KE of system / different reference frames question

[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.

 

[mender]

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're pretty good at this; some people take quite a few more posts to see that.

 

[mender]

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.

This is known as an assumed close.

 

[Humber]

You're pretty good at this; some people take quite a few more posts to see that.

He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.

 

[Humber]

This is known as an assumed close.

And you are known as a "troll", I believe?

 

[my_wan]

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.

How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.

 

[mender]

And you are known as a "troll", I believe?

The black pot speaketh!

 

[mender]

He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

 

[Dale]

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.

None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?

 

[my_wan]

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.

 

[Humber]

How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.

That is the point. They should be the same, but aren't.

Case 1
v_{i,c}=p_{f,c}/m= 10 \ m/s Initial velocity of car
v_{f,c}=p_{f,c}/m= 0 \ m/s Final velocity of car

v_{i,e}=p_{f,e}/M= 0 \ m/s Initial velocity of Earth
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s Final velocity of Earth

Obviously, when the car is at rest, the final velocities of car and Earth should be the same,
and -10e-21m/s, if the Car has accelerated the Earth in the opposite direction.
Total change for the car = 10m/s + 10e-21m/sCase 2;
v_{i,c}=p_{f,c}/m= 0\ m/s Initial velocity of car
v_{f,c}=p_{f,c}/m= -10 \ m/s Final velocity of car

v_{i,e}=p_{f,e}/M= -10m/s \ m/s Initial velocity of Earth
v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s Final velocity of Earth

This makes no sense at all.

 

[Humber]

Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.

I don't need to do that. The error is quite clear.
The change is 1e-18J in the first case and agrees with KE =p^2/2m, but not the revised calculation. The error there, is rather obvious.

 

[Humber]

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

You are trolling, and have nothing to contribute, but simply want to avoid the consequences for your ddw cart, when it is proven that there is no ground energy.
I have done that by two different means, and now you want to spread your bluff denial, and emulation of Spork, to here.

 

[mender]

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

 

[rcgldr]

there is no ground energy.

DDWFTTW like devices rely on the difference in speed between two media. Which media that KE is extracted from depends on the frame of reference.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.

 

[Humber]

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

Is that the topic of the OP? Again, you are trolling, and exposing yourself as a zealot.

 

[mender]

No; trolling would be to evade answering questions, even simple ones, to keep a discussion from reaching its inevitable (to most people at least) conclusion.

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

 

[Humber]

DDWFTTW like devices rely on the diffrence in speed between two media. Which media that KE is extracted from depends on the frame of reference.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.

Is that the topic of this thread? And the torpedo, as the reference states, was directly powered by on-shore steam engines. That has nothing at all to do with relative motion.

 

[Humber]

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

mender;3692808

 

[mender]

Back to the topic then: answer Dalespam's question.

None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?

 

[mender]

You can continue to show that you have no control at all of the zeal that forces you to make a fool of yourself.

Come on now, stay on topic and answer the question!