SRL Archive - NOTA

The bullet falls (mostly) straight to the ground.

You would shoot a hole in the back of the train.
In your reference frame (on the train) the bullet would be appearing to move 2000km/h to the left and hit the back of it at 2000km/h, to someone standing outside of the train the bullet would appear stationary, and hit the back wall of the train at 2000km/h as the train "catches up" to it. Boring without relativistic speeds :(

From: senrath
The bullet falls (mostly) straight to the ground.

"Muzzle speed" of 2000km/h, therefore it leaves the gun at 2000km/h regardless of how fast the gun itself is moving

The Mythbusters did it with an aircannon that launched a ball at the same speed as the truck they were driving. It fell straight down.

It would just drop to the ground, in the way you drop a tennis ball.

You guys are ignoring reference frames. Standing outside the train thats what it would look like but the back of the train would hit it, as the train is going 2000 km/h and would reach the bullet before it dropped

http://en.wikipedia.org/wiki/Frame_of_reference

Um, the picture has no "back" for the bullet to hit. The guy is on a flat rail-car.

Then s/back/nonexistant back/ for my above posts

You are both right because there is a picture depicting one view, but it also mentions the word 'you'. According to picture it would go down. From the point of 'you', if you had superman vision, you would see the bullet go away from you, heading past the back of the platform you are standing on, and you would see it hit the tracks when you are about 400m away, and if you didn't have superman vision, you would just see you shooting a gun :p.

From: senrath
The bullet falls (mostly) straight to the ground.

if the gun had a range of 2000km, that would happen. If, it had a range of 4000km, after one second, it would be 2000km from the origin.

Range doesn't matter, because it has an effective velocity of 0 km/hr.

From: jakeyboy29
if the gun had a range of 2000km, that would happen. If, it had a range of 4000km, after one second, it would be 2000km from the origin.

With a muzzle velocity of 2000km per >>>>HOUR<<<<, please tell me how?

From: Killerdou
With a muzzle velocity of 2000km per >>>>HOUR<<<<, please tell me how?

per hour, quite.

I'm not a gun expert, but from a physics perspective I would imagine that range is how far a bullet can go before gravity pulls it too far off course. An effect of gravity that is too much may be considered 6 inches, but I will give you an example applicable to the current question, ignoring wind resistance etc.

Let's say that the height of the gun is 10 feet above the ground, or 3 meters. With an acceleration of gravity of -9.8m/s^2, we can plug into the shortened equation
Distance = gravity*time*time/2
-3=-9.8*t*t/2
t=0.782460796
So the bullet (or anything else dropped from 3 meters with no air resistance and upward or downward force) will hit the ground in about 3/4 of a second. Where on the ground depends on the sideways velocity at release time relative to the ground, which in this case, is 0. After that 0.78 seconds from the shot, the train traveling at 2000*1000/60/60 m/s will be 434.700442 meters away. So if you were to shoot that gun from a standing position, the bullet would hit the ground about 400 meters away from you (assuming you shot it parallel to the ground). If you wanted to hit a target, and the bullet falling more than 10cm below where your laser sight is pointing would be unacceptable, the range would be about 79 meters.

From: Harry

get off 4chan already

It's the same thing as if the train would stop, the bullet would be shot, and then the train continued on… The speed of the gun doesn't affect the speed with which the bullet will fly.

Actually marpis, it does. Because the bullet is already traveling with a velocity of +2000 km/h, so the force from the charge drops it to 0 km/h, rather than go to -2000 km/h.

This has been 'proved' so many times before…
The bullet will fall to the ground just as the only force pulling it is gravity (2000km/h to the right direction - 2000km/h to the left = 0km/h in either of them)

From: marpis
It's the same thing as if the train would stop, the bullet would be shot, and then the train continued on… The speed of the gun doesn't affect the speed with which the bullet will fly.

This^

You who mention the mythbusters are surely forgetting that real life is not like physics in the least. In the show, there is wind resistance, in a physics problem, it does not seem that there is.

Using the logic that you have decided to take, standing on a train going 3000km/h would push the bullet through the back of the gun. I personally see no sense in that.

I personally cannot throw a baseball 100mph, but I know that if I were to throw a baseball off the back of a car that was traveling 100mph, it would not go in the negative direction after I threw it, so explain to me how a bullet is different?

