

Simple bouncing ball puzzle, with $50 prize - sswam
http://sam.ai.ki/ball.html

======
bravura
edit: I thought this over more, here is a new answer.

Spoiler: _it bounces an infinite amount of time in a finite space._

Actually, the horizontal distance it travels is proportionately decreased with
the decreased time in the air. If the ball is now in the air for k<1 times as
much time, it goes to the right a factor of k times as much. I forget but I
think these recursive series converge to a finite sum, they definitely do if
k<0.5. So the height and length of each bounce decays exponentially, and _it
bounces an infinite amount of time in a finite space._

\---

Old, wrong answer.

Spoiler: It keeps going to the right to infinity, just at a height that is
vanishing and asymptotically approaching zero.

Each time the ball bounces, it loses 40% of its vertical kinetic energy. But
the problem statement ("the ground is flat, and each part of the ball’s path
is a parabolic arc. Don’t consider friction, atoms, relativity, quantum,
etc!") indicates that the horizontal energy doesn't change, even though the
figure would suggest otherwise. So it will keep going to the right at the same
rate.

~~~
rbabich
Of the responses so far, this one is closest to being correct. Rather than
"asymptotically approaching zero," however, the height of the bounce will
quickly converge precisely to zero. Assume that the previous bounce (up and
back down) took time t. Then the ball will stop bouncing after time
t/(1-sqrt(0.6)) ~ 4.4t. After that, the ball will simply continue moving
("rolling") to the right. This follows from summing the geometric series 1 +
sqrt(0.6) + sqrt(0.6)^2 + sqrt(0.6)^3 + . . . , where 0.6 = (1 - 0.4) is the
ratio of the height of the next bounce to the current bounce, and we take the
square root since height and time are related by h = 1/2 at^2.

Incidentally, for anyone who has a ping pong ball handy, this is very close to
what happens in real life.

Edit: To clarify, it's the parent's "old, wrong answer" that's closer to being
correct. btilly (below) also has it right.

~~~
greenlblue
No, you are making the same mistake as some of the other people. You are
summing a geometric series so you are saying the ball bounces infinitely often
and the time it takes for the ball to bounce infinitely often is blah. But if
it bounces infinitely often then it never stops to roll on the floor because
if it did stop and roll on the floor in a finite amount of time then you
wouldn't have an infinite series to sum which would mean that the potential
energy in the horizontal direction would be zero in a finitely many bounces
which contradicts the problem statement and part of your original reasoning.

~~~
paulofisch
The series is infinite, but the sum of the series can still be finite. These
two things are not at odds.

