KE of system / different reference frames question

[Dale]

To start with the Earth at 10m/s relative to the car, is a totally different physical situation.

Here is where you are flat out wrong. It is merely a different (but equally valid) description of the same physical situation. You are denying the principle of relativity. This has been a cornerstone of physics since Galileo's time:
http://physics.ucr.edu/~wudka/Physics7/Notes_www/node47.html
You are literally centuries out of date in your thinking.

In the first case the calculation is done with the car having 50kJ, which after 10s, is transferred to the battery.

Yes.

In the second, it's already there, at t = 0.
Makes no sense.

This is incorrect. In the second the Earth has a huge amount of KE, of which after 10s, 50 kJ is transferred to the battery and 50 kJ is transferred to the car's KE.

These are both correct descriptions of the same physical situation. Momentum and energy are conserved in both cases, and Newton's laws are obeyed in both cases.

 

[Dale]

It's your notation i = initial, c = car, e= earth. There is one transfer of 10000kg.m/s to the momentum, p_i,e. It makes no difference what that is. The resultant KE depends on the Earth's mass.

So prove it, mathematically. If you are going to make claims, then you should be able to back them up with clear and unambiguous derivations, as I have done.

 

[Humber]

I figured this was your core error, you don't understand the meaning of conservation laws. Conservation laws don't mean that the quantities are fixed, regardless of frame. They mean that once you pick a frame, the total quantities will stay the same in that frame.

And none can be created.

If you change frames, the quantities change. That's how conservation laws work. Do you get this, or not?Wrong, the changes are frame independent, but that doesn't "follow" from anything, that statement, and only that statement, is the conservation law. Think about this as long as it takes.

Yes it does. There are two objects. The ground and the car. Momentum is conserved, so what one gains, the other loses, and that is entirely frame independent, as the total remains the same and pe - pc = 0 or p = -p

No, not if you think that p is a change in momentum, which is how you are using it.

When all is transferred Δp is p. 100% of p is p.

Doesn't it concern you that you do not get the correct answers, and you conclude that all the experts are wrong, but when they tell you what you are doing wrong, you just claim you are right? I can tell you right now, you will never learn anything that way. Is it all right with you to never learn anything?

Oh, right.

Notice where once again you associate p^2/2m with energy changes. Wrong.

Oh right.

p= mv ( I assume you know that)

p² = m²v²

p²/2m = m²v²/2m = 1/2mv² = KE.

 

[Dale]

Btw, I think that Huber's continued avoidance of the question is because he recognizes that the derivation is correct, but he feels that he would lose face to admit it.

 

[D H]

This nonsense has gone on long enough. Thread locked.