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Capiert
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8 hours ago, Capiert said:

What kind of Energy?
It's NOT chemical, thermal, NOR potential etc.
So what KIND of energy is that?

A change in KE means the energy is no longer in the system, or has been added to the system. Where that energy went or where it came from (i.e. the kind of energy) is not accounted for in these examples. It’s not required if these equations apply.

8 hours ago, Capiert said:

If a particle had lost that KEd
 then that particle has decelerated
 & is now moving slower,
 than it was (before the loss).
But it was lost to another particle
 (forced to accelerate,
 E.g. Newton's 2nd Law).

Look at the example I gave of the head-on collision. The KE is not lost to another particle even though there was an acceleration. There is zero KE after the collision. Both particles lost KE. Nothing gained KE.

KE is not generally a conserved quantity. It’s only conserved in the special case of elastic collisions.

8 hours ago, Capiert said:

How then can you claim
 that their (kinetic) Energy
 was NOT destroyed (=annihilated)?

I didn’t claim that. I claimed the opposite: KE is not conserved

8 hours ago, Capiert said:

NOT always.

On 8/4/2022 at 7:33 AM, swansont said:

And for an object moving at some constant speed, it's kinetic energy is decidedly NOT zero. For an object not starting from rest, this does not give the object's KE (=KEf). Which makes your equation wrong.  

 

I did not write KE (=KEf). You must have edited the quote, which is grossly dishonest.

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11 hours ago, swansont said:

A change in KE means the energy is no longer in the system, or has been added to the system.

That'( i)s how you (would) do it.
E.g. Just so you do NOT have to bother.

But why should I (try to) believe you.

You have given me NO proof,
 with your inability
 NOT to bother (attitude).

Are all scientists so lazy
 (like Minkowski hinted
 about Physicists)?
(Surely NOT!)

Why should (some of the kinetic) energy leave that system?
(& I DON'T accept warm, acoustic, excuses either.)
(I'd like to see (simple) tangible measurements.)

11 hours ago, swansont said:

Where that energy went or where it came from (i.e. the kind of energy) is not accounted for in these examples.

Oh! Abracadabra (then).
(It's a mystery!)

That sounds like a boring disinterest in science
 e.g. trying to know.

A half hearted attempt
 to throw a few things together.

You either: know; or (else) you DON'T,
 & you obviously DON'T,
 because you give me useless excuses.

Sorry, other people can try to be more thorough.
You DON'T even give the effort.

If you (were to) say:
 those answers can NOT be found;
 then there must be a reason why.

(Oh we are too feeble,
 (at) attempting, (to) zero_speed, =zero results.
It's more difficult that c.
)

Disinterest is NO excuse.

You also avoid commenting
 (up)on the (=my)
 initial_kinetic_energy KEi
 (perhaps because you habitually evade it
 by subtracting it away).

My syntax includes KEi.
(What is your syntax,
 if mine is NOT an extended (syntax)?)

All 3 (named KEs)
 KEf=KEi+KEd
 are "kinetic_energies"
((meaning) NOT your "the"(what? _unknown),
NOT mentioned f form) syntax).

You can clearly see that
 (they are kinetic_energies)
 in my syntax "KE"
 with a subscript.

There is NO difference:
 meaning a KE is a KE,
 whatever its subscript is.

The KEd can (equally)
 accelerate a(ny) mass
 from zero (speed)
 to a (new) final_speed vf,
 which would finally have
 its own KEf(new)=KEd
 equal to that kinetic_energy_difference KEd.

I DON'T see why you try
 to sell a KEdistinction
 (away) from any other KEsubscripted
 just because you do NOT know
 what (else) KEd is
 (or could be).

E.g.
Even though you only want KEf
 to be "the" (only) kind of KE (possible).

It is absurd to say: KEd
 is NOT a kinetic_energy
 simply because it is NOT "the" final_KE
 KEf=m*(vf2-0)/2
 which uses the mass m multiplied
 by half the final_speed squared vf2
 but "subtracted by zero(_squared)"!

