Author Topic: Shooting an arrow out the back of a moving train  (Read 2936 times)

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Offline EPAIII

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Re: Shooting an arrow out the back of a moving train
« Reply #25 on: January 17, 2025, 05:41:16 am »
Bingo!

Before shooting the arrow both the train and the arrow have kinetic energy because both are moving at 100 MPH with respect to the ground.

Now, every action has an equal and opposite reaction. So when the arrow is shot it loses it's kinetic energy with respect to the ground. And, due to that equal and opposite reaction, the train's kinetic energy is increased by exactly the same amount.

The arrow then drops to the ground because it has no horizontal velocity with respect to the ground. And the train is going faster. The total kinetic energy of the overall system remains exactly the same, neglecting the inevitable losses to heat due to friction of course.

Please spare me the nit picking. This is not intended to be a complete explanation of what happens but it does cover the broad strokes.



You have accelerated the train by stopping a moving arrow and dropping it on the ground.
Paul A.  -   SE Texas
And if you look REAL close at an analog signal,
You will find that it has discrete steps.
 

Online coppercone2

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Re: Shooting an arrow out the back of a moving train
« Reply #26 on: January 17, 2025, 07:26:50 am »
you shoot a neutronium tipped arrow from the back of a train car
 

Offline Andy Chee

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Re: Shooting an arrow out the back of a moving train
« Reply #27 on: January 17, 2025, 09:43:57 am »
Another relevant Mythbusters experiment is the elevator drop, and whether jumping before impact would soften the blow.
 

Offline CirclotronTopic starter

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Re: Shooting an arrow out the back of a moving train
« Reply #28 on: January 17, 2025, 11:34:35 am »
Bingo!

Before shooting the arrow both the train and the arrow have kinetic energy because both are moving at 100 MPH with respect to the ground.

Now, every action has an equal and opposite reaction. So when the arrow is shot it loses it's kinetic energy with respect to the ground. And, due to that equal and opposite reaction, the train's kinetic energy is increased by exactly the same amount.

The arrow then drops to the ground because it has no horizontal velocity with respect to the ground. And the train is going faster. The total kinetic energy of the overall system remains exactly the same, neglecting the inevitable losses to heat due to friction of course.

That's the same conclusion I came to.
Okay, so now I understand it is like this - before the arrow is fired it is moving at the speed of the train so it has kinetic energy. As it leaves the bow it slows down to zero speed with respect to the ground so it’s kinetic energy also falls to zero. This kinetic energy is transferred into the RECOIL of the bow, then into you, then into the train. So the train and everything in it gets accelerated by the kinetic energy that was in the arrow while it was moving with the train.
 

Offline EPAIII

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Re: Shooting an arrow out the back of a moving train
« Reply #29 on: January 21, 2025, 08:37:00 pm »
Evil minds think alike!

My apology, I didn't read all the responses before answering.

Seriously, it is the first year mechanics class in physics explanation. It leaves out all the absurd factors like the speed of light and the curvature nature of the gravitational field, relativity, etc. and probably still predicts the results to at least 10 places, probably 12 or more.

Did I say first year mechanics class? Probably in the first few weeks, before the first pop-quiz. Elementary my dear Watson, elementary!



Bingo!

Before shooting the arrow both the train and the arrow have kinetic energy because both are moving at 100 MPH with respect to the ground.

Now, every action has an equal and opposite reaction. So when the arrow is shot it loses it's kinetic energy with respect to the ground. And, due to that equal and opposite reaction, the train's kinetic energy is increased by exactly the same amount.

The arrow then drops to the ground because it has no horizontal velocity with respect to the ground. And the train is going faster. The total kinetic energy of the overall system remains exactly the same, neglecting the inevitable losses to heat due to friction of course.

That's the same conclusion I came to.
Okay, so now I understand it is like this - before the arrow is fired it is moving at the speed of the train so it has kinetic energy. As it leaves the bow it slows down to zero speed with respect to the ground so it’s kinetic energy also falls to zero. This kinetic energy is transferred into the RECOIL of the bow, then into you, then into the train. So the train and everything in it gets accelerated by the kinetic energy that was in the arrow while it was moving with the train.
Paul A.  -   SE Texas
And if you look REAL close at an analog signal,
You will find that it has discrete steps.
 

Offline BrianHG

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Re: Shooting an arrow out the back of a moving train
« Reply #30 on: January 21, 2025, 09:52:56 pm »
These 2 videos seem to hit the point:

Shooting a Nerf Gun Backwards While Driving At The Bullet's Speed Forward


Running Off A Vehicle Backwards At The Same Speed That It Is Driving Forward!
 

