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Is the rebound rate (or ratio) of a bouncing ball actually constant?

Edit: To clarify, I am considering a given ball and whether it has a different coefficient of restitution depending on the height dropped from (or velocity at impact).

It’s quite typical for high school mathematics textbooks to ask questions about a bouncing ball for the topic of geometric sequences. At this level of education, the maximum height of a bouncing ball is modelled as a geometric progression or exponential function. For instance, $h_{n}=h_{n-1}\times r=h_0\times r^n$, where $h_n$ is the maximum height after the $n$th bounce after it is dropped from a height of $h_0$, and $0< r < 1$ is the rebound rate (coefficient of restitution).

But is $r$ constant with respect to height? Why or why not?

If $r$ is not constant, how do other factors besides gravity and idealised elastic behaviour influence the $(h_0, r)$ relationship? For a typical ball dropped from a typical height, what association should we expect to see? How about in a vacuum to isolate deformation effects?

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  • $\begingroup$ Do you have access to a physics book? $\endgroup$
    – David White
    Commented Mar 22, 2021 at 16:08
  • $\begingroup$ I do not have access to any undergraduate or higher physics textbooks. $\endgroup$
    – lukejanicke
    Commented Mar 22, 2021 at 22:56
  • $\begingroup$ Here's a good place to start: scienceabc.com/pure-sciences/… $\endgroup$
    – David White
    Commented Mar 22, 2021 at 23:22
  • $\begingroup$ So ‘rebound rate’ is formally referred to as ‘coefficient of restitution’. Then the question could be, is the coefficient of restitution constant for a given ball? And if not, how does height in particular (or velocity on impact) affect the coefficient of restitution? I hadn’t considered sound, but in a vacuum that would be zero along with aerodynamic forces. $\endgroup$
    – lukejanicke
    Commented Mar 24, 2021 at 5:43
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    $\begingroup$ The "why" may well involve characteristics of the material that the bouncing object is made of. Answering that question to your satisfaction is going to be difficult. $\endgroup$
    – David White
    Commented Mar 31, 2021 at 18:46

1 Answer 1

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No, it's not constant. It decreases with the drop height.

https://www.researchgate.net/profile/Khairul-Ismail-6/publication/260991737_Coefficient_of_restitution_of_sports_balls_A_normal_drop_test/links/57a021d808aec29aed214c06/Coefficient-of-restitution-of-sports-balls-A-normal-drop-test.pdf?origin=publication_detail

enter image description here

Why? Because the coefficient of restitution depends not only on the material of the object but also on its shape. So when you increase the speed, some object might deform a lot at the impact so can't restitute the kinetic energy the same way than at lower speed.

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  • $\begingroup$ So if you approximate $R(h)=R_0 - R'h$ you get to solve $h_{n+1} = h_n(R_0-R'h_n) = R_0h_n - R'h^2_n$. $\endgroup$
    – JEB
    Commented Aug 4 at 19:43
  • $\begingroup$ Note that the linked paper defines the COR ($e$), in terms of $v_{n+1}/v_n$ (velocity), which ofc is square-root of height, so the OP's $r=e^2 $. $\endgroup$
    – JEB
    Commented Aug 4 at 19:51

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