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/sci/ - Science & Math


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2373293 No.2373293 [Reply] [Original]

sup /sci/

I'm a scientist out of college for a while now, but I'm not a physicist, so my understanding of mechanics is only so-so. Anyway, the Newton's cradle has always stumped me, and the fact that my first physics teacher way back in high school couldn't answer my questions really didn't help. Anyway, here goes.

Two obvious requirements are conservation of momentum and conservation of energy. Simple enough. If we have a two-ball cradle (equal masses for the two balls), and pull back one ball, we can calculate its velocity right before the collision. That leaves us with two unknowns--the two velocities after the collision--and with two equations (CM and CE), we can easily show the familiar result: the second ball gains the velocity of the first, and the first comes to rest.

Well, what about a cradle with three or more balls? Pulling back one ball, we can again calculate its velocity just prior to the collision, but now we're left with n > 2 unknowns. Because we still only have two equations, the system is underspecified.

I would think that the principle of least action should give us some simple, Newtonian-mechanical way of explaining why only one ball flies out (rather than, say, two forward and one backward), or if we pull two balls back, why two fly out, and so forth.

Anybody with a better background in physics than I have the answer? in b4 trolls

>> No.2373310

Assume elastic collisions, treat each pair separately.

>> No.2373315

>>2373310
A little more explanation...?

>> No.2373329

>>2373315
The amount of force put in by swinging the first ball is the amount of force that will propel the last ball. If you swing 2 balls into it, then there is only enough force to propel 2 balls out the other side at the same rate of swinging.

>> No.2373341

>>2373329
The amount of force? I don't think you know what you're talking about.

>> No.2373358

Because it is only with one ball that satisfies both conservation of energy and conservation of momentum. for instance if there are 3 balls 1kg each, with one going say 10 m/s, it would satisfy conservation of momentum that 2 of them go 5 m/s. however it would not satisfy conservation of energy, .5*1*(10^2) before, .5*2*(5^2). notice how they don't equal? you can prove it mathematically that it only works with one ball moving, but that is the general explanation of why

>> No.2373365

>>2373341
Not at all.

>> No.2373368

you are a scientist? You have a Ph.D in a science field? You actively participate in original research? You have published in a peer-reviewed journal???

>> No.2373451

>>2373315
He's saying to imagine there's a small space between the balls so that only two collide at a time. Then treat each collision between two balls using conservation of momentum and energy as you already described.

>> No.2373485

>>2373358
Please don't reply if you haven't actually stopped to work out the equations. I assure you, three balls is underspecified with just conservation laws.

>>2373368
Yes, yes, yes, yes.

>>2373451
What if there are five balls and I pull back two? What is the "pair" when balls 1, 2, and 3 collide?

>> No.2373516

>>2373485
First ball 2 collides with ball 3, bringing #2 to a stop and setting #3 moving. Then #1 collides with #2 and #3 collides with #4, and so on...

>> No.2373538

>>2373516
Again, please don't comment if you haven't actually written out the conservation equations. If you do, you'll see why I am posting.

>> No.2373563

>>2373538
Two balls colliding isn't underspecified.

>> No.2373720

>>2373563
All right, smart guy. Let's go back to the example of five balls, and we pull two.

When the two balls hit the three that are at rest, we have to include the "hidden" collision of the first ball with the second ball. This just has the effect of giving the second ball twice its original velocity.

Now it collides with the third ball, and transfers all of its velocity (twice the original) to it.

The third then does the same to the fourth, and the fourth to the fifth, and oh snap! The fifth emerges with twice the velocity of the first two. This is not what happens.

Don't treat the two initial balls separately, you say? Fine. Treat them as one big ball with twice the mass, and CM tells us that ball three leaves with twice the original velocity, but CE tells us that ball three leaves with sqrt(2) times the original velocity.

Thanks for nothing. Anybody want to actually tackle this problem by writing out the equations, instead of just assuming that you know how to solve it?

