Poss. Symmetry Break?

[John doe]

https://www.google.com/url?sa=t&source= ... wfSnpDKJPw

The is what the mythbusters had to say...

[ME]

The heads-on collision is an inelastic one (by design), the coefficient of restitution will be zero.
The Sum Total of the kinetic energy of that collision will just be 2*(0.5*m*v^2), and not miraculously become 1*(0.5*m*(2*v)^2

When one would track a single car with speed 'v' then its optical velocity becomes 0, while the other car would be doubled.
But when they collide the first car would suddenly get some speed 'v', while the other will half its speed and would also be 'v'.
All energy should be dissipated into forms of heat, and yet -from the tracking perspective- they both move, and one could assign some false energy label.

I guess the only way to calculate what happens in such moving frame of reference is by momentum, or otherwise temporarily step out of that frame of reference.

[Wubbly]

Here's a spreadsheet of MrV's collision from 3 different reference frames:
1) reference frame Stationary at center of mass,
2) reference frame moving at 10 m/s (same velocity as mass 1),
3) reference frame moving at 20 m/s (10 m/s greater than velocity of mass 1).

Momentum is conserved in each reference frame.
100 Joules of energy is lost in each reference frame.

You can't take the 100 J of "leftover energy" from reference frame 2 and transfer it back to reference frame 1, or the 400 J of "leftover energy" from reference frame 3 and transfer it back to reference frame 1.

So.. the train / platform paradox i understand, and is resolved by a deceleration of the train's momentum, in proportion to the accelerated mass's gained momentum, conserving the net total momentum.

That seems reasonable.

This is resolved by the fact that the train itself has lost momentum equal to the 10 J the 1 kg has gained from the static frame.

This sounds very odd. Momentum and energy are never equal. You can't take momentum and create energy. Momentum and energy are two very different concepts. One is a vector and the other is not. Only a momentumist would mix and match them.

[daanopperman]

Hi Mr V ,

What would happen in the case of a mass accelerated vertically , colliding and transferring it's KE to PE to a second mass , and then drops down vertically under the influence of gravity to the starting point .

[MrVibrating]

Mr V .. I've run sims of your problem scenario before. Don't have them now since my laptop was stolen.

I decided I had to find out how much 'Work' could be done from each scenario. That could even be compressing a spring in between.

IIRC it was identical for each reference frame at (in your case) 100 J.

F x d = 100 J.

LOL yes obviously - i don't for a moment suspect that there really is another 100 J here.

I'm just trying to resolve the discrepancy. The two masses do have a relative velocity of 20 m/s, as measured from either's rest frame, at which a 1 kg mass has 200 J.

So for me, at least, the question remains - why isn't this measure valid? Dumb question, i know. I'm sure it'll click soon..

[MrVibrating]

The bottom line, it seems to me, is that two identical 10 kg/m/s momentums, invested in identical masses, always have a value of 50 J each relative to Earth (or whatever common frame), but the total possible energy value of that momentum between the masses themselves depends upon their relative vectors, and any changes they undergo. So it can be anything from zero to 200 J..

Why stop at 200 J? Take your two masses approaching each other at 20 m/s, and measure them from a reference frame that is moving at 10000 m/s relative to the first mass, and your energy levels jump up considerably. Momentum would be conserved in the 10000 m/s reference frame, and you would still loose 100 J of energy in the 10000 m/s reference frame after the collision. You don't have to wonder where all the "unaccounted" energy went to. It's relative to the reference frame that you're using.

Sure we could take any external frame, and the energies of each mass would be modified relative to this new reference velocity.

But here we're just using the internal frames. Two masses on a collision course. Neither is accelerating, so we have a closed system of two non-inertial frames.

The collision is going to involve just these two masses, and we're only measuring the energy from their own reference frames.

Not some arbitrary external frame.

The relative velocity from either rest frame is 20 m/s. So each mass sees the other approaching with 200 J.

We could consider an equidistant observer as an external FoR, but his intercession has the effect of halving the apparent system energy, removing 100 J. Yet we wouldn't consider this an arbitrary frame since he's now very much part of the system (assuming he's not killed in the collision).

So what makes it interesting to me is that it's only dealing with the most basic, internal FoR's, not arbitrary external ones.

[MrVibrating]

Here's a spreadsheet of MrV's collision from 3 different reference frames:
1) reference frame Stationary at center of mass,
2) reference frame moving at 10 m/s (same velocity as mass 1),
3) reference frame moving at 20 m/s (10 m/s greater than velocity of mass 1).

Momentum is conserved in each reference frame.
100 Joules of energy is lost in each reference frame.