~~~
greenlblue
Ya, and what are the terms in the series representing? Is it air time of each
bounce? Your calculation makes it clear that it is. So you are saying you are
calculating the air time for _infinitely_ many bounces, the key word here is
_infinite_ , i.e. the ball bounces up and down, up and down infinitely often.
So if the ball bounces infinitely often how can it stop and roll on the floor
because _infinitely_ many bounces means not stopping after finitely many
bounces and rolling on the floor. Your calculation for the time is confounding
two things, air time and bouncing. You can calculate the air time assuming
infinitely many bounces but then you can't go and claim that the ball stops
bouncing after t = whatever because you calculated t = whatever assuming the
ball never stopped bouncing and then after the calculation went back and
changed your assumption, that is a logical fallacy if I ever saw one.

~~~
mkarmac
1) The ball stops bouncing in a finite _amount of time_.

2) The ball bounces an infinite _number of times_.

3) This is not a contradiction, no more than the idea that a projectile passes
through an infinite number of spatial points in finite time. Please go and
read about geometric series and Zeno's paradox on wikipedia, as another
commenter has already suggested.

------
greenlblue
So I did some math starting with 100J of potential energy and summed a
geometric series to see how long it would stay in the air if we had infinite
time to sum the geometric series and it was some finite number. Then I assumed
the initial velocity in the horizontal direction was 1 m/s so if we had
infinite time the ball would travel some finite distance because it would
bounce infinitely often and in every such bounce it would be in the air some
amount of time and would travel at a speed of 1 m/s in that amount of time.
But this assumes that the ball doesn't roll along the floor when it comes down
and immediately bounces up as soon as it hits the floor which is unrealistic
because it means I'm assuming infinite impulse on each bounce because the
momentum of the ball changes and I'm assuming this takes 0 seconds. So the
best I was able to do was give an upper bound on the total distance the ball
would travel if the universe were to last forever. It's a lot like Zeno's
arrow or is it Achilles and the tortoise.

------
drallison
Any horizontal component to the motion can be ignored. The only thing of
interest is the vertical motion of the ball in a gravitational field.

There is an elastic collision between the ball and the flat surface. Unlike an
inelastic collision, energy is not conserved. Energy is lost when the ball
hits the surface and rebounds; it shows up as heat in the ball and at the
surface at the point of collision. Assuming the surface is very hard, most of
the heat goes into the ball.

At some point the kinetic energy remaining is not enough to lift the ball
against gravity so the ball does not get lifted off the surface. If you follow
the center of mass of the ball, it continues to oscillate, compressing and
expanding elastically until the remaining kinetic energy is expended as heat.
As the size of the oscillations get smaller and smaller you eventually reach a
scale where the idealized model of the ball begins to fail; at that point,
things become complicated.

------
hardy263
I love how there's so many calculations with energy and time. Here's my
offering with just concept.

The first thing they teach you in projectile motion in grade 12 physics is
that the horizontal component is independent from vertical component. That
means, if you ignore friction, and you throw a ball in an arc, you'll find
that the horizontal speed is linear! This surprises many people, since it's
not very intuitive. You would expect the horizontal speed to be quadratic or
non-linear, which is not the case. If the ball loses 40% of its _vertical_
energy, it means that it'll just keep bouncing, but at lower and lower
heights, but the horizontal speed is continuous. In fact, if given the initial
velocity and angle, you could calculate the horizontal distance traveled by
the ball for any point in time.

------
barrydahlberg
The little boy picks it up and throws it again?

~~~
cperciva
There is no air resistance in this world, hence no air. The little boy
asphyxiated shortly after the first time he dropped the ball. :-)

~~~
barrydahlberg
Hah! Let's hope it wasn't the little boy on his home page then...
<http://sam.ai.ki/>

------
retube
The ball has horizontal energy as implied by the parabolic arc. Assuming no
friction the ball's horizontal velocity will be constant, hence it will just
roll away at whatever initial horizontal velocity it had.

~~~
cperciva
_just roll away_

Technically, it will _slide_ , since without friction it can't obtain any
angular momentum.

~~~
retube
yes, you are correct: the ball will not "roll", it will continue in a straight
line at constant v and ang. momentum, whatever it had initially.

------
TamDenholm
I'm no maths/physics geek so I'm just guessing here, but I'd assume that it
loses energy at a far more rapid rate and maybe bounces once more and then
just rolls along the ground.

My reasoning is based on the fact that while the ball absorbs 40% of the
energy on the higher bounces there is a minimum amount of energy it absorbs
and on the smaller bounces the energy left meets the minimum amount of energy
the ball will absorb and then simply stop bouncing.

Like i said this is pure conjecture and I may not have even articulated it
right.

~~~
cperciva
Generally things tend to get more elastic at low energies, not less; so I
think your 'minimum amount of energy the ball will absorb' hypothesis is
wrong. However, in the real world, there are other factors which drain energy
from bouncing balls -- air resistance, for example.

But this is all irrelevant, since the problem is explicitly in a non-
physically-accurate world.

------
cperciva
This is indeed a very nice puzzle.

~~~
sswam
thanks :)

------
burgerbrain
I'd love to know how many people have swamped him with the correct answer so
far. Any junior high physic student could answer this.

