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Simple bouncing ball puzzle, with $50 prize

sam.ai.ki

21–30 of 58 posts

Re: Simple bouncing ball puzzle, with $50 prize

#21
post #10

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.

Re: Simple bouncing ball puzzle, with $50 prize

#22
post #12

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 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…

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.

Re: Simple bouncing ball puzzle, with $50 prize

#23
post #8

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.

Re: Simple bouncing ball puzzle, with $50 prize

#24

Earlier quoted context omitted.

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 po…

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.

Re: Simple bouncing ball puzzle, with $50 prize

#25
post #22
post #12

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 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…

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 th…

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.

Re: Simple bouncing ball puzzle, with $50 prize

#26
post #21
post #10

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…

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.

Re: Simple bouncing ball puzzle, with $50 prize

#27
post #22

Earlier quoted context omitted.

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 th…

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…

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

Re: Simple bouncing ball puzzle, with $50 prize

#28

Earlier quoted context omitted.

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…

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

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

Re: Simple bouncing ball puzzle, with $50 prize

#29
post #12

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 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…

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 d…

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?

Re: Simple bouncing ball puzzle, with $50 prize

#30
post #21

Earlier quoted context omitted.

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…

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.

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