MrVibrating wrote: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?
Assuming a "head on" collision...
If the collision is inelastic so that the masses "stick" together, as viewed from the original reference frame where each was moving at 10 m/s toward one another, both come to a stop and all 100 J of kinetic energy is lost in the collision.
The momentum for the complete system was, of course, 0 kg*m/s both before and after the collision since the equal masses were moving with the same speed but in opposite directions. +10-10=0
If you are viewing the same collision from a reference frame in which one mass is at rest and the other is moving at 20 m/s, though, then the initial momentum of one mass is 0 kg*m/s and the other is 20 kg*m/s for a total system momentum of 20 kg*m/s. 0+20=+20
The total momentum has to be the same after the collision, so we will have 2kg stuck together and moving at 10 m/s. That means that in this reference frame the masses will still have a 100 J of kinetic energy after the collision.
100 J is lost in the collision no matter from which reference frame it is viewed.
I don't believe in conspiracies!
I prefer working alone.
From the external frame, each mass has 50 J and the system's net energy is 100 J.
But from the rest frame of either mass, the other is closing at 20 m/s.
And at 20 m/s, a 1kg mass has 200 J.
So, by applying the standard metric of KE - 1/2mV^2 - each mass in their own rest frame sees the other approaching with twice as much energy as the net system, as measured from an equidistant observer.
Internally, there's double the energy visible from the outside.
It appears to be another example of creation ex-nihilo, from nothing more than the exponential relationship of KE to linear P.
I originally noticed it in a rotating frame - a pair of contra-rotating rotors.
Linear or angular, the worst part is that we have 200% OU from a system with zero net momentum (since the two velocities are equal and opposite).
So... from either internal frame there's 200 J, while externally there's arguably zero, but certainly no more than 100 J.
We can dismiss as unhelpful the contradictory notion that the net system momentum and energies are actually zero - after all there is gonna be a collision, and it's gonna involve at least 100 J. Nonetheless, the pilot of each mass is bracing for a 200 J impact.
Common sense says the anomaly must be a measurement error, yet the measurement in question is simply 1/2 * 1 kg * 20 m/s squared: so, 20 * 20 = 400, and half of that is 200. It's the same, standard metric we're uising for the external measurement, so equally as valid.
The anomaly arises because energy squares with velocity, instead of summing linearly. Yet the only way i can resolve it is to assume that the observer frame is preferential (ie. the universe's 'preferred frame'), which contravenes the most fundamental tenets of relativity, not to mention our own predicates here.
FWIW i don't think the gain is real or can be measured - i'll run some sims later to investigate further - but i can't see where i could be going wrong with such basic principles..!?
Furcurequs wrote:
The total momentum has to be the same after the collision, so we will have 2kg stuck together and moving at 10 m/s. That means that in this reference frame the masses will still have a 100 J of kinetic energy after the collision.
Precisely my problem - if they're stuck together then it's a perfectly inelastic collision and 100 J has been dissipated, yet we still have another 100 J floating about. Apparently.
Clearly something's wrong. It just looks like the thing that's wrong is.. everything we know; our most basic energy measures, our very predicates... things ostensibly too simple and fundamental to be prone to such abuses..!?
I didn't follow this completely, but as far as I know Conservation of Kinetic Energy only works in a single reference frame, and why should it: as the velocity of the reference can be changed arbitrarily the v² (now relative to what) becomes meaningless.
Mr.V wrote: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?
I think Furcurequs explained it well.
I think you understand case 1.
In case 2, the reference frame is moving with one of the masses. After the collision, the reference frame is still moving. The two masses are stopped from an external reference frame, but from the reference frame that is still moving, they have a non-zero velocity. The extra 100J comes from the fact that your reference frame is still moving away from the "stopped masses".
Case 2 before collision:
reference frame is moving at 10 m/s with mass 1.
KE1 = 1/2 mv² = 0 J
KE2 = 1/2 mv² = 1/2 (1 kg) (20 m/s)² = 200 J
KE tot = 200 J
Case 2 after collision:
reference frame is still moving at 10 m/s, but now away from the masses.
