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Elastic collision?


Capiert

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  11 hours ago, Capiert said:

If I multiplied force by time, instead

 I would get momentum,

 which I would prefer (& trust more).

 

Swansont suggested (from my Fiv thread)

OK. Solve some problems involving thermodynamics with conservation of momentum. Go ahead. Try it. Or even a block sliding down an inclined plane with friction. Or an elastic collision, without involving energy.

That, too, has limits on practicality.

 

Here goes.

A few assumptions are necessary.

 

Collision takes time:

 

While colliding

 the 2 balls are:

 in contact with each other

 for a specific duration time (amount);

 & (thus) decelerating

 (to wrt a (united) non_elastic common_speed u´=(mom1+mom2)/(m1+m2)

 (& 1 specific point in time),

 that they reach=attain;

 & (then (they)) continue decellerating)

 til they disconnect.

 

After disconnection:

 No further acceleration occurs

 (because there is no affecting cause (e.g. elastic force, e.g. Hooke's force Fs=-x*k)

 when NOT (elastically) in contact, any more

 (=contact is missing).

 

(Then) at that (disconnected) point in time afterwards,

 the speed( velocitie)s are final.

v1´=2*u´-v1

v2´=2*u´-v2.

---

What do we know?

 

To summarize:

 

Collision takes time:

The 2 balls connect,

 indent (compressing, & decelerating),

 til effectively stopping

 wrt to a common=mutual non_elastic speed u´=(mom1+mom2)/(m1+m2);

 & then continue to decelerate while undeforming

 til they disconnect.

Assuming a linear acceleration equivalent,

 each phase ((elastic) compression & decompression)

 have the same amount of speed change vd1.

Thus the total speed change (for mass1) is 2*vd1=v1´-v1;

 & half the total (speed_changed (difference), for mass1)

 vd1=(v1’-v1)/2

 is an average.

 

In other words:

 mass1’s (initial) speed (velocity) v1

 (e.g. moving from left to right (is positive polarity)),

 decelerates

 to the (non_elastic) mutual (masses’) speed (velocity) u´

 ((unfortunately) here indicated as the addition

 of its (negative) (speed_difference) value vd1 ((that is) lost))

 v1+vd1=u´ (compression);

 & mass1 continues decelerating (from the mutual (masses’) speed u´)

 ((with) the other (negative) (speed_difference) value vd1 ((that is) lost))

 u´+vd1=v1´ (decompression, =decompressed)

 til its (=mass1’s) final_speed v1´.

 

 

So I’ve split the deceleration (e.g. of mass1)

 into 2,

 by using the

 (doubling method (syntax))

 (2)*vd1,

 (of the speed_difference vd1;

 instead of simply after-before=final-initial speeds)

 so I can easily (=conveniently) pinpoint

 the non_elastic (collision’s) time_point equivalent.

 

E.g. For identical masses (when we let m1=m2)

 we can (equivalently) observe if & when the speeds (velocities) reverse.

 

So let vd1=(v1´-v1)/2 be half of mass1’s (decelerated)
 (final minus initial) speed (velocities’) difference.

 

Rewritten:

u´=v1+vd1 (compression result), &

u´=v1´-vd1 (uncompression), add both equations gives

2*u´=v1+v1´, /2=divide both sides by 2, gives

u´=(v1+v1´)/2

Again

vd1=u´-v1, &

vd1=v1´-u´.

Significant is (that)

 mass1’s final speed (velocity) is

v1´=2*u´-v1, for after the elastic collision,

 & using the mutual (united) non_elastic speed (velocity)

 u´=(mom1+mom2)/(m1+m2)

 for the total mass mt=m1+m2 (or mass_sum ms=mt),

 where

 mass1’s initial momentum is

 mom1=m1*v1,

 & mass2’s initial momentum is

 mom2=m2*v2.

 

The same can be done for mass2:

u´=v2+vd2 (compression result), &

u´=v2´-vd2 (decompression), add both equations gives

2*u´=v2+v2´, /2=divide both sides by 2, gives

u´=(v2+v2´)/2

Again

vd2=u´-v2, &

vd2=v2´-u´.

Significant is (that)

 mass2’s final speed (velocity) is

v2´=2*u´-v2, for after the elastic collision,

 & using the mutual (united) non_elastic speed (velocity)

 u´=(mom1+mom2)/(m1+m2),

 where

 mass1’s initial momentum is

 mom1=m1*v1,

 & mass2’s initial momentum is

 mom2=m2*v2.

 

Thus mass1’s final speed (velocity) (after elastic collision) is

v1´=2*((mom1+mom2)/(m1+m2))-v1, or

v1´=2*((m1*v1+m2*v2)/(m1+m2))-v1;

&

mass2’s final speed (velocity) (after elastic collision) is

v2´=2*((mom1+mom2)/(m1+m2))-v2, or

v2´=2*((m1*v1+m2*v2)/(m1+m2))-v2.

 

Is that something like what you want Swansont?

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I have taken the liberty of removing all of your text modifications, which take poor formatting and make it even harder to decipher (I mean, you say you can't type a proper way because of a crappy keyboard, but you can add colored text? Stop making things harder to understand. Just stop.)

Another shortcoming is your use of non-standard terminology and symbols. I can't decipher what you are trying to do.

19 hours ago, Capiert said:

  11 hours ago, Capiert said:

 

(Then) at that (disconnected) point in time afterwards,

 the speed( velocitie)s are final.

v1´=2*u´-v1

v2´=2*u´-v2.

