What's the best answer to "how many bounces until the ball stops"?

Question

A problem question that has been coming has the form:

• How many complete trips between two plates moving towards each other can you make? (Given that you are moving between them at a constant velocity, until they both collide, and thus the space between them becomes 0)
• How many bounces does a ball reaching a height 0.6 of its previous one make until it stops? (Dropped from rest... at an initial height...)

There are similar questions such as how much TIME it takes until it stops are simpler to answer since a geometric series can be formed. Is there a way to deal with number of travels/bounces/etc.?

Thanks for any input you may provide!

Edit: I am aware of Zeno's Paradox. Is there a mathematical work-around for it?

Answer 1

This is the simulation of bouncing ball with the co-efficient of the restitution $~\epsilon=0.9$

you start from the height $~h_0~$ with zero velocity .

Section 0

$$y(t)=h_0-\frac{g\,t^2}{2}\\ v=-g\,t\\ y(t)=0~\Rightarrow~,t_0^2=\frac{2\,h_0}{g}\\ v_{01}=-\sqrt{2\,g\,h_0}$$

when the ball hit the floor it reach the velocity $~v_{01}$ for the next section the start velocity will be $~v_{01}\mapsto -\epsilon\,v_{01}~$ where $\epsilon~$ is the co-efficient of the restitution.

Section 1

$$y(t)=v_{01}\,t-\frac{g\,t^2}{2}\\ \frac{dy}{dt}=0~\Rightarrow~,t_m=\frac{v_{01}}{g}\\ \text{the max. height is}\\ h_1=y(t_m)=\frac 12 \frac{v_{01}^2}{g}\\ \text{with}~ v_{01}^2\mapsto 2\,g\,h_0\,\epsilon^2\\ h_1=h_0\,\epsilon^2$$

Section 2 you start from $h_1~$ with velocity zero, thus you can use the calculation of section 0, and so on, you obtain the next maximal height $y_2=h_0\,\epsilon^4$

the n'th max height is:

$$h_n=h_0\,\epsilon^{2\,n}~,n=1\ldots$$ from here $$n=\frac 12 \frac{\ln\left(\frac{h_n}{h_0}\right)}{\ln(\epsilon)} +1$$ plus one because we have to count the "section 0"

Verification

$h_n=20~,~\frac{h_n}{h_0}=\frac{20}{100}$

$\epsilon=0.9$

$n=\frac 12 \frac{\ln\left(\frac{20}{100}\right)}{\ln(0.9)}+1=8.6=8$

this is also what you see in the simulation, the ball bounces eight times until it reaches 20% of the start height

Answer 2

On the $i^{th}$ impact, as ball comes down and hits, we can consider the very instant of first contact with the floor, while the ball is still a sphere and hasn’t started to deform yet.

It has a velocity of $v_{d,i}$, where $d$ is for downward and $i$ is the $i^{th}$ impact. Assuming little wind resistance and therefore that friction loss is from the ball not having a perfectly elastic collision with the floor:

$$v_{d,i+1} = v_{u,i}$$

We also know that it will rebound to:

$$h_{i+1}=0.6~ h_i$$

The height and potential energy reduce 40% each time.

Every time the ball hits, it’s surface is deformed. The displacement $d$ into the ball (ie how much closer to the floor than $r$ the $com$ of the ball gets) will not be linear, ie $F \neq -kx$. The exponent will be higher than $1$. For example it could be $F=-kx^2$

The energy imparted is:

$$E = mgh_i = \int_{0}^d F dx ~ =(\tfrac{1}{2}mv_{u,i}^2)$$

Although we won’t use the last expression. Each time $h$ and $E$ go down a factor of 0.6. Whenever we get to $h<d$, ball will not go high enough to leave the surface of the floor. This can be estimated by:

$$h_N = h_0 0.6^N ~,~ \text{until} ~ h<d$$

What’s needed now is the force for small displacements of a sphere onto a surface, when all the energy via work is imparted.

We could compress a ball and measure displacement vs force. We would seek the distance $d$ and condition where:

$$ \int_{-d}^0 F dx = mgd$$

Because work put into rebound from rest ($\int_{-d}^0 F dx$) equals the ending potential energy at the apex of ($mgd$). Because we have only rebounded a height equal to the displacement, the surface of the ball does not leave the ground and bouncing is over: We bounce until $h_N=d$.

Barring experiments to get that data, let’s just guess that it is $0.1r$ (the ball displaced a tenth of a radius and let go does not spring back enough to leave the floor surface). This guess will not affect results all that much.

If we start out $K$ diameters above the floor and drop it, then it will bounce until $h$ goes from $2Kr$ to $0.1r$

$$2Kr ~0.6^N = 0.1r \implies -N \ln(0.6) = \ln(20K)$$

The general equation would then be:

$$ N = -\frac{\ln(20K)}{\ln(0.6)} ~,~ \text{for }~ h_0 = KD$$

If we start at twenty diameters above the ground:

It will bounce Twelve Times ($-\tfrac{\ln400}{\ln0.6}$)

It may not seem like many bounces, but 40% loss per bounce is a lot. Finally, it is not that sensitive to our guess. Without an $F(x)$ measurement, even if the $0.1r$ is off by a factor of three, which I doubt (between $0.03r$ and $0.35r$), $N$ will only change by two. So 12 is a pretty good guess. (If instead of $20D$, the initial height is $\frac{20D}{0.6}$, it’ll increase by one bounce, and for $12D$, it’ll decrease by one, etc.)