The initial_kinetic_energy
 KEi=m*(vi2-0)/2
 is also a kinetic_energy
 (just like KEf is)
 because its
 half the initial_speed squared vi2
 but is also "subtracted by zero(_squared)"!

The universal
 KE_difference formula
 KEd=m*(vf2-vi2)/2
 is the most universal
 KE "definition"!
There you can (=may)
 use any reference(_frame) speed
 vref=vi
 (below c, that)
 you want
 to be your reference speed
(e.g. at rest, when identical
 to the initial_speed vi).

If I have 7 oranges (analogy KEf)
 & subtract 4 (oranges, analogy KEi),
 then I would expect out, 3 (oranges, analogy KEd)
 like any reasonable thinking person.
NOT grapefruits or "lemons"! (or other hogwash).
That's only common sense,
 which seems to be missing here
 (in (what some people call) science).

You DON'T (even) have a clue where the energy has gone,
 you DON'T know what it is (e.g. called, other than "difference")
 & yet want to be called scientists.
(& you want me to believe you?)

Modern physics is like modern art,
 anything goes,
 even junk.
All that matters is who (e.g. what ego) has the say.

11 hours ago, swansont said:

It’s not required if these equations apply.

I suspect you mean,
 the initial_speed terms
 (e.g. KEi)
 are subtracted anyway;
 (&) so why bother.
?

11 hours ago, swansont said:

Look at the example I gave of the head-on collision. The KE is not lost to another particle even though there was an acceleration. There is zero KE after the collision. Both particles lost KE. Nothing gained KE.

Taken from a different perspective,
 of: if the 2 masses are on Earth
 & the Earth is rotating
 let'( u)s say vi=~1 [km/s]
 eastwards

 ((just) to keep things simple,
 at where they are
 on the Earth's surface);

 then they are still moving
 ~1 [km/s]
 although they appear
 to you
 as at rest.

(& that KEd did NOT leave the system.)

What seems (as) "at rest"
 is an optical delusion
 (of) for both: observer;
 & an object,
 having the same (=identical) speed.

(E.g. Even though they are separated
 by a distance d.)

In reality (e.g. the universe),
 (we know)
 everything is moving.

Meaning NOTHING is (really) static (=at rest, with zero speed).
(Everything has a speed difference
 wrt some other (moving) object (reference, frame).)

Your "choice"
 of reference(_frame)

 (e.g. of same speed
 as the observed object)

 (help) determines
 whether you want
 to be: deceived ((in)to think(ing): )
 an object is at rest (when it has the same speed
 as the reference_frame);

 or NOT!

11 hours ago, swansont said:

KE is not generally a conserved quantity. It’s only conserved in the special case of elastic collisions.

I didn’t claim that (**KE is destroyed=annihilated). I claimed the opposite: KE is not conserved

I did not write KE (=KEf**). You must have edited the quote, which is grossly dishonest.

**(Sorry! (Yes) "I") Modified.
(Or do you mean your quote is dishonest? Which I (rather) doubt,
 in preference for the former.)

How else should I add (extra) comments
 of mine
 into your text?

I bracketed it,
 to distinguish it from your original text.

Dishonesty was NOT intended;
 only clearness of the discussion
 (was intended),
 (before getting lost again
 in(to) confusion
 between syntaxes).

Should I use different brackets?

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10 hours ago, Capiert said:

That'( i)s how you (would) do it.
E.g. Just so you do NOT have to bother.

But why should I (try to) believe you.

 

You can check in any college 1st semester physics textbook and confirm that this is the mainstream view.

 

10 hours ago, Capiert said:



You have given me NO proof,
 with your inability
 NOT to bother (attitude).

It’s a basic definition.

I’m giving you information, and the benefit of my expertise. Not bothering would mean not responding at all.

10 hours ago, Capiert said:

Are all scientists so lazy
 (like Minkowski hinted
 about Physicists)?
(Surely NOT!)

Nobody here owes you any of their time.

10 hours ago, Capiert said:

Why should (some of the kinetic) energy leave that system?
(& I DON'T accept warm, acoustic, excuses either.)
(I'd like to see (simple) tangible measurements.)