Offline IanB

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Re: Shooting an arrow out the back of a moving train
« Reply #31 on: January 21, 2025, 11:18:51 pm »
Before shooting the arrow both the train and the arrow have kinetic energy because both are moving at 100 MPH with respect to the ground.

An important message to really take home from all the discussion above, is that kinetic energy is not a "property" of an object. You cannot know the kinetic energy of an object. You can only measure the kinetic energy of an object (strictly speaking the change in kinetic energy) relative to some external frame of reference.

For example, if you are inside the moving train, with all doors and windows blacked out, radio signals blocked, on a perfectly smooth track with no noise and vibration, and no acceleration, then there is no way to tell whether the train is moving or not. In that case, your frame of reference is the floor you are standing on, and the arrow in the bow has zero kinetic energy in your frame of reference. If you shoot the arrow from the bow, then it gains kinetic energy according to its new speed, and that is all you can say about it.

As a further illustration, consider that everyone reading this is currently traveling at a speed of at least 67,000 mph, and yet we don't think we have a kinetic energy corresponding to that speed. (Although, relative to something not moving at 67,000 mph, that is a LOT of kinetic energy.)
 
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Offline Analog Kid

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Re: Shooting an arrow out the back of a moving train
« Reply #32 on: January 21, 2025, 11:26:36 pm »
Heh; in my younger days when I'd get high (420) and take a walk, I'd sometimes imagine that, instead of me moving along my path, I was stationary and my footsteps were revolving the entire Earth beneath me. Amazing how effortless that was!

Pretty kewl illusion if you can manage it ...
 

Offline Andy Chee

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Re: Shooting an arrow out the back of a moving train
« Reply #33 on: January 22, 2025, 12:53:02 am »
Heh; in my younger days when I'd get high (420) and take a walk, I'd sometimes imagine that, instead of me moving along my path, I was stationary and my footsteps were revolving the entire Earth beneath me. Amazing how effortless that was!

Pretty kewl illusion if you can manage it ...

Something like this?

 

Offline Analog Kid

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Re: Shooting an arrow out the back of a moving train
« Reply #34 on: January 22, 2025, 01:07:38 am »
Something like that.
Never had a totally natty hat like that, though.
 

Offline SiliconWizard

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Re: Shooting an arrow out the back of a moving train
« Reply #35 on: January 22, 2025, 01:30:30 am »
It's all relative.

For a bit more elaboration, you can refer to this discussion: https://physics.stackexchange.com/questions/51220/kinetic-energy-with-respect-to-different-reference-frames
« Last Edit: January 22, 2025, 01:45:17 am by SiliconWizard »
 

Offline IanB

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Re: Shooting an arrow out the back of a moving train
« Reply #36 on: January 22, 2025, 03:41:25 am »
Unfortunately, that StackExchange discussion only deals with elastic collisions. An arrow hitting wood is an inelastic collision. But the theory still works if we consider conservation of momentum, and that the change in kinetic energy is what matters, before and after, rather than the absolute value of kinetic energy.

For example, let's suppose two lumps of dough, each weighing 1 kg, collide, and after the collision they stick together. Not sure how or why this might happen, but whatever. To an outside observer, each lump of dough is travelling at 5 m/s.

Case 1: From the outside observer's frame of reference, each lump of dough has a kinetic energy of 0.5 x (1 kg) x (5 m/s)² = 12.5 J, for a total of 25 J. After the collision, the combined lump of dough is stationary, so it has a kinetic energy of 0 J. The change in kinetic energy during the collision was 25 J − 0 J = 50 J.

Case 2: From the perspective of a fly sitting on one dough ball, it is stationary, but the other dough ball is approaching at -10 m/s. So the total kinetic energy before the collision is 0 J plus 0.5 x (1 kg) x (-10 m/s)² = 50 J. Assuming the fly didn't get squashed, it will find the combined lump of dough after the collision is now moving backwards* at -5 m/s. So the new kinetic energy is 0.5 x (2 kg) x (-5 m/s)² = 25 J. Therefore the change in kinetic energy is 50 J − 25 J = 25 J. The change in kinetic energy is 25 J, the same as before.

* Why it's moving backwards at 5 m/s is conservation of momentum: We have (1 kg) x (0 m/s) + (1 kg) x (-10 m/s) = (2 kg) x (? m/s), therefore the new velocity = -5 m/s to balance the equation.
« Last Edit: January 22, 2025, 02:36:44 pm by IanB »
 
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Offline .RC.

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Re: Shooting an arrow out the back of a moving train
« Reply #37 on: January 22, 2025, 11:00:54 am »
This is an easy one to experiment with yourself and someone reliable.   Just use a slow moving vehicle say under 10kph and jump from it preferably from the rear so you do not get run over.   It you just slowly step off onto the ground you will fall in the direction the vehicle is travelling.   If you jump off in the opposite direction the vehicle is travelling you will cancel out your momentum with respect to the ground.

