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Author Topic: Should the ball bounce off the wall at the same angle of impact?  (Read 5769 times)

Offline marcelo

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Hi,

This has been a long standing question about basic physics phenomenon. While playing billiard or just with any ball and you throw a ball (no top spin or anything) against a surface what angle is supposed to follow in its trajectory off the surface?

In internet i see this is a common question among game programmers games that want to do realistic physic simulations, and many analogies to light reflection are also invoked as part of the answers. However,  is the collision of a body equivalent to a light beam shone against a surface? My feeling is that not necessarily.  I am not interested in the similarity to light, but on the actual explanation for the mechanical bouncing and the bouncing angle.



ps. to the webmaster. My previous questions in the forum are indexed, but after clicking on them nothing comes, so they seemed archived somewhere else in the wesite?  Sorry about not closing my previous topics, but I didn't know how to do it in one case and I still had one counter argument to bring.


 

Offline RD

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... While playing billiard or just with any ball and you throw a ball (no top spin or anything) against a surface what angle is supposed to follow in its trajectory off the surface?

If itís a 3-dimensional system, rather than one confined to a horizontal plane, (e.g. squash rather than billiards), then the angle of incidence does not necessarily equal the angle of reflection because of the energy lost via inelastic collision affects the trajectory.
« Last Edit: 03/01/2012 03:22:10 by RD »
 

Offline MikeS

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Presumably, the ball will also be rotating after the bounce despite no rotation before.  I imagine this would cause some deviation off the predicted course (in comparison to a reflected photon).
 

Offline marcelo

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Ok, thanks for that first rounds, but I think I need to precise a bit more this:

I see that the "imperfect" quality of the system or the fact that is  3d may affect the outcome, but let's assume for a moment that all conditions are "optimal", perfectly elastic ball, no absortion of energy by the wall, no deformation, no 3d tricks (we can always "frame" the bouncing along a single 2d plane) , etc, etc.

so in this ideal system,  why should the ball necessarily bounce off the wall at the same angle of impact? then I would be happy to discuss any departure from the ideal system  ( I guess I am aiming to thermodynamical/energetic considerations, but I don't have a good answer )
 

Offline MikeS

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For every action there is an equal and opposite reaction?
 

Offline CliffordK

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In an "Ideal" system, it is simply a vector problem.

Consider a 2 dimensional vector with a vertical "wall" at the Y axis.

Your vector will have a horizontal component and a vertical component of its direction when it encounters the wall.

The horizontal component reverses in direction, while the vertical component is unchanged.  And, thus the angle is reflected.



Now...
If you are playing billiards, then spin might be an interesting concept.  If the ball is simply rolling when it encounters the bumper.  It would tend to continue rolling in the same direction...  at least for a brief period of time, before the roll is changed to the new direction of motion.  And, thus one might expect a slightly wider than ideal angle.
 

Offline marcelo

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[SOLVED]

Ok, yes
The vectorial diagram makes it crystal clear that only the perpendicular component should get a reversal in direction and of the same magnitude (if ideal of course) . thanks guys!
 

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