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SWF
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 Collision With Row Of Spheres   « on: Jan 29th, 2003, 5:41pm » Quote Modify

(Warning, a good understanding of physics is required)

Four solid spheres floating in space far from the earth are aligned in a row contacting one another with their centers all on the same line. A fifth sphere moving along the common line approaches from the left and impacts on the left end of the row. Assume perfectly elastic collisions and all spheres have the same mass and diameter. Also assume that the spheres are isolated from external influences such as gravity and friction.

After the collision, what determines how many balls are moving and in which direction?  What information and principles of physics are needed to figure this out?
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #1 on: Jan 29th, 2003, 8:27pm » Quote Modify

Actually, having seen a "desk clacker" in action also provides the answer without having to resort to any physics! (Ahh - the value of experiment over theoretical calculation!)
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James Fingas
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 Re: Collision With Row Of Spheres   « Reply #2 on: Jan 30th, 2003, 11:29am » Quote Modify

Icarus,

The question as it stands is a little more general than the desk clacker. Specifically, the initial velocity of the balls is not given. Of course, the initial conditions will affect how they move after the collision. Also, static charges of the balls and their magnetic polarization could have an effect. We could also have the "electron whiplash" effect coming in to play.

Only if we assume that they are all stationary and electrically and magnetically inert, do we get the desk clacker phenomenon.

To derive this, we must simply consider conservation of energy and conservation of momentum. I don't think you can consider the clacker itself as a proof
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #3 on: Jan 30th, 2003, 6:59pm » Quote Modify

on Jan 30th, 2003, 11:29am, James Fingas wrote:
 The question as it stands is a little more general than the desk clacker. Specifically, the initial velocity of the balls is not given.

?? Initial velocity with respect to what? With respect to each other (since they are specified as the only things around) the first 4 have zero velocity, and the fifth's initial velocity is along the common line. Only the actual speed of approach is not specified, but I assume that it is not intended to be relativistic. Anything short of that should follow the same pattern as that famous clacker. While you might not agree that charges and magnetic polarization are "external influences", I have to think that SWF is aiming at the inert situation, where the only aspects of the spheres coming into play are their elasticity and gross physical properties.

As for "proving" anything - see my post on another thread (I don't remember which one and am too lazy to go look it up) for my views on science, experimentation and "proof". However, I do believe the clacker correctly models the situation sufficiently to guide you to the answer.

This does not mean SWF has not posed a good question for any serious physics devotee. My only point is that a simple experiment shows what the behaviour is. Of course, it does not explain WHY that is the behaviour. Which is what SWF is really asking.
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SWF
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 Re: Collision With Row Of Spheres   « Reply #4 on: Jan 30th, 2003, 7:34pm » Quote Modify

Perhaps, I should add some more, since I wasn't precise enough to trigger confusion.

Suppose the approaching ball starts with velocity +V and the other 4 balls start at rest.  After the collision, the ball which was originally moving rebounds with velocity -V/3 (i.e. moves in opposite direction from which it came), two balls are stationary, and two balls from the other end move onward with velocity 2V/3.  Show how this can or cannot occur.

Please assume gravitational attraction, magnetic interaction, electrostatic repulsion, eddy currents, quantum tunneling and other bizarre phenomena are negligible.
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BNC
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 Re: Collision With Row Of Spheres   « Reply #5 on: Jan 31st, 2003, 12:54am » Quote Modify

on Jan 30th, 2003, 6:59pm, Icarus wrote:
 As for "proving" anything - see my post on another thread (I don't remember which one and am too lazy to go look it up) for my views on science, experimentation and "proof".

I remember it (obviouslly a hop point for you!)
it in the Walking On A Stretching Rubber Band thread, post dated Jan 23rd.

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James Fingas
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 Re: Collision With Row Of Spheres   « Reply #6 on: Jan 31st, 2003, 9:57am » Quote Modify

Icarus,

I guess my word "proof" was a bit ambiguous. Of course I am only asking you to prove it within the set of axioms that you hold dear. We're not breaking new scientific ground here, just making deductions from already-proposed scientific principles. These deductions can be made with perfect accuracy (even if they don't match the real world).

SWF specifies the initial positions of the four balls aligned in a row, but doesn't provide the initial velocities of the four balls. We can assume that they're zero, but I figure I might as well be nitpicky...
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #7 on: Jan 31st, 2003, 4:06pm » Quote Modify

I was being nitpicky about the use of "proof" with reference to experimental results. (See the link BNC so graciously supplied to understand my views on this, and yes, it is a hot button issue for me as I am constantly having to cringe when I hear people call their pet theories "scientific fact, not theory", a phrase that demonstrates a fundamental misunderstanding of the nature of science. It's bad enough when Joe Blow says this, but when practicing scientists say it...).

