Mechanical Symmetry?

[Capiert]

KE is not a conserved quantity. Why would you expect it to always "add correctly"?

(Assuming that's not a rhetorical or stacked question, intended to enlighten:)

 

I expected energy should add instead,

but found that it didn't (quite correctly, sometimes).

 

With (supposed) conservation of energy,

I expect to "account" for the (the system's (total)) energy,

(mass included),

(expecting) not more, nor less.

 

Simple addition, is my (natural) expectation.

 

That KE is not a "conserved" quantity,

(in my view, doesn't bother me, &)

does (NOT) exclude it (=KE)

from (a trade_off exchange, as)

being (some sort of) energy

(that can be accounted (for)).

 

(KE is only 1 piece of apparatus I used

in the energy equation (example).

I could just as easily use PE instead,

or as a part.)

 

E.g. If energy does not add (correctly, or simply),

then how does it (really) behave

so that I can account for it (=energy).

 

Have I made myself clear?

(If not maybe we must needle in,

to find what you are looking for

from me,

if that exists.?)

 

KE is obviously a variable

during collision,

the same as (for) momentum

(is also a variable

during the collision).

 

I'm interested in before (status) & after (results)

& the parts of (adding to) the total.

 

If I give (in as input) each of the initial masses & speeds, (then)

I want (a way) to (accurately) calculate their final speeds (& polarity) (out, as output).

 

Is that illogical?

 

If so, perhaps you can see a flaw in my reasoning

& be so good as to show me. e.g. why.

 

That my excel table works (=functions) relatively well

convinces me

(as a 1st approximation).

 

In view of this sommer's findings,

I suspect improved rigor

might (help) get the bugs out

(in the trouble shooting phase).

 

The derivation here

is only an approximation

(with relatively good success).

 

That's all.

 

P.S. Although KE is stated in the equation,

please drop the K.

The equation represents the (system's total) energy E

(although it only used momentum & KE

to get it (=E)).

Edited December 29, 2016 by Capiert

[swansont]

(Assuming that's not a rhetorical or stacked question, intended to enlighten:)

I expected energy should add instead,

but found that it didn't (quite correctly, sometimes).

With (supposed) conservation of energy,

I expect to "account" for the (the system's (total)) energy,

(mass included),

(expecting) not more, nor less.

Simple addition, is my (natural) expectation.

That KE is not a "conserved" quantity,

(in my view, doesn't bother me, &)

does (NOT) exclude it (=KE)

from (a trade_off exchange, as)

being (some sort of) energy

(that can be accounted (for)).

(KE is only 1 piece of apparatus I used

in the energy equation (example).

I could just as easily use PE instead,

or as a part.)

E.g. If energy does not add (correctly, or simply),

then how does it (really) behave

so that I can account for it (=energy).

 

You have to look at all possibilities. Collisions make noise - sound represents energy lost from the kinetic energy. The particles can deform - work is done in doing that. They can heat up.

& after the collision

they are stuck together

 

This is known as a totally inelastic collision and is the condition under which the maximum amount of KE is transformed into other forms

[Capiert]

You have to look at all possibilities. Collisions make noise - sound represents energy lost from the kinetic energy. The particles can deform - work is done in doing that. They can heat up.

 

This is known as a totally inelastic collision and is the condition under which the maximum amount of KE is transformed into other forms

Thank you for your reply,

but what you are suggesting is guesswork.

Should I ignore my (logical) math?

(It doesn't lie, if done correctly, e.g. don't break any rules for algebra, equations.)

It's telling me things you aren't.

Why should I believe in guesswork*

when the math to follow has NOT been exhausted, yet?

Your proposals for sound (trivial), deformation, & heat,

have not helped me in the past,

why should they now?

Sorry, but rejected.

They are NOT the way I deal

with the major part of this problem.

The task was to find the end speeds.

The inelastic collision was setup

so that it did NOT matter

whether the 2 masses (e.g. doe or wet clay globs) were together

or NOT, e.g. barely touching,

thus deformation & heat could be eliminated.

