Capiert

Thank you Strange,

I'm surprised to notice how distant physics is

from physical reality.

That's a paradox for me,

why get physical at all?

Or should I better ask,

are you dealing physical or not.

Still, it's informative.

--

Thank you Swansont for your sturdy words.

At least I (now) know how things will go, in advance.

I expected you would be glad to help (model),

but it seems I was wrong,

the intent is destructive instead.

I still appreciate the honest words.

 

I think you also mean,

there is only an inward force

for rotation?

 

But I'd also appreciate

if you could answer my question

about gravity (in general);

not just Newton's gravity

(based on the greek's pull),

if that is possible?

(I suspect you will agree

gravity in general is also a pull?)

 

--

 

Thank you Function for your noble efforts,

I'm sorry they didn't fullfill what I needed.

--

 

Cheers, Lord Antares.

Thanks for your answers Strange.

But I'm still not clear on this (modeling).

E.g. Either something is, or is not.

Is gravity a force, or NOT?

Assuming GR is newest (theory)

(then (now) gravity is not a force?)

You've got me there,

the mass in a rotating object

has non_linear acceleration,

which keeps rotating.

 

What I'm trying to figure out (in that question)

is the word net.

I assumed, centrifugal & centripetal accelerations balanced for circular motion.

If you're asking me, is the string tight in Newton's twirling water bucket pair experiment?

The answer is yes.

& that tightness seems (rather) constant for a given speed, but varies with speed.

 

But can rotation decrease (apparent) weight?

And where did I say it is otherwise?

..everything that spins exhibits the centripetal force which keeps the less massive objects spinning around the larger mass..

I said the centripetal force keeps the objects in orbit around the larger mass. Where did I give the reversed explanation?

In the word "spin" connected with centripetal, instead of centrifugal.

But some things fly apart, e.g. Roche limit,

when rotated too fast.

I suppose they are no longer spinning, if they have broken apart,

but centripetal force does NOT keep them together.

(=The smaller mass(ives) do NOT stay bound

to the more massives.

So your statement was wrong (for me, sometimes), implying (perhaps) things you did not notice.)

The centripetal law does NOT seem to dominate (for all cases).

The major (not minor)=general rule seems to be centrifugal, instead.

 

Also, no, according to general relativity, gravity isn't a force, so I'm not sure that you think you are on to.

? What? Please finish your statement.

This has been told for years.

 

Imagine you have two balls on a perfectly straight stripe of cloth held by the two edges. If you put your finger into the middle to dent the cloth, the balls will start moving towards each other. So this is not an attractive force as such, but you could say the net effect is attractive, if that makes sense to you.

 

 

Naturally, that makes sense to me.

Such examples are the best.

You are not saying anything new.

Say it this way:

We've been told for years

that centrifugal force is a force,

but now it's NOT!

Take it 1 step further.

Gravity is also NOT a force.

Well then, if that (force concept) was wrong before,

then is the "direction" of gravity still correct

(considering the force concept

was previously wrong)?

I'm only questioning the integrity of physics

considering it was wrong before.

That's nothing new.

#2 Function

 

1.

Sounds great!

We've got a "constant" G,

with a very exact value.

I haven't a clue where it came from
but (I guess that's the way Newton set up the formula),

it proves everything anybody wants to know

(except me).

 

2.

The acceleration we "exert?" on the earth

is "opposite" & equal (=the same).

(I suspect you mean "pressure" instead, when both earth & you (=we) touch?)

 

3.

"So surely the force is attracting"

seems to be based on Newton,

who "assumed" it's attractive.

That's NO proof for me,
that's a preference,

that's an assumption.

(It does NOT exclude 1 direction.)

#3 Lord Antares

 

I cannot quite follow you.

Everything that is spinning tends to centrifuge (outward).

It's the atomic bonds

that keep things together,

& pulls things in(ward) centripetally,

while they (=the (large) objects) spin.

A spinning centrifuge sends the more massive outwards, NOT inwards.

