Why do two balls launch from the right in a Newton's cradle instead of just one?

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In summary, the conversation revolved around the famous Newton cradle and the question of why two balls launch from the right when two balls are released from the left. The answer lies in the conservation of momentum and energy, which dictate that two balls will be knocked in order to satisfy both laws. The conversation also touched on the idea of an "ideal" universe and the role of dispersionless signal propagation in explaining the behavior of the Newton cradle. It was also proposed that gluing the two motive balls together or using a single ball with twice the mass would result in a different outcome. However, the conversation concluded that there are more than just two conserved quantities at play in this system, making it difficult to predict the exact result.
  • #1
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TL;DR Summary
Newton's cradle
That question came from one of the students in my classical mechanics class, it took me unprepared at first but was able to give the correct answer in like half a minute. They liked it a lot and it initiated a useful discussion about laws of nature extending even to some particle physics concepts like Compton scattering. So I wanted to share it here also... Tutors don't give the answer so quickly please :smile:

The famous Newton cradle: You have several balls hanging. You release one from left, it hits the balls and the rightmost one launches. Everybody has seen this. Now, another case. You release 2 balls together from the left and you observe that 2 balls launch from right. The question is WHY? Why not a single ball from right but two to be more clear...
 
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  • #2
Conservation of momentum and energy are why two balls are knocked. If the final was only one, it would go higher than the initial two. The PE is transferred to KE to maintain velocity.
 
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  • #3
osilmag said:
If the final was only one, it would go higher than the initial two.
And why can't this happen? There is no conservation law for velocity.
 
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  • #4
fresh_42 said:
And why can't this happen? There is no conservation law for velocity.
Not per se, but there is a conservation of energy, for which velocity is the key component here. If the mass changes you couldn't satisfy both conservation of momentum and energy at the same time because of the V2 term in KE.
 
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  • #5
I predict a different result if the 2 'motive' balls are super-glued together.
 
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  • #6
Dullard said:
I predict a different result if the 2 'motive' balls are super-glued together.
Why? Asking, not arguing.
 
  • #7
Ok, for momentum and energy arguments here is a configuration which conserves both momentum and energy... :smile:
243954
 
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  • #8
Nugatory said:
Why? Asking, not arguing.

I may be confused, but:
There are are several results which satisfy conservation laws (1 ball really high, 2 balls same height, 3 balls less high...). The number of balls 'excited' is the result of the number of motive impacts.
 
  • #9
Dullard said:
I may be confused, but:
There are are several results which satisfy conservation laws (1 ball really high, 2 balls same height, 3 balls less high...). The number of balls 'excited' is the result of the number of motive impacts.
You have to solve both: ##2m\vec{v}=m\sum_k \vec{v}_k## and ##m\vec{v}^2=\frac{m}{2}\sum_k \vec{v}_k^2##
 
  • #10
The "Newton's cradle" page on Wiki provides some interesting points, in particular the importance of the (small) separation between the balls.
 
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  • #11
I like to think about those sort of questions by asking myself what would be the case in an "ideal" universe. Here, what I mean by an ideal universe is the one in which all objects are perfectly rigid, zero elasticity so the interactions are instantaneous (and impulses are represented by dirac delta functions you might say). In such a universe (avoiding all sorts of complications regarding elasticity, pressure waves, etc...) the only way I can work my way through this one is to study the pairwise inteactions.

Quoting wikipedia:
"... when two balls are dropped to strike three stationary balls in a cradle, there is an unnoticed but crucial small distance between the two dropped balls, and the action is as follows: The first moving ball that strikes the first stationary ball (the second ball striking the third ball) transfers all its velocity to the third ball and stops. The third ball then transfers the velocity to the fourth ball and stops, and then the fourth to the fifth ball. Right behind this sequence is the first ball transferring its velocity to the second ball that just stopped, and the sequence repeats immediately and imperceptibly behind the first sequence, ejecting the fourth ball right behind the fifth ball with the same small separation that was between the two initial striking balls..."

Still not sure about the "small seperation" argument, and curious what would happen in my ideal universe if there was zero seperation. I can't convince myself that there will be one and only one path the system could take...