Velocity is a relative concept. Things do not have a velocity by themselves - they only have a velocity with respect to something else. Usually in physics, with respect to some convenient Frame of Reference. But in this case, we are told that there is nothing else around but the 5 balls. The only convenient Frame of Reference is one anchored on the 4 balls. It was not necessary for SWF to specify a velocity for them. Their velocity is 0. If SWF had said, for instance, "the 4 spheres are moving along their common axis at a speed of 50 m/s, and the 5th is following even faster", I would still throw away the worthless 50 m/s number and examine the problem from the frame of reference with respect to the balls - velocity 0. (If I were to go ahead and examine the problem mathematically - I haven't tried to yet!)
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SWF
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 Re: Collision With Row Of Spheres   « Reply #8 on: Jan 31st, 2003, 8:06pm » Quote Modify

Is somebody going to try to give an answer to this question, or will this turn into a debate as to which complications to include.

True, the original post didn't specify initial velocities, and as written, the four contacting spheres could all have different initial velocities (both magnitude and direction).  If somebody wants to include that possibility, be my guest.  However, being that nobody has given a correct answer to the case with the four contacting balls initially at rest, maybe beginning with the simple case is best.
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #9 on: Feb 1st, 2003, 7:53pm » Quote Modify

Well, as James has said, conservation of energy and of momentum are definitely required. I don't see that anything else is needed. You can view the collision as being a series of 4 collisions with no intervening time. Since there is no tranverse motion, forces, or contact angles involved, the problem can be treated in 1 dimension.

For two spheres of equal mass, one viewed at rest and the other approaching with a velocity V, conservation of momentum gives

V = V1 + V2

Where V1 and V2 are the ending velocities of the two spheres. Conservation of Energy likewise gives

V2 = V12 + V22

Squaring the first equation, subtracting the second, and dividing by 2 gives

V1V2 = 0

So one of V1 and V2 is zero and the other is V (since their sum is V). The only way the standing ball could remain still and the moving ball remain moving after the collision would be for the moving ball to pass through the standing one. Since this is impossible, the moving ball comes to a dead stop and the standing ball shoots off with velocity V.

Number the balls 1 through 5 with 1 initially moving, approaching at V.
1 collides with 2. 1 stops, 2 shoots off at V, but immediately collides with 3. 2 stops (without actually having gone anywhere), and 3 shoots off at V, but immediately collides with 4. 3 stops, 4 shoots off at V, but immediately collides with 5. 4 stops, and 5 shoots off with velocity V.

So the net effect: 1 comes to a dead stop touching 2, and 5 heads off with the same velocity V as 1 had at the start.

Exactly the behavior one sees in a clacker.
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SWF
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 Re: Collision With Row Of Spheres   « Reply #10 on: Feb 2nd, 2003, 10:32am » Quote Modify

Icarus, that does demonstrate how one ball coming out of the other end does not violate conservation of energy or momentum.  Please see my post from 1/30/2003. Is this there more than one way that energy and momentum can be conserved without the spheres passing through one another, and if so, is something else needed to figure out which situtation will occur?  Is the result I describe on 1/30/2003 possible?  How, or why not?
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #11 on: Feb 2nd, 2003, 11:51am » Quote Modify

Actually what I did does not show that all the velocity passing to ball 5 is a solution. It shows that it is the solution. The full picture allows no other.

The problem with the scenario you (SWF) described is that while the beginning and ending momentums and energies are the same, it violates conservation of momentum and energy in the individual collisions. If you look at the problem as a single collision, then you apparently have two equations in 5 unknowns (V is a "known"):

V = V1 + V2 + V3 + V4 + V5

V2 = V12 + V22 + V32 + V42 + V52

This has an infinitude of solutions, because it does not model the situation well enough.