In reference to the person shooting the bullet, it would move away from them. An explosion (the bullet firing) creates motion, and the motion carries the bullet out of the barrel, if it were not meant to move, the bullet would stay lodged in the barrel, no?

From: 3Garrett3
This^

You who mention the mythbusters are surely forgetting that real life is not like physics in the least. In the show, there is wind resistance, in a physics problem, it does not seem that there is.

That has no impact on the situation, thanks.

From: 3Garrett3

Using the logic that you have decided to take, standing on a train going 3000km/h would push the bullet through the back of the gun. I personally see no sense in that.

Who the heck said that? I didn't.

From: 3Garrett3

I personally cannot throw a baseball 100mph, but I know that if I were to throw a baseball off the back of a car that was traveling 100mph, it would not go in the negative direction after I threw it, so explain to me how a bullet is different?

No, if you matched it perfectly, it would only go down.

From: 3Garrett3

In reference to the person shooting the bullet, it would move away from them. An explosion (the bullet firing) creates motion, and the motion carries the bullet out of the barrel, if it were not meant to move, the bullet would stay lodged in the barrel, no?

It does leave the barrel. But it stays at the exact point it was at when the trigger was pulled.

From: senrath
Actually marpis, it does. Because the bullet is already traveling with a velocity of +2000 km/h, so the force from the charge drops it to 0 km/h, rather than go to -2000 km/h.

but the bullet isnt traveling at - anything

From: jakeyboy29
but the bullet isnt traveling at - anything

The - just means it's going to the left in the picture, while the train is going to the right. It doesn't mean that the bullet is going backwards.

From: senrath
Actually marpis, it does. Because the bullet is already traveling with a velocity of +2000 km/h, so the force from the charge drops it to 0 km/h, rather than go to -2000 km/h.

the bullet isnt moving anywhere untill it comes out the barrell. (on it's own) So, as soon as the bullet is shot, it is going to be moving 4000km/ph away from the train

everyone just give up, trolls trolling trolls trolling trolls.
Take over 9000 university physics courses and l2 frame of reference.

I feel dirty just posting in this thread…

From: jakeyboy29
the bullet isnt moving anywhere untill it comes out the barrell. (on it's own) So, as soon as the bullet is shot, it is going to be moving 4000km/ph away from the train

If the bullet isn't moving, then it's not gonna be in the gun, which IS moving at a rate of +2000 km/h.

(on it's own)

It's still moving, though, and that's what matters.

but is it though?

Yes.

but is it though?

Yes. You're just being purposefully obtuse.

&#8206;&#8206;&#8206;

but am i though?

but seriously, im really not sure about this; you very well might be right, i'll ask a teacher tomorrow

From: senrath
That has no impact on the situation, thanks.

Who the heck said that? I didn't.

No, if you matched it perfectly, it would only go down.
It does leave the barrel. But it stays at the exact point it was at when the trigger was pulled.

I'm saying

1) Comparing real life experience (Mythbusters) to this physics problem isn't realistic, there are many factors, so the people who said that it was tested are incorrect in using that reasoning.

2) It may not have been you, but whoever is saying that the train is going +2000 so the -2000 cancels to be 0, correct? So if the train is going 3000 to the right, the bullet only goes -2000 to the left. So 2000 cancels out, and we are left with 1000 to the right, which means you are saying that the bullet would turn around?

3) From the point of view of the person firing the gun, the bullet will move away from them, right? You should be able to understand that. It wont just fall to the ground at their feet. Use the baseball example again, if I simply throw the ball up in the air it will move away from me, and that is with what would be considered a positive velocity. If baseballs go with the motion of the car wherever you're from, I'm sorry, but in Canada, the baseball will move away from the person, hence the bullet will move away from the person when fired, not fall at his feet.

From: 3Garrett3
I'm saying

1) Comparing real life experience (Mythbusters) to this physics problem isn't realistic, there are many factors, so the people who said that it was tested are incorrect in using that reasoning.

It was tested, and found accurate. There are factors present in the real world that aren't in the physics problem, but that doesn't matter. Because the physics problem has been shown to work already, and then was shown to actually work in real life.