>> No.2373753

-sage design

>> No.2373756

>>2373720
> When the two balls hit the three that are at rest, we have to include the "hidden" collision of the first ball with the second ball. This just has the effect of giving the second ball twice its original velocity.

No, that's wrong. #1 and #2 are initially moving at the same velocity. They don't collide, and if they did, the result would not be what you stated. The first collision would be between balls #2 and #3.

The pair approach gets the expected result (you can see this easily because there are always two balls moving at v and three at 0), although I'm not sure if this is a valid way of working out the problem. But there might be some way of justifying it. Let me think about that...

>> No.2373762

>>2373538

op, the conservation equations were designed so that you could take pleasure in proving them to be innacurate, thus aquiring sage status by causing a massive paradigm shift in the scientific community at which point you will likely become aware of a new scientific community who were previously not heard from.

>> No.2373775

>>2373293
>principle of least action

Yes, you can use it to calculate everything about the system.

What is your problem exactly?

>> No.2373790

>>2373293
Your problem is that you assume conservations, which isn't first princple at all.

Start off with least action, then derive conserved quantites. You can also derive the equations of motion for each ball using the least action.

It all works out at the end.

>> No.2373793

I don't know if you can actually treat this as one simultaneous collision instead of a series of collisions. When pulling multiple balls back, my intuition would say the one of the balls pulled back transfers its energy to the ball on the other rend, then the second ball pulled back does the same thing a few milliseconds later, etc. It seems like a simultaneous collision to us, but they should actually be distinctly separate collisions. Somebody more familiar with solid mechanics could probably give a better explanation.

>> No.2373795

>>2373775
Show, using equations, how the principle of least action exactly specifies the following system:

Newton's cradle, five balls of equal mass, two are pulled back to some height and released.

In other words, show (using equations) how the principle of least action requires that exactly two balls emerge from the collision, and not some other number (e.g. with some moving backward).

>> No.2373806

>>2373790
You know, everyone says things like this, but no one (Google it, you'll be surprised) actually bothers to demonstrate that it is so. If it's so simple that you all can brush it off with a single sentence, then demonstrate it with equations.

>> No.2373811
File: 31 KB, 265x350, 70870871094641.jpg [View same] [iqdb] [saucenao] [google]
2373811

>>2373762
>implying that those shitty conserved quantities work in general

lrntoNoether

>> No.2373815

>>2373793
I actually think this is a rather difficult problem and involves the propagation of shock waves through the "stationary" balls. I'm looking for someone to either give me the simple, correct answer, or to explain why the answer is too complicated to present here.

>> No.2373816

If you want to get sophisticated, what you can do is model the interaction between balls as a springlike force.

>> No.2373832

>>2373816
Tell me more...

>> No.2373843

The only way that you can conserve both momentum and energy is to conserve both mass and velocity.

IOW:
m1.v1 = m2.v2
(1/2).m1.v1^2 = (1/2).m2.v2^2

The only solution is m1=m2, v1=v2. And it's velocity that's preserved, not speed. If all balls have the same mass, you end up with exactly one ball moving at the same velocity as the original. This can only be the ball at the other end.

>> No.2373846

>>2373832
There isn't any more until I or you or someone else sits down with a piece of paper and tries to work it out this way.

>> No.2373856

>>2373843
Thanks, gen-ass. How about you read the OP? We're talking about systems with three or more balls: two conservation equations, three unknown velocities after the collision. Underspecified.

>> No.2373859
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2373859

>>2373806
>actually bothers to demonstrate that it is so

Why would they? The methodogy and actual physics involved is fucking trivial. It is first year physics, and not a novel way of teaching. It is not an "example" problem.

The calculation for the equations could be considered tedious. No one wants to calculate that shit, unless they really have to.

If you try calculating it, I will guide you. But I aint doing the whole fucking thing for you. Start by writing your energies and constraints.

>> No.2373864

In this thread; COMPLEXITY.

Read up on it if you're into physics.
People who assume "equations are the language of the Universe" haven't seen the language of the Universe.

