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Let's say that a ball dropped from $1 \ \text{m}$, covers $\frac{1}{n^2} \ \text{m}$ for every $n$ bounce, i.e $\sum_{n = 1}^{\infty} \frac {1}{n^2}$ so the total distance will be $\frac{π^2}{6}$, right? But, when I tried to calculate the time, it got a bit messy.

enter image description here

(Please forgive me if I am wrong or if I made any dumb mistake, I am not as knowledgeable as a college guy. I am just a typical 9th grader.)

Qmechanic
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Not_CarlFriedrichGauss07
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    It seems there is some info missing from your problem, or at least it is not clear. Once the ball bounces, what do you assume ? I guess there is some restitution coefficient which reduces the energy of the ball after each bounce. Do you assume it is such that the ball travels 1m, then $1/4$m, then $1/9$m ? If so I think this is some weird physics, as typically you would expect an exponential decay of the height.

    In any case, it is not difficult to devise situations where you cross a finite distance in infinite time, so it is definitely possible.

     – Frotaur May 31 '22 at 13:33
  • umm, lemme rephrase it, suppose, there are 2 infinite mass walls, one at the bottom, one at the top, and the top wall moves down to 1/n^2th distance , when the ball touches the bottom after nth bounce (no air resistance, super ideal) ....does it make sense now....? – Not_CarlFriedrichGauss07 May 31 '22 at 13:47
  • "In any case, it is not difficult to devise situations where you cross a finite distance in infinite time, so it is definitely possible" , wait !! realllyy !!! , so like is the avg velocity 0 ? as its lim t --> infinity of c/t = 0 ? (where c is some constant) !! that's awesome !! can you give me some examples of that ? – Not_CarlFriedrichGauss07 May 31 '22 at 13:51
  • For the second comment : Take the following curve, followed by a particle : $s(t)=1-1/(t+1)$, for $t\in (0,\infty)$. In infinite time, the particle will go from $s=0$ to $s=1$. You can verify by taking the time derivative that the velocity is never 0. – Frotaur May 31 '22 at 13:54
  • As for the second comment, it is still quite a complicated setup, because if you include gravity the balls trajectory will be quite involved. Disregarding that, if there is no energy loss then the ball will continue bouncing around the two walls until they close together. This will however take finite time, as the time is dictated by the distance traveled by the ball (wall advance at each bounce). So your computation of the time taken does not depend on $g$ anymore, and thus your argument breaks down. – Frotaur May 31 '22 at 13:56
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    "()=1−1/(+1), for ∈(0,∞). In infinite time, the particle will go from =0 to =1. You can verify by taking the time derivative that the velocity is never 0 " asymptotic nature of the function helps here !!! , it is true and it's so fascinating !!! to see this thank you soo much @Frotaur !!! it's really awesome to learn about these !! – Not_CarlFriedrichGauss07 May 31 '22 at 14:00
  • Let us continue this discussion in chat. – Frotaur May 31 '22 at 14:01

4 Answers4

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This is correct and not too surprising. Each time it bounces, the ball loses some energy. But with the hypotheses in OP, the ball always has some mechanic energy left, so it is possible that it never stop bouncing. Performing the calculations as in OP, we find out that this is indeed the case. As @Frotaur points out in the comments, for a more realistic system where the fraction of energy lost at each bounce is constant, then the height of the bounces decreases exponentially and the process stops in finite time.

SolubleFish
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    If we assume a restitution coefficient $(1-\epsilon)$, as it is usually done, then the successive bounce heights decay like : $h_0=1$, $h_1=(1-\epsilon)$, $h_2=(1-\epsilon)^2$... $h_n=(1-\epsilon)^n$. In this case you also always have some energy left, and yet the whole movement is completed in finite time. So you answer is incorrect. – Frotaur May 31 '22 at 13:37
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    Thank you for pointing this out ! I corrected my answer. I agree that the physics of the energy loss here are rather curious, but you could engineer a situation where this would happen. – SolubleFish May 31 '22 at 13:40
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    I removed the downvote accordingly ! – Frotaur May 31 '22 at 13:56
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Here is a diagram of the bounces of a ball:

enter image description here

It seems that the ball will require an infinite number of bounces to come to rest. Intuition says that it will take an infinite amount of time for this to occur. Analysis however, shows that in this model, the ball can make an infinite number of bounces in a finite amount of time. The reason is that the time between bounces decreases as the bounce height decreases.