~~~
cperciva
I'm more curious to know how many people have sent in _incorrect_ answers. My
immediate guess disagreed with my mathematics.

~~~
btilly
My immediate guess agreed with my mathematics. But I have the advantage that a
variant of this problem had occurred to me some 20 years ago, and I worked it
out.

In fact, pick up a bouncy ball and drop it on a flat surface. You can observe
the fact that bounces become smaller and more rapid, and then they stop
bouncing entirely. There may be some residual vibration that is not apparent,
but it sure seems to act like this toy model of the situation says it should.

~~~
cperciva
_In fact, pick up a bouncy ball and drop it on a flat surface. You can observe
the fact that bounces become smaller and more rapid, and then they stop
bouncing entirely._

Indeed, I've done that -- but my intuition told me that it was stopping due to
friction, not due to the exponential decay of an infinite number of bounces.
:-)

It's oddly disappointing to realize that the model actually acts more or less
the same as the real world for once.

~~~
btilly
_Indeed, I've done that -- but my intuition told me that it was stopping due
to friction, not due to the exponential decay of an infinite number of
bounces. :-)_

But your intuition was correct! Without friction the bounces would be
perfectly elastic and wouldn't go into exponential decay! :-)

------
pero
Where's the spoiler?

~~~
btilly
Here is a spoiler.

Energy = force * distance

After each bounce you are left with 60% of its energy, so it comes back up 0.6
times as high as the previous bounce.

Distance falling in time t is proportional to the square the time, so each
bounce takes sqrt(0.6) times as long as the previous bounce did.

Thus the timing of the bounces forms a geometric series. It is well known that
the sum of such a geometric series is 1/(1-r). In this case r = sqrt(0.6)
which is roughly 0.774596669241483 and so from the time it first hits the
ground to the time it it finishes bouncing is approximately 4.43649167310371
times as long as the time for the first full bounce. But we didn't start with
a full bounce, we dropped the ball. Thus we start with a half-bounce, followed
by a full bounce that takes 2 * sqrt(0.6) times as long, followed by the rest
of the sequence. This works out to be 7.87298334620742 times the time it took
to initially fall to the ground the first time.

Hopefully I haven't made any silly mistakes. If I have, correct the error and
the general analysis is correct.

~~~
greenlblue
Same mistake as everyone else. Geometric series means infinitely many bounces
and infinitely many bounces means it never stops bouncing. Everyone is making
the same logical fallacy.

~~~
btilly
I refer you to Zeno's paradox for an example of how a geometric series can
allow an infinite number of things to happen in a finite time.

In this case the time taken forms a geometric series, and the total time taken
is the sum of that geometric series. Which means that, for the same
mathematical reasons that let Achilles catch the tortoise, it stops in finite
time.

~~~
greenlblue
I am familiar with Zeno's paradox and all other things Zeno and infinite
series summing to finite things but there is fallacy here that nobody seems to
get. Yes the time taken is indeed a geometric series but if the time taken is
a geometric series, an infinite one at that, then that means there are
infinitely many bounces, no? So if there are infinitely many bounces how can
you claim the ball stops bouncing? You are confounding two things, air time
which is indeed finite because it forms a geometric series, and the number of
bounces. You can not have an infinite geometric series to calculate the time
and only have finitely many bounces.

~~~
sswam
The ball bounces infinitely many times, in a finite period of time, and then
stops! :) You can model this completely, i.e. describe height as a function of
time. You can know exactly where the ball is at any particular time, and it
follows a continuous curve. So it does make sense as a model, although it is
perhaps a bit mind-bending.

------
sswam
Congratulations to everyone who figured it out, and especially to Glenn from
Alaska who was the first to email me with the correct answer, and wins $50. In
his words:

"The ball makes an infinite number of progressively smaller bounces in a
finite amount of time, and then proceeds to slide (roll?) along the ground at
a constant speed."

I was just looking for "infinite number of bounces in a finite amount of
time/distance".

I think it's a nice puzzle, because it illustrates Zeno's 'paradox', with a
simple model of an everyday occurrence. The answer makes sense, but it is not
obvious unless you understand limits. I thought of this puzzle while playing
with a pool cue, you can really hear/feel them bouncing faster and faster.