All your proposals can be delt with

after the basis has been (correctly) established,

NOT before.

 

* P.S. I know we must make assumptions somewhere,

but ..(see above).

Edited December 30, 2016 by Capiert

[Bignose]

Thank you for your reply,

but what you are suggesting is guesswork.

If you did some reading, you'd find that the study of impacts is very sophisticated. See Werner Goldmsith's text Impact, available very cheaply as a Dover title. 416 pages of not-guesswork.

 

Stronge's Impact Mechanics is also highly regarded.

 

Why don't you take a peak at some of the existing literature before trying to reinvent it yourself and see if it addresses your concerns?

Edited December 30, 2016 by Bignose

[Capiert]

I can't believe it,

you guys are ignoring the math (that is presented)

& diverging to other things.

Can we stay on topic?

 

It's not that I don't appreciate

the helpful suggestion, (thank you for the tip)

but it's very frustrating

not to complete this

before hopping on the next band wagon.

Edited December 30, 2016 by Capiert

[swansont]

Thank you for your reply,

but what you are suggesting is guesswork.

Should I ignore my (logical) math?

 

 

 

 

Yes, we should ignore your math. It's wrong.

 

Momentum in physics uses the symbol p

 

You wrote that

(p1+p2)^2=(p3)^2

 

But momentum is a vector, not a scalar.

 

You wrote

"The final kinetic energy

KE3#KE2+KE1

is NOT the sum of the 2, but instead

KE3~(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2)."

 

But that's wrong, too. KE is not conserved, as I have stated a number of times. There IS NO FORMULA to find the kinetic energy in an arbitrary collision.

 

You can send two particles of identical mass toward each other, at the same speed. The momentum of the system is zero. If the collision is totally inelastic, the combined object will be at rest — there will be no kinetic energy left. There's no violation of any law here, because KE is not a conserved quantity. Total energy is conserved.

[Bignose]

I can't believe it,

you guys are ignoring the math (that is presented)

& diverging to other things.

Can we stay on topic?

I'm not ignoring math. Please see the texts I referred to. There is a lot of math in both of them. I just don't think there is any reason to type it all here when you should be capable of reading those texts yourself.

[Capiert]

Yes, we should ignore your math. It's wrong.

 

Momentum in physics uses the symbol p

 

You wrote that

(p1+p2)^2=(p3)^2

 

But momentum is a vector, not a scalar.

I'm sorry, you've lost me there.

I don't follow what you're trying to tell me.

All I know for that is

speed (vector) has a direction (e.g. polarity, for 1D);

& mass (scalar) doesn't.

You wrote

"The final kinetic energy

KE3#KE2+KE1is NOT the sum of the 2, but instead KE3~(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2)."

 

But that's wrong, too. KE is not conserved, as I have stated a number of times. There IS NO FORMULA to find the kinetic energy in an arbitrary collision.

Why then, can an (=my) excel sheet find it?

Maybe I'll have to name the quantity something else other than KE?

 

You can send two particles of identical mass toward each other, at the same speed. The momentum of the system is zero. If the collision is totally inelastic, the combined object will be at rest there will be no kinetic energy left. There's no violation of any law here, because KE is not a conserved quantity. Total energy is conserved.

Thanks for the reply.

But let me get this straight.

Do you mean the energy is NOT zero after the collision?

Edited December 30, 2016 by Capiert

[Mordred]

maybe wiki will help. Perhaps you might see Swansont is correct. Considering he has a Ph.D in particle physics. Which deals with elastic and inelastic collisions. He might just have a far greater understanding than you do. Just saying...

 

https://en.m.wikipedia.org/wiki/Inelastic_collision

 

"An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction."

Edited December 30, 2016 by Mordred

[Capiert]

maybe wiki will help. Perhaps you might see Swansont is correct.

https://en.m.wikipedia.org/wiki/Inelastic_collision

"An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction."