Another one that catches people out is say you have two tractors joined by a piece of rope pulling against each other, each tractor can pull five tonnes.   How much strain is on the rope?

A:       0 tonnes
B:       5 tonnes
C:      10 tonnes
D: 12.75 tonnes
E: There is no rope.
« Last Edit: January 22, 2025, 11:04:15 am by .RC. »
 

Offline EggertEnjoyer123

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Re: Shooting an arrow out the back of a moving train
« Reply #38 on: January 23, 2025, 02:31:16 am »
If same speed for train and shooting arrow, then yes, if you are a tree, then for you the arrow will just fall like a drop of rain.

The arrow already has kinetic energy before you shoot, because of the moving train.  Then when the arrow is shoot with a bow, all the extra energy from the bow is used to defeat the speed of the train.



But if you want to break physics, take a voltage source, and charge a capacitor through a resistor.  Then disconnect all, and calculate how much energy is left in the capacitor, how much energy was spent by the resistor, and how much energy was taken in total from the voltage source.

You'll discover that half of the energy apparently is missing.  Where energy?  :o



The internet might tell you that the missing energy went into outer space, dissipated as electromagnetic fields.

Now, do the same, charge a capacitor to a certain voltage then disconnect all, except this time, use a current source instead of a voltage source.  If you count the energy again, you'll probably expect half of the energy is missing again, dissipated as electromagnetic fields, just as before when you charged from the voltage source.

Well, no.  When you charge the capacitor from a constant current source, no energy will be missing.  Why?  ;D

You could theoretically have a current source connected to a capacitor with no resistance, and no rules are broken. There is no infinite voltage or current in this scenario. The capacitor voltage would just increase linearly. With the voltage source, infinite current would flow through the capacitor if there were no resistance, which is impossible, so there must be some kind of loss in the circuit. If we assume the voltage source is 1V and the capacitor is at 0V, then charging the capacitor from 0V to 0.1V forces you to waste at least 90% of the power from the voltage source (the remaining 0.9V is dissipated or radiated across something). With the current source, charging the capacitor from 0V to 0.1V costs pretty much no power, because power is I * V and V is almost zero. If the current source is 1A, then at the beginning no power is supplied by the current source because 1A * 0V = 0. At the end, the current source is only providing 0.1W (1A * 0.1V). If the voltage source provides 1A, then it must produce 1W, which is a lot higher than what the current source provides.
 

Offline Andy Chee

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Re: Shooting an arrow out the back of a moving train
« Reply #39 on: January 23, 2025, 02:44:05 am »
Another one that catches people out is say you have two tractors joined by a piece of rope pulling against each other, each tractor can pull five tonnes.   How much strain is on the rope?

A:       0 tonnes
B:       5 tonnes
C:      10 tonnes
D: 12.75 tonnes
E: There is no rope.

Mythbusters also has a similar episode, testing the statement that two cars crashing head-on at 50mph is the same as a single car crashing into a stationary brick wall at 100mph.

 

Offline BrianHG

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Re: Shooting an arrow out the back of a moving train
« Reply #40 on: January 26, 2025, 12:22:23 am »
 

Offline electrodacus

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Re: Shooting an arrow out the back of a moving train
« Reply #41 on: February 03, 2025, 04:50:23 am »

Third paragraph is something that I've accidentally "discovered" by myself while trying to come up with a more intuitive explanation for the capacitor paradox.  I still don't understand why the capacitor paradox doesn't happen any more when charging from a constant current source instead of a constant voltage source.


I think the original problem was answered.

Regarding the capacitor paradox (there is no paradox).

In your example charging a capacitor trough a series resistor from a constant voltage source half of energy is found in capacitor and the other half was dissipated by the series resistor.
There is nothing missing. That half you think is missing had just increased the temperature in the room you performed the experiment.

Online coppercone2

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Re: Shooting an arrow out the back of a moving train
« Reply #42 on: February 03, 2025, 05:40:17 am »
this is why a educational lab needs a 50 amp power supply, super caps and smoke
 

Offline SiliconWizard

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Re: Shooting an arrow out the back of a moving train
« Reply #43 on: February 03, 2025, 05:50:36 am »
Or you need to destroy vehicles just to illustrate basic laws of physics.
 

Online coppercone2

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Re: Shooting an arrow out the back of a moving train
« Reply #44 on: February 03, 2025, 04:53:34 pm »
you wont hear someone going 'well uh... lemme try to remember what that guy wrote on a chalk board 20 years ago..."
 


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