=The less massive remain near the middle (NOT outwards).

Your explanation sounds reversed (=backwards) to me (if the less massive go outwards).

 

E.g. Freeze drying: the (lighter) water surfaces as ice (is not the cause of spinning).

 

What I'm onto?
Wiki states gravity is NOT a force (said bluntly),
I found that in their centrifugal force page.
Relativity confirms that,

(but I'm not interested in its (=Einstein's) reason (=assumption) being: volume moving in a curve,

by "multiplying" distance with time (as a factor), instead of "dividing" (for speed, as a quotient);

because that's too (spastic=abstract) wasteful & imprecise, for my fantasy).

 

Claiming "foolishness" is NO proof for me,

that's NOT what I'm looking for.

 

I need something to bite on. e.g. tangible.

 

=At present with all these assumptions,

you're still guessing.

If yes please prove it.

All I observe is,

gravitational acceleration "g is only an acceleration".

That does NOT imply (only) "pull" to me.

I get hints, it's (=the acceleration is) the opposite (direction), instead.

Please clarify.

 

Wiki's "Centrifugal force", says,

"fictitious forces, like gravity, pulls.."

 

If they are NOT real forces,

then are they in the right direction, at all?

E.g. Gravity is NOT a force?

 

Wiki gives no reference to [1]:

"The centrifugal force

is an outward force (which I recognize as, only "acceleration", of mass)

apparent in a rotating reference frame;

it does NOT exist
when measurements are made
in an inertial frame of reference[1]."

I find the (Wiki) statements admirable, & courageous (against ridicule).
But I don't know what the status quo is. (..But can (sure) guess.)

 

 

So you are proposing something new. Including a unit vector with KE. You could write the term as sqrt(mE). And then sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation. That would work, because it's conservation of momentum. p = sqrt(2mE), and the sqrt(2) is divided out, being common to all terms. This is nothing new. IOW, it adds nothing, and is not a new insight.

 

What you had was

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

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

 

With a vector, that's still just applying conservation of momentum. (It's wrong by a factor of two, however. KE = p²/2m)

 

What you've succeeded at is a more complicated way of doing something that physics already does.

Is this following formula correct? (Yes?)

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

Is it not conservation of momentum? (No it is com?)

 

Aren't these 2 following equations equal?

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

(Yes?)

 

Why then should that #19 formula be wrong?

"Unit vector" is another name for "normalized vector".

Thank you Sensai for the link & (excellent) explaination.

But I don't understand the need for the redundant complexity

with the hatted i,j,k syntax.

I thought the direction was already implied

in x,y,z.

Why is that redone with hats (on or off, pun; & primes ' )?

 

E.g.

(I guess (the unit vector is an intermediate bridge (=tool))

for converting to circular coordinates,

using Pythagorus's rule r^2=(x^2)+(y^2), + etc.)

Defining (direction) twice, seems backwards to me. (E.g. not progress.)

I isolate bidirection, with polarity + or -, e.g. 2D x,y.

3D is definitely a challenge, to get the twist right; which Lorentz did 1904 (GR).

Maybe you can say some more to clear my confusion?

But that's not my problem.

I'm confused with the #19 "equation"

whether that formula (which now has the common factor 2 cancelled)

is valid

because (according to standard algebra rules)

if you asked me I would say yes;

but coming to this website forum

everybody has said no it (=that (algebra) formula) is wrong.

Can you understand how I feel?

I'm getting mixed up & confused.

Would somebody please clarify that for me?

Either confirm; or reject it (=the formula in question)

& show me why it's wrong

& will fail,

so I can clean up things on my side.

One error was treating KE as a vector when it isn't, as I explained.]

What is (the formal name for) KE called after you use the unit vector with it. "KE vector"? Please give it an appropriate name.

& what is the operation used? (Dot product? or multiply?)

 

The other is that KE = p²/2m, as I also explained, and which you acknowledged a few posts back ("Finally recognized. I'm very happy.")

 

Do you not recognize this mistake anymore?

