# How Fast is it Going?

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This question may seem simple but for a non-physicist like myself it is sometimes hard to grasp. I'm sure this is beginner's stuff.

I have 2 guns, when fired the bullets leaves the barrel at the same speed relative to where I am standing. I put a target 100 yds away, mount one gun on a vehicle moving towards it and one gun stationary. The mounted gun is preset to fire at the target when it is 100 yds away and is synchronized with the firing of the stationary gun, also at the target. My feeling is that both bullets should hit the target at the same time.... is this correct?

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No, they won't. The speed of the bullet from the gun mounted on the moving vehicle will be faster (relative to the stationary target) than the other bullet, by the speed of the vehicle. Consequently the time taken will be less.

PS : The speeds here are low enough that relativistic corrections are meaningless, so this thread should be moved to classical physics (not that SR isn't a classical theory).

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If I changed the guns to flashlights then the light from them hits the target at the same time.....correct?

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I believe so. Things behave different at the speed of light.

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I believe so. Things behave different at the speed of light.

Light itself behaves differently.

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If I changed the guns to flashlights then the light from them hits the target at the same time.....correct?

Yup.

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why does light behave differently ?

in lay please .. thanks

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why does light behave differently ?

in lay please .. thanks

It's not that light behaves differently, as the understanding that the approximation we use at low speeds (compared to c) does not work at all at speeds near c.
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It's not that light behaves differently, as the understanding that the approximation we use at low speeds (compared to c) does not work at all at speeds near c.

I disagree. Light always travels at c for any inertial observer. That is different behavior than for massive particles.

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What I was trying to highlight is that the algebra of velocity addition (which seems to be the crux of the discussion) does not change only when you consider light - you start to see deviations from Newtonian behavior as the velocities get closer and closer to c.

I disagree. Light always travels at c for any inertial observer. That is different behavior than for massive particles.
No doubt, it is - I should have worded my previous post differently. In theory though, if I considered a massive particle (a neutrino, say) traveling at a speed that's arbitrarily close to c in some inertial frame S, its speed will remain arbitrarily close to c in any other frame S'.
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This question may seem simple but for a non-physicist like myself it is sometimes hard to grasp. I'm sure this is beginner's stuff.

I have 2 guns' date=' when fired the bullets leaves the barrel at the same speed relative to where I am standing. I put a target 100 yds away, mount one gun on a vehicle moving towards it and one gun stationary. The mounted gun is preset to fire at the target when it is 100 yds away and is synchronized with the firing of the stationary gun, also at the target. My feeling is that both bullets should hit the target at the same time.... is this correct?[/quote']

It could go either way. If the vehicle is substantial (massive enough with gun mounted solidly) then the vehicle fired bullet should arrive first. If the vehicle is ultralight then the ground fired bullet may arrive first. Somewhere between you get your "tie".

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It could go either way. If the vehicle is substantial (massive enough with gun mounted solidly) then the vehicle fired bullet should arrive first. If the vehicle is ultralight then the ground fired bullet may[/b'] arrive first. Somewhere between you get your "tie".

I think the word is 'semantics'.

I mean, if you really want to get into details, you could look at the first condition stipulated in this experiment:

I have 2 guns, when fired the bullets leaves the barrel at the same speed relative to where I am standing.

and conclude that the bullets MUST hit at the same time no matter what they're mounted on.

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I think the word is 'semantics'.

I mean' date=' if you really want to get into details, you could look at the first condition stipulated in this experiment:

and conclude that the bullets MUST hit at the same time no matter what they're mounted on.[/quote']

That crossed my mind. LOL. But I assumed, from the way it was written, that the guns were equivalent on the ground prior to the experiment. You could assume a certain mass of the vehicle also which would make the recoil equivalent/negligibly different. But you have to assume it and somebody reading this physics forum may not know the difference.

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But I assumed, from the way it was written, that the guns were equivalent on the ground prior to the experiment.

That assumption is correct, that's what I meant when I typed the question. If you don't mind I would like to reword the question with a slightly different scenario.

A pitcher winds up to throw a baseball. He releases the ball and the Jugs gun records the velocity. At same time a baseball is fired from a stationary pitching machine and the velocity is also recorded. The points of release are equidistant from the target and the recorded velocities are equal in both instances.

By the answers given to my previous question am I correct to conclude that the baseballs will arrive at the same time but that the thrown baseball depended upon the forward arm motion of the pitcher and without it would not have equalled the velocity of the machine's baseball?