Thank you.

Maybe I've labled (=named) things wrong?

What is it called when the final masses barely touches,

continuing to travel both at the same speed,

but did NOT collide or bang together?

Neither elastic nor inelastic?

Edited December 30, 2016 by Capiert

[swansont]

Thanks for the reply.

But let me get this straight.

Do you mean the energy is NOT zero after the collision?

The energy is not zero. But one form, the kinetic energy, is zero.

 

maybe wiki will help. Perhaps you might see Swansont is correct. Considering he has a Ph.D in particle physics.

Atomic physics. But (as you know) this is first-semester physics. High school students are able to learn this.

[Capiert]

The energy is not zero. But one form, the kinetic energy, is zero.

I think my formula gives that zero KE.

(Please note or remember, it uses KE polarity, to give the values in excel. It's 1D.)

[Mordred]

Thank you.

Maybe I've labled (=named) things wrong?

What is it called when the final masses barely touches,

continuing to travel both at the same speed,

but did NOT collide or bang together?

Neither elastic nor inelastic?

All collisions are elastic or inelastic.

 

If the two masses touch its a collision even if its so instantaneous as to appear otherwise.

Atomic physics. But (as you know) this is first-semester physics. High school students are able to learn this.

Thanks for some reason I always thought it was in particle physics.

[Capiert]

If you did some reading, you'd find that the study of impacts is very sophisticated. See Werner Goldmsith's text Impact, available very cheaply as a Dover title. 416 pages of not-guesswork.Stronge's Impact Mechanics is also highly regarded.Why don't you take a peak at some of the existing literature before trying to reinvent it yourself and see if it addresses your concerns?

Hi, I think you've (almost) stated the reason why I'd prefer to simplify impacts

instead of reading tons of liturature.

It's very sophisticated, & time consuming.

But I want a toy that will tell me the answers, instantly;

without all the hassels.

That's why.

Best

Regards

& thanks for the tips.

They might become (very) useful

later if or when I can get my hands on them

if more improvements are needed.

But maybe we could do a 1D test.

Give me a task of givens

& unknowns

that my excel sheet should calculate.

You must surely know some tricky 1's

that bluff the average student.

We first have to establish

if it's a task that it can solve.

E.g. elastic collision. (non_relativistic).

You are aware that Swansont's proposal (noise, deformation, heat etc)

won't be delt with.

I'm curious how different my results

will be from yours.

& naturally the reasons why.

Is that fair?

[Mordred]

This happens all the time in speculation. People always question existing physics. They go about trying to reinvent it instead of taking the time to understand the existing physics.

 

End result, this methodology never works.

 

Its always best to understand the existing physics before you try to change it. Particularly in classical physics.

Take the time to understand inelastic vs elastic collisions. The same rules apply whether its relativistic or not.

 

Here

 

http://bolvan.ph.utexas.edu/~vadim/Classes/2014s/collisions.pdf

Edited December 30, 2016 by Mordred

[swansont]

I think my formula gives that zero KE.

(Please note or remember, it uses KE polarity, to give the values in excel. It's 1D.)

KE is a scalar.

 

The equation I quoted above takes the square root of the KE, so you'd be getting an imaginary value with a negative KE, but never zero.

[Bignose]

Hi, I think you've (almost) stated the reason why I'd prefer to simplify impacts

instead of reading tons of liturature.

It's very sophisticated, & time consuming.

But if your simplifications are wrong and predict the wrong things, then you've made things worse. The authors of those books didn't make them hundreds of pages long just because they like writing. The books are that long to handle the problems properly. Simplifications are fine, up until the points when they aren't anymore. At the very least, like Swansont said, the beginnings of this stuff is covered in high school level physics. You need to understand that -- and how supremely successful it is -- before you try to come along and introduce simplifications or other changes to it. And then you need to show us how your simplification make even better predictions than what we have now. This will require extraordinary evidence as what we have now has been, well, supremely successful. Your inability to understand it is not a good enough reason to change it -- when again it has been used for a very long time and has been very successful.