When you consider all the collisions separately, there are 3 more variables (S2, S3, S4 - the intermediate velocities of the 3 center balls between their 2 collisions), but there are also 2 equations for each collision: a total of 8 equations in 8 unknowns:

V  = V1 + S2,   V2  = V12 + S22
S2 = V2 + S3,   S22 = V22 + S32
S3 = V3 + S4,   S32 = V32 + S42
S4 = V4 + V5,   S42 = V42 + V52

This has a unique solution that does not have balls passing through other balls:
V1 = V2 = V3 = V4 = 0
S2 = S3 = S4 = V5 = V

Now you could argue that since Balls 2, 3, 4 never actually move, they could not have a non-zero velocity between collisions. My answer is that they do have the velocity, but for 0 time. How can this be? - IT CAN'T - not in real life. But we are not dealing with a real situation here - that was established with the introduction of perfect elasticity. In a real world situation the behavior is much more complex. The collisions take a short amount of time to complete and overlap each other in time. When ball 1 strikes ball 2, they both deform, then spring back into shape. Some energy is lost to pressure waves passing through the balls, which spread out and reflect around until they finally drop into the general motion noise of the molecules, which we call "heat". Ball 2 is pushed into ball 3, deforming at the other end. Ball 3 pushes back even as it is pushed into ball 4. This back push stops the minute motion of ball 2, just as 4's back push stops 3's, and 5's back push stops 4's motion. The result is that except for the small amount of energy lost to heat, the momentum and energy of ball 1 is passed down the line to ball 5.

So my description above is only an approximation of the real world situation. How can I be sure my approximation is adequate? By the only way you can justify any physics model of a real world situation - by experiment! In this case the experiment is the clacker, which demonstrates quite closely the velocity transfer I described. This is also the real reason I have to reject the -V/3 + 2(2V/3) scenario - I don't see this behaviour in experiments.
 « Last Edit: Feb 2nd, 2003, 12:11pm by Icarus » IP Logged

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 Re: Collision With Row Of Spheres   « Reply #12 on: Feb 3rd, 2003, 2:59am » Quote Modify

Can't this be done using Galilean relativity (which anyone could deduce knowing the earth is moving and physics still works) and symmetry?

Number the spheres 1-5, 1 being the one that is moving in at a velocity of V relative to the others.

Have 2 observers A & B. A is moving at V/2 relative to the spheres 2-5, whereas B is stationery.

So the first impact from A:

Sp#1 has relative velocity of V/2
Sp#2 has relative velocity of -V/2

Since this collision is symmetrical and those spheres are "perfectly elastic" it follows that after the collision

Sp#1 would have relative velocity of -V/2
Sp#2 would have relative velocity of V/2

So translating back to B gives the Sp#1 as stationery and Sp#2 is going on at V

A similar arguement applies as in Icarus' proof so Sp#1-4 all become stationery and Sp#5 flies off at V.
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Chronos
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 Re: Collision With Row Of Spheres   « Reply #13 on: Feb 3rd, 2003, 5:11pm » Quote Modify

The point is, you do need more than just conservation of energy and momentum.  You also need to assume some things about the balls and how they're arranged and how they're interacting.  For instance, I could, in principle, spot-weld balls 4 and 5 together, and still satisfy the conditions of the original post.  But if I did that, then the behaviour coming out of the collision would be completely different (although I'm not sure offhand exactly what the end result would be).  If we assume that the balls can only interact by a normal contact repulsive force, then we can model it with infinitesimal spaces between the balls, and get the solution Icarus found.  But there are other assumptions which could be made.
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #14 on: Feb 3rd, 2003, 7:45pm » Quote Modify

One can also argue that "sphere" means a small furry rodent in the antediluvian language of the Mutar tribe of Far Fandango, and that by placing them in space, SWF is guilty of the most atrocious attitudes towards our animal friends.

Or, one could stop coming with spurious objections and simply look at the problem as offered!

on Jan 30th, 2003, 7:34pm, SWF wrote:
 Please assume gravitational attraction, magnetic interaction, electrostatic repulsion, eddy currents, quantum tunneling and other bizarre phenomena are negligible.
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SWF
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 Re: Collision With Row Of Spheres   « Reply #15 on: Feb 3rd, 2003, 8:17pm » Quote Modify

I am claiming there is way to have the one ball rebound at -V/3 and two balls 2V/3.  This does not involve welding, gluing, brazing, taping, velcro or joining the balls together in any way.
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Icarus
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 Re: Collision With Row Of Spheres   « Reply #16 on: Feb 4th, 2003, 4:49pm » Quote Modify

Alright - there is one aspect of the spheres that I have overlooked, and that I personally would agree is not spurious (i.e. does not require an uncommon interpretation of the question, and does not ignore SWF's statements concerning what is involved here). That would be if the spheres are spinning. If they are just barely in contact, then any frictional interaction would be infinitesimal, so they could be considered "floating in contact" as the puzzle says.

In this case when they actually collide, you can generally expect them to pick up some tranverse motion, and fly off in various directions depending on the orientations and speeds of the rotation.