(just a crazy idea: the indeterminacy in quantum mechanics... could it be because the systems are similar to my "ideal" universe where things and rules are just "simple"...)
 
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  • #12
fresh_42 said:
You have to solve both: ##2m\vec{v}=m\sum_k \vec{v}_k## and ##m\vec{v}^2=\frac{m}{2}\sum_k \vec{v}_k^2##
Yes, but you have 2 equations and 5 unknowns, so there will be multiple solutions. One such solution is shown by @erbahar in post 7. This is a valid solution, but I have never seen it happen, so there is something more than just conservation involved.
 
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  • #13
Dale said:
Yes, but you have 2 equations and 5 unknowns, so there will be multiple solutions. One such solution is shown by @erbahar in post 7. This is a valid solution, but I have never seen it happen, so there is something more than just conservation involved.
Newton #1?
 
  • #14
It's anything but trivial. The analysis with simple models shows that Newton's cradle is a system with (nearly) dispersionless "signal propagation" and that explains the observed behavior. The apparently simple explanation with energy-momentum conservation alone is of course wrong for the above discussed reasons. It only provides the "explanation" if you already assume the observation that you try to explain! For a very nice review, see

https://doi.org/10.1119/1.2344742
and references therein.
 
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  • #15
Dullard said:
I predict a different result if the 2 'motive' balls are super-glued together.
Has anyone tried it? Or replacing the 2 dropped balls with a single one of twice the mass?
 
  • #16
vanhees71 said:
https://doi.org/10.1119/1.2344742and references therein.
I haven't followed the references, because I presume it would have been mentioned in the article linked to if another significantly different explanation was contained therein.
My immediate reaction to the above discussion is that energy and momentum are not the only conserved quantities. There are five second-order differential equations in the ##n## variables ##\theta_i##, say, with a force law ##f(\theta_i-\theta_{i+1})## for the interactions between the balls, giving us the ##n## equations $$\frac{d^2\theta_i}{dt^2}+\theta_i+(1-\delta_{i,1})f(\theta_{i-1}-\theta_i)+(1-\delta_{i,n})f(\theta_i-\theta_{i+1})=0,\quad 1\le i\le n,$$ assuming the ##\theta_i## are small and scaled so the mass is 1. That system of equations has more than just two conserved quantities. The ##2n## initial conditions would fix the subsequent evolution, which is presumably not very sensitive to the precise force law insofar as real-world examples are typically not finely engineered.
 
  • #17
Yeah, with two motive balls, why not one ball twice as high?
 
  • #18
DaveC426913 said:
Yeah, with two motive balls, why not one ball twice as high?
russ_watters said:
Not per se, but there is a conservation of energy, for which velocity is the key component here. If the mass changes you couldn't satisfy both conservation of momentum and energy at the same time because of the V2 term in KE.
 
  • #19
Does that ... answer my question? :frown:
 
  • #20
DaveC426913 said:
Yeah, with two motive balls, why not one ball twice as high?

One key point is that if a ball (moving) impacts another ball (at rest) of equal mass, then the first ball stops and the second moves with the velocity/momentum/energy of the first.

The height of the ball, therefore, only determines this momentum.

If, however, a more massive ball impacts a less massive ball, then the larger ball will continue with some of the momentum.
 
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  • #21
Yes, if momentum ##mv## is conserved and kinetic energy ##\frac{1}{2}mv^2## then mass cancels out, and an increased velocity cannot be both: linear and quadratic.
 
  • #22
Good Lordie, Isn't it amazing just how complicated even such a very "simple" application of the laws of Mechanics can be??
 
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  • #23
fresh_42 said:
Yes, if momentum ##mv## is conserved and kinetic energy ##\frac{1}{2}mv^2## then mass cancels out, and an increased velocity cannot be both: linear and quadratic.
That works for a two ball collision where you have two unknowns and two equations. But with 5 balls you have 5 unknowns and still just two equations. So there are an infinite number of solutions.
 
  • #24
Dale said:
But with 5 balls you have 5 unknowns and still just two equations.

But you do have constraints, and those constraints will provide additioonal equations (only one thing happens). For example, a ball cannot overtake another ball.
 
