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Gaussian Gun toy

Interesting. I don't get why the end ball goes flying off at speed though. From the looks of it, the loose ball rolls towards the magnets, which then move back towards the loose ball (along with the magnetised balls), hit it... and then the end magnetised ball goes flying off?
 
I think it has to do with properties of the magnets themselves. They're "neodymium." There's another toy on the site, which they call "Magnetic Torpedoes," made of the same stuff, that apparently also does unexpected things...

http://www.grand-illusions.com/torpedo.htm

But I don't really know.

Any physicists out there who wanna try and explain this?
 
Originally posted by Malenfant:
Interesting. I don't get why the end ball goes flying off at speed though. From the looks of it, the loose ball rolls towards the magnets, which then move back towards the loose ball (along with the magnetised balls), hit it... and then the end magnetised ball goes flying off?
The way it works (If I understand correctly) is that the Ep of the loose ball is transferred by kinetic transfer when it impacts the mangnet; that energy transfer is near-linear or linear drop-off and quite high, relatively, as the loose ball has been accelerating down an Ep slope until impact; it's acquired a big bunch of energy by the rolling downhill*. The end ball receives a DIRECT KINETIC kick; the Ep slope is, courtesy of the inverse-square law, 1/16th that of the point of impact. (4 times the distance from bottom of slope: the magnet). If even only 1/.2 the kinetic transfers, that should produce a considerable kick.

*the hill and slopes are not physical slopes, but rather virtual slopes created by the magnetic field pulling the balls into it.

Edit: Ep = Energy, Potential.
 
Friends

I am bemused.

I have studied that little clip perhaps 20 times, it appears that the 5th ball does indeed accelerate to contact, at contact
0000MM<-0
becomes not
0000MM0
but rather
0<-->000MM0,
with ball 1 (far LHS) gaining the original Ek of the ball 5. So much is expected, it is what you would have for the simpler case of non-magnetic balls.

What is so interesting is the momentum exchange of the RHS group
000||0
displacing to the right at a small fraction of the velocity of the ball 1 moving now to the left, indicating a momentum transfer. I don't see how this can happen.

The real mystery solved though, is what I am buying my 4 year old for Christmas, well, maybe I mean what I'm getting my INNER 4 year old for Christmas - and I do not intend to share this little toy, for a while anyway - but I will play with same and advise the panel.

Clearly, further study is advised.

sojourner
 
At this time I can only think of two explanations for the behavior of the rest of the stack when the end ball shoots off.

1) The rest of the stack is recoiling from the impetus given to the end ball, or

2) As the first ball rolls down towards the stack, it becomes inductively charged by the magnets (a metal object moving through a magnetic field) and attracts the rest of the stack.

I don't think the second one is right, so it's either the first one or there's something else going on.
 
Ok, I've got some of these at work. I'll give it a try on Monday and see what happens.

I think Aramis is right though. As the slow moving ball enters the strong magnetic field it has quite a high amount of magnetic potential energy, which is converted to kinetic energy at the moment of impact.
Look at the recoil of the stack, equal and opposite. The energy must be coming from the magnets.
 
as the hammer ball approaches the magnets , the force increases due to proximity causing it to accelerate and gain momentum . the bullet ball , however ,is held by a lesser force as is it further from the magnets due to the balls in between .does this play a role?

maybe????

(my physics a-level was a loooong time ago)
 
THat's exactly what I said, hirch.

The binding force at hammer strike is assigned an arbitrary force of 1. the binding force at starting position of bullet ball is 1/16 that of the binding force upon the hammer at impact. The hammer has not only whatever inbound force gained from starting state (IE, hand applied) plus the accelleration time and distance, which gains force AND ACCELLERATION as the ball closes, to impact.

Lets assume these are 1m balls rather than 1 cm (makes the calcs easier), and the overall system is 4x1mballs, 2x1m magnets, and 1x1m ball starting at 5m. We'll assume 1m/s accell at magnet.
</font><blockquote>code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">Sec Di Vi ?v ?d
1 5 0 0.04 0.02
2 4.98 0.04 0.04032193029 0.06016096515
3 4.9198390349 0.08032193029 0.04131409198 0.10097897628
4 4.8188600586 0.12163602228 0.04306370296 0.14316787375
5 4.6756921848 0.16469972523 0.04574126541 0.18757035794
6 4.4881218269 0.21044099065 0.04964445255 0.23526321692
7 4.25285861 0.2600854432 0.05528892057 0.28772990349
8 3.9651287065 0.31537436377 0.06360414547 0.34717643651
9 3.61795227 0.37897850924 0.07639665328 0.41717683588
10 3.2007754341 0.45537516252 0.09760893846 0.50417963175
11 2.6965958023 0.55298410098 0.13752076885 0.6217444854
12 2.0748513169 0.69050486983 0.23228760619 0.80664867292
13 1.268202644 0.92279247601 0.62175987473 1.2336724134
14 0.03453023063 1.5445523507 838.68918352 420.88914411
15 -420.85461388 840.23373587 0.00000564593 840.23373869</pre>[/QUOTE]Di= Distance, Initial for that second
Vi= Vector, initial for the second.
?v= Delta-Vector (Change in vector)
?d= change in distance. (Vi+(?v/2))

So, it hits during the 15th second, with a vector of about 1.5m/s... if we assume 50% transfer... that's 0.75m/s, with an initial Ep to overcome of 1/16=0.625 m/s to overcome...

Now, the aceelerations on the real one are FAR higher ( more than 9.8 m/s/s, as it can hold against 1G...), the distances relative are far smaller, and the relative accelerations are the same.
 
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