You can't take the 100 J of "leftover energy" from reference frame 2 and transfer it back to reference frame 1, or the 400 J of "leftover energy" from reference frame 3 and transfer it back to reference frame 1.

So.. the train / platform paradox i understand, and is resolved by a deceleration of the train's momentum, in proportion to the accelerated mass's gained momentum, conserving the net total momentum.

That seems reasonable.

This is resolved by the fact that the train itself has lost momentum equal to the 10 J the 1 kg has gained from the static frame.

This sounds very odd. Momentum and energy are never equal. You can't take momentum and create energy. Momentum and energy are two very different concepts. One is a vector and the other is not. Only a momentumist would mix and match them.

The accelerated mass has gained 1 kg/m/s at the expense of the train's momentum.

So if the mass has gained 1 kg/m/s, the train has been decelerated by 1 kg/m/s (assuming it's coasting without losses).

The energy value of this transferred momentum is vastly unequal - the train has lost a tiny fraction of the 10 J gained by the accelerated mass on board, due to the vastly different inertial momentums and respective velocity changes..

So if you could accurately measure the system's momentum and energy from the station platform, the former remains constant, and equally divided, but the latter does not - the train loses 1 kg/m/s and nanojoules of KE, while the 1 kg mass gains 1 kg/m/s and 10 J of KE.


The tables you've given are great (tho quite unecessary - this is not what i was considering looking at in sims), however example 3 is an arbitrary external frame, and example 2 confirms the validity of the 200 J measure from either rest frame.

Obviously we can't 'take' that 100 J leftover and use it. That was never my point.

My point was simply that the internal frames aren't arbitrary. I chose values of 1 and 10 precisely to obviate any need for number crunching - everyone can do the maths in their head, so it's not a 'maths' problem as such.. just a curio i guess.

But the question remains - why can't we 'take' the extra 100 J from the internal For? I mean, suppose we place a deformable medium at the collision site, per S'Gravesande's famous experiements - would it receive 100 J or 200 J of deformation? What if we place it in one of the mass's rest frames?

[Furcurequs]

Only just noticed something silly - i'm sure this must've come up here before, but would appeciate some help resolving it:

- two identical 1 kg masses on a collision course, each travelling at 10 m/s

- from the external frame, each mass has 10 kg/m/s of momentum, and 50 Joules of KE

- from the rest frame of either mass, the other is approaching with 20 kg/m/s and 200 J

..so momentum's conserved, but where's that extra 100 J come from?

Two identical cars are in the same scenario as above. They crash head on. What is the energy of the collision ? Both cars deform the same extent.

Is this the same as one car traveling the same speed colliding with a concrete wall ?

I believe in the youtube link posted, the Mythbusters crashed same sized cars into a wall at 50 and 100 mph. They also did a head on collision with two others each going 50 mph.

The damage to each car during the head on collision was equivalent to that of the car crashing into the wall at 50 mph and not the one hitting the wall at 100 mph.

You are probably already aware of this. I'm just trying to summarize what it showed so that I could add the following.

It would have been nice if they had also crashed a car going 100 mph into a stationary car. Had they done this, the damage should have been about the same as in the head on collision they actually did do, but both cars would have been moving approximately 50 mph immediately after the collision.

The car moving 100 mph, of course, had 4 times the kinetic energy of the one moving at 50 mph. Most all of that energy, then, went into the destruction of the car when it was crashed into the wall. Only about half that energy would have gone into destroying the two cars if it had hit a parked car. ...though not exactly since there would be some interaction with the earth through the tires.

[MrVibrating]

Missed that link, will go back and check it, but sounds like it speaks directly to the problem..


The simplest embodiement of the problem would look something like this:

- the two identical masses are launched into orbit, using spring launchers (or gas guns) instead of rockets (which would complicate things) - so their energies are precisely apportioned at launch.

- the launchpad straddles the north pole, each launcher on either side an equal distance from the earth's axis

- upon launch, the pilots of each projectile can see each other's crafts as they're shot upwards, and that they have the same energy relative to the ground

- they settle into opposing orbits, and will collide on the other side of the planet, over the south pole.

- as they approach the south pole and their imminent collision, the pilot of each craft looks down and sees that they're still travelling at equal velocity relative to the ground

- but looking forward, they see each other aproaching at twice their ground velocity. If each is travelling at 100 m/s, their relative closing velocity must be 200 m/s

- finally, suppose we have an accurate instrument cluster in the nose-cone of one projectile, and we remain in its reference frame throught the experiment

The telemetry from the on-board instruments should confirm an impact energy proportionate to that closing relative velocity - ie. four times the energy of either projectile relative to the ground, and twice their net energy.