For example, this. You don't explain what the primed variables are, though it seems they are after the collision. But you have a variable u. What is this mystery variable? And where does this equation come from? 

 

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30 minutes ago, swansont said:

I have taken the liberty of removing all of your text modifications, which take poor formatting and make it even harder to decipher (I mean, you say you can't type a proper way because of a crappy keyboard, but you can add colored text? Stop making things harder to understand. Just stop.)

My development history is very complicated

 with various syntaxes & attempts

 (I often can NOT decide which is best),

 I tried to give you a good cut

 otherwise you would have seen ~5x too much.

Last minute (insertions) changes, were added to clarify.

(I doubted your chances to understand (well) without (them).)

Quote

Another shortcoming is your use of non-standard terminology and symbols. I can't decipher what you are trying to do.

If you tell me what you do understand

 (put things in your own words),

 maybe we can take it from there.?

Quote

For example, this. You don't explain what the primed variables are, though it seems they are after the collision. But you have a variable u. What is this mystery variable? And where does this equation come from? 

" common (=mutual) non_elastic speed u´=(mom1+mom2)/(m1+m2); "

 

u' is the (equivalent, united) speed (velocity) after a non_elastic collision.

mom1+mom2=momt, total momentum momt=

before=after

m1*v1+m2*v2=(m1+m2)*u´, (non_elastic).

Yes (you are right) prime ´ is (for) after the collision.

(I assumed you would recognize them.)

Edited by Capiert
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19 hours ago, Capiert said:

   Assuming a linear acceleration equivalent,

 each phase ((elastic) compression & decompression)

 have the same amount of speed change vd1.

Which stems from conservation of energy, which you are not allowed to assume. It certainly doesn't come from conservation of momentum.

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1 hour ago, swansont said:

Which stems from conservation of energy,

 in your mind (maybe?). I.e.

Which can stem also from conservation of energy.

(It might stem from coe (in your thinking); but it does NOT have to (in mine)!)

Where (from what I said)
 did you get the idea that
 linear acceleration
 & speed change
 are suppose to be (according to you) exclusively from energy?

I only stated linear acceleration (equivalent)
 & speed change (difference).

I said nothing about energy (there)!
You did!

You are using your mind's memory
 & trying to project it onto what I said
 when I said nothing of the sort.

I do not mind,
 if you attempt to recognize things in your (own energy) terms.
But please avoid the distortion, & explain what you mean.

E.g.
Do you mean
 that linear acceleration (also) is not possible
 with (just only, some) Newton's force F=m*a
 & some speed_difference v=vf-vi? (a=v/t).

Or do you mean Hooke's law F=-x*k?

Quote

which you are not allowed to assume.

I can't follow you, (there, yet).

I'm studying motion
 (from point A to point B).
It's that simple.

I don't see the error
 (nor a potential error).

Please clarify.

Quote

It certainly doesn't come from conservation of momentum.

I never said it (=linear acceleration a=v/t; & speed_difference v=vf-vi) did, either.

Perhaps that is why (normal) people find it difficult
 to discuss with scientists?

Please help. (Signed -Troubled. You lost me at the bakery.)

21 hours ago, Capiert said:

Assuming a linear acceleration equivalent,
 each phase ((elastic) compression & decompression)
 have the same amount of speed change vd1.

I've assumed an equivalent,
 & said (=set) it (=acceleration, for a speed_difference) to be linear,
 so that I can use it (=half the speed_difference) twice
 as 2 halves.

To define a (virtual) maximum_compression('s)
 point_in_time.

I.e. A virtual maximum
 of elastic force,
 where both (elastic) forces are (maximum,) opposite & equal.

We can not know all the details (of acceleration & elasticity);
 but we can equate (virtual speeds) to them, with an equivalent.

 

 

P.S. Caution

Error (to the Software (Bug) team):

While editing,
 marking text,
 & then pressing the backspace (key)
 (=attempting to delete the text),
 ejects me out from the thread
 onto another page (somewhere else).

Edited by Capiert
It bugs me that (typing the) CR (key, automatically) splits my phrases (apart) so much (between the lines).
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3 hours ago, Capiert said:

It bugs me that (typing the) CR (key, automatically) splits my phrases (apart) so much (between the lines).

That is what the CR key is for. Adding blank lines (the clue is in the name). If you don't want so many blank lines, then don't press the CR key so often. 

And here is another, related, tip: How about NOT pressing CR in the middle of a line as well ! That might make your posts look slightly less idiotic, even if it doesn't improve the content.

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7 hours ago, Strange said:

That is what the CR key is for. Adding blank lines (the clue is in the name).

?
CR=
Carriage =buggy, horse & carriage, old_fashioned time westerns.
&
Return.
Where does blank_line come into place there?

Quote

 If you don't want so many blank lines, then don't press the CR key so often. 

And here is another, related, tip: How about NOT pressing CR in the middle of a line as well ! That might make your posts look slightly less idiotic, even if it doesn't improve the content.

Thanks for the tips.
But it looks like I need to use: Shift+CR, instead of CR.

I.e. (I have to do more (keys), to get less (spacing)!

Very logical! ?

It still doesn't work right using a tablet. (As you can see.)

Edited by Capiert
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7 hours ago, Capiert said:

CR=
Carriage =buggy, horse & carriage, old_fashioned time westerns.
&
Return.
Where does blank_line come into place there?