From: 3Garrett3

2) It may not have been you, but whoever is saying that the train is going +2000 so the -2000 cancels to be 0, correct? So if the train is going 3000 to the right, the bullet only goes -2000 to the left. So 2000 cancels out, and we are left with 1000 to the right, which means you are saying that the bullet would turn around?

That was what I said.

And the bullet simply wouldn't leave the barrel, because the explosion wouldn't be able to overcome the momentum it already has.

From: 3Garrett3

3) From the point of view of the person firing the gun, the bullet will move away from them, right? You should be able to understand that. It wont just fall to the ground at their feet. Use the baseball example again, if I simply throw the ball up in the air it will move away from me, and that is with what would be considered a positive velocity. If baseballs go with the motion of the car wherever you're from, I'm sorry, but in Canada, the baseball will move away from the person, hence the bullet will move away from the person when fired, not fall at his feet.

No, the bullet will NOT move away from the person. The person will move away from the bullet. There is a HUGE difference.

From: senrath

No, the bullet will NOT move away from the person. The person will move away from the bullet. There is a HUGE difference.

Frame of reference.
Both shooter and bullet are moving at the same speed to different directions.

Movement always needs a static point where we compare positions. In this case the static point would be the point where the bullet was shot. Both bullet and shooter move from this spot with the same speed to different directions.

I don't even think this is anyhow interesting situation. :rolleyes:

From: senrath
And the bullet simply wouldn't leave the barrel, because the explosion wouldn't be able to overcome the momentum it already has.

It would leave the barrel, but it would have a speed of 0 km/h. The reason is because there's a lot of gas inside the barrel, forcing the bullet out, but since the muzzle speed is the same as the train it will just fall to the ground. (EDIT: Read: Ground. It will not hit the train floor because the train has already moved a couple of meters before the bullet even get's close to the floor)

It may well be a bit late, but I'd have to agree with senrath.

For this problem I'd assume we're ignoring things like air resistance, standard high-school level questions like this tend to ignore these, hence we're dealing with this from a logical-perspective rather than real-life physics.

We've got four major objects within this problem, the person, the gun, the bullet and the train.

Obviously, the train's going at +2000km/h, and as such is having to exert a force upon the person to propel them at +2000km/h who consequently provides the necessary force for the gun and bullet to be travelling at +2000km/h.

I think this is where most get caught up - it's seen that because the bullet is stationary within the gun that it's velocity is 0 km/h. However, the person (who has a force applied by the train) applies a force on the gun, which applies a force on the bullet to produce it's subsequent velocity of +2000km/h. It isn't stationary, but from a relativistic view of the person on the train it is. (Just like the gun is).

The moment the gun fires, the gun exerts a force which produces a resultant velocity of -2000km/h on the bullet. Ignoring recoil from the gun, the person and the train are unaffected and remain moving at a constant +2000km/h.

However, the gun has exerted the -2000km/h force on the bullet and as such the bullets new velocity is : -2000 + 2000 = 0. (In the perfect circumstances).

Within this logical world, the bullet will drop to the ground in the same location as it was in the gun, as there is no force present to alter it's velocity and consequently it's location anymore.

As has been said though, the train, gun and person will continue to travel at +2000km/h and hence move away from the bullet, but the bullet itself won't move at all.

From: Aequix
… force of -2000km/h …

I don't like how you call a velocity a force

From: Brain
I don't like how you call a velocity a force

Aww just ignore that. He's right!

/thread

From: Brain
I don't like how you call a velocity a force

Indeed, my mistake - the gun exerts a force on the bullet which produces a resultant velocity of -2000km/h. Thanks for pointing that out! :)

Thanks for the agreement Zyt3x! :) It's important to point out the small issues with terminology though, not good to get bad habits.

But wouldn't there be an extra +2000km/h velocity due to the bullet leaving the gun? Sorry, I haven't taken physics. :( But as far as I know, there's some law about equal and opposite reactions. Sorry if I'm totally wrong (I probably am). :p

Interesting though. :)

From: i luffs yeww
But wouldn't there be an extra +2000km/h velocity due to the bullet leaving the gun? Sorry, I haven't taken physics. :( But as far as I know, there's some law about equal and opposite reactions. Sorry if I'm totally wrong (I probably am). :p

Interesting though. :)

You're not wrong as such, you're right about the law. It's Newton's Third Law of Motion upon which if a body #1 exerts a force on body #2, body #2 will exert an equal and opposite force on body #1.

