The problem of one tube and two balls on a plane

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The discussion revolves around a physics problem involving a tube and two balls, where participants express confusion about the setup and significance of the tube's infinite mass and its placement on a plane. Key points include the assumption that the tube is finite and closed at both ends, which affects the motion of the balls inside it. Participants debate the implications of the tube's orientation and whether it is stationary or moving, as well as the conditions under which the balls can collide. There is a consensus that the question lacks clarity and detail, particularly regarding the initial conditions and the nature of the collisions. Overall, the conversation highlights the complexities of the problem and the need for clearer definitions to facilitate understanding.
  • #31
phinds said:
Then he should say that, not bring in an infinity.
I said that the mass of the tube is infinite in order to prevent people from calculating the tube's rotation when considering the problem.I didn't expect that it would lead to everyone's misunderstanding. I'm sorry.
 
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  • #32
jbriggs444 said:
Let us consider a simpler and illustrative scenario.

Suppose that we have a rectangular area. Like an ideal air hockey table. We place a single ideal puck on this table. We launch the puck from one of the corners at a randomly chosen angle. We ask whether it will ever bounce back to its starting corner again.

For almost any launch angle we choose, the period for puck bounces between the short walls will not be a rational multiple of the period for puck bounces between the long walls. So there is no time after the start at which an exact integer number of round trips will have occurred for both directions simultaneously.
Your understanding of the problem is completely correct. When there is only one ball, I understand your answer. But when we need to consider the collision of two balls, is the above answer still valid?
 
  • #33
crazy lee said:
Your understanding of the problem is completely correct. When there is only one ball, I understand your answer. But when we need to consider the collision of two balls, is the above answer still valid?
In post #1, q1 is given as "can" the state return exactly, not "will" it.
Which is it? Showing that it can is easy.
 
  • #34
crazy lee said:
First of all, I have tried to solve this problem
SAYING that you have tried to solve the problem is NOT trying to solve the problem. You have to show us what you have tried.
 
  • #35
haruspex said:
In post #1, q1 is given as "can" the state return exactly, not "will" it.
Which is it? Showing that it can is easy.
I'd like to ask whether the following problem is easy to solve. Given that the initial states of two small balls in a tube are unknown constants, find the specific position of the center of mass of the small balls in the tube when the two small balls collide, as well as the velocities of the small balls at that time and the specific time of the collision. Thank you. I want to ask about this specific problem first.
 
  • #36
phinds said:
SAYING that you have tried to solve the problem is NOT trying to solve the problem. You have to show us what you have tried.
phinds said:
SAYING that you have tried to solve the problem is NOT trying to solve the problem. You have to show us what you have tried.
Firstly, this is not a homework problem. Secondly, I've been out of school for more than 20 years, and I've pretty much forgotten all the English and physics I learned back then. Finally, for me, just accessing this website is already a challenge. I would really appreciate it if you could understand me.The most crucial thing is that I haven't found any effective and simple train of thought.
 
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  • #37
I'd like to ask whether the following problem is easy to solve. Given that the initial states of two small balls in a tube are unknown constants, find the specific position of the center of mass of the small balls in the tube when the two small balls collide, as well as the velocities of the small balls at that time and the specific time of the collision. Thank you. I want to ask about this specific problem first.I communicate with everyone through AI translation, so there may be some translation errors. Thank you all for your help.
 
  • #38
In fact, it's still a bit difficult to discuss physical problems with others in a language without having a good command of that language. Besides, I'm not familiar with some of the rules on this website. I have seriously thought about this problem, but I haven't come up with any simple and effective ideas. Even if I take pictures of my notes and upload them, it won't be of any help to you, and the notes are all written in Chinese. So I hope some friends can be understanding. Thank you all.
 
  • #39
Since the objective is to make things simple, we should add these restrictions:
1) The balls track along the floor of the tube - so they follow a strictly 1-dimensional path.
2) When the balls collide with each other, they are not only elastic, but they reverse direction without deforming at all.
3) At least one of the balls has a non-zero starting velocity.

It has already been mentioned in this thread that without end stops, there is no hope of ever returning to the original state. So ...

4) There are end stops and when the balls collide with and end stop, the collision in not only elastic, but the ball reverses direction without deforming.

Initial conditions:
L: The length of the track (inside of the tube) as measured from one collision contact point to the other.
D: The diameter of each ball.
F = L-D: The range of travel for the balls. Positions will be measured from -D/2 (collision contact point 1) to F+D/2 (collision contact point 2).
P1: The initial position of ball 1 (from 0 to F-D).
P2: The initial position of ball 2 (from D to F).
V1: The initial velocity of ball 1.
V2: The initial velocity of ball 2.