I had the impression you were old enough to remember typewriters. Possibly even their invention. But apparently not. 

7 hours ago, Capiert said:

But it looks like I need to use: Shift+CR, instead of CR.

How about not doing either. Why the bizarre need to break your text into short unreadable lines?

Just stop it. 

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On any modern electronic device you only need to hit CR at the end of a paragraph. The software does the line wraps for you.

On 8/25/2018 at 9:24 AM, Capiert said:

 in your mind (maybe?). I.e.

Which can stem also from conservation of energy.

No, in physics. They are linked. 

On 8/25/2018 at 9:24 AM, Capiert said:

(It might stem from coe (in your thinking); but it does NOT have to (in mine)!)

Then come up with your own self-consisten theory. Otherwise, learn mainstream physics.

On 8/25/2018 at 9:24 AM, Capiert said:

Where (from what I said)
 did you get the idea that
 linear acceleration
 & speed change
 are suppose to be (according to you) exclusively from energy?

The changes you propose do not come from conservation of momentum. So how do you justify the speed changes that you propose?

On 8/25/2018 at 9:24 AM, Capiert said:

I only stated linear acceleration (equivalent)
 & speed change (difference).

I said nothing about energy (there)!
You did!

Just because you didn't say it does not mean it's not the underlying principle. You gave certain speed changes in "compression" and "decompression". Where does the justification for that come from? What is the physics?

On 8/25/2018 at 9:24 AM, Capiert said:

 

Do you mean
 that linear acceleration (also) is not possible
 with (just only, some) Newton's force F=m*a
 & some speed_difference v=vf-vi? (a=v/t).

Or do you mean Hooke's law F=-x*k?

I didn't mention any specific force. But we know that a collision can end up with the two objects sticking together, in which case your assumption about this decompression, and the associated math, does not hold. If you can't justify it, you can't use it.

On 8/25/2018 at 9:24 AM, Capiert said:

 I.e. A virtual maximum
 of elastic force,
 where both (elastic) forces are (maximum,) opposite & equal.

But you can't assume it is elastic in order to show that it's elastic. That's circular reasoning, i.e. invalid logic.

 

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16 hours ago, Strange said:

I had the impression you were old enough to remember typewriters.

The meaning is also non_sense for typewriters.?

How old do you think I am? At least 5.

16 hours ago, Strange said:

Possibly even their invention. But apparently not. 

How about not doing either. Why the bizarre need to break your text into short unreadable lines?

For me it looks neater,
 to help me correct my (text) errors.

16 hours ago, Strange said:

Just stop it. 

I I capt'n

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24 minutes ago, Capiert said:

For me it looks neater,

To everyone else, it makes it look like the text is written by a four year old. 

P.s. your pathetic attempts to justify your immature writing style is much more interesting than your delusional ideas about physics. 

Edited by Strange
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10 hours ago, swansont said:

On any modern electronic device you only need to hit CR at the end of a paragraph. The software does the line wraps for you.

I don't want line wraps.
This website('s software) & my PC have too many bugs (to fix).
I want to see (small) phrases so I can correct faster.

 

Let's try again please (for your next comment),
 against what was my wrong choice of words (Assuming vs set)
 (e.g. instead of trying to be diplomatic,
  (now) I have to boss around, telling what to do).

Capiert: (Do NOT assume; but instead)
 s
et a virtual (linear acceleration) equivalent,
 so each phase (of (elastic) compression & decompression)
 have the same amount of speed change vd1,
 as
(only) a (math) construct.
(vd1 is my variable because I created it,
 so I can say what I want to do with it. Basta.)

If you don't like that setup,
 then set mass1's total speed difference to 2*vd1=v1'-v1,
 & then divide by 2
 (to get vd1).
That's all I'm doing.
Now I (still) don't see where that (setup) involves coe
conservation of energy.?
I've only stated the initial & final speeds of mass1.

Swansont: Which stems from conservation of energy, which you are not allowed to assume.

Capiert: Ok, so I've now stopped assuming (that); & started defining with "set" (verb) instead.

Swansont: It (=Capiert: the (virtual) linear acceleration?)
certainly doesn't come from conservation of momentum.

Capiert: in your mind (maybe?).

(It does NOT come from COE (now) either.)

I.e.

Which (now) does NOT stem from either conservation of energy, nor momentum.
(Instead of: could also (stem from coe).)

I hope that now improves the situation better, for your comment:

10 hours ago, swansont said:

No, in physics. They are linked. 

Please explain (now what you mean, by linked to (?).
Can you still disagree, if yes then with what?
(E.g. It's doubtful (now) that the vd1construct (definition) stems from either coe or com, at all!
So which are you linking?
Do you mean linear_acceleration &/or speed_difference (vd1) are exclusively derived from coe?

10 hours ago, swansont said:

Then come up with your own self-consistent theory.

I'd be happy to,
 but I can't always see my mistakes,
 until you help make me aware of what you do NOT understand.
I've attempted (stating) so often that I can not always remember whether you've got it, or not.

10 hours ago, swansont said:

Otherwise, learn mainstream physics.

The changes you propose do not come from conservation of momentum.

Nor neither from coe.

10 hours ago, swansont said:

So how do you justify the speed changes that you propose?

Those virtual_speeds (& their linear change(s)) are a(n abstract) math construct (tool), for a virtual equivalent.