My happyness was generally the summary for the whole post (icing on the cake, to top it off)

but with the detailed quotes (inserted later, that statement) seemed to refer only to the last quote,

instead of reinforced restatement (=getting better).

Generally you were recognizing the steps,

& the steps were going in a common direction,

climaxing finally with a correct formula,

instead of doubtful ones.

A day after I posted the formula #19

I thought I had (really) goofed!

But as the 2nd day rolled by

it occurred to me

that it (=that posted formula) still might work,

& so I then forgot about it along during the week.

That formula had a very complicated history,

& I had too separate myself from at least

half of its original development (stumbling on it)

after finding the approach with coe

shockingly failed on me,

& let me down.

I wasn't able to work on a recovery strategy

for more than a month, due to another job,

so I was in a terrible state (of) not knowing

what the score is.

So (later) I retried based on com instead

& got good results

except a few times

when coe seemed valid again

(but not com). Very puzzling.

Over the years I've noticed

about 5 errors

I can NOT explain

on other projects.

My indecision has gone on too long

so I thought I should post the formulas here

to get some help (with the math algebra (intended for the excel sheet))

& (try to) bring it (=the project) to an end, as finished.

(=Decided, & without errors.)

 

To answer your question:

 

I immediately recognized your #59 formulas

were missing the factor 2

because in #19 I had told our india friend (Sriman Dutta)

mom^2=2*m*E is pronounced "to me"

for quick identification (strategy).

That formula is only correct with the factor 2;

not without.

But that's not my problem.

I'm confused with the #19 "equation"

whether that formula (which now has the common factor 2 cancelled)

is valid

because (according to standard algebra rules)

if you asked me I would say yes;

but coming to this website forum

everybody has said no it (=that (algebra) formula) is wrong.

 

Can you understand how I feel?

I'm getting mixed up & confused.

The unit vector and the factor of 2 are separate issues. One is direction, the other is magnitude.

 

No, physics is doing what you intended to do. What you actually did was wrong.

Ok (=good, accepted). What is the actual mistake? The missing factor 2?

Meaning I'm not allowed to cancel the factor(s) 2?

I'm still confused there, needing orientation to help me.

Or do you mean other errors too? (Naturally, yes?)

Please show me to help me pinpoint them.

 

Thanks again.

Including a unit vector with KE.

You could write .. sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation.

That would work, because it's conservation of momentum.

p = sqrt(2mE),

and the sqrt(2) is divided out,

being common to all terms.

 

What you had was

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

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

 

With a vector, that's still just applying conservation of momentum. (It's wrong by a factor of two, however. KE = p²/2m)

Please clarify.

Above did you say (or mean) the formula above would work with a unit vector, because common 2 was divided out?

If it works, that nears saying it's right, it's not (completely) wrong, if it's working.?

 

What you've succeeded at is a more complicated way of doing something that physics already does.

If physics is already doing what I've done,

then it (=what I've "done" (in the formuls)) can't be wrong,

e.g. extracting KE from momentum.

(Because physics is suppose to be right.)

 

So we (=physics & me) are both either doing something right, or else something wrong.

(Because we get the "same" results.)

 

But then later you say "It's wrong by a factor of 2, however",

although the factor 2 seemed correctly cancelled.

 

So I'm still confused, what's right & wrong, & about that 2 (if it's ok=allowed to do cancelling so?).

 

Facit: What (or where) are the problems (=errors) if it is working?

 

Perhaps a few words (=sentences) about "unit" vector would help too.

I don't (really) know what is meant by "unit" (without an s for plural) other than e.g. kg, m, s units; or 1 (a single thing, e.g. radio).

 

Thanks in advance.

It (=?) was wrong. You can't just arbitrarily change what standard terminology means. It would be easier to recognize if you didn't misuse the language.

Please be more specific & show me the formula errors.

Your "It" (pronoun) does not pinpoint the individual errors, for me enough.

I'm still a little foggy (=vague) which formula (mine or yours) you were referring

in your previous post #58, #59

"It's wrong by a factor of 2, however."