[Sriman Dutta]

In an inelastic collision the sum total of kinetic energies of two bodies does not equal to the total kinetic energy of the system after collision. For such a collision, the equation KE1+KE2=KE3 does not hold true.

But momentum is conserved so that we get m1u1+m2u2=(m1+m2)v, where u1 and u2 are initial velocities and v is the final velocity of the whole system.

[Capiert]

KE is a scalar.

The equation I quoted above takes the square root of the KE, so you'd be getting an imaginary value with a negative KE, but never zero.

Wrong, or at least not quite.

I use Napier's syntax for negative polarity, so I can maintain tracking it.

I've been able to setup the excel sheet to observe KE polarity, from the speed direction.

Edited January 1, 2017 by Capiert

[swansont]

Wrong, or at least not quite.

I use Napier's syntax for negative polarity, so I can maintain tracking it.

I've been able to setup the excel sheet to observe KE polarity, from the speed direction.

 

Your equation is

KE3~(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2).

 

You are, in fact, taking the square root of a KE. This gives an imaginary number if the KE is negative.

 

You add the two values. There is no way to get zero from this equation, unless you are doing the math incorrectly.

[Capiert]

It's using an extended syntax which eliminates the imaginaries.

Is that wrong?

 

e.g. Let

v=-1 m/s

v^2=(-1 m/s)^2

Rooting gives

((-1)^2)^0.5=(-1 m/s).

 

Where is the imaginary value there?

Let each mass m=m1=m2=2 kg

v1=1 m/s, & v2=-1 m/s.

Edited January 1, 2017 by Capiert

[Strange]

It's using an extended syntax which eliminates the imaginaries.

Is that wrong?

 

 

Yes.

[Capiert]

KE1=m1*(v1^2)/2=2 kg*((1 m/s)^2)/2, & KE2=m2*(v2^2)/2=2 kg*((-1 m/s)^2)/2,

 

KE3~(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2)

KE3~(((2 kg*2 kg*((1 m/s)^2)/2)^0.5)+((2 kg*2 kg*((-1 m/s)^2)/2)^0.5)^2)/(2 kg+2 kg)

KE3~(((2 (kg^2)*((1 m/s)^2))^0.5)+((2 (kg^2)*((-1 m/s)^2))^0.5)^2)/4 kg

Isolate terms under each root sign

KE3~(((2^0.5) kg*((1 m/s)^2)^0.5)+(2^0.5) kg*(((-1 m/s)^2)^0.5)^2)/4 kg

KE3~(((2^0.5) kg*(1 m/s)+(2^0.5) kg*(-1 m/s))^2)/4 kg

KE3~((((2^0.5) kg*m/s)-(2^0.5) kg*m/s)^2)/4 kg

KE3~((0 kg*m/s)^2)/4 kg

KE3~0 kg*((m/s)^2)

 

 

 

 

Edited January 1, 2017 by Capiert

[Sensei]

Mass m vs speed_difference v "trade_off"

is very important for conservation of momentum (com).

Unfortunately conservation of energy (coe)

doesn't see (=conceptualize) things

the same way,

& instead (some sort of) a speed_squared

vs (single) mass relevance exists

(or should it?)!

E.g. How can both laws be valid,

when they seem to contradict?

Does (kinetic) energy really have

a mass vs speed_squared trade off? E.g. in collisions.

If NOT, then why do we use it (=(the) energy_construct)?

 

That's 4 questions.

Thanks in advance.

I think the best is to show it on examples.

 

Suppose so we have radioactive isotope, such as Uranium-238.

It has 92 protons, and 146 neutrons, total 238 baryons.

It decays the most often by alpha decay to Thorium-234 (It has 90 protons, and 144 neutrons, total 234 baryons), emitting alpha particle, Helium-4 nucleus.