>> No.2373870
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2373870

>>2373856
>Forgets to take into account the constraints of the system

>Thinks it underspecified

>> No.2373878

>>2373293
Does OP still need a simple, intuitive, and sound explanation for this?

>> No.2373895

>>2373846
I'm going to take a look at it once I get back from eating.

>> No.2373896

>>2373293
You have constraints on your system OP.
It isn't just conservation of mometum and energy. You have additional relationships between the Balls. You really need to do the Lagragian Formalism to understand.

http://en.wikipedia.org/wiki/Lagrange_multiplier#Handling_multiple_constraints

>> No.2373899

>>2373878
Seeing as no one has provided one, yes. No dismissive, one-sentence answers, please. I can come up with those on my own.

>> No.2373910

>>2373896
I'm familiar with the Lagrangian formalism and that's why I brought up the principle of least action, but I don't see what the other constraints would be here. Again, I think this problem is actually rather difficult to explain and as I mentioned in my OP, I have a serious problem with high school physics teachers using Newton's cradle as a demonstration. A demonstration of what, I ask?

>> No.2373927

>>2373293


If it helps, OP, consider what would happen if Newton's cradle just had one ball. That ball would, once it was started, swing back and forth, right?

And if something is in its way, such as your hand, the object would hit your hand with a force, transfering its velocity to your hand.

Since your hand isn't attached to the beam, it would go sailing off, and the ball would be at rest forever more, having lost all of its energy to your hand.

But, since the ball hits something that IS attached to the crossbeam, that object is forced to return, and when it does, it has lost very little of the energy with which it was struck, and so strikes the ball next to it.

Next ball strikes the next and so forth until a ball which CAN move (the one of the end) swings off, falls back, and the whole cycle starts over again.

It works the same way with two, because the amount of energy is enough to move two balls.

Follow?

>> No.2373957

>>2373856
>>2373870
Now that I re-read the OP's question, it seems like he's asking something other than what I thought he was. I thought he was wondering how to work out 3 balls pulled back = 3 balls flying, but apparently not. The balls in the middle are necessarily constrained to be stationary because the collisions are perfectly elastic. Incoming ball hits middle ball, which then hits the ball at the other end. For conservation of energy and momentum to apply, the middle ball HAS to be stationary after the collision. This means we can pretty much ignore the middle ball(s).

>> No.2373966

>>2373927
worthless answer

To prove to 90% of the responders here that the conservation equations are totally insufficient to solve a three-ball problem, here are actual numbers:

Three balls, pull only one back. Assume equal masses, and of course, elastic collisions. Assume mass of 1 and initial velocity of 3.

Conservation of momentum:

1*3 + 1*0 + 1*0 = 1*v1 + 1*v2 + 1*v3
3 = v1 + v2 + v3

Conservation of energy:

0.5*1*3^2 + 0.5*1*0^2 + 0.5*1*0^2 = 0.5*1*v1^2 + 0.5*1*v2^2 + 0.5*1*v3^2
4.5 = 0.5*(v1^2 + v2^2 + v3^2)

Everyone knows that the physical answer is v1 = 0, v2 = 0, v3 = 3, but there are clearly infinitely many other solutions. How about v1 = -1, v2 = 2.707, v3 = 1.293?

>> No.2373973

>>2373966
This seems to be the case of a mathematician trying to solve a physics problem. You can't think like a mathematician to do physics. You can think like a mathematician once you've got the equations, sure, but you need to think conceptually first.

>> No.2373979

>>2373973
Care to enlighten us?

>> No.2373982

>>2373910

agreed. OP try to build one of these. what you will end up with will match your conservation equations more to your liking. newton has simply pwned conservation with his design. and very eloquently if i may say so.

>> No.2373988

>>2373982
in during trolls

>> No.2373991

>>2373957

lol, u r silly =P

>> No.2374001

>>2373927
Clearly the same amount of energy is enough to move two balls, but it's also enough to move one ball \sqrt{2} times as fast. Why doesn't that happen?

Wikipedia says it's because the shock wave in the balls doesn't disperse, but I know nothing about that kind of physics.

