engineering subsection

[Janus]

Look at DrP's answer just above your comments.

 

The added energy is immeasurably small or the man could not hold the weapon's recoil. How strong do you believe the human spine to be?

The added energy is not immeasurably small, It is just enough to make up for the apparent "missing" KE. The change is velocity of the train and Earth caused by the recoil will be extremely small but is made up for by the large total mass of the pair.

The Human spine only has to resist the recoil from firing the bullet. This force will not be huge, and it is this magnitude of force it what it transfers to the train and the Earth. This transfer of force causes an acceleration of the train and Earth, albeit a extremely small one.

This acceleration, in turn translates into a change in KE of the train and Earth.

[RiceAWay]

The added energy is not immeasurably small, It is just enough to make up for the apparent "missing" KE. The change is velocity of the train and Earth caused by the recoil will be extremely small but is made up for by the large total mass of the pair.

The Human spine only has to resist the recoil from firing the bullet. This force will not be huge, and it is this magnitude of force it what it transfers to the train and the Earth. This transfer of force causes an acceleration of the train and Earth, albeit a extremely small one.

This acceleration, in turn translates into a change in KE of the train and Earth.

The recoil of a man's weapon is absorbed by the spinal column and supporting muscle tissue.

 

What "missing" KE are you talking about? As I said, the man could absorb and dissipate all of the energy and virtually none of the insignificant mass of the 300 grain bullet or the powder (energy) to drive it at about 3000 feet per second initial speed.

 

Are you suggesting that if a man standing on the equator shoots the same gun due east true that the Earth slows?

 

I did. It was not correct.

 

 

I'm sure that there are plenty of people here that want some of what you're smoking.

[Janus]

The recoil of a man's weapon is absorbed by the spinal column and supporting muscle tissue.

 

What "missing" KE are you talking about? As I said, the man could absorb and dissipate all of the energy and virtually none of the insignificant mass of the 300 grain bullet or the powder (energy) to drive it at about 3000 feet per second initial speed.

The "missing" KE was the KE Bender was referring to as seen by the person watching the train go by. From his view, the bullet goes from 2000 mph to zero and losses KE and momentum. Something has to account for this to keep the books balanced in terms of energy and momentum.

Are you suggesting that if a man standing on the equator shoots the same gun due east true that the Earth slow

By a small bit, Yes. If he wee standing on a friction free surface, upon firing the gun the Bullet would gain a West-East momentum. Momentum of the system must be conserved, so he will take on a small East-West change in momentum to keep the balance.

If he is standing fixed to the Earth, the recoil force is transferred through him to the Earth. Momentum must still be conserved, so the Earth has to make a East-West change in momentum. Since the momentum balance equation is

m₁v₁=m₂v₂ And the mass of the bullet is very small compared to the Earth, the change in velocity of the Earth will only need to be very, very small to balance.

Of course, this velocity change will only be temporary, as eventually the bullet will hit an object fixed to the ground or the ground and return its momentum to the Earth*. The only way to make it permanent is to fire the bullet at better than 11 km/sec so that it escapes the Earth completely, taking its momentum with it.

 

*Though to be strictly correct, not quite all of the momentum is returned. Some of the KE is converted to heat by passage through the atmosphere and upon striking the target/ground. This is then radiated away as IR. EMR has momentum also, which is proportional to its frequency. The IR radiated towards the East will be Doppler shifted to a slightly higher frequency and larger momentum, and that to the West to a slightly lower one with less momentum. This results in a slightly Easterly net momentum being being lost and carried away by this radiation. A minuscule loss, but a real one.

[DrP]

The "missing" KE was the KE Bender was referring to as seen by the person watching the train go by. From his view, the bullet goes from 2000 mph to zero and losses KE and momentum. Something has to account for this to keep the books balanced in terms of energy and momentum.

 

Surely this energy just comes from the gun powder explosion, no? What am I missing then?

[Bender]

 

Surely this energy just comes from the gun powder explosion, no? What am I missing then?

Yes.

 

Ignoring acceleration of the Earth and efficiency of gun powder to kinetic energy conversion:

- if the gun is stationary: the useful energy of the explosion goes to kinetic energy of the bullet

- if the gun is firing backwards on a riding frictionless train: the energy goes to acceleration of the train. Moreover, the KE the bullet had before also goes to the train

- if the gun is firing forwards on a riding frictionless train: the energy goes to the bullet. Additionally, the train slows down slightly and this KE from the train is transferred to the bullet.

[Function]

Why do we actually have to make things this complicated? If you stand on a platform and you see a train starts moving and a person walking in the opposite direction, you find it funny he isn't making any progress whatsoever in relation to your frame of reference, while he's gaining quite some metres in his frame of reference

 

Makes the discussion a bit easier: no question where the energy comes from or goes into.

Edited January 28, 2017 by Function

[Bender]

Why do we actually have to make things this complicated? If you stand on a platform and you see a train starts moving and a person walking in the opposite direction, you find it funny he isn't making any progress whatsoever in relation to your frame of reference, while he's gaining quite some metres in his frame of reference

 

Makes the discussion a bit easier: no question where the energy comes from or goes into.

The original question was already answered in the second post. Feel free to ignore anything that comes after .

[Janus]

 

Surely this energy just comes from the gun powder explosion, no? What am I missing then?

This energy is present whether or not you measure things from the train or from the ground. From the train its is simple to balance it out; The energy of the explosion is transferred to the bullet as KE since it is now moving a 2000mph with respect to the train.

From the ground, you have the same energy content of the explosion, but now the bullet goes from 2000 mph to 0, and loses KE. Adding the energy of the explosion to the bullet no longer works as that can only result in an increase in bullet energy because energy is a scalar not a vector, and in this frame it has lost energy.

 

So the question Bender posed was; "How do you account for energy conservation in this frame?" To answer this, you have to consider the effect of the powder explosion on the train, AKA the recoil.