Here is a specific case:
Can the balls ever return to their initial conditions:
P1: Any (from 0 to F-D).
P2: Any (from P1+D to F).
V1: 1.
V2: pi
 
  • #40
crazy lee said:
I'd like to ask whether the following problem is easy to solve. Given that the initial states of two small balls in a tube are unknown constants, find the specific position of the center of mass of the small balls in the tube when the two small balls collide, as well as the velocities of the small balls at that time and the specific time of the collision. Thank you. I want to ask about this specific problem first.
The answer to your question is obvious. If the initial states of the small balls in the tube are not known it is impossible, not just difficult, to find the specific position of the center of mass when the balls collide or at any time. Suppose I tell you that at noon I start moving from some unknown position moving with unknown but constant velocity. Would you be able to find my position and velocity some specific time later if I didn't tell you where I started, how fast I am moving and in what direction?

I understand that English is not your language but you need to formulate your question in a way that makes sense. The final velocities and positions depend on the initial velocities and positions. This means that final answers at a given specific time, say 2 minutes, can be found as numbers if the initial positions and velocities are given as numbers or at least as symbols corresponding to numbers. You cannot say that the initial state is unknown and expect to find the final state at some specific time.
 
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  • #41
kuruman said:
The answer to your question is obvious. If the initial states of the small balls in the tube are not known it is impossible, not just difficult, to find the specific position of the center of mass when the balls collide or at any time
There is a difference between the initial conditions being unknown vs. simply not provided. Even without knowing the initial conditions (or all of the initial conditions), one can describe under what conditions the balls will return to their initial state and under what conditions they will not.

As an example, in post #40, I do not specify the initial positions of the balls, but the question (will the balls return to their initial states) can still be answered.
 
  • #42
.Scott said:
Since the objective is to make things simple, we should add these restrictions:
1) The balls track along the floor of the tube - so they follow a strictly 1-dimensional path.
With the given relationship between tube width and ball width, it seems clear we need to allow two dimensions at least, and the intent may even be to allow three.
.Scott said:
4) There are end stops and when the balls collide with and end stop, the collision in not only elastic, but the ball reverses direction without deforming.
… and the end stops are flat discs normal to the tube axis.

Two simple cases to start with would be an initial contact at the centre of the tube, equal and opposite longitudinal velocities, equal or equal and opposite lateral velocities.
 
  • #43
crazy lee said:
I'd like to ask whether the following problem is easy to solve. Given that the initial states of two small balls in a tube are unknown constants,
Do you mean known but arbitrary, i.e. not special values? If so…
crazy lee said:
find the specific position of the center of mass of the small balls in the tube when the two small balls collide, as well as the velocities of the small balls at that time and the specific time of the collision.
Since the system is deterministic, yes, but the equations would be messy.
 
  • #44
haruspex said:
Since the system is deterministic, yes, but the equations would be messy.
No, with the limits I put on it, there is nothing messy.
Although, I will also assume no friction and no spin. Just two balls bouncing back and forth along a linear track.
 
  • #45
.Scott said:
No, with the limits I put on it, there is nothing messy.
Sure, but as I posted, I feel sure it is supposed to be 2D and maybe 3.
 
  • #46
haruspex said:
With the given relationship between tube width and ball width, it seems clear we need to allow two dimensions at least, and the intent may even be to allow three.

… and the end stops are flat discs normal to the tube axis.

Two simple cases to start with would be an initial contact at the centre of the tube, equal and opposite longitudinal velocities, equal or equal and opposite lateral velocities.
Your judgment in this post is completely correct. This is exactly what I meant to express.
 
  • #47
haruspex said:
Do you mean known but arbitrary, i.e. not special values? If so…
Yes, I mean that the initial state of the small balls is known, but it's not represented by definite numbers.
 
  • #48
.Scott said:
No, with the limits I put on it, there is nothing messy.
Although, I will also assume no friction and no spin. Just two balls bouncing back and forth along a linear track.
Thank you very much for your help. This is indeed a two-dimensional problem. A friend of mine previously compared my problem to ice hockey pucks moving on an air hockey table, and that metaphor is really quite appropriate.
 
  • #49
I would like to add three more questions to help everyone understand the original intention of my question. The direction of the length of the tube is set as the x-axis, and the direction of the width is set as the y-axis. The two small balls A and B in the tube are exactly the same, and the collisions are ideal collisions. Friction and spin are not taken into account. The initial state is known, but not represented by specific numbers. It is expressed by letters representing constants. The total energy of the two small balls is e. Then, after a sufficient number of collisions, the questions are as follows:

4. When the two small balls collide, record the motion state at the moment of the collision. If the component of the velocity of the small ball on the y-axis is positive, mark it as 1. Otherwise, mark it as 0. Then, for each collision, a set of two-digit binary numbers can be obtained. Does the change of this set of binary numbers follow a certain pattern? Is there any periodicity?