The equivalent_speeds are set to be linear (even though the real speeds are (probably) not.
That's why I originally (but apparently wrongly_)requested, to assume (=imagine): because it (is virtual, &) does NOT really exist. It only assists, to help (understand).

10 hours ago, swansont said:

Just because you didn't say it does not mean it's not the underlying principle.

If my statement is for: to set a variable, then the underlying principle is strictly mathematical, instead of physical.
Only you would imply, if you (tried) to jump the gun. E.g. jump to a conclusion, that is NOT validated.

10 hours ago, swansont said:

You gave certain speed changes in "compression" and "decompression". Where does the justification for that come from?

Elasticity.
Have you never seen a ball elastically deform
 as it bounces from a (hard wall or ground) surface?
Does not the ball's contact surface deform inward while ramming?
Does not the deformation begin to recompensate & undeform, upon (beginning) reversing direction?
Are you saying (=implying) bouncing takes no time?
I hope not. ?

10 hours ago, swansont said:

What is the physics?

Hooke's law F=-x*k.

 

Capiert:
Do you mean
 that linear acceleration (also) is not possible
 with (just only, some) Newton's force F=m*a
 & some speed_difference v=vf-vi? (a=v/t).

Or do you mean Hooke's law F=-x*k?

10 hours ago, swansont said:

I didn't mention any specific force.

Surely not,
 & neither did I.
I said that although we do NOT know the exact force (amount),
 we do know a point in time exists,
 when the 2 repelling forces are equal (as construct),
 even though we do NOT know exactly when.

10 hours ago, swansont said:

But we know that a collision can end up with the two objects sticking together,

Yes, & they are (called) a non_elastic collision.
But this thread is (only) about elastic collisions
 (& how to use the non_elastic collision (theory) for them (elastic collisions).
It's NOT just any collision,
 but instead (aimed at) a collision which does NOT stick together, finally.

If we are dealing with (finally) only elastic collisions,
 then I do NOT see how they will end up staying (=sticking) together permanently.
Do you?

10 hours ago, swansont said:

in which case your assumption about this decompression, and the associated math, does not hold.

Yes.
Reading through the lines (i.e. interpreting what I suspect you are trying to say):
 we know non_elastic collisions do NOT have the decompression of elastic_collisions.
Decompression takes place for sure in elastic collisions (as a bounce, bouncing).
But I think that is obvious. Isn't it?
& you got mixed up a bit.?

E.g. Elastic collisions can be equivalent to: non_elastic collision, (only) for an instant in time (=point in time);
 but non_elastic collisions can never really be equivalent to: elastic collision, because they (=the 2 balls) do NOT stay together, permanently, after.

10 hours ago, swansont said:

If you can't justify it

=?
I'll assume you're referring to decompression, & its (elastic collision) math
 but you inappropriately (=wrongly) tried to use that on non_elastic collision.?
It's possible to hi_speed photograph partially decompressed non_elastic collisions,
 if that's what you mean, instead?

10 hours ago, swansont said:

, you can't use it.

I didn't you did.

10 hours ago, swansont said:

But you can't assume it is elastic in order to show that it's elastic.

I didn't, I assumed non_elastic (briefly, that) continued to elastic.
Otherwise, I'm sorry I can't follow you there,
 that (=your false accusation) sounds like non_sense,
 to breed non_sense.

A collision is finally, either: elastic; or it is NOT.
A non_elastic collision can be made up of a mixture of both;
 but it is still finally (only) non_elastic.

Your elastic elastic, unfortunately:

10 hours ago, swansont said:

That's circular reasoning, i.e. invalid logic.

Which had nothing to do with me.
Or have I missed you?
Because you obviously mixed things up a bit.
If that's because of my sentence structure by saying something wrong, then I'm sorry.
It wasn't intended to go stray.

2 hours ago, Strange said:

To everyone else, it makes it look like the text is written by a four year old. 

P.s. your pathetic attempts to justify your immature writing style is much more interesting than your delusional ideas about physics. 

Thanks.

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6 hours ago, Capiert said:

E.g. Elastic collisions can be equivalent to: non_elastic collision, (only) for an instant in time (=point in time);
 but non_elastic collisions can never really be equivalent to: elastic collision, because they (=the 2 balls) do NOT stay together, permanently, after.

Errata: it sounds like chaos.

Rethinking that sentence's 2nd half I'm not (completely) happy with that word "never" & "equivalent", e.g. the statement's structure.

Please help me state that correctly. I.e. Help me say what I want, in view of what I am aiming at.

P.S. I'm: "stuck" (but not together, =scattered!), on that idea. "Getting lost in the sauce."

& can't say it right. (So I guessed "never"; but I'm NOT sure if you can say that (so), & (then) depend on it.)

Is it safe to say:

 An elastic collision is temporarily like a non_elastic collision, thus never permanently.

 But a non_elastic collision is never like a completed elastic_collision.?

That also sounds wrong because they both can begin similarly, but end up differently.

How should I say it?

Edited by Capiert
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8 hours ago, Capiert said:

This website('s software) & my PC have too many bugs (to fix).

I think it is entirely due to your incompetence.

8 hours ago, Capiert said:

I want to see (small) phrases so I can correct faster.

Why are short lines easier to correct? That is an idiotic excuse.

But you admit you are deliberately writing in an annoying and hard to read manner just yo make your life a little bits easier. That is a pretty offensive level of arrogance.

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1 hour ago, Strange said:

I think it is entirely due to your incompetence.