As a sanity check, suppose instead that there's a very tall mountain at the south pole... reaching all the way up into their orbital paths. So instead of colliding with each other, they strike the earth instead. Relative to the earth, they only have KE equal to their ground velocity. But in their frames, the mountain is a stationary obstacle... whereas the oncoming projectile is "oncoming", and not-at-all stationary. THEREFORE if the collision energy between the masses is only the sum of their respective KE relative to the ground, rather than 2x that value, then surely their KE relative to the mountain must also be 2x less than their sum total?

Can't have it both ways, surely? If two masses are at 10 m/s relative to the ground, on opposing vectors, then they're at 20 m/s relative to one another, and KE scales as half the square of velocity, not summing linearly.

For their energy to sum linearly, the 1/2mV^2 function would have to fail, instead being replaced by 1/4mV^2..

I'll have to watch this Mythbusters episode..

[Furcurequs]

The link to the youtube video was in John Doe's post above. It went through Google, though, and so had a bunch of Google stuff in it.

Here is the direct link:

https://www.youtube.com/watch?v=r8E5dUnLmh4

[John doe]

Only just noticed something silly - i'm sure this must've come up here before, but would appeciate some help resolving it:

- two identical 1 kg masses on a collision course, each travelling at 10 m/s

- from the external frame, each mass has 10 kg/m/s of momentum, and 50 Joules of KE

- from the rest frame of either mass, the other is approaching with 20 kg/m/s and 200 J

..so momentum's conserved, but where's that extra 100 J come from?

Two identical cars are in the same scenario as above. They crash head on. What is the energy of the collision ? Both cars deform the same extent.

Is this the same as one car traveling the same speed colliding with a concrete wall ?

So what is the relevance or what does it show?

I believe in the youtube link posted, the Mythbusters crashed same sized cars into a wall at 50 and 100 mph. They also did a head on collision with two others each going 50 mph.

The damage to each car during the head on collision was equivalent to that of the car crashing into the wall at 50 mph and not the one hitting the wall at 100 mph.

You are probably already aware of this. I'm just trying to summarize what it showed so that I could add the following.

It would have been nice if they had also crashed a car going 100 mph into a stationary car. Had they done this, the damage should have been about the same as in the head on collision they actually did do, but both cars would have been moving approximately 50 mph immediately after the collision.

The car moving 100 mph, of course, had 4 times the kinetic energy of the one moving at 50 mph. Most all of that energy, then, went into the destruction of the car when it was crashed into the wall. Only about half that energy would have gone into destroying the two cars if it had hit a parked car. ...though not exactly since there would be some interaction with the earth through the tires.

[Mark]

Textual summary:
http://scienceblogs.com/gregladen/2012/ ... ollisions/

[Furcurequs]

The fellow who wrote that blog post got it wrong about the 100 mph car. Some people in the comments section tried to explain it to him. The 100 mph car has 4 times the energy of one 50 mph car, so then it has twice the total energy of the two cars that were going 50 mph before the head on collision.

ETA: Maybe he just worded it oddly? I don't know. Maybe all he meant was that the energy of destruction in the head on was equal to two cars going 50 mph into the wall.

[Mark]

Sorry, thought I was being helpful. When I saw it I just skimmed thru it, didn't take time to analyze. My mistake.

:-/

[Furcurequs]

...but it's "scienceblogs"! How could they be wrong?!

...lol

No problem.

I was trying to give the blog poster the benefit of the doubt myself, thinking he might have just been a little vague in his explanation, but it seems he also explained the pendulum test that was in the original Mythbusters episode wrong, too.

Just make sure to read the comments section, then. Some people explain things correctly there. ...and remember that the car moving at 100 mph has 4 times the kinetic energy of the same mass car moving at 50 mph.

In the original episode, the Mythbusters also did a preliminary test where they used pendulums to crash into each other and deform clay. To get twice the speed of the pendulum bob at the bottom of the swing, you need to start it from 4 times the height. The blog poster said it was twice the height.

Based upon some other stuff I've read about the episode, I think the Mythbusters actually did try to do a 4 to 1 height ratio, though even they may have had a slight screw up. I may try to find the original episode to see.

I think they dropped the bob from the horizontal for the maximum speed, which means they should have dropped from an angle of about 41 degrees from the vertical for the minimum. According to comments I've seen, though, they may have mistakenly used the wrong angle - measuring 49 degrees from the vertical instead of 41. If they measured 49 degrees, it should have been from the horizontal.