[Professor Mega-Mind (OP)]

The basic idea of energy content depending on mass and speed is correct .
P.

[Kryptid]

The basic idea of energy content depending on mass and speed is correct .
P.

Stating it as mass multiplied by velocity, however, is incorrect and should therefore never be stated in that way.

[Bored chemist]

Note - I consider the "vacate" talk a form of running away .

We didn't so much ask you to g. we asked you to come back... when you had time to post something useful.

I stand by that fundamental logic ,

Fine. Go and apply it to the question I keep asking in the other thread.

I stated for the easy understanding of the uninformed !

Is that self referential, or is there some group who you wish to remain uninformed or what?

[alancalverd]

Momentum is a vector. Energy is not. Momentum is conserved.You need to alter the momentum of an object to make it move. You can't do this by jumping around inside it.

[Bored chemist]

..............Drifting in space .
Let me postulate a possible exception , the man in a box (drifting in space) .
Picture , if you will , a large steel box , floating in space .  In the middle stands an astronaut .  In his right hand he holds a 20lb sandbag , in his left a 20lb steel ball .  When his timer dings , he throws both objects at the opposing walls .  The sandbag hits the right wall , goes "splat" , deposits %55 of it's kinetic energy into the wall , %45 into friction heat , then slowly drifts back to the 'stronut .  The steel ball bounces off the left wall , deposit-ing %2 of it's kinetic energy into the wall , and reversing it's flight with %98 of it's initial kinetic energy intact .  It then strikes a sandbag mounted on the right wall , depositing %55 of that %98 KE into the right wall .  It then drifts back to the "stronut" . Of all the energy he mustered for his simultaneous throws , only %1 went into pushing the left wall , ~%55 went into pushing the right wall , the rest became omni-directional waste heat .  The result is that the box slowly drifts to the right , without ejecting any reaction mass .  Repeating the process would result in a small rate of acceleration .  This would definitely qualify as a reactionless drive .
A repetitive motion system , losing some of it's kinetic energy uni-directionally , experiences an effective "push" from the remaining  kinetic energy .  Energy conversion is the key .  E=MC2 is proven out .  Long live the "Epstein Drive" !
P.M.

OK, in the other thread you can now see the calculation which shows that an elastic collision transfers more  momentum than an inelastic one.

So this idea of the OP's is dead because it relies on "Of all the energy he mustered for his simultaneous throws , only %1 went into pushing the left wall , ~%55 went into pushing the right wall , " when the push against the left wall is actually stronger than the push against the right.

[Professor Mega-Mind (OP)]

Aaargh , my diaghram's still sore from the last yuck-fest !  Incredible how 1% becomes greater than 55% with your math , think I'll stick with 2 + 2 = 4 .  It works for me !
P.

[Bored chemist]

Aaargh , my diaghram's still sore from the last yuck-fest !  Incredible how 1% becomes greater than 55% with your math , think I'll stick with 2 + 2 = 4 .  It works for me !
P.

Thanks for clarifying that you find actual physics "Incredible".
That's why you get stuff utterly wrong.

You really need to either find out the difference between momentum and energy or stop posting about things where that difference matters.

(BTW, have you noticed that none of the grown-ups agrees with you?)

[Professor Mega-Mind (OP)]

The established Massive Wall effects are not my creation .  I'm thinking you are confusing a fast reflection for a heavy one .
P.

[Bored chemist]

The established Massive Wall effects are not my creation

Nobody said they were.
It's just that you don't know how to use them correctly.

I'm thinking you are confusing a fast reflection for a heavy one .

There is no difference.

[Professor Mega-Mind (OP)]

Doode ,
Your own formulas tell you that very hard ( glass ) objects rebound from a massive-plate with ~90% of their initial kinetic energy !  Trying to confuse , with scrambled-up exponential equations , does not change the fundamental kinetic energy imbalance !  Let’s take a glass ball : K.E. after rebound=90%.
Now Bean Bag : .002 freakin' % !
Even a daft wank_r can tell that the beanbag put a helluva hit on the wall , while the glass ball reflected back with little energy exchange !  BY THE BOOK , the 55%/45% split in K.E. means that 55 was transferred to the Massive Plate when they "bonded" , 45 became "molecular" momentum  ( heat ) , in that process .  Said heat is then radiated to the universe as EMR (photons) , with neglegible effect upon the momentum of anything ! 
 It's time to stop screaming " I do formulas better than you ! " , and acknowledge the creative-mechanical genius which birthed this triumph ! 
OK , 90% humbles .002% hugely !
P.M.

[Bored chemist]

rying to confuse , with scrambled-up exponential equations ,

None of the equations contains any exponential function.

Your own formulas tell you that very hard ( glass ) objects rebound from a massive-plate

They don't say anything about hardness.

ven a daft wank_r can tell that the beanbag put a helluva hit on the wall ,

Yes, but a clever one knows that , if the ball and the beanbag started with the same momentum (same mass, same speed) te ball transfers more momentum to the wall than the bag does.

I realise you can't understand this but it's a standard result in physics (and you even know about and accept it for photons), so you really should learn to accept it.

Transfer of momentum and transfer of heat are independent processes, so all this stuff is just dross.
.

