Simple bouncing ball puzzle, with $50 prize | Hacker News Reader

Simple bouncing ball puzzle, with $50 prize

sam.ai.ki

16 pointssswam53 comments

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.

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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.

It's true that the horizontal energy doesn't change and it keeps going to the right but it stays in the air for less and less on each bounce and travels a shorter and shorter distance on each bounce assuming the ball only goes forward when it bounces but you still didn't answer the question because if you sum the total time the ball will spend in the air you will get a finite number so the ball will travel a finite distance which is a much better description of what happens to the ball than just saying it keeps going to the right.

The problem statement says 'ignore friction'. Doesn't that imply the ball's initial energy (when thrown, presumably, but it doesn't matter) is conserved? How, then, will it travel a finite distance, since we are not told of any obstacles?

The problems with questions in imagined worlds is that sometimes it is hard to get the premises right. Even if you treat the problem as a thought experiment, it feels uncomfortable because:

- Ball bouncing with 60% energy remaining all the time is actually a premise, a rule, because we are not supposed to consider atoms or their interactions. There is no reason to it, it becomes a fact.

- Ball is a singular object. We don't have atoms, so we don't have to think things like "What happens when the height of the bounce gets shorter than the size of the atom?" The concept of the ball becomes a premise.

- We get a picture that shows the balls losing speed in direction x at each bounce, but since the question asks for us to judge "qualitatively", we will omit that. We cannot have observations for this problem, it is not the real world.

- There is no friction, or spin, so vertical speed cannot be transfered into horizontal speed and vice versa.

- The concept of bounce might be different, this question probably assumes it happens at 0 (instant) time without deforming the ball, since remember our ball is singular.

- The rest we can probably treat with Newtonian physics in Euclidean geometry, no air, interaction between ball and surface frictionless, etc.

But it feels uncomfortable, because I too can make up an imaginary world for myself, dress it like the real world and ask my question and hide the premises behind.

I'm not sure what you're trying to say. I understand you might feel 'uncomfortable' hiding electromagnetic forces and quantum mechanics behind Newtonian physics (how so?), but how's that relevant to the problem, or its answer?

Spoiler: That was my first guess.

This assumes that each bounce takes the same time to complete instead of also tending to zero.

If you expect that each bounce also takes a smaller amount of time to complete, a series of infinite bounces takes a finite amount of time and therefore a finite distance.

At the point in time and distance where the bouncing ends, the forward motion of the ball (unimpeded by friction) continues in a slide along the ground.

Ah but it takes infinite amount of time for the ball to stop and start rolling on the floor and infinite amount of time means the ball bounces infinitely often so the ball never stops to bounce and never rolls along the ground. It is easy to calculate how long the ball stays in the air on each bounce a formula from high school physics tells you that the potential energy of an object is m x g x h so if you know the potential energy then you know how high the ball is and if you know how high the ball is then you know how long it will take for it to fall to the ground but no matter how long you wait the ball will have some potential energy left so it will be bouncing no matter how long you wait.

The bounces are an infinite series, each with a time attached to them tending to zero.

Time itself is not a series, this is where the fallacy creeps in.

A infinite series can have a finite sum.

True but then in what order are you calculating things. In order to have an infinite series for the air time to sum you must assume there are infinitely many jumps so if there are infinitely many jumps then the ball never rolls on the floor by definition. But then if you say you sum the infinite series of time intervals and the ball stops at that time then you don't have infinitely many bounces because if there were infinitely many bounces the ball would not roll on the floor. So you are missing something somewhere.

http://en.wikipedia.org/wiki/Geometric_series

See also: http://en.wikipedia.org/wiki/Zenos_paradox (Hint: It's not really a paradox.)

This doesn't address the fact you are making a logical fallacy. You can't make a calculation assuming the ball bounces infinitely often and then after the calculation go back and say the ball stops bouncing because it invalidates your original calculation.

In addition to the fallacy that the ball will keep bouncing for an infinite amount of time, I'd like to add that the ball will never roll even when it stops bouncing, because of the frictionless environment, it will slide instead.

Unless of course it was rolling in the first place and I missed/misread it.

This is true but I was just looking for the 'infinite number of bounces in a finite time' part; after that it would slide or roll without bouncing, or sit still if there had been no horizontal component initially.

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.

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.

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

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.

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.

no rolling without friction.

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.

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.

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.

The little boy picks it up and throws it again?

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

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

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.

just roll away

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

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.

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.

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.

This is indeed a very nice puzzle.

thanks :)

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.

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

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.

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.

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! :-)

My experience has been that although the topics needed are covered in high school physics and math (limits), most people did not get it right.

Where's the spoiler?

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.

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.

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.

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.

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.

You don't give up, do you? An infinite series can have a finite sum. Yes, the ball bounces infinitely often, but it does that in a finite amount of time.

I'm not disagreeing with you. Yes, an infinite series can have a finite sum, no argument there but what I am arguing with is that people are saying the ball stops bouncing after time t = whatever. Everyone is confounding two things here, air time and bouncing. Yes, the air time is finite but the bouncing isn't. So you can't say it stops bouncing after time t = whatever if you calculate t = whatever by assuming infinitely many bounces which is what everyone is doing because they are summing a geometric series where the terms of the series represent the air time of each bounce.

There is a time t such that after time t the height of the ball's bounces is zero.

Personally, I call that "not bouncing".

Ok, now we are in agreement because my definition of "stops bouncing" was finitely many bounces.

Bzzt Still wrong.

If you have infinite bounces, and I ask you "Which number bounce in the series happens at exactly time X?", there is a time for X for which you will not be able to give an answer.

This is because there is a limit for the latest time at which bounces happen.

Yes, even with infinite bounces.

Whatever. I got it sorted out so go bother someone else.

I suspect this is what the intended solution to the puzzle is... Now for a follow up: Instead of assuming no friction, assume no slippage. What happens now?

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.

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