>> No.2374003

I think we can all agree that no one here can demonstrate the principle behind a Newton's cradle. Sigh.

>> No.2374004

>>2373979
Stop trying to solve for unknowns, and just look at why the balls are moving. They're moving because they're getting hit, to put it roughly. There's just enough energy transfered in the impact to move one ball from rest to the peak at which the other ball was lifted.

More balls don't move to half of that resting height because the impact is an all-or-nothing sort of an operation, in this instance.

>> No.2374011

>>2374004
So why do two balls come out when you swing two out of five? Why not one ball moving really fast? Seems like that would be more "all" than "nothing." P.S. Your answer is worthless. 0/10

>> No.2374016

>>2374004
>impact is an all-or-nothing sort of an operation

new physics discovered
"it works because it does"

>> No.2374028

this thread is so full of fail

if you don't know, don't act like you do

>> No.2374030
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2374030

>>2373966
Your equations don't model Newtons Cradel. Your equations model a fictionless point particle that breaks appart.

Your equations treat the interaction as it all occurs at one point! When in fact each ball collision have a different point of contact. DURRR.......! Additionally, your equations also don't account for the swiming or gravity.

>> No.2374034

>>2374030
>the swiming or gravity

>> No.2374035
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2374035

>>2374030

>> No.2374036

>>2374011
The energy of the leading ball, when there is another 'helping' to push it down is equal to that of lifting two balls.

That same force travels through the line of balls until there is enough of a decrease in the 'resistance' to the flow of the energy, so to speak, that it can lift balls again. As the 'flow' of this energy is nearly lossless, when two balls go down on one side, two balls go up on the other.

>> No.2374038

>>2374036
not a scientific argument

>> No.2374045

>>2374036
That's f'ing stupid. If there's some magical "flow" of energy, how come it can push out two balls at exactly the original velocity, but not three balls at a slower velocity? Are you postulating that there is a work function for emitting balls from a line of balls?

>> No.2374046
File: 15 KB, 400x320, facepalm2.jpg [View same] [iqdb] [saucenao] [google]
2374046

>>2374030
>swinging

fixed

>> No.2374047

>>2374038
What drives you to this conclusion?

>> No.2374051

>>2374046
the real problem is that swinging and gravity have nothing to do with the instantaneous velocities of the balls the moment after the collision

>> No.2374055

>>2374047
>'resistance' to the flow of the energy

>> No.2374058
File: 10 KB, 300x300, Telescope[1].jpg [View same] [iqdb] [saucenao] [google]
2374058

>>2374055
inertia is a property of matter

>> No.2374062

>>2373966
v1 = -1, v2 = 2.707, v3 = 1.293 wouldn't work because of the constraint that v3 >= v2 >= v1 in the final state. But one that would work is v1 = -1, v2 = 2, v3 = 2. Intuitively we can see that this from the fact that this is the solution when balls 2 and 3 are glued together.

>> No.2374065

>>2374045
Theoretically, it could push three balls out at slower velocities if two balls were attached to each other , and then had one more on the end.

The issue here is that the energy wants to continue in the direction it is traveling in, in this case sideways, and will continue to do so until it cannot travel in that direction any longer, as, in this case, the vessels carrying it have run out of links in the chain and must convert this kinetic energy back into potential.

>> No.2374067

>>2374062
That's a trivial point. Just change the indices:

v1 = -1, v3 = 2.707, v2 = 1.293

no problo

>> No.2374071

>>2374065
i really hope you're not a scientist

>> No.2374078

>>2374065
energy wants to flow in a direction? energy flows in a direction? energy flows?

>> No.2374081

>>2374071
The sad part? 95th percentile in scientific reasoning on the ACT, 90th and up percentile in all my physics classes to date, et cetera.

... I'm just awful at explaining the concepts.

>> No.2374084

necessary information to describe this system with 100% accuracy?

GO!

>> No.2374085

>>2374078
It's called a Watt.