KE tot = 1/2 mv² = 1/2 (1kg + 1 kg) (-10 m/s)² = 100 J
The net energy lost is still 100 J.
The "extra" 100 Joules of energy after the collision comes from your choice of reference frame (that is still moving away from the collided objects). The reference frame does not stop because your mass stopped. The reference frame keeps moving.
1) If Bruce lee punches someone with his 6" punch, as he is flying through the air do you think his opponent wonders where the energy came from ? Do you think the effect is any less real if it came from a "fictitious" force ? To his opponent does it really matter where the energy came from ? The effect is still the same.
For Every action There is an equal an opposite reaction. This is the law and it stands for every observer. I believe ( just thinking).
Once you have eliminated the impossible whatever remains however improbable must be the truth.
I understood your question the first time, but maybe I wasn't as direct as I could have been with my answer. Hopefully, Wubbly's explanation is more thorough than mine.
I should have pointed out this significant bit:
If you arbitrarily pick a reference frame, that means you have an "observer" with some sort of arbitrary motion. The kinetic energy, then, which this observer sees will depend upon his own arbitrary motion and so could itself be somewhat, erm, arbitrary.
Since there can be no actual physical interactions with our imaginary observer, however, the only relevant bit is how much energy is available to be harnessed or lost in an interaction with the actual masses being considered.
If we choose a reference frame in which the center of mass of the total system of moving masses is at rest (your original condition), then under this condition all the kinetic energy seen by our imaginary observer is available for use/loss - the 100 J in your example.
Last edited by Furcurequs on Tue Apr 26, 2016 6:42 pm, edited 1 time in total.
1/2mv^2 is an interesting concept for kinetic energy. A basic example is a bowling ball rolling down the lane. It's angular and linear momentum would be close. Thus 1/2 would be linear while the other half is angular.
1/2mv^2 seems to be more representative of it's linear momentum. This is because if it's linear velocity is less than 9.8 m/s, then there is an extra potential of 9.8 m/s acting on it. This is why a 1 kg weight has 9.8 netwons of force while at rest.
As such, a bowling ball at the top of a ramp has 9.8 newtons of potential KE. Once it starts rolling down the ramp, it's velocity plus the 9.8 newtons of force (gravity) would seem to be a more accurate determination of it's KE.
And in this, mv+a might be more representative of the actual KE of a body in motion or at rest.
This sparks the same question (directed to anyone, it just sparked) as with Pequaide, how are these similar? (perhaps I'm the one who doesn't understand)
The angular momentum of a solid ball is (2/5)*m*r²*w = (2/5)*m*r*v measured in [Joule second] or [kg m² / s].
The linear momentum would be m*v, measured in [Newton second] or [kg m / s]
Kinetic energy is not representative for both of them [Joules]
When it rolls just right, I guess we should use the combination (rotational kinetic energy+linear kinetic): 0.5*I*w²+0.5*m*v², (for a solid ball): 7/10*m*v² [Joules]
-is this all correct?-
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
It's obviously not a real gain and i don't expect a firm conclusion to suggest any different.
Still, i recently looked at KE gains from moving reference frames, concluding that the momentum of the donor frame is obviously slowed down by the transfer..
Eg: you're on a station platform, i'm on a train passing thru at 10 m/s, and i accelerate a 1 kg mass forwards, in the train's direction of travel, by 1 m/s. Aboard the train, i've only spent 1/2 J, but from the platform you see a 1 kg mass accelerate from 10 to 11 m/s - a rise of 10.5 J.
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.
So no problem there.
The problem here is that energy measurements in all reference frames have to remain equally valid. In this case, the masses could be following curved trajectories, having pushed off from a third mass (such as the Earth), or even from one another, if they were tethered to a common axis.
If we take the Earth as our static reference frame, both masses have 50 J each. On board each mass, they both see themselves as having 50 J relative to Earth. If they were launched side-by-side, then they initially had zero energy and momentum relative to each other.