5. Is the average velocity of the components of the velocity of the small balls on the x-axis and y-axis exactly equal to the ratio of the length to the width of the tube?

6. If the sum of the energies of the two small balls and the ratio of the length to the width of the tube are fixed, does it mean that regardless of the initial states of the two small balls, the motion states of the two small balls will eventually show similar periodic oscillations? What rules should these oscillations follow?

In fact, I want to obtain two sequences. The first sequence records the position of the center of mass of ball a each time it collides, and the second sequence records the velocity of ball a after the collision. However, I don't know how to calculate them. Or is there any other way to get the answers to the above six questions without calculation? Thank you all very much.
 
  • #50
haruspex said:
Sure, but as I posted, I feel sure it is supposed to be 2D and maybe 3.
Your judgment is completely correct. Besides, I made supplementary explanations to the question in post 50. Thank you.
 
  • #51
kuruman said:
View attachment 358524If the pucks are identical, we know that their CM at the time of puck-to-puck collision is at the point of contact. This allows mapping the region where the CM can be at any time. It is shown in blue in the figure on the right with the pucks at one of the extremal positions. The ratio of the tube width to twice the tube diameter is 7/9. To describe the motion, one can imagine the CM bouncing elastically inside the blue hexagonal area and then write some equations.
I made supplementary explanations to the question in post 50. I would be extremely grateful if you could take a look at it.
 
  • #52
kuruman said:
To describe the motion, one can imagine the CM bouncing elastically inside the blue hexagonal area and then write some equations.
The CM will not bounce elastically.

Total energy is conserved, of course. But that total energy can be seen to be divided into the bulk kinetic energy of the center of mass and the internal kinetic energy in the relative motion of the individual balls. The relative velocity of the balls is not constant. So the internal kinetic energy is not constant. So the center of mass energy is not constant.
 
  • #53
In post #1 you ask
crazy lee said:
1. After a sufficient number of collisions, can the positions and velocities of the two small balls return to the initial state simultaneously? Or can only the positions return to the initial state?
In post #50 you say
crazy lee said:
4. When the two small balls collide, record the motion state at the moment of the collision. If the component of the velocity of the small ball on the y-axis is positive, mark it as 1. Otherwise, mark it as 0.
What do you mean by "motion state"? Is it position and velocity or just velocity?

Also, if you consider what happens when two masses collide, you will understand that the velocity "at the moment of the collision" is not a well-defined quantity. The velocity before the collision is well-defined and so is the velocity after the collision. During the collision while the masses are in contact, the velocity of each mass is changing continuously from its "before" value to its "after" value. So what are we supposed to record, the velocity before or after the collision?
 
  • #54
crazy lee said:
I want to obtain two sequences. The first sequence records the position of the center of mass of ball a each time it collides, and the second sequence records the velocity of ball a after the collision.
That is not possible because the sequence keeps forking according to whether a state value is more or less than a threshold. E.g. at one angle it will strike the side of the tube next, at a slightly different angle it will strike the end of the tube next.

Try to answer the first question, which I interpret to mean "is there an initial state (positions and nonzero velocities) such that it will eventually return to that state?".

Next, suppose that just after the balls have collided ball A has velocity vector ##v_x\hat i+v_y\hat j##. What are the possible velocity vectors for A when next they are about to collide?
 
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  • #55
jbriggs444 said:
The CM will not bounce elastically.

Total energy is conserved, of course. But that total energy can be seen to be divided into the bulk kinetic energy of the center of mass and the internal kinetic energy in the relative motion of the individual balls. The relative velocity of the balls is not constant. So the internal kinetic energy is not constant. So the center of mass energy is not constant.
I agree that the relative velocity of the balls is not constant. However, the CM cannot be said to "bounce" elastically (or otherwise) until the collision is complete. Let me explain.

Shown below is a sequence of frames illustrating the collision of the CM with the wall. The balls have equal speeds ##v## along the horizontal. Velocity arrows are omitted to reduce clutter. The color code is: Lime green → moves to the Left; Red → moves to the Right. The CTC (center-to-center) separation between balls is ##d## and the red dashed vertical line is parallel to the wall at the point of closest approach of the CM.

The onset of the collision of the two-mass system is when the left mass hits the wall at ##t=0## as shown in the first frame; it is over when both masses have reversed direction and the information that one part of the system has collided with the wall is transmitted to the other part. This is shown in the last frame. The collision duration is ##\Delta t=\dfrac{d-2R}{v}.##
Balls in Tube_3.png


This is a simple 1D example with the common velocity in the CTC direction. For a tube of length ##L##, the period of horizontal oscillations is $$T_{\text{hor}}=2\times\left(\Delta t +\frac{L-2d}{v}\right)=\frac{2}{v}(L-2R)$$and is independent of the separation ##d##.