The initial condition of this problem is such that all four objects present have a velocity of +2000km/h. That's 2000km/h 'to the right'.

If we take some known formulae, v (velocity) = u (initial velocity) + a*t (acceleration * time) and F (Force) = m*a (mass * acceleration), we can determine that at an initial velocity of 0, to produce a velocity of -2000km/h on a 1kg mass over time 't', we need a force of :- F = (-2000 / t)N.

Due to this, the gun must exert a force of (-2000 / t) Newtons on the bullet in order to produce a velocity of -2000km/h as the question states. This force is equal to the force applied to the bullet initially to reach the velocity of +2000km/h and as such the two 'cancel' each other to produce an overall velocity of 0 km/h on the bullet.

After diverging a bit though, back to what you stated - the gun exerted that force of (-2000 / t) Newtons on the bullet, and from Newton's Third Law like you stated, this must therefore also be applied in the opposite direction, (2000 / t) Newtons, on the gun. This will have an effect, but on the gun and subsequently other bodies. (i.e Person). What it doesn't have an effect on though is the bullet as the equal and opposite force is exerted, by the bullet, on the gun. ;) (Also note that the force applied would be the equivalent of the force needed to create a 2000km/h velocity on an object with the same mass as the bullet, but when this opposite force is applied to the gun, the gun will have a different mass and as such the resultant acceleration and consequently velocity due to the force would be different to that on the bullet - this follows from the formula : a = F / m, and v = u + at.)

A lot of it is taking things from a 'perfect' perspective though, in reality we'd have all kinds of outside influences that would affect this, but the fundamentals still are in play.

Is force really a factor in this? And if so, wouldn't we need the masses to find the acceleration? Or is that equation just to show me something (I'm already aware of a few physics equations and some laws, I just haven't taken classes).

Also, in my mind I agree with you, and I think it makes perfect sense that the bullet would simply drop (and it would "move away from you" due to the time it'd take for it to fall); I'm just trying to see if you can disprove every aspect of the counter argument (it helps me learn :3).

Thanks a lot for explaining everything by the way. Your posts are definitely the ones I enjoy the most so far (out of the few you've posted). :)

If you consider the starting situation, we have four objects with a constant velocity and therefore an acceleration of 0.

This is fairly logical, but there's the equation : a = (v - u) / t, if velocity and initial velocity are the same, you can see a = 0.

Just to clarify, v = velocity, u = initial velocity, t = time, a = acceleration, F = force, m = mass.

So we have four bodies with a constant velocity.

Whenever a change in acceleration and therefore a change in velocity occurs, we need to have a force being applied to a body. This follows in that in order for a velocity change to occur, we must have an acceleration. (Be it positive or negative). This is from : a = (v - u) / t. If v =/= u, then a > 0 or a < 0. (Obviously not considering the time = 0 scenario! :))

As we now know acceleration is a necessity within a velocity change, we can see through the formula : F = ma, a = F/m, that we must have a force being active to produce that acceleration. So we now know we need a force to be present in order to have a velocity change. :) This is fairly standard and logical as any movement is as a result of a force being applied.

The equations are mainly to try and tie in a bit more logic / maths to the problem. Physics follows quite nicely in that if you follow mathematical equations and rules if you relate one to another it is accurate. (Providing you don't make a mistake with the maths. ;)). Thanks to that you can relate things like forces being applied on a body to resultant velocities and vice-versa. So it allows us to show that the velocity that the question's asking for is a result of a force being applied by the gun to the bullet as that's the only body interacting with it. :)

Haha, good to see it's making sense! ;) Playing Devil's Advocate is definitely one of the best ways to learn, and it helps make people like me double-check we actually know what we're saying! Glad you enjoy the posts too! :)

Just should clarify though - "(and it would "move away from you" due to the time it'd take for it to fall)", remember it's not the bullet that's actually moving, but the three other bodies (gun,person,train). ;) (Well, the bullet would move as gravity comes into playing producing a constant-downwards force of 9.8N and therefore acceleration, subsequent velocity increase etc, but that's vertical movement.)

HOLDDDDDDDD on. :p Sorry for continuing, I'm just trying to make complete sense of this (you're not doing a bad job, I just want to make sure I figured it out :)).

