KE of system / different reference frames question

[Dale]

There is a change of 100kJ for the Earth in the second case, but the car gains only 50kJ.
Does the remaining 50kJ go as heat?

If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.

 

[Humber]

Sure, I would be glad to clarify.

Differences in velocity are frame invariant in Newtonian physics. So you are right to be concerned about this.

Differences in velocity are frame invariant? No, velocity is relative, so is frame dependent.

In frame (a)
$v_{f,e}-v_{f,c}=10^{-21} \ m/s - 0 \ m/s = 10^{-21} \ m/s$

In frame (b) note the number of 9's behind the decimal point
$v_{f,e}-v_{f,c}= -9.999999999999999999999 \ m/s - (-10 \ m/s) = 10^{-21} \ m/s$

So the final velocity of the car relative to the Earth is the same in both frames.

No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, ...
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.

Just as the Earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the Earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.

No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

ETA: That is in contrast with the case where actual values are used, and the same method.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J

 

[Humber]

If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.

That's true, but I understood that it was idealised. The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth. Which does not happen in the first case.

 

[Ken G]

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth

This is incorrect, and will lead you astray. Changes in energy are not p²/2m, they are changes in p²/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus. Then a change in p²/2m looks like (p+dp)²/2m - p²/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp² terms.

 

[Humber]

This is incorrect, and will lead you astray. Changes in energy are not p²/2m, they are changes in p²/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus.

Momentum is frame independent, otherwise, there could be violations of conservation.
There would be no means of correcting after the fact, so violations simply don't occur.
That alone guarantees frame independence. If follows that changes are also frame independent.

Then a change in p²/2m looks like (p+dp)²/2m - p²/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp² terms.

That is not correct, because "change" in momentum dp/dt = Force. KE is not frame independent, and that is also accommodated.
KE = p^2/2m = m^2v^2/2m = 1/2mv^2.

Total energy = p^2/2mcar + p^2/2mearth is correct. If the energy change of the car were the same as the change in the ground, p^2/2mcar - p^2/2mearth = 0, then obviously the KERS could not store any energy.

 

[Dale]

No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

The quantity you are describing, $v_{i,c}+v_{f,e}$, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be traveling the next couple of days so I won't be able to do it myself.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

The fact that the result is independent of the mass of the Earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J

Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.

 

[Dale]

The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth.

What makes you say that?

 

[Humber]

The quantity you are describing, $v_{i,c}+v_{f,e}$, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be traveling the next couple of days so I won't be able to do it myself.

I did that. There are several problems, including independence of Earth's mass.
That is the result of simply assigning a velocity to the Earth that is the same as the car's.

The fact that the result is independent of the mass of the Earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

(Δp_c=1e4kg.m/s)
p_i,e = M.v
KE_i,e = p_i,e²/2M
KE_f,e = (p_i,e)²-Δp_c/2M
KE_i,e - KE_f,e = 100kJ

Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.

V_earth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
m_earth ~ 6e24kg (mass of the Earth)
p_i,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KE_i,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δ_pc= 10000kg.m/s ( The applied impulse from the car)
KE_f,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.

 

[Humber]

What makes you say that?

The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely. The actual reason is that there is no 100kJ

Total energy = p^2/2m_car + p^2/2_mearth
Applies to energy to and from the cart as the case may be.

 

[Humber]

When a car brakes, momentum is transferred from the car to the ground.
From the conservation of momentum;

Car gains momentum -p
Earth gains momentum p

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.

All of that energy will have come from the fuel of the F1 car. It could be batteries, and the result is the same.

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

There is no significant transfer of energy to or from the F1 and the ground, as a result of KERS.

 

[Dale]

Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

They may not be random, but they are not relative velocity either.

I did that. There are several problems, including independence of Earth's mass.

I don't know why you think that is a problem.