I gave you an example that shows it does. You are free to recreate it.

You can also drop a wet rag or raw egg on the floor. It will have some speed just before it hits the floor (i.e. it has kinetic energy) and then just after the collision, it won’t be moving, so its KE is now zero.

You’re working very hard to deny something so trivially easy to observe.

Why should the KE change? The real question is why wouldn’t it? TOTAL energy is conserved. Not any particular type of energy, unless you have specific circumstances. Energy can change from one type to another. Lots of devices take advantage of this - a generator takes mechanical energy and converts it to electrical. 

You want an example of kinetic energy converted to thermal energy, you can see this in a car’s brakes. They heat up as the car slows. Any object sliding along a surface with friction present will heat up. You can check this yourself.

 

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13 hours ago, Capiert said:

You DON'T (even) have a clue where the energy has gone,
 you DON'T know what it is (e.g. called, other than "difference")
 & yet want to be called scientists.
(& you want me to believe you?)

In some cases it’s difficult to track; you might not be measuring it. But in e.g. a completely inelastic collision, it doesn’t matter, since conservation of momentum applies and allows you to solve the problem. In a situation with friction it may not matter if you can calculate the amount of work done.

Some problems might just not be solvable without additional information. 

13 hours ago, Capiert said:

How else should I add (extra) comments
 of mine
 into your text?

You shouldn’t add text to anyone’s quote, since a quote implies that it’s an actual quote. i.e. what they said.

13 hours ago, Capiert said:

Taken from a different perspective,
 of: if the 2 masses are on Earth
 & the Earth is rotating
 let'( u)s say vi=~1 [km/s]
 eastwards

 ((just) to keep things simple,
 at where they are
 on the Earth's surface);

 then they are still moving
 ~1 [km/s]
 although they appear
 to you
 as at rest.

(& that KEd did NOT leave the system.)

Then there is no change in the motion. So what? It’s a different example than what I gave. 

(And this ignores that you can choose whatever convenient frame of reference you want. I chose one where the momentum is zero. But pick whatever frame you want; in a completely inelastic collision, the kinetic energy is smaller after the collision)

KE not being conserved means you can’t assume it is as a general principle. It doesn’t mean you can’t find individual examples where it doesn’t change. i.e. “not conserved” does not mean “must change”

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Posted (edited)
On 8/6/2022 at 3:18 AM, Capiert said:

That'( i)s how you (would) do it.
E.g. Just so you do NOT have to bother.

But why should I (try to) believe you.
 

On 8/6/2022 at 2:00 PM, swansont said:

You can check in any college 1st semester physics textbook and confirm that this is the mainstream view.

It’s a basic definition.

I’m giving you information, and the benefit of my expertise. Not bothering would mean not responding at all.

Nobody here owes you any of their time.
 

Yes, True. Sorry for my (sudden, shocked) reaction.
(Anything to do with Newton 3?)

It's difficult
(for me)
 to accept
 that you will ignore
 & deny the syntax,
 & thus undermine
 any (further) argument;
 making all rationality attempts seem pointless.

(That takes more than a few aspirin,
 with the feet up
 (to recover from).
)

It'( i)s similar,
 to (suddenly) declaring
 that the Earth's North Pole
 (which is known & documented for centuries,
 as found in the northern hemisphere,
 near, side by side,
 to the north (geographical) rotational_"axis")
 is suddenly (now, renamed to) a south (magnetic) pole
 because your compass needle
 is pointing to it (=the Earth's North Pole).

& everybody else's compass
is (now) doing the same thing (too=also)!
They are (all) "pointing to"
 (the North Pole).
(Which is NOTHING new!)

I mean, why was the North "Pole"
 named so
 in the 1st place
 (instead of axis)?

Obviously because
 magnetic loadstone
 orientated
 towards the Earth's 2 magnetic poles.

Nobody then knew what kind
 of magnetic pole
 each end of the loadstone had;
 other than to follow its (end's) orientation
 to the Earth
 (as reference).