>> No.2374086

>The issue here is that the energy wants to continue in the direction it is traveling in

I'm just going to toss my hat in with the other people who have pointed out that what's taking place is a series of collisions, not one simultaneous collision. As long as you're doing perfectly elastic collisions, there's a physical constraint on the middle balls to stay at 0 velocity.

>> No.2374088

>>2374081
I earned a perfect score on the SAT, and that's not a lie. You're not impressing me, and frankly, I don't care if you're in the bottom 10th percentile--if you're going to "explain" a physical phenomenon without making any attempt to use scientific language, you're not explaining anything.

So far, you've basically said that energy is flowing, energy likes to keep flowing in one direction, and when it runs out of balls to keep flowing into, it stops flowing. So...it stops when it does, because that's what it wants to do? How insightful.

>> No.2374092

>>2374086
Isn't gravity mostly taking care of that part?

>> No.2374096
File: 231 KB, 850x645, wtfamIdrinking.png [View same] [iqdb] [saucenao] [google]
2374096

>>2374086
Forgot my image

>>2374081
You're not right. You're not even wrong. Energy is a scalar, not vector.

>>2374085
gtfo

>> No.2374097

>>2374086
That doesn't answer why, when two out of five balls are swung, the fourth ball, which is by all accounts a "middle" ball, swings away with the fifth.

If it's as simple as using conservation equations on individual pairs of balls colliding, then it should be simple to type up here.

>> No.2374099

>>2374088
Then explain it better, and stop bashing my attempts. With so many apparently well-qualified people here, we should be able to answer this question. We are, after all, Anon.

>> No.2374100

>>2373846
>>2373832
>>2373816
So far I've worked out the two-ball case (with ball 2 at rest initially) using a springlike force between the balls. We already know what the result is, but one new result is the time it takes to transfer momentum from one ball to the next. For a force of <span class="math">F = k((x_1+r) - (x_2 - r))[/spoiler] I get a transfer time of <span class="math">\pi \sqrt{m/2k}[/spoiler].

I think from this one can argue that considering forces between the balls is enough, and one doesn't have to worry about the details of a shock wave passing through the balls. The reasoning is that the force between two halves of the same ball has a much higher k because it's compressing the middle of the ball. Whereas the force between two balls mostly compresses the region where the balls come into contact, which has a much smaller cross section and thus a much smaller k. This means that momentum transfer within a ball happens over a much shorter time scale than momentum transfer between balls, and can be safely ignored.

>> No.2374104
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2374104

>>2374051
>>2374051
FOR THE MILLIONTH FUCKING TIME!
YOU NEED ADDITIONAL CONSTRAINTS!

You will need to take in account the swinging and gravity in modeling the whole system. If you just model the interaction you need to to into account additional constraints as well. It's even in the fucking WIKI!

Under the further physics section:
http://en.wikipedia.org/wiki/Newton's_cradle

"There must therefore be an additional constraint at work. This additional condition relates to a shock wave, which has to propagate through the line of balls without dispersion."

CONSTRAINTS! CONSTRAINTS! CONSTRAINTS! CONSTRAINTS!
YOU NEED CONSTRAINTS FOR MOST PHYSICS! NOT JUST CONSERVATION OF ENERGY AND MOMENTUM!

>> No.2374107

i would say just building one and observing it means you are now capable of describing it with 100% accuracy. then you can play with the materials used, the length of the strings, and size of the balls to get whatever results you wanted. for instance, if you wanted ball 1 to collide with the other balls, causing them to all move slightly together while ball one bounced back off, i'm sure that could be done. you could get an infinite number of slightly different responses, but the mind bending result of newton's cradle would take some tinkering. i suppose decent math skill could reduce the number of attempts, guiding you to the balance you're looking for at an accelerated rate.

>> No.2374110

>>2374100
I appreciate your real attempt at solving this problem using math. So, what does the spring model tell us about the five balls, pull two back, case?

>> No.2374115

>>2374104
Welcome to the first post, asshole. Your answer to the OP's question ("What is the extra equation?") is "YOU NEED ANOTHER EQUATION". Thanks.

>> No.2374116

>>2374110
Ha, first I'm going to try the three balls, one ball pulled back case.