Didn't mine have a factor 2, but yours that you said were mine did not?

Won't com also work with the factor 2?

Or doesn't it (= factor 2) belong there, at all?

I don't understand the distortion. (=I'm confused, please clarify.)

E.g.

Quoting my formula wrong, makes it wrong

(unless typos, mistakes, etc caused that).

But then you can say it (=the formula that is supposed to be mine, but isn't) is wrong?

I don't follow.

 

Surely there must be a misunderstanding.

 

It looks like #19 could be wrong?

My error, sorry.

& Thanks.

 

Correction for #47

Note: vf & vi must be swapped

to eliminate (Sensai's #46, relativity) gammas,

Thus

 

final speed vf=c (NOT vi).

 

The initial speed vi

is NOT zero

& is the isotopes' speed.

The kinetic energy is KE=m*v*va

for mass m

difference speed v=vf-vi

& average speed va=(vi+vf)/2.

 

(I still need more time to prepare.)

So you are proposing something new. Including a unit vector with KE. You could write the term as sqrt(mE). And then sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation. That would work, because it's conservation of momentum. p = sqrt(2mE), and the sqrt(2) is divided out, being common to all terms. This is nothing new. IOW, it adds nothing, and is not a new insight.

It's missing the factor 2. I don't know why you are presenting something wrong 1st

(unless it's to show something near (=very similar) which is very encouraging & helpful);

but I am happy you are finally recognizing what I am presenting.

I had to finally state it as new,

because you rejected it previously

(wrongly?) stating mine was wrong.

It's very difficult to get you guys

to recognize what you already know.

I have been (wrongly?) accused of not understanding

the material.

 

What you had was

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

That's also missing the factor 2.

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

dito

 

With a vector, that's still just applying conservation of momentum.

What was the thesis #1?

(It's wrong by a factor of two, however. KE = p²/2m)

Finally recognized. I'm very happy.

 

What you've succeeded at is a more complicated way of doing something that physics already does.

Thanks for recognizing.

Capiert you must first understand what's scalar and what's vector.

Force and momentum are vectors because they involve scalar multiplication of vectors (in force we have the multiplication of scalar mass and vector acceleration, and in momentum we have multiplication of scalar mass and vector velocity).

Thank you very much for explaining vectors vs scalars.

That (simple) is how I saw it before I came to this forum,

but some of their crew has made me insecure

with further details

about dot & cross products.

Perhaps you would like to continue.

 

energy is not a vector.

Is not a scalar, multiplied by a scalar, a scalar?

& is not a scalar, multiplied by a vector, a vector?

With that I have every indication that KE is a vector,

unless you can show me otherwise.

 

Also you are not clear about squaring and rooting. Any integer squared "gives" a positive integer.

We don't have to accept every gift given. E.g. Trojan horse.

I distinguish between what things (really) are

& what they "can" become (virtual, equivalence), but do NOT have to be.

 

But, on extracting the square root we "put"

Oh! We make an exception, to try & fix the damage done.?

a plus-minus sign before the number whose square we have rooted. And if we extract the square root of a negative integer we "call" that an "imaginary" number. [math]\sqrt{-1}=i[/math]

That sounds like a lot of exceptions to me.

I guess nobody complains if it (still) works.

 

Energy does not have polarity (I don't know what you mean by polarity---I know polarity of magnets, but not this).

I think some are forgetting my goal

is to put this (formula) into an excel sheet

& that I am dealing with algebra.

A number line can be (for (all) practical purposes)

treated as a vector.

It has (both) minus & positive (sign) polarity,

you say "directions".

E.g. a battery is also labled positive & negative

with the same symbols "+" & "-"

for the A, B, (baby) C, AA, cells & 9 V block cells (etc).

"Swap poles" or "swap polarity" is a common expresion.

Please make a suggestion,

I'd like to explain things to kids too.

 

Bi_mention (etymology), you say "dimension",

& mention (=talk about) that it is a pair.

