Abacus

Here’s the deal. I drop a ball from rest onto the floor. It hits the floor and bounces back up to a height less than its original height before falling to the floor again. It does this repeatedly until it finally comes to rest on the floor.

Here’s the question. How much time elapses between when the ball is initially dropped and when it finally comes to rest on the floor?

Assume that the ball is initially dropped from rest at a height h = 1 meter from the floor. Assume that it is dropped in a uniform gravitational field with g = 10 meters/second^2. Assume that the ball is dropped in a vacuum (no air friction). Assume that with each bounce, the ball loses exactly one-half of its energy. Assume that the time the ball spends in contact with the floor during each bounce is vanishingly small (zero).

Undercurrent

It never comes to rest on the floor under those approximations. Even if it loses half its energy at each bounce, it still always has some positive energy to bounce back with. The bounces get exponentially smaller, eventually being swamped by random thermal vibrations, but never stop completely.

trunks2k

But wouldn't there be a point at which the energy of the ball leaving ground is not able to overcome the energy of the force exerted on the ball by gravity (m*G)? Sure, it will never COMPLETELY stop moving, but it is still no longer leaving the ground, which is what I believe the problem is asking.

simian

The total distance the ball should travel is something that can be calculated: it will bounce up 1/2 the original distance, then 1/4 the original distance, then 1/8 the original distance ....

Now, for the time, some arbitrary value will need to be assigned for when the ball is "at rest" - or there could be some nice calculus work done (don't as me I am merely an engineer, not a mathematician) - when the bounces get really small, each one will take such a small increment of time that it may be ignored.

Well, here is something that may be a starting point... Perhaps Goliath will see this - he may be able to answer it - & maybe tell me my assumptions are complete and utter crap....

Simian

Lobstrosity

Ok, let's see. Immediately after the nth bounce the ball has kinetic energy En = mg/2^n, which will take it to a height of 2^-n. The time needed to go from the floor to this height and back to the floor again is:

tn = 2 * sqrt((2^-n)/5) = 2/sqrt(5) * 2^(-n/2)

Total time is just the sum of tn from n = 0 to infinity - 1/2 to

I'd need to think for a moment about how to evaluate that power series, but I would bet it converges.

Ok, I thought about it and you can break it into two power series whose sums you know

sum tn from n = 1 to infinity can be written as the sum of ti where i are all the even numbers from 1 to infinity + sum of tj where j are all the odd numbers from 1 to infinity

What does this do for you? Well the first series of the sum is just a normal power series. The second series of the sum is a normal power series times 1/sqrt(2).

1 + x + x² + x³ + ... = 1/(1-x)

so the answer is (2 / sqrt(5)) * (1/(1-.5) + sqrt(.5)/(1-.5)) - 1/sqrt(5)

Edited to remove some horribly incorrect simplifications

Abacus

I took a power series approach to solving this problem. With my assumptions, the ball would bounce an infinite number of times, but in a finite amount of time. I guess it's this finite time I had in mind when I used the term "comes to rest."

Anyways, I'd be interested to see if anybody comes up with the same answer as I did.

Lobstrosity

I take it you did not come up with 1 second?

Edit: oops, in my above calculation I confused a sqrt(.5) with a sqrt(5). My simplification was incorrect. Hmm, it also seems I confused dividing by 1/2 with dividing by 2. Ok, so now my simplification is giving me:

(3 + sqrt(1/2))/sqrt(5)

Abacus

No. I got 2.61 seconds (rounded off). I'm not sure if that's right. I'm comparing your solution to mine to see who did something wrong.

Abacus

Alright, you're very close Lobstrosity. I think you want to take that sum from n=1 to infinity, rather than 0 to infinity. It is only after the first bounce that the ball goes through a complete cycle of up and down. To that sum, you will want to add the time for the first half-cycle (the initial drop).

Also, it's probably not necessary to split that sum into two sums. I used the formula for a geometric series:

sigma(n=0 to inf) ar^n = a/(1-r) ,|r| < 1 where a is the first value in the series and r is the ratio of the n+1th value to the nth value in the series.

Lobstrosity

Power series summation formulas range from n = 0 to infinity, which is why I took the sum starting from n = 0. I accounted for this, however, by subtracting the time for the first half of the cycle (since by adding To I was effectively counting this first drop twice). How did you manage to start from n = 1 while still using that power series relation? Also, the power series relation you used isn't valid. You can see yourself that it requires n to have only non-negative integral values. You need to split it up into two pieces (I believe) in order to make n be an integer. Otherwise, you would find that

sum(n = 0 -> inf) ar^n = a/(1-r) = sum(n = 0 -> inf) ar^(n/2)

when clearly this shouldn't be true.

Basically, you have the height is linearly proportional to energy. Energy is halving each time, so height is halving each time. Height relates to time like

h = (1/2)10 t^2

Because of that extra power of 2, you have that time is a power series in .5^(n/2), not .5^n