The idea that the ball executes an infinite number of bounces in a finite amount of time is perhaps the most interesting response to Zeno's paradoxes. We have experimental evidence that an infinite number of events can occur in a finite amount of time. Mathematically, we have use the fact that (for some cases) the sum of an infinite number of terms is a finite number.

Using the labels in the diagram, we show that the time from $t = 0$ to $t = t_1$ is given by

$$t_1 - 0 = \sqrt\frac{8h_1}{g}$$

In words, the bounce-to-bounce time is proportional to the square root of the bounce height.

Using the fact that $h_2 = g\cdot h_1$, we see that

$$t_2 - t_1 = \sqrt\frac{g8h_1}{g} = (t_1 - 0)\sqrt{g}$$

Adding up all the bounce-to-bounce times (from $n = 0$ to infinity) gives the total time to come to rest. The sum is $$T = (\sqrt\frac {8h_1}{g}) (\frac {1}{1-\beta})$$

$\beta$ is best measured by measuring the time for the second bounce and the first for the second bounce: $\beta$ is the ratio of those times. In this way, the time to come to rest is predicted from the times for the first two bounces.

Hope this helps.

Debanjan Biswas
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  • In Ops, question, he assumes $h_n = 1/n^2$. In that case your computation shows it indeed takes infinite time. You assume that $h_n \propto \beta^n$, which is what would physically happen (exponential decay of heights, as a ratio $\beta$ of energy is lost between bounces). But in op's question, we could interpret this as assuming the bouncing floor to be tiled with different materials, such as to obtain the series $h_n=1/n^2$, which indeed takes infinite time to complete. – Frotaur May 31 '22 at 14:26
  • Note that the smallest distance that can exist in nature is the plank length. The ball can't bounce more after covering a plank length distance in a bounce. – Debanjan Biswas May 31 '22 at 14:42
  • Sure, but we are talking about classical mechanics here. In any case even for finite n your time formula is incorrect if we want to follow OP's prescription for the heights. – Frotaur May 31 '22 at 15:08
  • @DebanjanBiswas “Note that the smallest distance that can exist in nature is the plank length.” – That’s a common misconception. Distances around the size of the Planck length may be impossible to measure, but there’s a big difference between being impossible to measure and not existing. – Tanner Swett Jun 01 '22 at 00:55
  • @TannerSwett Thank you for your comment. I was thinking a lot of about it for months. – Debanjan Biswas Jun 01 '22 at 01:36
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    But I think the kinetic energy of bounces should be completely converted into heat and will increase the temperature of the ball. Because the bounces will remain in the molecules of the ball vibrating the molecules. – Debanjan Biswas Jun 01 '22 at 01:39
  • In this site, we have limited ways of sending a message. You just deleted a post. I was working on an answer. If you undelete it, I will finish. Sorry it sounded like I was dismissive. People sometimes vote to close because they don't understand a question well enough. If they close, I cannot add a new answer. But I can edit an answer I already posted. So I post a placeholder. – mmesser314 Sep 21 '23 at 13:38
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My analysis is this. Roundtrip time for a ball in one iteration is: $$ t_n = S_n/\overline {v_n} ~,$$

Where $\overline {v_n}$ is average speed of ball in current bounce. Then according to problem statement,

$$ S_n \approx h_0/n^2$$,

Where $h_0$ is initial starting height. But, average speed in each iteration also decreases due to the fact that gravity has less and less time to speed-up the ball until collision to the ground, so let's assume that average ball speed in current bounce also decays as :

$$ \overline {v_n} \approx v_0/n^2 $$

Where $v_0$ is average ball speed in first bounce.

So total time taken of ball is this series :

$$ T = \sum_{n=1}^{\infty} t_n = \sum_{n=1}^{\infty} \frac {h_0/n^2}{v_0/n^2} = \sum_{n=1}^{\infty} \text {const} = \infty $$

Diverges.

Unless :

  1. Average speed changes in a different manner than distance traveled per bounce.
  2. In practice when ball kinetic energy will be a very small amount,- then dissipation forces such as air drag, ground friction, collision conversion to thermal energy ,- will take last ball energy in one go, leaving it at rest.
Agnius Vasiliauskas
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enter image description hereThank you soo much y'all ! I think this is the solution that I was looking for !

Not_CarlFriedrichGauss07
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