However, I don't see that this could explain SWF's (-V/3)+2(2V/3) scenario. Any interaction between spinning spheres would pick up some transverse forces, unless the spin axes are all the common axis through the centers of all the spheres. And in that case there is not coupling of the rotational energy and the kinetic energy, so it is back to the situation I have described for the kinetics.

Really, I don't see any Newtonian mechanical way for the (-V/3)+2(2V/3) outcome to arise, assuming unconnected perfect spheres.
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SWF
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 Re: Collision With Row Of Spheres   « Reply #17 on: Feb 4th, 2003, 7:40pm » Quote Modify

The spheres are not spinning, and I would classify this a classical mechanics phenomenon.  You could even have a desk clacker that does this sitting on your desk without fear of radiation or becoming 100 years younger than your twin.  However, after the second 'clack' the behavior may get ugly.  I will leave the explaination for the messy behavior as an easier side question.
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 Re: Collision With Row Of Spheres   « Reply #18 on: Sep 24th, 2003, 1:03am » Quote Modify Remove

Are the spheres all of uniform density?
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wowbagger
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 Re: Collision With Row Of Spheres   « Reply #19 on: Sep 24th, 2003, 4:10am » Quote Modify

on Feb 4th, 2003, 7:40pm, SWF wrote:
 You could even have a desk clacker that does this sitting on your desk

Unfortunately, I haven't, so I can't check by experiment whether my guess is sensible.

The only conserved quantity of classical mechanics not yet used is angular momentum. Since the spheres are arranged along a line, the total angular momentum has to be conserved. I'm not sure whether that's what SWF has in mind, but someone with access to a desk clacker could try to lift one sphere up a bit on one end, and the remaining four spheres a little less at the other end. If you then let go, all spheres will be along a straight line at every instant. Of course, they won't be moving along a line, but in a plane (thus, the total anglar momentum is still conserved).

I'm usually not exactly an experiment enthusiast, but I don't like to ponder this in great depth now, let alone calculate something. So if someone could be so nice to quickly check my proposal, I'd appreciate it.
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SWF
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 Re: Collision With Row Of Spheres   « Reply #20 on: Sep 24th, 2003, 8:37pm » Quote Modify

on Sep 24th, 2003, 1:03am, Guess wrote:
 Are the spheres all of uniform density?
Yes, each sphere has uniform density.

I do not have the equipment to try wowbagger's experiment, but can say that trick will not cause the behavior I am describing. Too much experience with the standard desk toy is partly responsible for the misconceptions about this problem. Note that the the orginal question was phrased as spheres floating in space to avoid the additional complexity of the pendulum motion (not that it is much of a factor).
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James Fingas
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 Re: Collision With Row Of Spheres   « Reply #21 on: Sep 25th, 2003, 5:29am » Quote Modify

I almost posted this yesterday, but managed to lose my post to that great bit-bucket in the sky.

The answer could be in the elasticity of the spheres. If some or all of the spheres compress during the collision, the answer could be different from the desk clacker phenomenon. But I'll have to come up with some way of analyzing the situation, to decide which sphere(s) must be more compressible than others to get the desired effect.
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aero_guy
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 Re: Collision With Row Of Spheres   « Reply #22 on: Sep 25th, 2003, 6:34am » Quote Modify

Wow, I must have missed this one when it originally came out.  I think James is right, though you do not mean elasticity, but compressibility (not interchangeable).  If the first two balls compress significantly, than they will not have fully expanded (complete kinetic energy will not have transfered) by the time the pair hits the next ball.  This causes all kinds of fun stuff to happen.  I imagine that by changing the compressibility of the balls you can get any potential solution to the set of five equations presented by Icarus earlier.  The reason his original answer would not be appropriate is that they are no longer seperate interactions.
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James Fingas
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 Re: Collision With Row Of Spheres   « Reply #23 on: Sep 25th, 2003, 10:04am » Quote Modify

From civil and materials engineering:

Modulus of Elasticity, measured in Pa (or kPa, MPa, and often GPa for stiff materials) is defined as the stress divided by the strain in the elastic region of the stress/strain curve. In this region, no energy is absorbed in deforming the material, and it returns to its original shape once the stress is removed.

So there.
 « Last Edit: Sep 25th, 2003, 10:06am by James Fingas » IP Logged

aero_guy
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 Re: Collision With Row Of Spheres   « Reply #24 on: Sep 25th, 2003, 10:26am » Quote Modify

yeah, that is the definition I normally use, but it seemed that people were confusing "perfectly elastic" with non-deformable.  Maybe I misinterpreted.

Anyway, I have come up with a computer model that is giving some interesting results.  More later.
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