But then their trajectories curve (due to gravity, say) and they end up on a collision course, still carrying all their initial momentum. From either's frame, both still have 50 J relative to Earth.. but relative to one another they now have 200 J, simply because they converged on equal opposite vectors, doubling their effective velocity, and KE squares with velocity.
So is the 200 J combined impact energy real or not? Where's the slowed net system momentum (the 'train') - it can't be the Earth, since that's the common frame of both masses, from which the net energy remains 100 J. Both masses are decelerated equally in the collision, or bounce back with equal momentum if it's fully elastic, and adopting either as a rest frame we find that it is actually accelerated by the collision, not decelerated (because it was already considered static)..
It looks to me like the 'gain' is simply due to re-apportioning our given system momentum - assigning the momentum to one frame that we take from the other, investing it instead in a higher relative velocity and thus greater energy value of the available momentum - akin to that 'momentum transfer' concept that Fletcher started..
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.
But here, both masses push off against the Earth (or each other, even), so the 'train' here is the Earth, and its momentum can remain constant (the masses could've been launched from opposite poles, for instance), thus precluding any deceleration of the Earth's momentum.
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..
This is doing my nut in.. on the one hand i can't believe it's a valid measurement, but on the other, all energy measures from all frames must be equally valid with regards to their relative velocities..
If you're on the launchpad, both masses definitely only have 50 J. But if you're on board either and staring at a head-on collision, there's definitely a 1kg mass approaching at 20 m/s, with 200 J of energy. Both measures must be true and valid in their respective conditions, yet they seem irreconcilable.. what's to stop us harvesting the 200 J (by compressing a spring, say) and letting the masses fall back to earth carrying double their initial PE?
Maybe i should focus on the rotary example - two counter-rotating rotors, torqued against a stator bound to the Earth. That's gotta simplify the problem...
This sparks the same question (directed to anyone, it just sparked) as with Pequaide, how are these similar? (perhaps I'm the one who doesn't understand)
The angular momentum of a solid ball is (2/5)*m*r²*w = (2/5)*m*r*v measured in [Joule second] or [kg m² / s].
The linear momentum would be m*v, measured in [Newton second] or [kg m / s]
Kinetic energy is not representative for both of them [Joules]
When it rolls just right, I guess we should use the combination (rotational kinetic energy+linear kinetic): 0.5*I*w²+0.5*m*v², (for a solid ball): 7/10*m*v² [Joules]
Mr.V wrote: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.
This sparks the same question (directed to anyone, it just sparked) as with Pequaide, how are these similar? (perhaps I'm the one who doesn't understand)
The angular momentum of a solid ball is (2/5)*m*r²*w = (2/5)*m*r*v measured in [Joule second] or [kg m² / s].
The linear momentum would be m*v, measured in [Newton second] or [kg m / s]
Kinetic energy is not representative for both of them [Joules]
When it rolls just right, I guess we should use the combination (rotational kinetic energy+linear kinetic): 0.5*I*w²+0.5*m*v², (for a solid ball): 7/10*m*v² [Joules]
-is this all correct?-
After double checking your math, I'd say that looks correct.
A rolling ball, of course, has both translational and rotational energy.
For such a "pure roll" situation, we can see that approximately 29% of the energy, then, is in the rotation.
ETA: It, of course, is different for other geometries.
Before there was Walter Lewin, we had Julius Sumner Miller!
This is a nice video:
"Julius Sumner Miller: Lesson 12 - The Strange Behavior of Rolling Things"
Thanks Furcurequs, sometimes those things need a double check (I don't use this stuff outside PM-research)
It's much fun to see Sumner present his stuff, I don't know if that would make him a great teacher.
MrVibrating wrote:The problem here is that energy measurements in all reference frames have to remain equally valid.
But only valid within their own reference.
When there is a moving frame of reference with velocity V the total momentum is changed by (m1+m2)*V for that reference frame.
The energy change is complex.
When both masses are equal (to make it easier) then reference frame V adds (m*V*(V+v1+v2)) to the Total Kinetic Energy.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