There was NO other reason
 for naming it "pole"
 (instead of axis)
 other than that it was (& still is)
 a "magnetic"_pole!
 

Clear cut!

Now some (so_called) genius
 comes around
 with his (own) compass
 in his hands
 (seeing is believing)
 (ignoring that it (=the compass's_needle)
 is only "pointing"
 to the north(ern) direction,
 & (=but) says "his" needle('s end)
 must be North (magnetism)
 (when it is NOT!),
 because its pointing
 "to" the Earth's north (magnetic) pole.

I mean how far does narrow_minded egoism have to get.

You get 1 (raving) idiot
 (ignoring the consistency=continuum
 with the rest of the concept, &) 
 messing (=mixing) things up;
 & then every other (stupid) idiot jumps on the bandwagon, too
 (as ja_sayers);
 instead of saying NO!.

(Stunned with astonishment,
 which boardered upon stupification
 (..they left the forest).
-Jules Verne.

NOT seeing the forest
for (=because of) the trees.)

That is only an example
 of inconsistent logic.
 
In my book
 the Earth's magnetic pole
 is in the Northern hemisphere
 (as it should be)
 (NOT the southern);
 & I have seen maps so, too.

I see NO reason
 to change that,
 for the crackpot idea
 that it should be south,
 just so some egoist
 can (hi_jack the show,
 &) gain all the attention
 as a "new" breakthru
 just to be different=elite.

I mean which came 1st
 the chicken
 or the egg?
Obviously, "its" egg:
 the proprietary owner('s right(s)).
I.e. Historical sequence,
 squatter's rights.
Credit where it belongs.

But getting back
 to tracking KEd.

On 8/6/2022 at 4:45 PM, swansont said:

In some cases it’s difficult to track; you might not be measuring it. But in e.g. a completely inelastic collision,

You must (surely) know,
 (at least by now),
 that:
 KE does NOT work
 for NON_elastic collisions
 because
 KE's mass m is NOT correctly proportioned
 to the speed vf.
The mass will change (amount)
 (becoming larger)
 (instead of stay(ing) the same).
That is a strictly=pure(ly) math error
 (of WRONG proportioning).

Mass*Energy m*E
 helps correct that error
 (of WRONG proportions);
 but (it seems) you wish
 to ignore that
 (proportioning error;
 & m*E,
 altogether).

Kinetic_Energy on its own
 is (inferior) NON_SENSE!
 (because it'( i)s only (an intermediate,
 a) part
 of a term
 of m*E);
 that's why it (=KE) will NOT work generally
 for (exchanging to) different (sized) masses
 (such as (in)
 e.g. NON_elastic collisions).

On 8/6/2022 at 4:45 PM, swansont said:

it doesn’t matter, since conservation of momentum applies and allows you to solve the problem.

Oh yes it does matter,
 because you ignore COE is fake
 by (still=continuing) declaring energy
 is a conserved quantity
 (although NOT completely).
It either is, or it is NOT.
Energy has a flaw,
 & breaks down sometimes
 in some cases, e.g. NON_elastic collisions;
 & you refuse to admit
 that weakness
 is a (math) ERROR!

Your general (conservation) laws
 rely completely
 on only COM (conservation of momentum);
 & NEVER COE (conservation of energy),
 because COE is corrupt!
 (an incomplete distortion)!

You misleed people
 (like me)
 into believing
 conservation of energy
 (is general)
 when it can NOT be (e.g. completely),
 inspite of your limitation warnings.
It's like saying, I'm a bit pregnant;
 but NOT completely.
I'm NOT (pregnant)
 when my mass has increased
 e.g. in an inelastic collision.

(Partially (conserved) is NO excuse for the real McCoy!, COM.)


m*E is a unified form(ula)
 of both COE & COM.

Why need to choose
 anything else?

1 formula
 will now finally do.

On 8/6/2022 at 4:45 PM, swansont said:

In a situation with friction it may not matter if you can calculate the amount of work done.

Some problems might just not be solvable without additional information. 