Fine. Descartes also thought, that therefore he was; or was that is?

Quote

Why are short lines easier to correct?

I can see irregularities sooner.

Quote

That is an idiotic excuse.

I often don't write like you, starting with 1, 2 or 3 words & fill up the sentence from a very abstract vague impression.
I often don't know how to say things, & have to program the sentence.
Is that better?

Quote

But you admit you are deliberately writing in an annoying and hard to read manner just to make your life a little bit easier.

Yes the laptop overheats & hangs, sometimes loosing info so I have to be faster.
It's a little chaotic.
My laptop doesn't like it (getting hot).

Quote

That is a pretty offensive level of arrogance.

Trying to correct (typing) errors to make it easier for others to read including me?
If you didn't notice, I also corrected your nonsense here.

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13 hours ago, Capiert said:

 

Let's try again please (for your next comment),
 against what was my wrong choice of words (Assuming vs set)
 (e.g. instead of trying to be diplomatic,
  (now) I have to boss around, telling what to do).

Capiert: (Do NOT assume; but instead)
 s
et a virtual (linear acceleration) equivalent,
 so each phase (of (elastic) compression & decompression)
 have the same amount of speed change vd1,
 as
(only) a (math) construct.
(vd1 is my variable because I created it,
 so I can say what I want to do with it. Basta.)

No, the wording is not the problem. You have to justify why you would "set" things this way. It has to depend on some physics.

Quote

 I hope that now improves the situation better, for your comment:

No. You're still making this up out of thin air, in order to get the answer you want.

Quote

 Elasticity.
Have you never seen a ball elastically deform
 as it bounces from a (hard wall or ground) surface?
Does not the ball's contact surface deform inward while ramming?
Does not the deformation begin to recompensate & undeform, upon (beginning) reversing direction?
Are you saying (=implying) bouncing takes no time?
I hope not. ?

Not all objects will return to their original state. But in the context of collision (since that's what we're discussing), whether or not the objects deform does not affect the conservation of momentum.

What does affect the behavior is to the extent that KE is conserved. But you claimed that you did not like conservation of energy. And I challenged you to solve various problems using momentum. Conservation of momentum does not distinguish between elastic and inelastic collisions.

 

Quote

 

Yes.
Reading through the lines (i.e. interpreting what I suspect you are trying to say):
 we know non_elastic collisions do NOT have the decompression of elastic_collisions.
Decompression takes place for sure in elastic collisions (as a bounce, bouncing).
But I think that is obvious. Isn't it?
& you got mixed up a bit.?

No, not mixed up. Elastic vs inelastic is a determination based on conservation of KE. To make assumptions based on it is to invoke circular logic, which is invalid. The tool you get to use, based on your stated preference, is conservation of momentum.  

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On ‎2018‎ ‎08‎ ‎27 at 6:15 PM, swansont said:

No, the wording is not the problem. You have to justify why you would "set" things this way. It has to depend on some physics.

I thought linear_acceleration is a physics concept dealing with speed, distance, per time;
 & has the advantage that we can predict them well (when linear).
I don't see how we can know the real exact speeds while colliding.
That's the reason why I chose to use a linear acceleration (equivalent).
So I attempted to convert what we know (physically) into a (virtual) linear format (called equivalent) for the surety.
Are you saying there is (absolutely) no physics there? (Maybe you're right.?)
Not even the initial & final conditions?
(It's back_engineering, & I want answers (so I apply math).)

Quote

No. You're still making this up out of thin air,

Well, I did NOT get it out of a physics book
 if that's what you mean,
 because they did NOT have it.

So I sat down & tried to figure out how I could solve it (a momentum conversion).
So that led me to the elasticity of Hooke's law of the 2 balls
 & that their (motional) energy (or also momentum) could be (temporarily) stored into their elastic deformation displacement
 similar to springs.
I must admit: that perspective, led me to question whether Hooke's law was a kind of energy (or also momentum) storage or not;
& the whole perspective (development) began wrt energy (instead of momentum).
But my intention was to put (=convert) things from an energy perspective to a momentum perspective.
How they are (or can be correctly) proportioned (to each other) seems long obvious to me now.
Energy is a well known concept in physics, but you say I am not allowed to use it.
As far as I know, your physics does not use a spring_loaded momentum storage. Does it?
Is such an equivalent possible, or conceivable?

Strangely we both get similar answers (with 2 different methods: yours & mine).
How is that possible?

Quote

in order to get the answer you want.

Yes, it's true I want the correct momentum answer
 (Ewert's impulse experiment baffles me the most);
 it would be foolish to want a wrong answer.
Or do you think otherwise?

Quote

Not all objects will return to their original state. But in the context of collision (since that's what we're discussing), whether or not the objects deform does not affect the conservation of momentum.

So I guess you mean momentum is conserved for both (=either elastic or non_elastic collisions).?

Quote

What does affect the behavior is to the extent that KE is conserved.

You mean, how much KE is conserved?
That sounds similar to my concept as to how much momentum_squared can be saved.

Quote

But you claimed that you did not like conservation of energy.

Yes, I said I didn't like energy (in its naked format, with single masses; instead of squared masses);
 but when as 2*m*E (e.g. ruffly the momentum_squared format)
 I know where to place energy in the equation
 for correct proportions (or proportioning).

Quote

And I challenged you to solve various problems using momentum.