 

U-238 -> Th-234 + He-4 + 4.27 MeV

 

Decay energy, energy released by this reaction is approximately 4.27 MeV.

To convert from electron volts to Joules (and back), we multiply or divide by e=1.602176565*10^-19.

4.27 MeV = ~4270000 eV * 1.602*10^-19 = 6.84*10^-13 J

 

But where does this 4.27 MeV come from?

It's difference between mass of parent isotope, and daughter isotope with alpha.

Mass of U-238 with electrons is 238.0507882 u.

Mass of Th-234 with electrons is 234.043601 u.

Mass of He-4 with electrons is 4.0026 u.

 

Subtract them 238.0507882 u-( 234.043601 u + 4.0026 u) = 0.0045872 u

1 u (a.m.u. mass atomic unit) is approximately 1.66*10^-27 kg and 931.494 MeV/c^2

0.0045872 u * 931.494 MeV/u = 4.27 MeV energy released.

 

This energy is taken by either newly created Thorium-234 atom, and Helium-4 nucleus. That's simply sum of kinetic energies of newly created particles (assuming there is no gamma photons emitted).

But Thorium is 234.043601 u / 4.0026 u = ~58.5 times more massive than Helium-4 nucleus!

 

Conservation of momentum (which is initially 0 prior decay), without special relativity corrections is

 

[math]m_{th}*v_{th}=m_{he}*v_{he}[/math]

 

Reverse equation:

 

[math]\frac{m_{th}}{m_{he}}=\frac{v_{he}}{v_{th}}[/math]

 

 

Conservation of momentum (which is initially 0 prior decay), with special relativity corrections is

 

[math]m_{th}*v_{th}*\gamma_{th}=m_{he}*v_{he}*\gamma_{he}[/math]

 

Reverse equation:

 

[math]\frac{m_{th}*\gamma_{th}}{m_{he}*\gamma_{he}}=\frac{v_{he}}{v_{th}}[/math]

 

Kinetic energy in special relativity is:

 

Thorium-234:

[math]E.K._{th}=m_{th}*c^2*\gamma_{th}-m_{th}*c^2[/math]

 

Helium-4:

[math]E.K._{he}=m_{he}*c^2*\gamma_{he}-m_{he}*c^2[/math]

 

And we know that their sum is 4.27 MeV:

 

[math]E.K._{th} + E.K._{he} = 4.27 MeV[/math]

 

Are you now able to calculate velocity (and thus kinetic energy) of newly created Thorium-234 and Helium-4 nucleus from what I presented above.. ?

 

Helium-4 taking the largest part of decay energy in form of kinetic energy will be passing though medium, and decelerating, and losing its kinetic energy, which will change to different kind of energy, f.e. heat.

 

Total energy of Uranium prior decay, is exactly the same as total energy of newly created Thorium-234 and Helium-4 plus their kinetic energies after decay.

[math]m_{u}c^2 = m_{th}c^2\gamma_{th} + m_{he}c^2\gamma_{he}[/math]

Edited January 1, 2017 by Sensei

[Capiert]

I don't think you can do it in reverse, because of the heat released,

or something else, I don't know yet.*

But please give me time, to check.

(It's a neat & fascinating task.)

 

3 typos distract me:

we divide (or multiply) by e (to get Joule from MeV, how can you do either?)

Helium.. passing through (a) medium..

e.g. heat (not f.e.?)

 

Please correct,

if you want I can erase my correction request (if possible) after you are done.

 

*The (1st) problem (=difficulty, obstacle) I see (coming) is the approximation KE~m*(v^2)/2,

(had) allowed simple rooting (giving no imaginaries);

(but) which is no longer possible

with KE=m*(((v^2)/2)+vi*vf), speed_difference v=vf-vi,

using initial_speed vi=c (NOT zero anymore), final_speed=vf,

for dealing with relativity to exchange (=eliminate, replace) your gammas accurately.

Edited January 1, 2017 by Capiert