I think that is the reason
 why I am probing (=e.g. asking)
 what kind of energy KEd is.
I'( a)m asking for that additional info.

I suspect we have something (new?) to learn
 with the correct answers.

At least from
 that endeavour
 we might
 develop more ability
 e.g. simpler formulas,
 e.g. shortcuts.

(Why hinder the progress?)

On 8/6/2022 at 4:45 PM, swansont said:

You shouldn’t add text to anyone’s quote, since a quote implies that it’s an actual quote. i.e. what they said.

OK. Good advice.

On 8/6/2022 at 4:45 PM, swansont said:

Then there is no change in the motion.

NOT true.

v1i=(1000-1=~999) m/s
v2i=(1000+1=~1001) m/s

v3f=~1000 m/s

On 8/6/2022 at 4:45 PM, swansont said:

So what? It’s a different example than what I gave. 

(And this ignores that you can choose whatever convenient frame of reference you want. I chose one where the momentum is zero. But pick whatever frame you want; in a completely inelastic collision, the kinetic energy is smaller after the collision).

To make things simple,
 let'( u)s put your example
 in a car (or train, frame)
 (that is)
 moving (at only e.g.) +2 [m/s] (eastwards;
 instead of 1 [km/s]);
 & watch the (masses') speeds
 from (wrt) the (Earth's surface) ground.

Mass 1 (initial):
m1i=1 [kg]
v1i=3 [m/s]
mom1i=m1*v1=1 [kg]*3 [m/s]=3 [kg*m/s]
KE1i=m1*(v12-0)/2=1 [kg]*(3 [m/s])2/2=4.5 [kg*m2/s2]=4.5 [J].

Mass 2 (initial):
m2i=1 [kg]
v2i=1 [m/s]
mom2i=m2*v2=1 [kg]*1 [m/s]=1 [kg*m/s]
KE2i=m2*(v22-0)/2=1 [kg]*(1 [m/s])2/2=0.5 [kg*m2/s2]=0.5 [J].

Mass 3 (final):
m3=m1+m2=1 [kg]+1 [kg]=2 [kg]
v3=? [m/s]
mom3=mom1+mom2=3 [kg*m/s]+1 [kg*m/s]=4 [kg*m/s]
v3=mom3/m3=4 [kg*m/s]/2 [kg]=2 [m/s]
KE3=m3*(v32-0)/2=2 [kg]*(2 [m/s])2/2=4 [kg*m2/s2]=4 [J].

The total final kinetic_energy
 KETf=KE3=4 [J].

The total initial kinetic_energy
 KETi=KE1i+KE2i=4.5 [J]+0.5 [J]=5 [J].

The Energy defect (e.g. neutrino=(math_)error! KEd4)
 is the initial_total energy
 minus the final_total energy
 KEerror=KETf-KETi=4 [J]-5 [J]=-1 [J] lost!
 =-20%
of (the total) initial(_energy)
 (for that (NON_elastic) example).

Things look different(ly)
 with mass*Energy m*E,
 because the defect
 is m4*KE4=~-2*mom1*mom2.

But
 there is a big difference
 between m*E=mom*moma=mom2*(va/v)
 versus
 mom2=mom*mom=m*E*(v/va).
 

The catch there
 is the average_speed
 va=(vi+vf)/2
 needs the initial_speed vi;
 which is a speed
 you ignore
 by subtracting it out.
E.g. By (typically) setting
 the initial_speed vi
 to be your reference_frame's
 speed.
Thus it (vi) is gone from the picture.
Anything that has the same speed
 as vi
 is declared
 as "at rest"
 (when it's NOT really
 when wrt any other speed than vi).
"At rest" is the deceiving(_speed) exception,
 of really being (=having) the "same speed".

But you (already) know that.
I'm NOT telling you anything new.

What bugs me is the partial truth.
(The biggest lie,
 is only half the truth.
-A banker.)
E.g. The exceptions.

On 8/6/2022 at 4:45 PM, swansont said:

KE not being conserved means you can’t assume it is as a general principle.