Yes, I squared the (before after collision) momentum equation.
That eliminated the polarity (sign) problem,
  because some vector_speeds get squared.

Quote

Conservation of momentum does not distinguish between elastic and inelastic collisions.

That's why I believe com can be used for both;
 but that an elastic collision undergoes a temporary non_elastic collision, briefly;
 & it has a time_point,
 even although we cannot know it.

Is that the wrong way to look at it?

Quote

No, not mixed up. Elastic vs inelastic is a determination based on conservation of KE.

I think you are trying to say KE determines elasticity (elastic_collision).?

Quote

To make assumptions based on it is to invoke circular logic, which is invalid. The tool you get to use, based on your stated preference, is conservation of momentum.  

Yes but I must get from 1 system (KE) to the other (mom), securely.
I do not know how to proceed (otherwise) if those are your restrictions.
(We know the conditions when KE works; & we can know the conditions when momentum functions correctly too.
I'm looking for a seamless bridge between them both, & believe I have found it.) 
I was interested in delivering a conversion (equality)
 (e.g. using Hooke's law Fs=-x*ks);
 but you want me to do it without using an existing (energy) physics.
But to get there (to wrt mom_squared) I had to (or did), use your energy observations.

It was an end to a means (if that's what you mean?).
On the way, I became discouraged with energy('s reliability), (so much) that I wanted to leave it,
 in favor of momentum.

Is Hooke's law not acceptable enough?

Edited by Capiert
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1 hour ago, Capiert said:

Energy is a well known concept in physics, but you say I am not allowed to use it.

You said you didn't trust it.

Quote

avoiding energy as much as possible   with the suspicion   that it is corrupt

So don't blame me. I challenged you to solve certain problems without it, and rely on momentum, which you "trust completely"

You could have agreed with me, but you didn't — you accepted the challenge.  

Quote

As far as I know, your physics does not use a spring_loaded momentum storage. Does it?

That depends on what you mean by spring-loaded momentum storage. 

Momentum is conserved when there is no net external force on a system. One does not "store" momentum.

 

Quote

So I guess you mean momentum is conserved for both (=either elastic or non_elastic collisions).?

Yes. As long as the net external force is zero, momentum will be conserved.

Quote

 I think you are trying to say KE determines elasticity (elastic_collision).?

Elasticity determines whether KE is conserved, because KE is, in general, not a conserved quantity.

Quote

Yes but I must get from 1 system (KE) to the other (mom), securely.

Yes, you do. You need to apply conservation of KE. So isn't it rather silly to have accepted a challenge where you claim you don't?

Quote

I do not know how to proceed (otherwise) if those are your restrictions.
(We know the conditions when KE works; & we can know the conditions when momentum functions correctly too.
I'm looking for a seamless bridge between them both, & believe I have found it.) 
I was interested in delivering a conversion (equality)
 (e.g. using Hooke's law Fs=-x*ks);
 but you want me to do it without using an existing (energy) physics.

I don't want you to. I want you to drop your silly notion that energy is "corrupt"

Stop pretending that this limitation comes from me. You are the one claiming that you don't want to use the concept. All you have to do is realize you were wrong.

 

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12 hours ago, swansont said:

You said you didn't trust it.

Yes, til you (helped improve that situation).

Quote

So don't blame me. I challenged you to solve certain problems without it, and rely on momentum, which you "trust completely"

Can I say it so?: I can trust energy when it is properly (isolated, &) incorporated in the momentum_squared equations.

Quote

You could have agreed with me, but you didn't — you accepted the challenge.  

I guess I didn't recognize what you really wanted. In fact I know.

Quote

That depends on what you mean by spring-loaded momentum storage. 

That sounds like a possibility although you end up saying no.

What example do you have in mind?

Quote

Momentum is conserved when there is no net external force on a system.

That's so (=too) elegantly said that I'll need an example to picture it.

Quote

One does not "store" momentum.

I have to accept that. (But have difficulty & can't (quite).)

But isn't every moving thing a storage of momentum?

Non_recognizable (as static), when our surroundings are moving at our same speed.?

Quote

Yes. As long as the net external force is zero,

When is that ever possible?

(According to you, net gravity is from the universe's core.)

But drop that example, & try, e.g. 2 weights balanced on a balance.

Is potential momentum possible? (Similar to the potential energy PE=m*g*h=Wt*h.)

mom=m*v

v=((vi^2)+2*h*g)^0.5-vi

 where we have

 initial_speed vi

 height h

 free_fall acceleration g.

I.e. The mass multiplied by the rooted 2*h*g (part).

?

Quote

momentum will be conserved.

Elasticity determines whether KE is conserved, because KE is, in general, not a conserved quantity.

Yes, you do. You need to apply conservation of KE. So isn't it rather silly to have accepted a challenge where you claim you don't?

I'd be more than happy to to agree with you there.

But you did not answer all of my post, leaving things: mom^2 (instead of KE), & Hooke's law.

So it's difficult to orientate.

Indirectly then, I have to assume they mean nothing for you here,

 & are thus an inadequate basis.

You have brought me so far that (thru) the various threads I could incorporate Energy correctly into my momentum formulas.

Thanks for helping me correct them.

But if you do not comment on points (which I've based) adequately enough,

 then they are left in my head to rumble around, causing further disturbance.

Quote

I don't want you to. I want you to drop your silly notion that energy is "corrupt"

At present I can trust the momentum_squared formula,

 & in that (equation) is where I find (accurate) energy relations.

