3 ms·

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 t

by bravura 16y ago

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.

greenlblue 16y ago
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.
- mfukar 16y ago
  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?
  - utoku 16y ago
    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.
    - mfukar 16y ago
      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?
cperciva 16y ago
Spoiler: That was my first guess.
paulofisch 16y ago
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.
- greenlblue 16y ago
  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.
  - paulofisch 16y ago
    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.
    - greenlblue 16y ago
      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.
deleted 16y ago
[deleted]
rbabich 16y ago
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 16y ago
  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 16y ago
    The series is infinite, but the sum of the series can still be finite. These two things are not at odds.
    - greenlblue 16y ago
      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.

by bravura 16y ago

greenlblue 16y ago

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.

mfukar 16y ago
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?
- utoku 16y ago
  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.
  - mfukar 16y ago
    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?

cperciva 16y ago

Spoiler: That was my first guess.

paulofisch 16y ago

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.

greenlblue 16y ago
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.
- paulofisch 16y ago
  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.
  - greenlblue 16y ago
    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.

deleted 16y ago

[deleted]

rbabich 16y ago

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 16y ago
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 16y ago
  The series is infinite, but the sum of the series can still be finite. These two things are not at odds.
  - greenlblue 16y ago
    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.