Ok, I can acknowledge,
 that KE is only 1 kind of energy
 (among [m]any);
 (but it's (=KE is) NOT the big cheese(y principle)!)
Energy (in general)
 is that (general concept)
 (even if it has its weaknesses).

On 8/6/2022 at 4:45 PM, swansont said:

It doesn’t mean you can’t find individual examples where it doesn’t change. i.e. “not conserved” does not mean “must change”

IOW:
It means, you can find individual examples
 where KE does NOT change.
(E.g. identical masses?)
I.e. “not conserved” means “might change”;
 but does NOT have to change.

I sure hope
 that will sink in(to my head).

Maybe you could (please) give me
 an example
 of that,
 (so I can remember better,
 instead of (me) falling
 into the same old hole)?

I'( woul)d like
 to present
 you the m*E example;
 but (can NOT do it properly yet,
 because) you'( ha)ve
 gotten me off track
 by (=into) zeroing
 my KEf & KEi
 away from (my original concept)
 KEd=m*v*va.

E.g. Versus mom2=m*(m*v*v).

v#va.

Which means
 I (still) have to give it some thought,
 as to the consequences.

(=I need some time.)

Edited by Capiert
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4 minutes ago, Capiert said:

Oh yes it does matter,
 because you ignore COE is fake
 by (still=continuing) declaring energy
 is a conserved quantity
 (although NOT completely).
It either is, or it is NOT.
Energy has a flaw,
 & breaks down sometimes
 in some cases, e.g. NON_elastic collisions;
 & you refuse to admit
 that weakness
 is a (math) ERROR!

That you don’t understand it is not something that matters. Physics doesn’t claim that any individual form of energy is conserved. Only the total energy. 

Nobody cares that you don’t like it. It works. People do thousands upon thousands of successful experiments that use the concepts of physics.

Quote

when my mass has increased
 e.g. in an inelastic collision.

Mass doesn’t increase in a collision. The mass at the beginning of the example is the same as the mass at the end. In a completely inelastic collision, the two masses are moving at the same speed.

Quote

 

NOT true.

v1i=(1000-1=~999) m/s
v2i=(1000+1=~1001) m/s

v3f=~1000 m/s

 

That’s different than what you said, which was just two particles moving at 1000 m/s

But OK, then calculate the kinetic energy before and after the collision. Is it the same?

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Posted (edited)

 

1 hour ago, swansont said:
  Quote

when my mass has increased
 e.g. in an inelastic collision.

1 hour ago, swansont said:

Mass doesn’t increase in a collision. The mass at the beginning of the example is the same

(amount (of mass))

1 hour ago, swansont said:

as the mass at the end.

But it (that mass) is distributed differently
 (into the (mass) terms)
 so that it is only (1 mass, now e.g.) together.

& that ((different) togetherness)
 (now) has a different affect
 on the math answer.

1 hour ago, swansont said:

In a completely inelastic collision, the two masses

 are now 1 total mass
 m3=m1+m2
 which is now larger
 than either 1
 of its initial components
 (e.g. either: m1; or m2)
 &

1 hour ago, swansont said:

are moving at the same speed.

If mass is squared,
 then its parts
 produce different (math, number, value) results
 compared to its (=the total mass) sum squared.

 m32=(m1+m2)2
 m32=m12+m22+2*m1*m2   <---But look at that!

m32#m12+m22

Mass squared does NOT simply add
 only the squared terms;
 because the (NON_squared term) +2*m1*m2
 is missing!

I hope you get my drift.
(What I am "trying" to tell you.)

Such naive math (just) can NOT work, (sometimes).

Edited by Capiert
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32 minutes ago, Capiert said:

But it (that mass) is distributed differently
 (into the (mass) terms)
 so that it is only (1 mass, now e.g.) together.

& that ((different) togetherness)
 (now) has a different affect
 on the math answer.

They’re moving at the same speed is all.

What’s the physics of this “togetherness” that has a “different affect”?

Actual physics says nothing about this, of course. Mass and speed - existing physics variables - are all you need to describe what’s going on.

