KE of system / different reference frames question

DaleSpam

No, the initial relative verlocity of the earth wrt the car would be [itex]v_{i,e}-v_{i,c}[/itex] the final relative velocity would be [itex]v_{f,e}-v_{f,c}[/itex]. Your quantity [itex]v_{f,e}+v_{i,c}[/itex] is not a relative velocity of anything. See the link I posted.

Sure I do, Newton's laws are the basis for all of my calculations. That and the standard definitions for the various quantities.

What makes you think it is an error? A small rock and a big rock fall at the same acceleration despite the difference in the mass. Not everything is a function of every mass.

If you are trying to calculate the change in the KE of the car then you are missing a factor of 1/2. Otherwise I don't know what you are trying to calculate.

DaleSpam

I don't know if you are being deliberately irritating or you really can't grasp something this basic. I have said this multiple times already:

Earth loses 100 kJ KE
Car gains 50 kJ KE
Battery gains 50 kJ electrochemical energy
All energy is accounted for.

There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm.

The KERS recovers 50 kJ of electrochemical energy from the KE of the earth and the car gains 50 kJ of KE from the KE of the earth. The earth loses 100 kJ of KE.

Why don't you ponder that a bit until you get it.

Humber

They are different on each case.

And the change in momentum that you said you did not make, though it is quite clear that you did.

The primary error remains. The result of the elaborated calculation that eliminates the Earth's mass in the process is, and always is;

iΔpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

Humber

In the first case, the car will be stationary, with the rest state KE, and with a known amount of charge in the battery.
The above says that in the second case, there will be both charge in the battery and kinetic energy. A clear violation of both the conservation of momentum and energy.
All the result of one arithmetic error, that has been flagged to your attention.
Δpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

Your calculation says you have made that error.

Yes, I agree. Your calculation confirms that you have made that error.

jduffy77

I should clarify something which most people interested in physics will not be familiar with. What Humber means when he says "a calculation confirms an error" is not that an error was made in the traditional sense, like incorrect math or a logical inconsistency. He simply means that the result conflicts with his preconceived expectation of what the result should be. So it is wrong.

Humber

The correct solution to this problem, when the juvenile runs out of excuses, and recognizes his error, relies on nothing more than the conservation of momentum, which is a law quite independent of frames.

There are only two bodies involved: the car and the Earth so, a very simple case. When the car slows, it changes momentum;

Car gains momentum -p
Earth gains momentum p

p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

The total energy is therefore;
p²/2m_car+ p²/2m_earth

Because m_earth is very large ,~6e24kg, the second term is negligible and can be approximated to the first term.

Initially, all of that energy will have come from the fuel of the F1 car, and in part, like the consumer counterpart, KERS is intended to reduce fuel consumption by returning energy that would otherwise be lost.

In a conventionally braked vehicle, the KE of the car's mass is dissipated as heat in the brakes. KERS does not affect momentum transfer to the ground, but stores the KE that is otherwise lost to heat.

That energy may be transferred to a flywheel;
The angular momentum L, of a particle, and the moment of inertia, I
L = r.p = r.mv
L =I.ω
KE = 1/2.I.ω²

In the given example of a 1000kg vehicle braking to a stop from 10m/s, and where the mass of the Earth is 1e25kg, the energy transferred to the ground is;

(100000kg.m/s)^2/2e25kg
= 5e-18J

There is no significant transfer of energy to or from the F1 car and the ground as a result of KERS. The conservation of momentum says that result is independent of frame.

Humber

The following calculation shows that there is negligible transfer to the ground, and is in agreement, also producing 5e-18J

[itex]p_{i,c}=m v_{i,c}=10000 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ[/itex]

[itex]p_{i,e}=M v_{i,e}=0 \ kg \ m/s[/itex]
[itex]f_e=-f_c=1000 \ N[/itex]
[itex]\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s[/itex]
[itex]p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s[/itex]
[itex]v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s[/itex]
[itex]KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ[/itex]
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J[/itex]



All that remains, is to show that result is independent of frame.
Frame independence for the result can be shown by the introduction of an observer, moving relative to the Earth and Car at velocity u.

m = mass of car

ΔKE(car)

= (−m u − p)^2/2m − (−m u)^2/2m

= (m^2 u^2 −2m u·p + p^2)/2m − (m^2 u^2)/2m

= −u·p + p^2/2m


M = mass of Earth.

ΔKE(earth)

=(−M u + p)^2/2M − (−M u)^2/2M

= (M^2 u^2 + 2M u·p + p^2)/2M - (M^2 u^2)/2M

= u.p + p^2/2M

The total energy is once again

p^2/2m + p^2/2M

DaleSpam

This is true, due to Newton's 3rd law.

This is false, in general.

If some object gains momentum Δp then [itex]p_f=p_i+\Delta p[/itex]

Dividing both sides by the mass m we get [itex]v_f=v_i+\Delta v[/itex]

Substituting that into the expression for the object's final KE we obtain
[itex]KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m [/itex]
[itex]\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m [/itex]

This is different from what you wrote because of the [itex]v_i \Delta p[/itex] term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the [itex] \Delta p^2/2m [/itex] term drops out, but the [itex]v_i \Delta p[/itex] term does not.

DaleSpam

That is correct.

DaleSpam

What exactly are you claiming that I said? Please use the quote feature to show exactly where I said I did not make a change in momentum. If you cannot find where I said it then I didn't say it, you only inferred it. Given this conversation so far, you probably inferred wrongly.

DaleSpam

Yes, that is correct, 50 kJ charge in the battery and 50 kJ KE in the car.

It is not a violation of either conservation of momentm or conservation of energy.

DaleSpam

I love the arrogance of the ignorant. Keep it coming Humber, it is very amusing.

Again, this is false, in general. See my post 59 above. This is your key error, so please spend some time studying the derivation.

Humber

KE_earth =p^2/2M_earth
= (10000kg.m/s)^2/2*1e25
= 5e-18J

DaleSpam

Which works since [itex]v_i=0[/itex] for the earth in frame (a). It does not work in frame (b) where [itex]v_i \ne 0[/itex] for the earth.

Humber

KE = p²/2m = m²v²/2m = 1/2mv²

When you finish falling over yourself and the arithmetic churn, you can stop.

DaleSpam

Do you agree with the derivation in post 59?

If yes, then do you understand how it was your key mistake?

If no, then which specific step of the derivation do you believe is in error, and what do you believe is the error?

Humber

You correct it. I am not going to help you chase your tail.