Magnets and collissions

[Experimenter]

I am building a little experiment to demonstrate the impact of metal objects. It consists of the following two configurations:

Magnets used are grade N38 NdFeB rare earth magnets.

$RM$ - Cylindrical magnet that is 1 1/2" long x 1/2" diameter laid lengthwise
CM - 1/2" cube magnet
B - 1/2" diameter steel ball

Configuration 1:
$RM$-B-B-B

Configuration 2:
CM-B-B-B

A ball will be rolled to impact with the left side as the magnet draws it in. In theory, the impact should be transferred to the right most ball to the right of the magnet, causing it to disconnect from the group of balls to its left and roll to the right. The question is: Which of the two configurations will produce a faster propelled ball leaving the system on the right hand side. Assume that the impact force of the ball hitting the left hand side of each respective magnet is identical.

Thanks

The Experimenter

[Experimenter]

I should also add that the magnets are secured by paper tape so as not to move towards the ball being rolled from the left and that motion is restricted (by a track for example) to be only in the left and right directions.

Experimenter

[Experimenter]

Just so there is no confusion

[chroot]

You have presented nowhere near enough information for any of us to have any idea what will happen.

- Warren

[arcnets]

Originally posted by Experimenter
Assume that the impact force of the ball hitting the left hand side of each respective magnet is identical.I think this assumption is problematic. No way to really ensure this, IMO...

[On Radioactive Waves]

Still waiting for my magnets I just ordered... I do know one possibility though, they shatter!

[Experimenter]

Originally posted by chroot
You have presented nowhere near enough information for any of us to have any idea what will happen.

- Warren

What information is missing? I thought I presented all physical characteristics.

Thanks

[Experimenter]

Originally posted by arcnets
I think this assumption is problematic. No way to really ensure this, IMO...

One way to do this is by starting with a ball at rest to the left of the magnet. The starting distance from the magnet can be adjusted so that the force of impact is equal in both scenarios.

Thanks

Experimenter

[Experimenter]

Originally posted by On Radioactive Waves
Still waiting for my magnets I just ordered... I do know one possibility though, they shatter!

Shattering and chipping is a real problem. One way I thought of to minimize this possibility is to use a tiny steel washer to spread the force of impact from a single point on the face of the cylinder or cube, to a small circular area.

[arcnets]

Originally posted by Experimenter
The starting distance from the magnet can be adjusted so that the force of impact is equal in both scenarios.Can it? I doubt that.

You're talking about collison here, so force may not be the proper quantity at all. Momentum might.

[Experimenter]

I am referred to the energy transfer to the right most sphere and the initial velocity with which it starts rolling to the right.

[arcnets]

Originally posted by Experimenter
I am referred to the energy transfer to the right most sphere and the initial velocity with which it starts rolling to the right.
That ball is made of steel, right? Plus, it's in the magnetic field, isn't it? So it will be slowed down while rolling away. By many effects, one of which is eddy currents. So how do you want to determine its initial velocity?

[Experimenter]

Originally posted by arcnets
That ball is made of steel, right? Plus, it's in the magnetic field, isn't it? So it will be slowed down while rolling away. By many effects, one of which is eddy currents. So how do you want to determine its initial velocity?

I only need the initial velocity at departure time (i.e. the time when it separates from the ball next to it.

Experimenter

[arcnets]

I guess the most exact method to measure this would be one that uses Doppler effect - like when the police measure the speed of a car.

Maybe some stroboscopic method, or even video, will do as well.