32 minutes ago, Capiert said:

If mass is squared,

But it’s not. There’s no m^2 in kinetic energy or momentum equations. You’re just making up a problem.

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  • 2 weeks later...
Posted (edited)

 

On 8/7/2022 at 11:51 PM, swansont said:
  On 8/7/2022 at 11:22 PM, Capiert said:

But it (that mass) is distributed differently
 (into the (mass) terms)
 so that it is only (1 mass, now e.g.) together.

& that ((different) togetherness)
 (now) has a different affect
 on the math answer.

On 8/7/2022 at 11:51 PM, swansont said:

They’re moving at the same speed is all.

Well done!

But if the proportionality
 to distance
 is a (correct) mass_squared;
 instead of (only mass) NOT squared,
 then you have a definite difference
 that can NOT be pushed_away (=denied!)
 (& hid under the carpet).

On 8/7/2022 at 11:51 PM, swansont said:

What’s the physics of this “togetherness” that has a “different affect”?

The numbers (resulting) for mass_squared
 versus only mass (NOT squared).
Btw. That ("squared" versus NOT_squared (mass))
 is math;
NOT physics.
Math_Physics at the most.

On 8/7/2022 at 11:51 PM, swansont said:

Actual physics says nothing about this, of course. Mass and speed - existing physics variables - are all you need to describe what’s going on.

Quite right.
But with what (=which?) method
 you choose,
 to describe,
 determines your outccome.
Choose momentum
 & then you are probably (quite) right.
Choose Energy
 & then it can go wrong
 when mass changes
 in the collision.
Choose mass*Energy
 & then you might get it right (again),
 when dealing with changing mass.
The choice is (really) yours (NOT mine).
(I'( woul)d call that bias.)

On 8/7/2022 at 11:51 PM, swansont said:

But it’s not.

..What?
(Please explain yourself.
"NOT" what?)

On 8/7/2022 at 11:51 PM, swansont said:

There’s no m^2 in kinetic energy or momentum equations.

Quite right.
Mass_squared
 does NOT exist in either
 of those (mom or Energy) equations,
 (& because they are inferior, e.g. limited).

Mass_squared is a new concept
 from Ewert (1996)
 & it does seem to work right=correctly,
 e.g. better,
 such as in problems
 where the mass changes.

& the cool (=neat) thing
 about it (=mass_squared)
 is m*E,
 mass*Energy (m*E)
 is VERY similar
 to (only) Energy
 so you hardly (=barely)
 need to change a thing.
It'( i)s very easy.
Simply multiply mass
 by Energy
 & you might (=should) get
 the correct answers.
That'( i)s over_simplified
 of course.

On 8/7/2022 at 11:51 PM, swansont said:

You’re just making up a problem.

What sort of problem
 am I making up?
I see definite problems
 with Energy
 that can NOT be over_come
 (otherwise).
Dark_Energy is a real eye_sore.

But the so_called
 energy loss
 in an inelastic (=NON_elastic)
 collision
 is most prevalent (=obvious).

& I am NOT making that 1 up.
YOU are!

The Energy NEVER left the system.

& I can tell you where it is.

On 8/7/2022 at 10:03 PM, Capiert said:

The total final kinetic_energy
 KETf=KE3=4 [J].

The total initial kinetic_energy
 KETi=KE1i+KE2i=4.5 [J]+0.5 [J]=5 [J].

The Energy defect (e.g. neutrino=(math_)error! KEd4)
 is the initial_total energy
 minus the final_total energy
 KEerror=KETf-KETi=4 [J]-5 [J]=-1 [J] lost!
 =-20%
of (the total) initial(_energy)
 (for that (NON_elastic) example).

I (wrongly) said that you "lost" -1 [J];
 but that is wrong
 only because you guys condition me
 into your same bad habits.
(& I fell for it,
 as naive as I am.)

You haveN'T lost a thing!

