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If you were to drop a ball onto a surface from a height $h$ and the ball collided with the surface and then rebounded, would the ball then travel a distance of $eh$ back up from the surface where $e$ is CoR? If so why?

Also, just generally, is the CoR a constant in this scenario? I.e, is the ratio of separation speed to approach speed the same even though these speeds are most likely decreasing? If so, why?

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    $\begingroup$ en.wikipedia.org/wiki/Coefficient_of_restitution says that $COR=\sqrt{h/h_0}$, ball will bounce to a height of $COR^2*h_0$ $\endgroup$ – aaaaa says reinstate Monica Mar 24 '15 at 0:03
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    $\begingroup$ @aandreev Do not answer in comments. This is not how stack exchange works. If you wish to supply an answer, use the "Your Answer" box below, not the "add comment." $\endgroup$ – PipperChip Mar 24 '15 at 0:08
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    $\begingroup$ i comment due to the fact that answer is googlable $\endgroup$ – aaaaa says reinstate Monica Mar 24 '15 at 2:06
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The ball will rise again to height 'h' if the surface is absolutely smooth and there is no air friction (elastic collision).

The ball will rise to 'eh' if the collision is inelastic. This is because some fraction of initial velocity is lost due to rough surface etc.

The ratio i.e COR is 1 when it is the former case and < 1 when it is the latter. COR is the ratio between velocity after impact and velocity before impact and is a constant if surfaces, frictions and other things are fixed in the problem.

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The answer by RGJ is not exact: the ball will always travel back up $eh$.

The only difference is that:if the collision is inelastic $e<1$, if the collision is elastic $e=1$,

But as correctly said by angel, it is almost impossible that this may happen and the coefficient of restitution will be at most 0.99......

This because $e$ is the ratio between the two values of the speeds

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  • $\begingroup$ Why will the ball alaways travel up $eh$? $\endgroup$ – tey yreryt Mar 24 '15 at 16:33
  • $\begingroup$ ask angel, he might be able to explain. I can't because it is so obvious $\endgroup$ – user76123 Mar 27 '15 at 5:44
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The answer by RGJ is not exact: the ball will always travel back up eh.

The only difference is that:if the collision is inelastic $e<1$, if the collision is elastic $e=1$,

But as correctly said by angel, it is almost impossible that this may happen and the coefficient of restitution will be at most 0.99......

This because e is the ratio between the two values of the speeds

This answer somewhat explains it, but it left some confusion.

Why will the ball always travel up eh?

This is because of conservation of momentum. Let's ignore air resistance. The ball cannot come up higher than it was dropped because then it would get momentum from nowhere. The perfect scenario in which all the momentum is transferred from the ball to the wall back to the ball is $e = 1$. e is basically a measure of efficiency, how well the surface can "give back" momentum. Momentum can be lost to heat, "squishyness", etc. therefore in real life, $e < 1$.

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