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Posted

We've all seen Newton's Cradles.

 

Pull one Ball back, and let it go.

 

It goes 1-Right, 1-Left; 1-Right, 1-Left.....And so on.

 

Nothing strange yet.

 

Take two Balls at one time.

 

It Goes 2-R, 2-L; 2-R, 2-L.....And so on.

 

Once again, nothing surprising.

 

Pull back three Balls at one time.

 

It goes 3-R, 2-L; 3-R, 2-L; 3-R, 2-L.....

 

How does the Cradle "Know" or "Remember" when two Balls strike it from the Left, that those two Balls were launched by three Balls?

 

Why doesn't it go:

 

3-R, 2-L; 2-R, 2-L; 2-R, 2-L ?

 

It isn't a matter of force. You can contrive to give two Balls twice or three times their traditional momentum.....

 

And they still won't kick out three Balls?

 

It does not work:

 

2>>R, 3-L; 2-R, 3-L.....

 

Or any other permutation that you can imagine.

 

.....Saxon Violence

Posted

i'm pretty sure on a neuwton cradle if you pull back 3 balls, 3 balls will come out on the other side; ie. 3-R 3-L 3-R 3-L.

if you don't get this; could you post a youtube video of your cradle giving a different result? cause that basically violates the laws of physics.

(momentum in = momentum out)

Posted

:) Hi Saxon – welcome to hypography, and thanks for the video, and a physics question I can actually answer!

 

We've all seen Newton's Cradles.

...

Pull back three Balls at one time.

 

It goes 3-R, 2-L; 3-R, 2-L; 3-R, 2-L.....

No, as Phillip though, and the video shows, it goes 3R, 3L, 3R, 3L ...

 

or, to ASCII art draw it perhaps more clearly, it goes

 

OOO-> OO / OO OOO-> / OO <-OOO / <-OOO OO / OOO-> OO / ...

 

The physics of this cool old desk toy is pretty simple. Because the system is fairly elastic – not much of its kinetic energy is converted into sound, heat, etc. – it must satisfy the 2 equations for conservation of momentum and conservation of energy:

 

[math]P = \sum_{i=1}^n M_i \overrightarrow{V_i}[/math]

 

and

 

[math]2E = \sum_{i=1}^n M_i \left( |V_i| \right)^2[/math]

 

Where [imath]M_i[/imath] is a ball i’s mass, [imath] \overrightarrow{V_i} [/imath] it’s velocity, and [imath]|V_i|[/imath] its speed (the scalar component of its velocity), and P and E constant.

 

I’ll leave it as an exercise to the reader to prove that only one possible “permutation” is possible for a Newton’s cradle with equal mass balls and no tricky stuff to sick them together, etc. It’s not a hard proof/derivation, and one you need to do whenever you write any sort of physics-based motion simulator with elastic collisions, such as my old “Newtonian bowling” sim, where you can see the solution written in an actual program.

Posted

I will add my two cents on this subject as I had such a toy when I was a child.

 

It behaved exactly as CraigD described in the previous post.

 

I did in essence your experiment.

 

When I let go 1 ball, 1 ball came out other side.

 

When I let go 2 balls together, 2 balls came out other side

 

When I let go 3 ball together, 3 balls came out other side

 

When I let go 4 balls together, 4 balls came out other side.

 

This was the demonstration of the Conservation Laws as CraigD described.

 

Nothing Magical at all.... ;)

 

maddog

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