 

But maybe I have selected the wrong word "polarity", e.g.

the ends of a pole (is 3D thick, as a real physical example,

for common speach

& explaining)

instead of (a virtual zero thickness) number line.

An object has 8 Joules of kinetic energy. What direction is it moving? Where is the "polarity" information?

If I gave you that task (instead)

(with my syntax technique)

then I (& you) could say (for a 1D number line)

"to the right".

But since you are giving me the task

it can NOT be said

whether it is going

to the right or the left,

it's ambiguous

because you are not observing polarity so,

& have thrown away that (option, possiblity) ability.

 

Is that a satisfactory answer?

KE is always a positive value.

Yes, but you do NOT use that "+" symbol (for 360 degrees),

as a compromise indicating 2 crossed negative symbols i.e.

1 horizontal & the other vertical.

Instead, you drop it completely.

 

You've seen how the rooting operation was cancelled out

with the squared operation

in #45.

 

Can you tell me why that (double operation, as a "do & undo" should not work, or) is not allowed?

 

At present what I've shown you in #49 works.

 

Please show me a(n algebra) math law broken.

 

There is no direction information in it.

That's not true.

The speed has polarity.

 

I've only made polarity observations

& reported them.

 

No one will prevent you from closing your eyes, if you want.

No one will force you to observe.

 

KE2~m2*((v2)^2)/2=2 kg*((-1 m/s)^2)/2=(-)*(-)*1 kg*((m/s)^2).

Have I forgot anything in that rigor?

--1 kg*((m/s)^2).

Negative speeds have 2 negatives in their KE.

"You" can keep them or get rid of them.

I'm only telling you they are there,

something you already know.

If you get rid of them

then you turn the equation

into an ambiguous thing

& can no longer extract the original polarity.

It's thus your choice,

what you want

& want to do.

Not mine.

You can "assume" they are the same,

but I don't.

Thus I can not (completely) agree with you,

when you say KE is always positive.

 

Basically you're making stuff up and wondering why it's not working.

Newton made things up too. I.e.

They were "new" equations meaning nobody else knew about them.

The rest is an attempt to explain how the things work. (Self defining.)

But I think you are guessing when you say that "wondering why"

because with your help

I have narrowed in on the problem.

(But I still don't know why, the algebra rule is not allowed.

It's in a different place not yet mentioned.)

When you square a negative number, the result is a positive number.

When you root a squared number what do yo get?

When you square a rooted number,

you get a positive.

The other (way around) is NOT guaranteed.

There is a sequence priority.

 

(E.g. The egg must come before the chicken,

otherwise no chicken, e.g. minus or error.)

When "I" (do a) square (operation) & then reverse that operation with rooting,

then I expect (to get) the original (as NOT modified).

That is reversable math. (Otherwise that is NOT reversible math, by definition.)

For each operation, an anti_operation should (exist &) be available.

E.g.

Addition vs subtraction

multiplication vs division

squaring vs rooting.

 

Am I demanding too much (rigor)?

 

"You" (are perhaps a little more careless in that respect, if I may say, &) generalize & make the approximation:

that a squared minus is positive.

 

I however distinguish

between subtraction

vs negative polarity.

For me they are NOT identical.

Negative polarity is equivalent

to 180 degrees rotated, into the opposite direction.

The root of negative polarity is 90 degrees.

Polarities are multiplied to add angles.

(-1)*(-1)=(--1)

180 degrees+180 degrees=360 degrees.

360 degrees is (to) the (angle) position as 0 degrees (starting place),

but they do NOT mean (exactly) the same thing.

 

Does that help explain that technique?

 

The polarity (symbolic for angle) is treated like a unit

so it is multiplied by the quantity (=amount)

like you do with length, for meter e.g. 1 m.

I could have written it (=negative polarity) so

(1-)*(1-)=1*1*(-)*(-).

It is simply another syntax (unit).

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.

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)

 

 

 

 

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.

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.

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?

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.)

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?

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?