Take a look at
 (conservation of momentum, COM)
 mom1+mom2=mom3.
That'( i)s a perfectly balanced equation.
Left_side (is mom1+mom2) is "before" the collision
 & right_side (is mom3) is "after" the collision.
Square both sides
 & it is still balanced (=perfect)!
 (mom1+mom2)2=mom32.
But that produces
  mom12+mom22+2*mom1*mom2=mom32.
That +2*mom1*mom2 belongs to "before" the collision (=left_side
of the equation);
 NOT "after" the collision (right_side of the equation).
Again:
That +2*mom1*mom2 belongs to the "before"_ side
 of the collision equation;
 NOT the "after"_side of the collision equation.
Putting the 2*mom1*mom2 on the "after" side
 of the collision equation
 means it "needs" to be NEGATIVE.
I.e. -2*mom1*mom2.
NON_elastic collisions
 might seem
 to have lost energy
 on the (heavier) result;
 but they have NOT lost a thing
 (in that sense);
 you need only observe the math
 to correctly determine
 what has happened.

& (thus) you physicists (have) preach(ed) an untrue gospel (=NONSENSE!)

Sorry!

But more SORRY for your (misled) followers.

Edited by Capiert
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8 hours ago, Capiert said:

But if the proportionality
 to distance
 is a (correct) mass_squared;
 instead of (only mass) NOT squared,
 then you have a definite difference
 that can NOT be pushed_away (=denied!)
 (& hid under the carpet).

...

The numbers (resulting) for mass_squared
 versus only mass (NOT squared).
Btw. That ("squared" versus NOT_squared (mass))
 is math;
NOT physics.
Math_Physics at the most.

You got that right: it's math, not physics. So why bother with it? It's just math you made up, and has no applicability to physics.

Quote

But with what (=which?) method
 you choose,
 to describe,
 determines your outccome.
Choose momentum
 & then you are probably (quite) right.
Choose Energy
 & then it can go wrong
 when mass changes
 in the collision.
Choose mass*Energy
 & then you might get it right (again),
 when dealing with changing mass.
The choice is (really) yours (NOT mine).
(I'( woul)d call that bias.)

You choose the method that applies to the problem. It comes from understanding physics and the experience of having solved similar problems many, many times. You can evaluate and see what conservation laws can be applied, and what other equations can be applied.

If you think you can solve a problem with mass*energy, go ahead and show that it works: derive the formula from known physics, and show that gives the correct answer. Tell us when it applies and when it doesn't, so others can test it (and make sure that whatever example you give isn't correct by accident)

Quote

Mass_squared is a new concept
 from Ewert (1996)

That's not a good enough citation.

8 hours ago, Capiert said:

I (wrongly) said that you "lost" -1 [J];
 but that is wrong
 only because you guys condition me
 into your same bad habits.
(& I fell for it,
 as naive as I am.)

You haveN'T lost a thing!

Take a look at
 (conservation of momentum, COM)
 mom1+mom2=mom3.
That'( i)s a perfectly balanced equation.
Left_side (is mom1+mom2) is "before" the collision
 & right_side (is mom3) is "after" the collision.
Square both sides
 & it is still balanced (=perfect)!
 (mom1+mom2)2=mom32.
But that produces
  mom12+mom22+2*mom1*mom2=mom32.
That +2*mom1*mom2 belongs to "before" the collision (=left_side
of the equation);
 NOT "after" the collision (right_side of the equation).
Again:
That +2*mom1*mom2 belongs to the "before"_ side
 of the collision equation;
 NOT the "after"_side of the collision equation.
Putting the 2*mom1*mom2 on the "after" side
 of the collision equation
 means it "needs" to be NEGATIVE.
I.e. -2*mom1*mom2.
NON_elastic collisions
 might seem
 to have lost energy
 on the (heavier) result;
 but they have NOT lost a thing
 (in that sense);
 you need only observe the math
 to correctly determine
 what has happened.

& (thus) you physicists (have) preach(ed) an untrue gospel (=NONSENSE!)

Sorry!

But more SORRY for your (misled) followers.

momentum is not kinetic energy. They are distinct concepts.

You can check that KE is lost because you can calculate the KE for each object, and the values do not match. As in the example I gave. 

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