Well first of all, using a = (v-u)/t, what would t be? Because it doesn't work, does it?

a = (v-u)/t
0 = 2000/t

Now what? You could multiply both sides by t, but then it'd still be 0 = 2000, which doesn't work?

And if acceleration is zero, then both force and mass would equal zero, wouldn't they?

Sorry if I'm wrong, but now that I put in the numbers it doesn't make sense. :<

From: marpis
Frame of reference.
Both shooter and bullet are moving at the same speed to different directions.

Movement always needs a static point where we compare positions. In this case the static point would be the point where the bullet was shot. Both bullet and shooter move from this spot with the same speed to different directions.

I don't even think this is anyhow interesting situation. :rolleyes:

Nope. Only the shooter moves from that spot (besides vertical movement). The bullet WAS moving towards to this spot, before it was shot. The expansion of the gases stopped this movement when it got to that spot. You might be thinking of what i luffs yeww mentioned, equal and opposite reaction. This does not mean equal and opposite speed, it means equal and opposite force. It doesn't mean equal and opposite acceleration either, since the mass of the bullet is not the same as the mass of the gun. The expansion of the gas applies an equal force to the right, but has a meaninglessly small effect on the massive gun/shooter/train. When the bullet is in the chamber, it has a speed of 0 relative to the gun, and when the gas applies the force to it it gets an acceleration of about 1500000m/s^2. Due to that acceleration, by the time it gets to the end of the 10cm barrel (where the gas escapes and no longer provides the force), it has reached 2000*1000/60/60 m/s to the left relative to the gun. If the barrel was crazy long so that the bullet was in the barrel for more than a fraction of a second, after the first second in the barrel the bullet would be traveling 1500000m/s (to the left relative to the barrel), and after the next second it would be 3000000m/s, and so on until it left the barrel. Now relative to the spot on earth where the bullet was shot/'photo' was taken, when the bullet was in the chamber it had a speed of 2000*1000/60/60 m/s to the right. Again with a crazy long barrel, relative to the spot, after the first second it would have a speed to the right of 2000*1000/60/60-1500000 m/s. As each second goes by, another 1500000m/s is subtracted, until it gets to 0, at which point it starts going left relative to that spot. This ends when the force is longer being applied. In the case of the regular barrel with the given parameters, it has had enough time to subtract 2000*1000/60/60 m/s, making it 0 relative to the spot.

From: Zyt3x
It would leave the barrel, but it would have a speed of 0 km/h. The reason is because there's a lot of gas inside the barrel, forcing the bullet out, but since the muzzle speed is the same as the train it will just fall to the ground. (EDIT: Read: Ground. It will not hit the train floor because the train has already moved a couple of meters before the bullet even get's close to the floor)

Correct, assuming the the persons arm is 5 feet above the train floor, the person will have moved about 300 meters away by the time bullet has dropped 5 feet.

So what I said about

a = (v-u)/t
0 = 2000/t

and

And if acceleration is zero, then both force and mass would equal zero, wouldn't they?

? :(

Sorry - had to sleep! ;)

The formula is :- Acceleration = Velocity - Initial Velocity / Time.

Based on what you're putting in, they're contradictory values. If acceleration is 0, then velocity and initial velocity will be the same over any period of time as there's no acceleration to change the velocity.

Therefore, we have : if a = 0, 0 = 0 / t, which holds true. :)

Ahh. Alright. After I reread what I said I realized what I did. :) Thanks a lot! :D