BY THE BOOK , the 55%/45% split in K.E. means that 55 was transferred to the Massive Plate when they "bonded" , 45 became "molecular" momentum  ( heat ) , in that process .  Said heat is then radiated to the universe as EMR (photons) , with neglegible effect upon the momentum of anything ! 
 It's time to stop screaming " I do formulas better than you ! " , and acknowledge the creative-mechanical genius which birthed this triumph ! 
OK , 90% humbles .002% hugely !
P.M.

[Professor Mega-Mind (OP)]

Howdy ,
 P=m•v also prove me out !  It says that if the hard ( glass , steel , etc. ) ball bounces back at 90% of the striking speed , then it retains 90% of it's starting MOMENTUM !  That only leaves 10% to be transferred to the Massive Plate !  The soft beanbag,  however , transfers ~70% of it's momentum to the Massive Plate .  This is because it has a much longer contact time , spread out over a much larger contact area .  This means it gives a much more extensive push to the wall overall , transferring much more momentum and kinetic energy in the process . 
 You need to drop the photon physics for this , and use a Table of Coefficient of Restitution for different materials .  That will back up my Reply#49 , and highlight the difference between lightwave reflection, and bulk matter rebound .  Right now , all I'm hearing is a Scottie dog yipping out a bunch of put-downs, and misrepresentations !  Stop misleading the readers , and use honest reasoning and basic math to make your points ! Fingers stuck in ears , while head shakes violently , and mouth spews nah-nah-nah-nah , does not impress anyone .  Neither does pretending that you've somehow "won" your argument , it just looks "jejeune" !
P.M.

[Kryptid]

It says that if the hard ( glass , steel , etc. ) ball bounces back at 90% of the striking speed , then it retains 90% of it's starting MOMENTUM !

90% of its starting speed? I thought you said that it bounces back with 90% of its kinetic energy? Velocity and kinetic energy are not the same thing nor do they change at the same rate.

[Bored chemist]

P=m•v also prove me out !  It says that if the hard ( glass , steel , etc. ) ball bounces back at 90% of the striking speed , then it retains 90% of it's starting MOMENTUM !  That only leaves 10% to be transferred to the Massive Plate !

No.
As I explained, momentum is a vector quantity like velocity.
If a ball is rolling along a table from right to left I need to push on it (leftward) to stop it.
When I stop it, it has zero momentum and it has transferred its original momentum to me (and, unless i'm on a skateboard or something the momentum is also transferred through me to the world).
And, if I want it to reverse its original movement, I need to push it again (leftward) just as hard as I had to to stop it.

So, to get it to change direction I have to add twice as much momentum as I did to merely stop it.

Well, in the same way, bouncing off a big heavy thing also has to supply twice as much momentum as hitting the heavy thing and stopping.
The change of momentum is twice the momentum that the ball originally had.
And, since momentum is conserved, if the "wall" or whatever bounces the ball back, the change in its momentum (like that of the ball) is twice that of the incoming ball.

How did you come to the conclusion that science has been getting it wrong all this time?

[Bored chemist]

Right now , all I'm hearing is a Scottie dog yipping out a bunch of put-downs, and misrepresentations !  Stop misleading the readers , and use honest reasoning and basic math to make your points ! Fingers stuck in ears , while head shakes violently , and mouth spews nah-nah-nah-nah , does not impress anyone .  Neither does pretending that you've somehow "won" your argument , it just looks "jejeune" !

Are you looking in the mirror?

[Professor Mega-Mind (OP)]

Good one Scooby !

[Bored chemist]

Good one Scooby !

Thanks for clarifying your ability to hold a rational discussion.

[Bored chemist]

Good one Scooby !

Have you forgotten that you are expected to respond to valid criticism of your posts, or are you just trolling?

[Professor Mega-Mind (OP)]

I no work hear , I jest play hea-yar !
Besides , it's tit-for-tat I believe .
.....................Argument .
We are at an impasse because you erroneously assume that you must impart energy in order to reverse a glass ball's direction . That is false; a Massive Steel Wall  will reflect the ball very well , without any energy input at all .  Momentum vector may reverse , but kinetic energy content will remain almost the same as before rebound . Your refusal to acknowledge the C.of R. tables is both disturbing , and disheartening .  "Valid Criticisms" would encompass the accepted scientific and technical inter-relationships and values , and involve them with full accuracy and  context .  Omission and mis-representation only convince me that that conversation is not worth pursuing .  Now then , convincing ARITHMETIC if you please !
P.M.

[Bored chemist]

We are at an impasse because you erroneously assume that you must impart energy in order to reverse a glass ball's direction .

I made no such assumption.
You Are wrong- as usual.
Would you like to try again?

Your refusal to acknowledge the C.of R. tables is both disturbing , and disheartening . 

It's not that I don't acknowledge the table of them.
The fact is that the values do not actually matter much.
Materials exist with CoR near zero (such as porridge or velcro); and materials exist with CoR near 1 (such as hard steel or glass).
I even used examples of such materials in my calculations.
So, the data in the table was never in question.

So, once again, the problem is not with me.
The problem is that you don't understand what the issue is. You think the problem is somehow going to change if I acknowledge the existence of a table of data.
But I never questioned or doubted that tabel, did I?

Go on, show where I said that the table was wrong.


I pointed out that you do not understand the data, but that's a separate issue.