Quote

Stop pretending that this limitation comes from me. You are the one claiming that you don't want to use the concept.

If I have a seamless formula, then I will not hesitate (& be more than happy) to use it.

e.g. mom_squared.

Quote

All you have to do is realize you were wrong.

I've been wrong on (so) many things.

But that's not the important thing for me,

 getting it fixed=corrected is.

If I have to make the effort to just believe

 (like christians often demand)

 (then that naive belief) is simply not enough.

It won't stay (in), & that's valid for my technical mind.

I must address the issues that make it (=my mind) tick so.

(Me) evading those issues won't help me.

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23 minutes ago, Capiert said:

 What example do you have in mind?

I don't have an example in mind. Why would I? It's your claim! I don't know what you mean by the phrase.

Quote

That's so (=too) elegantly said that I'll need an example to picture it.

F = dP/dt

i.e. Newton's second law of motion

Quote

 But isn't every moving thing a storage of momentum?

It depends on what you mean by storage, which is something you still haven't clarified.

Quote

Non_recognizable (as static), when our surroundings are moving at our same speed.?

Then we are at rest with respect to our surroundings, and have no momentum.

Quote

When is that ever possible?

Collisions are one example. Neglecting external forces acting during collisions is usually a reasonable assumption.

Quote

(According to you, net gravity is from the universe's core.)

Whenever you try and summarize what I have told you, it invariably comes out as wrong. I have no idea where this bit about net gravity came from, or what it means. 

Quote

But drop that example, & try, e.g. 2 weights balanced on a balance.

Is potential momentum possible? (Similar to the potential energy PE=m*g*h=Wt*h.)

mom=m*v

v=((vi^2)+2*h*g)^0.5-vi

 where we have

 initial_speed vi

 height h

 free_fall acceleration g.

I.e. The mass multiplied by the rooted 2*h*g (part).

?

If they are balance, they are not moving. There is no momentum. if they are falling, then you can find the motion either by an energy balance, or with Newton's second law and the definition of acceleration (depending on what information you have and what you want to know) 

Quote

I'd be more than happy to to agree with you there.

But you did not answer all of my post, leaving things: mom^2 (instead of KE), & Hooke's law.

You keep mentioning Hooke's law, and I see no connection with any of our discussion. Hooke's law apples to springs. 

 

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4 hours ago, swansont said:

I don't have an example in mind. Why would I?

Your 1st sentence saying "It depends" gave an inkling of hope,

 that neither yes nor no were completely right.

Thus I concluded a yes possibility also exists.

Quote

 It's your claim! I don't know what you mean by the phrase.

I hope now you do.

Quote

F = dP/dt

Ok. What I usually write as F=mom/t (so I can make the connection for myself, & don't forget.)

F=m*v/t.

Is my notation wrong?

Quote

i.e. Newton's second law of motion

Force will change a mass's speed (velocity).

But I still see the acceleration

a=F/m

of a mass m as the observable.

Quote

It depends on what you mean by storage, which is something you still haven't clarified.

Momentum changed is like a bank account.

What goes into the mass m (as a speed change differencee v=vf-vi)

 can come out.

If the speed was increased,

 then decreasing the speed

 can return the mass to its original speed.

How we invoke & do the (speed) change is another story.

But I think I'm off topic there,

 momentum_squared is the better concept

 instead of momentum

 & kinetic energy, there.

Quote

Then we are at rest with respect to our surroundings, and have no momentum.

We have no momentum wrt to each other, true.

But I don't think we can say that for any other reference,

 because we are moving thru the universe (no doubt).

I mean surely there is an underlieing inherent momentum for every mass,

 based on let us say the universe's center

 (from where you (=physicists) say the big bang happened).

Quote

Collisions are one example. Neglecting external forces acting during collisions is usually a reasonable assumption.

Ok. That makes things easier. i.e. less to think about.

Quote

Whenever you try and summarize what I have told you, it invariably comes out as wrong. I have no idea where this bit about net gravity came from, or what it means. 

I guess according to the big bang (hypothesis, since I doubt we'll ever have a time machine or viewer to the past to prove it),

 everything is suppose to be flying away from the universe's center

 & being also gravitationally attracted (according to you physicists).

That looks like a radial acceleration to me,

 e.g. a non_balanced force (setup)

 in ruffly 1 direction

 out from the core (center).

Quote

If they are balance(d), they are not moving.

Does that mean the earth on which the balance is fixed

 is not turning.

Surely not.?

Stationary is a static (d)illusion.

No motion does not exist in the universe,

 everything is moving.

Stationary (or static, at rest) only means both (the object (e.g. mass),

 & its reference sytem) are moving at the "same" speed.

To say they do not move (at all) is absurd.

Or do you disagree?

Quote

There is no momentum. if they are falling, then you can find the motion either by an energy balance,

Math or instrument machine?

Quote

or with Newton's second law and the definition of acceleration (depending on what information you have and what you want to know).

Ok.

Quote

You keep mentioning Hooke's law, and I see no connection with any of our discussion. Hooke's law apples to springs. 

 I said the balls deform like a spring,

 in order to decellerate (elastically, during the collision).

That deformation needs time,

 (even if it's very quick,

 & the 2 balls are very hard,

 like steel or glass.

Both materials still have (hard) elastic properties).

I hope you understand me now.?