>> No.2374119

you know, with the number of worthless answers in this thread, you'd think this was /b/

>> No.2374123

Hahahaha, I've seen the explanation to this. I remember it now. My mad teacher in High School had an explanation, it was beautiful, as Science is meant to be.

The solution lies NOT in the eq. but in the concepts of how the balls move. The ones in the middle crash back and forth with EQUAL velocities. Hence for conservation of energy, it is the equal amount of balls getting moved.

It's brilliant. Simply brilliant.

>> No.2374127
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2374127

>>2374123
BRILLIANT

>> No.2374135

>>2374116
Post your answer to wikipedia, if you come up with anything decent. The article sucks right now. It basically admits that it doesn't know and mumbles something about shock waves.

>> No.2374136

Its all gravity carrying the balls as they hit one another. You could have 20 balls and It'd still be the same. I'm in physics next year and even I can figure this shit out....

>> No.2374137

>>2373293
Good luck, OP. I've done what I can at two in the morning.

>> No.2374138

>>2373966
We should consider also that the physical answer "everyone knows" is wrong or an approximation. We all know that if you pull back a large enough number of balls on a large enough Newton's cradle, that eventually you no longer have a perfect n balls coming out at an identical velocity. We can chalk that up to the collisions being not quite elastic, but the real reason for the discrepancy might be that the answer "everyone knows" is a tad bit off.

>> No.2374139

>>2374137
it's one in the morning, idiot

no wonder you didn't solve the math

>> No.2374142

>>2374138
That's a really interesting experiment. I think you might see a small gap develop between some of the "stationary" balls.

>> No.2374143

>>2374097
Sure it does. The first ball pulled transfers its momentum through the first stationary ball - the pulled ball comes to a stop, while the previously stationary ball now has all the momentum and energy transferred to it. It doesn't go very far, however, before it runs into the second stationary ball, and the same thing happens. This keeps going until the last ball is hit, and since it has no other balls to smack into, it goes flying. Following closely behind it is the ball that goes flying from the second pulled ball, whose "shockwave" has been closely following that of the first impact, but at the same speed, so it never catches up.

>> No.2374153

>>2374119
/sci/ has always been 90% useless kids looking for homework help

>> No.2374154

>>2374123
That's basically
>>2373310
>>2373451
>>2373516
>>2373756
but the potential problem with that solution is that it assumes that balls are not touching (ungay case). But Newton's cradle may also be set up so that the balls are leaning into each other slightly. In that case, I'm not sure if the solution is valid.

>> No.2374156
File: 60 KB, 750x600, 1266749743357.jpg [View same] [iqdb] [saucenao] [google]
2374156

>>2374115
KE = 0 for all masses not on the ends.
p = 0 for all masses not on the ends.

ShockWaves/CompressionWaves are perfectly (near perfectly) transferred by the balls. This is because of the material the balls is made out, as well as there shape.

Did OP even bother to read the fucking wiki? WTF guy!

\thread

>> No.2374157

>>2374143
WINNER

>> No.2374161

>>2374154
should say
>not initially touching
The balls do touch when they collide, of course.

>> No.2374166

>>2374161
kevin?

>> No.2374168

>>2374166
No.

>> No.2374173

>>2374168
david?

>> No.2374179

fuckin newtons cradle, how do they work

>> No.2374182

>>2374154

If You, my dear fellow citizen of Earth, had known chemistry as well as I do, which I must assume in this case isn't the case, You, my dear, would have maintained enough knowledge of physical objects and their complex interaction, to know, that in fact, yes, I said 'fact', they never really, and by that I do mean really really, touch.

tl;dr: the balls don't touch

>> No.2374196

>>2374182
By "touching" I really mean that the balls are compressing each other, and there is a force between the balls. What's happening at the atomic level may be relevant in determining whether it is gay, but isn't relevant for solving the problem.

>> No.2374238

>>2374196

They aren't compressing each other. Get it? They CANNOT merge into behaving like one object. You reductionist person. Get it? They are seperate.

