BALL BOUNCING INFINITELY to find distance covered

[abrahamrenns]

ok, the qn is..
a ball is dropped from H height. it bounces with restitution e. homework much distance will it cover before it comes to rest..? surely, it is not infinity if i am not mistaken, since the bounces become infinitesimally small and then it tends to zero.. thnx in advance for any help..

 

[xlines]

ok, the qn is..
a ball is dropped from H height. it bounces with restitution e. homework much distance will it cover before it comes to rest..? surely, it is not infinity if i am not mistaken, since the bounces become infinitesimally small and then it tends to zero.. thnx in advance for any help..

What did you try to solve your problem?

If on every bounce ball loses some fixed percentage of it's momentum, then you should calculate path traveled depending on starting momentum and sum it up. I am pretty sure you should get geometric sum.

 

[GRDixon]

ok, the qn is..
a ball is dropped from H height. it bounces with restitution e. homework much distance will it cover before it comes to rest..? surely, it is not infinity if i am not mistaken, since the bounces become infinitesimally small and then it tends to zero.. thnx in advance for any help..

You have more or less rediscovered Zeno's paradox. But you're instincts are correct. The key is that the bounces become infinitesimally small, and no matter how great a finite number of bounces (finite amount of time) the distance traveled with always be finite.

 

[abrahamrenns]

can some one please take the pains to actually calculate it? it wudnt be difficult for a few here atleast.. i am not able to do it..

 

[xlines]

can some one please take the pains to actually calculate it? it wudnt be difficult for a few here atleast.. i am not able to do it..

Well, we do have assignments forums, but here you go. First, what distance will travel ball with upwards initial velocity v' ?

v(t) = v' - gt

it will reach max. height in t' = v'/g . Total distance traveled in one jump will be

d(v') = 2*1/2*g*t'^{2} = v'^{2}/g

After each bounce starting velocity is lowered by factor \alpha . So total distance is

D = H + v'^{2}/g + (\alpha v')^{2} /g + (\alpha^{2} v')^{2} /g + ... = H + v'^{2}/g (1+\alpha^{2} + \alpha^{4} + ... )

Substitution \beta = \alpha^{2}
allows you to calculate infinite geometric sum.

Use energy conservation to link starting height H with v' .

 

[HallsofIvy]

That's getting more complicated than necessary!

The original problem told us that "it bounces with restitution e" which means that if it was dropped from height h, then it bounces up to height eh on the first bounce, e(eh)= e^2h on the second, etc. Because it goes the same distance up as down, the total distance will be h+ 2eh+ 2e^2h+ \cdot\cdot\cdot, almost a geometric series. We can make it a geometric series by adding and subtracting h: -h+ 2h+ 2eh+ 2e^2h+ ... where everything after the first "-h" is a gemetric series with first term 2h and constant ratio e. That total distance is
-h+ \frac{2h}{1- e}= \frac{-h+ +eh+ 2h}{1- e}= \frac{1+e}{1- e}h

 

[Phyisab****]

Yep it's a nice problem to work out.

 

[Saw]

The original problem told us that "it bounces with restitution e" which means that if it was dropped from height h, then it bounces up to height eh on the first bounce, e(eh)= e^2h on the second, etc. Because it goes the same distance up as down, the total distance will be h+ 2eh+ 2e^2h+ \cdot\cdot\cdot, almost a geometric series. We can make it a geometric series by adding and subtracting h: -h+ 2h+ 2eh+ 2e^2h+ ... where everything after the first "-h" is a gemetric series with first term 2h and constant ratio e. That total distance is
-h+ \frac{2h}{1- e}= \frac{-h+ +eh+ 2h}{1- e}= \frac{1+e}{1- e}h

That is a nice solution. And if you wanted to fine-tune it... would it be possible to take into account that with every bounce, the ball gets deformed, less elastic and the coefficient of restitution diminishes?

 

[xlines]

That's getting more complicated than necessary!

The original problem told us that "it bounces with restitution e" which means t... total distance is

English is not my native and, to tell you the truth, I thought e plays role something like my \alpha. I too learned something today! Thanks for clearing that up! :)

 

[Zeus0]

It's about right except that on rebounce it's not eh but e^{2}h.
by definition e = \frac{v_2}{v_1} = \frac{\sqrt{2gh_2}}{\sqrt{2gh_1}} or e = \sqrt{\frac{h_2}{h_1}} . Look at the wikipedia entry at http://en.wikipedia.org/wiki/Coefficient_of_restitution" Meaning the finale answer is D = \frac{1+e^2}{1-e^2}h . I also took the liberty to calculate the time it would take T = \frac{1+e}{1-e}\sqrt{\frac{2h}{g}}

I am more curious about weather or not real superball or bouncy balls rebounce an infinite number of time . It is likely not because of imperfections they may have or because of other processes in the way . I would estimate a real superball bounces between 20 to 50 times before resting.