(It seems you (also) did not register squared_momentum.? Have I said something wrong? Are you still trying to figure out what that means (because you have not commented on it so I assumed you are not sure). Please ask that I can try to fill in the blanks. Sometimes the mind automatically blocks things out too fast to notice.)

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21 minutes ago, Capiert said:

based on let us say the universe's center

 (from where you (=physicists) say the big bang happened).

There is no centre of the universe in the Big Bang model.

22 minutes ago, Capiert said:

I mean surely there is an underlieing inherent momentum for every mass,

Nope.

22 minutes ago, Capiert said:

I guess according to the big bang (hypothesis, since I doubt we'll ever have a time machine or viewer to the past to prove it),

There is an overwhelming amount of evidence for this Big Bang theory (which is why it is a theory and not a hypothesis). It will never be "proved" because science doesn't really "prove" anything.

23 minutes ago, Capiert said:

everything is suppose to be flying away from the universe's center

Nope.

23 minutes ago, Capiert said:

That looks like a radial acceleration to me, e.g. a non_balanced force (setup) in ruffly 1 direction out from the core (center).

Nope.

But I am not at all surprised to find out that you don't know what you are talking about.

 

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13 minutes ago, Capiert said:

I hope now you do.

I don't see any explanation from you, so your optimism is unwarranted.

13 minutes ago, Capiert said:

Ok. What I usually write as F=mom/t (so I can make the connection for myself, & don't forget.)

F=m*v/t.

Is my notation wrong?

Yes, in a subtle way that will cause issues in some problems. But acceptable for most cases.

13 minutes ago, Capiert said:

 Force will change a mass's speed (velocity).

But I still see the acceleration

a=F/m

of a mass m as the observable.

Momentum changed is like a bank account.

What goes into the mass m (as a speed change differencee v=vf-vi)

can come out.

If the speed was increased,

 then decreasing the speed

 can return the mass to its original speed.

That's trivially true.

13 minutes ago, Capiert said:

How we invoke & do the (speed) change is another story.

But I think I'm off topic there,

 momentum_squared is the better concept

 instead of momentum

 & kinetic energy, there.

That would depend on the circumstance. You seem to be focused on there being one concept to apply, and that's simply shortsighted. There are a number of concepts that might apply in a physics problem. 

13 minutes ago, Capiert said:

We have no momentum wrt to each other, true.

But I don't think we can say that for any other reference,

 because we are moving thru the universe (no doubt).

I mean surely there is an underlieing inherent momentum for every mass,

No, in fact that opposite is true. There is no inherent motion. You are free to choose any inertial reference frame, and physics will work in that frame. IOW, you are free to choose a frame where you are at rest, and you are free to choose a frame in which you are moving at some constant velocity.

We generally choose the frame where the solutions are the simplest to come by.

13 minutes ago, Capiert said:

 based on let us say the universe's center

The universe doesn't have a center.

13 minutes ago, Capiert said:

 (from where you (=physicists) say the big bang happened).

Ok. That makes things easier. i.e. less to think about.

I guess according to the big bang (hypothesis, since I doubt we'll ever have a time machine or viewer to the past to prove it),

 everything is suppose to be flying away from the universe's center

That's an incorrect summary of the big bang, but then, how did the big bang enter the conversation? We were talking about collisions.

13 minutes ago, Capiert said:

 & being also gravitationally attracted (according to you physicists).

That looks like a radial acceleration to me,

The salient question is how big is it? If it's imperceptibly small, it can be ignored.

13 minutes ago, Capiert said:

 e.g. a non_balanced force (setup)

 in ruffly 1 direction

 out from the core (center).

Does that mean the earth on which the balance is fixed

 is not turning.

Surely not.?

In most cases, the rotation of the earth will not have any effect that you can measure.

13 minutes ago, Capiert said:

 

Math or instrument machine?

Math

13 minutes ago, Capiert said:

  I said the balls deform like a spring,

But not all balls deform like a spring, so this is a bad assumption.

To assume deformation like that is to assume that energy is conserved, because an ideal spring is a device that conserves mechanical energy.

13 minutes ago, Capiert said:

 in order to decellerate (elastically, during the collision).

That deformation needs time,

 (even if it's very quick,

 & the 2 balls are very hard,

 like steel or glass.

Both materials still have (hard) elastic properties).

I hope you understand me now.?

Yes. You want to assume mechanical energy is conserved but do it in a way that doesn't directly invoke the conservation. But that doesn't mean you aren't using the concept.

IOW, you can't assume the collision is elastic, because "elastic" and "conservation of KE" are equivalent statements, and you said you could do this without invoking energy concepts, as you don't trust them.

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20 minutes ago, Strange said:

There is no centre of the universe in the Big Bang model.

Then center of the big bang explosion as the reference instead.

Quote

Nope.

There is an overwhelming amount of evidence for this Big Bang theory (which is why it is a theory and not a hypothesis). It will never be "proved" because science doesn't really "prove" anything.

Well then I guess we can give up.

Quote

Nope.

Nope.

But I am not at all surprised to find out that you don't know what you are talking about.

You knew it (excluding your assumption of James Watt's horsepower history).

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4 minutes ago, Capiert said:

Then center of the big bang explosion as the reference instead.

The Big Bang is not an explosion and there is no centre. Maybe you should start a thread to ask questions about what the Big Bang model actually says, instead of making up nonsense (as usual).

5 minutes ago, Capiert said:

Well then I guess we ca give up.

Why? Because you have an unrealistic view of science?

 

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