Similarly, it can be shown that if the CTC is horizontal, but the common velocity is vertical, the period of vertical oscillations for a tube of width ##W## is $$T_{\text{ver}}=\frac{2}{v}(W-2R)$$as the balls bounce up and down without colliding with each other.

I stop here. This is a live homework problem after all.
 
  • #56
kuruman said:
In post #1 you ask

In post #50 you say

What do you mean by "motion state"? Is it position and velocity or just velocity?

Also, if you consider what happens when two masses collide, you will understand that the velocity "at the moment of the collision" is not a well-defined quantity. The velocity before the collision is well-defined and so is the velocity after the collision. During the collision while the masses are in contact, the velocity of each mass is changing continuously from its "before" value to its "after" value. So what are we supposed to record, the velocity before or after the collision?
In Question 4, the state of motion refers to the velocity of the small balls before the collision, specifically at the instant just before the collision. If the component of the velocity along the y-axis is positive, it is marked as 1, and if it is negative, it is marked as 0. Then, the velocities of the two small balls at the instant just before each collision can be represented by a set of two-bit binary numbers. I would like to ask if there is any pattern in the change of this set of two-bit binary numbers. Thank you.
 
  • #57
kuruman said:
I agree that the relative velocity of the balls is not constant. However, the CM cannot be said to "bounce" elastically (or otherwise) until the collision is complete. Let me explain.

Shown below is a sequence of frames illustrating the collision of the CM with the wall. The balls have equal speeds ##v## along the horizontal. Velocity arrows are omitted to reduce clutter. The color code is: Lime green → moves to the Left; Red → moves to the Right. The CTC (center-to-center) separation between balls is ##d## and the red dashed vertical line is parallel to the wall at the point of closest approach of the CM.

The onset of the collision of the two-mass system is when the left mass hits the wall at ##t=0## as shown in the first frame; it is over when both masses have reversed direction and the information that one part of the system has collided with the wall is transmitted to the other part. This is shown in the last frame. The collision duration is ##\Delta t=\dfrac{d-2R}{v}.##
View attachment 358733

This is a simple 1D example with the common velocity in the CTC direction. For a tube of length ##L##, the period of horizontal oscillations is $$T_{\text{hor}}=2\times\left(\Delta t +\frac{L-2d}{v}\right)=\frac{2}{v}(L-2R)$$and is independent of the separation ##d##.

Similarly, it can be shown that if the CTC is horizontal, but the common velocity is vertical, the period of vertical oscillations for a tube of width ##W## is $$T_{\text{ver}}=\frac{2}{v}(W-2R)$$as the balls bounce up and down without colliding with each other.

I stop here. This is a live homework problem after all.
Thank you for typing so many words and drawing the diagrams. After studying for a long time, I finally understood it. You gave a special example and explained the oscillation periods of the center of mass of the two balls in the horizontal and vertical directions. Thank you.
 
  • #58
haruspex said:
That is not possible because the sequence keeps forking according to whether a state value is more or less than a threshold. E.g. at one angle it will strike the side of the tube next, at a slightly different angle it will strike the end of the tube next.

Try to answer the first question, which I interpret to mean "is there an initial state (positions and nonzero velocities) such that it will eventually return to that state?".

Next, suppose that just after the balls have collided ball A has velocity vector ##v_x\hat i+v_y\hat j##. What are the possible velocity vectors for A when next they are about to collide?
I also know it's quite difficult. So is it possible to calculate the first three terms or the first term of this sequence? Besides, I'd like to ask you, how can I find the post I made two years ago? Which icons or texts should I click on? I also want to ask, can a post be modified after it is published? That is, attach the modified content to the end of the original post instead of directly modifying the original post. Thank you.
 
  • #59
In fact, this problem of two balls in a tube is the first one in a series of problems. It is a problem that has been repeatedly simplified from other problems. This problem of two balls in a tube has two simple variations. For the first variation, with all other conditions remaining the same, another identical small ball is added into the tube. The question is whether the speed of the small ball in the middle is slightly slower than that of the other two small balls, or vice versa. For the second variation, the width of the tube is widened to 2.1 times the diameter of the small ball. The question is what is the probability that when the two small balls meet in the length direction of the tube, they pass through each other through the gap and reach the other end of the tube. It seems that it will be even more difficult to figure out these two problems.
 
  • #60
crazy lee said:
In fact, this problem of two balls in a tube is the first one in a series of problems.
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