(Δp_c=1e4kg.m/s)
p_i,e = M.v
KE_i,e = p_i,e²/2M
KE_f,e = (p_i,e)²-Δp_c/2M
KE_i,e - KE_f,e = 100kJ



V_earth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
m_earth ~ 6e24kg (mass of the Earth)
p_i,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KE_i,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δ_pc= 10000kg.m/s ( The applied impulse from the car)
KE_f,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.

This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.

 

[Dale]

The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely.

The remainder goes into the battery. Where else would the energy that goes into the battery come from?

 

[Tea Jay]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

;)

 

[Humber]

They may not be random, but they are not relative velocity either.

The quantity you are describing, vi,c+vf,e, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.

I don't know why you think that is a problem.

Then you don't have a case for your calculations.

This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.

(b) $p_{i,c}=m v_{i,c}= 0 \ kg \ m/s$
$\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s$
$p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s$
$v_{f,c}=p_{f,c}/m= -10 \ m/s$
$KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ$
$KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ$

$p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s$
$f_e=-f_c=1000 \ N$

This is the impulse momentum of the car
$\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s$

Here is where the momentum is decreased.
$p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s$

$v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s$
$KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ$
$KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J$

The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.
The result is indenpendet of the Earth's mass because it is the equivalent of

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

 

[Humber]

The remainder goes into the battery. Where else would the energy that goes into the battery come from?

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.

 

[Humber]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

;)

Infinitesimal it is. With the given numbers, 5e-18J and a velocity change of 1e-21m/s. That is well wide of the 100kJ claimed, and of the 100W of the o.p.
In reality, the Earth is not an infinitly rigid sphere, and will dissipate energy, just as it does for Earthquakes and nuclear weapons tests.

 

[Dale]

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.

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

Then you don't have a case for your calculations.

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

The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.

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.

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

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.

 

[Dale]

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

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.

Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.

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]

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

They are different on each case.

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.

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

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.

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]

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.

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

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

Your calculation says you have made that error.

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.

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

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

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

 

[Dale]

When a car brakes, momentum is transferred from the car to the ground.
From the conservation of momentum;

Car gains momentum -p
Earth gains momentum p

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

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

This is false, in general.

If some object gains momentum Δp then $p_f=p_i+\Delta p$

Dividing both sides by the mass m we get $v_f=v_i+\Delta v$

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

This is different from what you wrote because of the $v_i \Delta p$ 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 $\Delta p^2/2m$ term drops out, but the $v_i \Delta p$ term does not.

 

[Dale]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

That is correct.

 

[Dale]

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

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.

 

[Dale]

The above says that in the second case, there will be both charge in the battery and kinetic energy.

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

A clear violation of both the conservation of momentum and energy.

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

 

[Dale]

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.

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

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

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]

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 = p^2/2mcar
The Earth gains KE = p^2/2mearth

This is false, in general.

If some object gains momentum Δp then $p_f=p_i+\Delta p$

Dividing both sides by the mass m we get $v_f=v_i+\Delta v$

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

This is different from what you wrote because of the $v_i \Delta p$ 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 $\Delta p^2/2m$ term drops out, but the $v_i \Delta p$ term does not.

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

$p_{i,e}=M v_{i,e}=0 \ kg \ m/s$
$f_e=-f_c=1000 \ N$
$\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s$
$p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s$
$v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s$
$KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ$ Change in Earth's KE;
$KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J$

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

 

[Dale]

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

Which works since $v_i=0$ for the Earth in frame (a). It does not work in frame (b) where $v_i \ne 0$ for the earth.

 

[Humber]

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.

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

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

So in the first reference frame the energy for charging the battery comes from the KE of the car which goes down by 50 kJ. The Earth gains a negligible amount of energy.

KE = p²/2m = m²v²/2m = 1/2mv²

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

 

[Dale]

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]

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?

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

 

[Dale]

It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.

 

[Humber]

It is correct. Do you agree or not?
It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.

You have now made 4 unforced errors, and contradicted your own result.

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.