From: Boreas
Nope. Only the shooter moves from that spot (besides vertical movement). The bullet WAS moving towards to this spot, before it was shot. The expansion of the gases stopped this movement when it got to that spot. You might be thinking of what i luffs yeww mentioned, equal and opposite reaction. This does not mean equal and opposite speed, it means equal and opposite force. It doesn't mean equal and opposite acceleration either, since the mass of the bullet is not the same as the mass of the gun. The expansion of the gas applies an equal force to the right, but has a meaninglessly small effect on the massive gun/shooter/train. When the bullet is in the chamber, it has a speed of 0 relative to the gun, and when the gas applies the force to it it gets an acceleration of about 1500000m/s^2. Due to that acceleration, by the time it gets to the end of the 10cm barrel (where the gas escapes and no longer provides the force), it has reached 2000*1000/60/60 m/s to the left relative to the gun. If the barrel was crazy long so that the bullet was in the barrel for more than a fraction of a second, after the first second in the barrel the bullet would be traveling 1500000m/s (to the left relative to the barrel), and after the next second it would be 3000000m/s, and so on until it left the barrel. Now relative to the spot on earth where the bullet was shot/'photo' was taken, when the bullet was in the chamber it had a speed of 2000*1000/60/60 m/s to the right. Again with a crazy long barrel, relative to the spot, after the first second it would have a speed to the right of 2000*1000/60/60-1500000 m/s. As each second goes by, another 1500000m/s is subtracted, until it gets to 0, at which point it starts going left relative to that spot. This ends when the force is longer being applied. In the case of the regular barrel with the given parameters, it has had enough time to subtract 2000*1000/60/60 m/s, making it 0 relative to the spot.


Correct, assuming the the persons arm is 5 feet above the train floor, the person will have moved about 300 meters away by the time bullet has dropped 5 feet.

The bullet starts off with an explosion so we can assume that the change of speed and direction is instant.

From: marpis
The bullet starts off with an explosion so we can assume that the change of speed and direction is instant.

Sure but do you see how I was dragging it out to show the change in speed is -2000 (to the left), which when added to the initial +2000 (to the right) makes 0?

From: Boreas
Sure but do you see how I was dragging it out to show the change in speed is -2000 (to the left), which when added to the initial +2000 (to the right) makes 0?

It might be my english but I still can't see how the movement of the train affects the bullet's location compared to the place where it was shot. Mehh, I quit.

In the pic, the shooter is pointing to left, and the train is going to right. That just means the bullet would travel same speed (2000km/h) the instant trigger hits the bullet setting it off.

Reason? Bullet heading the opposite direction only appears to be going 4000km .h because one object heads left, the other to the right. The speed of both objects leaving the point when the bullet first got shot from the gun is 2000km/h. From the shooters perspective, it would look like the shot is leaving at 4000km/h, but in reality, the bullet is still travelling at 2000km/h.

Bullets in general move because the explosion in the barrel presses against the gun and the bullet (and the bullet being much less massive, has a much greater acceleration). Bullets are not being pushed away from some spot in space, they are being pushed away from the gun. If the gun stays in some spot in space, then it appears that the bullet is moving from the place it was shot, but this is just a coincidence. If the gun is moving, then the place the bullet was shot from is moving.

Take a look:
http://dsc.discovery.com/videos/mythbusters-vector-vengeance-high-speed.html
if IP is blocked: http://www.youtube.com/watch?v=33JBhnr9lds 3:24

eehhh when the bullet is fired, something comes out of its end no? wont that build up in the gun :o? and have some sort of effect, (im no physicist btw :D )

Well I just watched the video. It makes sense to me. :) Did you watch it, hack? (If anything, it should be more exact for a gun, since it doesn't take time for the bullet to get up to 2000km/h

From: i luffs yeww
(If anything, it should be more exact for a gun, since it doesn't take time for the bullet to get up to 2000km/h

Everything takes some amount of time to accelerate…except for some massless particle…

But for the bullet to go from 0-2000km/h from a gun would take less than a cannonball to go from 0-whatever km/h if it uses air (which is I think what they used), wouldn't it?

link to mythbuster please, i'd like to see this.

From: i luffs yeww
But for the bullet to go from 0-2000km/h from a gun would take less than a cannonball to go from 0-whatever km/h if it uses air (which is I think what they used), wouldn't it?

That would depend on the amount of force applied in each case and if those forces were equal to each other, I believe.

From: Main_Ftw
link to mythbuster please, i'd like to see this.

From: Boreas

Take a look:
http://dsc.discovery.com/videos/mythbusters-vector-vengeance-high-speed.html
if IP is blocked: http://www.youtube.com/watch?v=33JBhnr9lds 3:24

in the thread already

Imagine going on a marry go a round and throw a bommerang and it has the possibility of dropping dead on the floor.

We should have a discussion subforum in homework forum, or put topic like these in the general section.

It would hit you in the face and kill you.



(Joking :) ) but this thread gave me an interesting read on physics even though I don't take it yet, I'm glad I read this.