Are you seriously trying to troll us all or is this the ignorance of your brain? You can never ever learn physics without knowing chemistry. Never. Not possible. Go home.

TL;DR: THE BALLS ARE NOT ONE OBJECT BUT INDIVIDUAL OBJECTS READ SOME EFFIN' PHILOSOPHY ON THE SUBJECT, LIKE COMPLEXITY THEORY. capslock i am this mad

>> No.2374251

>>2374143
YES, this is correct. This all boils down to the fact that the collisions don't happen instantaneously, since the shock wave from each collision travels at a finite speed in the metal balls!

>> No.2374258

>>2374251
But that shock wave takes much less time to get through a ball than the time it takes for momentum to transfer from one ball to another. The collisions do occur simultaneously.

See
>>2374100

>> No.2374272

>>2374258
Come on now. How do you think momentum is being transferred? It's clearly being transferred by the wave propagating through the ball (ie, moving it): the "shock wave" and collisions are one and the same.

>> No.2374290

>>2374272
Yes, but your picture of how the shock wave behaves isn't accurate. The shock wave will bounce around many times within ball n, transferring a little momentum to ball n+1 each time in the small contact region, before ball n+1 starts moving with any appreciable speed.

>> No.2374297

>>2374258
The collisions cannot occur instantaneously. The flow of information cannot be faster than the speed of light in a vacuum.

>> No.2374306

>>2374297
Nobody thinks the collisions occur instantaneously. What I think is that there will be a force between ball n-1 and ball n at the same time there is a force between ball n and ball n+1. Simultaneous, not instantaneous.

>> No.2374309

>>2374306
Sorry, it cannot be simultaneous*, (mistype before), because the flow of information cannot occur faster than the speed of light in a vacuum.

>> No.2374317

>>2374306
And because I may not have explained that very well: The momentum takes some time to transfer from one ball to the other. The time interval there's a force between ball n-1 and ball n, and the time there's a force between ball n and ball n+1 is not identical obviously because of the finite speed of sound, but the intervals will be about the same, only offset by a comparatively small amount of time.

>> No.2374404

>>2374116
Hmm... The spring model tells me that the first ball should bounce back with velocity -v/3 and the other two go forward with velocity 2v/3. I can show calculations if anyone wants to check them. Obviously that's not what really happens, so there must be a problem with the model. Either the force between balls isn't actually linear in displacement, or we need to actually consider the behavior of the shock wave (which would be rather complicated if you do it correctly).

>> No.2374407

>>2374135
And Wikipedia probably isn't a good place to post this sort of stuff. Original research policy and all. Might be interesting to see if there's some stuff in the scientific literature about this problem, though.

>> No.2374538

From the Wikipedia article:
>The intermediate balls appear stationary. In fact, the cradle continues to work even if the intermediate balls are physically clamped still. This is also counter intuitive - transmitting motion without moving.

This is interesting and was something I didn't know. Has anyone here tried this? Does it work exactly as before, or is only some fraction of the momentum transferred? In any case, it certainly suggests that the detailed behavior of the shock wave is important in understanding the thing properly.

>> No.2374588

lrn2thermodynamics

energy can not be created or destroyed.

so we can quite easily calculate the energy made by a ball drpping from a specific height now all we need to do is calculate air resistance which is so minimal you can neglect it and the energy lost to sound when the ball hits the next (have no idea how to do this but sure it can be done. Now we have the amount of force with which the ball on the other end is catapulted up.

>> No.2374611

>>2374104
You need no additional constraints. If you regard each collision as a collision between two balls, then conservation of energy and momentum perfectly predicts the final velocity of all the balls.

You can't just ad hoc apply a certain amount of energy and momentum to the whole system without making some assumptions, but with a couple assumptions, you can do so. For example, the assumption that balls never bounce backwards. But it suffices to just work out the whole thing as a series of 2-ball collision to work out everything. In the 3-ball case, conservation laws require the entire energy and momentum to be transferred from the 1st to the 2nd, and then from the 2nd to the 3rd.