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A pitcher winds up to throw a baseball. He releases the ball and the Jugs gun records the velocity. At same time a baseball is fired from a stationary pitching machine and the velocity is also recorded. The points of release are equidistant from the target and the recorded velocities are equal in both instances.
If the velocities are the same, the balls are thrown and fired at the same time, and they are equidistant from the target, then they will arrive at the exact same time.

The motion of the pitcher's arm does not factor in as something you add to the velocity, if that's what your wondering. The motion of his arm is actually what is accelerating the ball to it's initial velocity (right when it leaves his hand). Thus, as long as the velocity of the ball as it leaves the barrel is equivalent to the velocity of the ball as it leaves the pitchers arm is the same, and they are at an equal distance from the target, they will arrive at the same time.

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If I changed the guns to flashlights then the light from them hits the target at the same time.....correct?

i cant understand what happens to the light, like physically, if one bullet (in the scenario at the beginning) is known to hit first, why doesnt one light hit first, does it slow down?? or what. can someone please try to explain. thanks

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Ok, that's what I figured. However, if the pitcher was running toward the target when he released the ball then would that momentum have been necessary for the pitch to equal machine's velocity? In other words without the forward run his arm could not have reached the velocity of the machine's ball by itself? I am just trying to get the original thread question's answer straight in my head.

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However, if the pitcher was running toward the target when he released the ball then would that momentum have been necessary for the pitch to equal machine's velocity?
I'm not sure what you're asking. Could you rephrase it? We assumed that the velocity of the balls were the same. If the pitcher is running, then you would have to add the arm velocity + the velocity of his body to get the final velocity of the ball.
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i cant understand what happens to the light, like physically, if one bullet (in the scenario at the beginning) is known to hit first, why doesnt one light hit first, does it slow down?? or what. can someone please try to explain. thanks

The nature of the universe

Nothing can travel faster than c

This is not to say that NOTHING is different about the light...

The light from a flashlight that is moving toward the target will be blueshifted (have more energy) compared to the light from a flashlight that is stationary (relative to the target) But they will arrive at the target at the same time.

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Ok, that's what I figured. However, if the pitcher was running toward the target when he released the ball then would that momentum have been necessary for the pitch to equal machine's velocity? In other words without the forward run his arm could not have reached the velocity of the machine's ball by itself? I am just trying to get the original thread question's answer straight in my head.

I think you are complicating things unnecessarily. You should stick with your original scenario...

2 guns that, when stationary, and side by side, and fired at the same time, release bullets with identical velocities that hit a target at the same time.

Now leave one gun where it is (100 yards from target) and put the other gun solidly on a heavy mount on a track that moves the gun toward the target. Set the experiment up so that the moving gun will fire when it is 100 yards from the target. Simultaneously, the stationary gun fires.

The moving gun's bullet will hit the target first.

velocities are addable (is addable a word?)

So if you have a gun that fires bullets at 100 yards per second, and put it on a track that is moving at 30 yards per second in the same direction as you are firing, the final velocity of the bullet, when fired, is 130 yards per second (relative to a stationary target).

When the velocities you are dealing with approach the speed of light, you can't just add them anymore, the math gets a little more complicated.

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I'm not sure what you're asking. Could you rephrase it? We assumed that the velocity of the balls were the same. If the pitcher is running, then you would have to add the arm velocity + the velocity of his body to get the final velocity of the ball.

Thanks, you answered my question. Sorry to confuse, I just meant that if the pitcher had ran prior to releasing ball and both balls hit target at same time then His velocity plus arm would add up to machine's velocity...that's all. I'm ok with it now...thanks a lot.

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I think you are complicating things unnecessarily. You should stick with your original scenario...

2 guns that' date=' when stationary, and side by side, and fired at the same time, release bullets with identical velocities that hit a target at the same time.

Now leave one gun where it is (100 yards from target) and put the other gun [b']solidly[/b] on a heavy mount on a track that moves the gun toward the target. Set the experiment up so that the moving gun will fire when it is 100 yards from the target. Simultaneously, the stationary gun fires.

The moving gun's bullet will hit the target first.

velocities are addable (is addable a word?)

So if you have a gun that fires bullets at 100 yards per second, and put it on a track that is moving at 30 yards per second in the same direction as you are firing, the final velocity of the bullet, when fired, is 130 yards per second (relative to a stationary target).

When the velocities you are dealing with approach the speed of light, you can't just add them anymore, the math gets a little more complicated.

Thanks also....

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i cant understand where the extra velocity given by the train goes ... i thought at first that the light accelerates upto C and would have just started at the speed of the train, but that cant be correct cause i recently learnt that C is a constant. So where does that extra velocity go ... i dont understand how blueshift ties into that either, can speed be converted into energy ? i didnt think it could

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