Perfectly inelastic, rigid collision, vel. ever 0?

[wvguy8258]

This likely represents a physical impossibility or contradiction, but I'd like to know that or, if not, understand what is going on at a deeper level.

Let's say we have a perfectly rigid and immovable wall. We have a perfectly rigid (doesn't compress) ball moving toward it at 10 m/s. The two meet and the collision is perfectly inelastic and so the ball moves off in the opposite direction afterward. I would assume it would reach 10 m/s given I believe an inelastic collision involves no loss.

Now, since the ball changed directions, it makes sense that the ball at some instantaneous point in time had a velocity of 0. However, since both objects in the system are incompressible, it is hard for me to imagine the ball not doing the following:

1. losing all velocity in the original direction instantaneously (so I suppose infinite negative acceleration)
2. at the same moment begin moving in the opposite direction

I know that if the velocity changes sign it must at some point be zero, however I can't picture it like a series of still shots where I can intuitively get a grasp of that ball being still at all. It is obviously easier to do this with usual collisions (tennis ball meets wall).

I'm thinking my problem is: 1. no experience with this type of system as it doesn't exist and so my intuition can't handle it and 2. I believe the moment at which velocity is exactly zero would be infinitely short and my mind doesn't deal with infinitely small time quantities well.

Any help? Thanks, if I'm asking an old question, let me know. Didn't find it on a search here.

Seth

 

[fluidistic]

My thoughts:

Let's say we have a perfectly rigid and immovable wall. We have a perfectly rigid (doesn't compress) ball moving toward it at 10 m/s. The two meet and the collision is perfectly inelastic and so the ball moves off in the opposite direction afterward. I would assume it would reach 10 m/s given I believe an inelastic collision involves no loss.

Fine.

I know that if the velocity changes sign it must at some point be zero

No. The velocity function needs not to be continuous. The velocity of the ball when it hits the wall isn't defined (with your assumptions above), maybe it has to see with Dirac's delta; I am not 100% sure here, but it's definitely not 0 m/s.

 

[wvguy8258]

My thoughts:
Fine.




No. The velocity function needs not to be continuous. The velocity of the ball when it hits the wall isn't defined (with your assumptions above), maybe it has to see with Dirac's delta; I am not 100% sure here, but it's definitely not 0 m/s.

Thank you. So, I'm not totally crazy ;)

So, a position versus time graph (position is in one dimension) for the ball would be linear but with a peak at the wall's location. Therefore, if we take the 1st derivative of the function to find instantaneous velocity, this derivative is not defined at the abrupt peak (a witches hat shape). Is my interpretation correct? This is very fascinating to me, thank you. -Seth

 

[fluidistic]

Thank you. So, I'm not totally crazy ;)

So, a position versus time graph (position is in one dimension) for the ball would be linear but with a peak at the wall's location. Therefore, if we take the 1st derivative of the function to find instantaneous velocity, this derivative is not defined at the abrupt peak (a witches hat shape). Is my interpretation correct? This is very fascinating to me, thank you. -Seth

I think so.

 

[PhanthomJay]

This [strike] likely [/strike] represents a physical impossibility or contradiction

Scratch the word 'likely'. Rigid bodies are non existent. It's like asking what happens when I travel at the speed of light. Both are physically impossible. You might want to assume, instead, that the objects are 'nearly' rigid, and that the collision is 'nearly' totally elastic. Gotta get rid of those infinities.

 

[wvguy8258]

Just a thought experiment.

 

[DrewD]

elastic, not inelastic