That means energy is conserved, and is the contrary of what you hope to say. One rocket at 2V has 4 times the KE of one at V. So the collision will not be the same.
The problem Carteers have has nothing to do with frames,
but a simple schoolboy error that has been made into a theory. TAD have been told countless times they have made this error, and now Prof Rhett has told them again, and with an in-your-face example from Mythbusters.
KE is quadratic with velocity, so KE =1/2mv^2 simple and linear addition is not possible.
The TM is a monumental error. Velocities may be added linearly, and that is what is meant by them being relative: there is no dependence upon some absolute frame or magnitude of the velocity ( for speeds of interest to us).
Velocity is not speed. When making the transform, one may add or subract speed
or velocity, but the latter may not be treated as speed.
If the arithmetic result changes sign an that is simply accepted, error may result.
It is expected that the reader of the text knows how to handle such math, and when the use of RMS may be a way to do that for example.
The TM is a monumental error. Taking the constant speed from from the wind, brings it to zero, but taking it from the ground, makes it negative. That is a reversal of the relationship of wind and ground that does not exist.
Dr Camp says that the making the road a belt that is moving backwards, is an invalid transform and that is the reason why.
Humber,
I believe I did understand the statement-"
As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity."
What the reference means concisely is:
For a closed system, Pick an IRF, specify all the vector velocities of all the entities in question before the event in question and compute the Energies using those velocities. After the event occurs, COE states the the numerical sum of the Energies of all varieties will be the same as the numerical sum of all Energies before the event in question.
Specifically since there is only "the event", all observers will observe the same event and their physics rules will be the same from any IRF. When one IRF is translating linearly wrt to any other, the arithmetic details will differ but the answers will be the same.
Specifically, it is an indication to beginning physics students that if the results from their calculations are different when viewed from two different IRFs, that they have made an error in understanding or their calculations.
One should not assume that some of the posters here do not understand that for KE calculations the translation should be in the "v" part of 1/2mv*2 before the calculation is executed and that this is a vector velocity for more than simple treatments. To state this concisely, pick reference frame, compute all KEs and PEs wrt to that frame. To translate to any other frame, translate all vector velocities and recompute all KEs and PEs wrt the the 2nd frame if required. Most posters understand that the velocity and momentum are vectors and that KE (and energy in general) is a scalar computed from the vector components wrt to an arbitrary but defined IRF.
One salient note:
One should not assume that one knows the educational level of the poster one is addressing unless stated by the poster and their competence with physics and mathematics. Their quality of posts should speak for them in that regard. One should not assume that, because the posters reduce the complexity of an issue to a low level (while retaining the cogent features), that the posters are necessarily incapable of working the issue at a much more sophisticated treatment level.
I believe from the quantity of posts and the information contained, I am satisfied that I understand the disconnect with your level of understanding and my own. I think I will sit this out for a while also.
When you have the "Top Secret" cart ready, let's see it. It must be subtle indeed as you mention it quite a while back.
Regards,
WindGrins