  • #25
Vanadium 50 said:
But you do have constraints, and those constraints will provide additioonal equations (only one thing happens). For example, a ball cannot overtake another ball.
The configuration in post 7 also satisfies those constraints.
 
  • #26
This is not a simple situation. See, e.g., Matthias Reinsch, "Dispersion-free linear chains," Am. J. Phys. 62, 271-278 (1994) and references therein. A good article at a lower level was in The Physics Teacher, 35, Oct. 1997 by J.D. Gavenda and J.R. Edgington. Hope this helps.
 
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  • #27
In another current thread on the subject I modeled the three ball case as three springs.
See post #11 at https://www.physicsforums.com/threads/conservation-of-momentum-elastic-collisions.972238
As has been noted, the "perfect" Newton's Cradle result requires a small separation so that no three balls are in contact simultaneously. On the other hand, I suspect my model exaggerates the discrepancy by ignoring the time it takes for the shock wave to traverse a ball.
 
  • #28
Two assumptions:

1. The balls ARE touching.
2. The spheres are PERFECTLY rigid.

Then it seems we have to conclude that those assumptions are so oversimplifying for this problem that it can not be solved?
 
  • #29
erbahar said:
2. The spheres are PERFECTLY rigid.
Not a good assumption. That leads to infinite forces.
Such idealisations - inextensible strings, perfect elasticity etc. - are only valid as asymptotic limits. E.g. if we let the spring constant of a string be k, solve, let k tend to infinity and find the solution converges, then we can argue we have the solution for a string of negligible extensibility.
In the present case, that is what I did at the link I posted. The end result does not depend on the value of k, so represents the limit as the spheres tend to perfect rigidity.
 
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  • #30
Dale said:
The configuration in post 7 also satisfies those constraints.

Which is why I said "for example".
 
  • #31
Dale said:
That works for a two ball collision where you have two unknowns and two equations. But with 5 balls you have 5 unknowns and still just two equations. So there are an infinite number of solutions.

Well, it really isn't a linear momentum problem. The balls are on strings and are therefore, pendula. However, it's not necessary to go through anything tedious because there is symmetry about the middle point of where the strings attach. For example, if there are 5 balls, then the line along the string of the middle ball is the symmetry axis. The initial and final motion should be symmetric about that axis. If there were 4 balls, the symmetry would be the line between (and parallel to) the two middle strings.
 
  • #32
bobob said:
The balls are on strings and are therefore, pendula
Any displacement of the angle of the string during the period of interest can be made arbitrarily small.
bobob said:
The initial and final motion should be symmetric about that axis.
It is tempting to assume that, but as I show at the link in post #27 it is not true. What you can say is that the solution should be time-reversible.
 
  • #33
Vanadium 50 said:
But you do have constraints, and those constraints will provide additioonal equations (only one thing happens).
"Only one thing happens" doesn't necessarily mean that modelling it as a single collision will yield a unique solution, even with all appropriate constraints.
 
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  • #34
erbahar said:
Two assumptions:

1. The balls ARE touching.
2. The spheres are PERFECTLY rigid.

Then it seems we have to conclude that those assumptions are so oversimplifying for this problem that it can not be solved?

I wonder whether the issue isn't whether the balls are touching but that the bonds inside a sphere are different from the "bond" between adjacent touching spheres. Certainly if you applied a force in the opposite direction, then two touching balls would come apart easily; whereas, it would be hard to split a ball.

You could try the experiment with the wires not quite vertical, so that the balls are held together slightly by gravity.

I wonder whether it would work the same with cubes? That would give more symmetry for for shock wave along its line of propagation. Then one could argue that the cubes would not be perfectly smooth, and this would again create a delay in the shock wave at the point of contact.

Finally, if you drop two balls, then assuming this delay the experiment breaks down nicely into an impact from the first ball that results in one ball being ejected at the far end; followed almost immediately by a second impact that results in a second ball being ejected.

If, however, a single larger ball is dropped, then there would be theoretically a geometric sequence of reducing shockwaves, which would result in all the other balls being ejected, each with less momentum than the ball before.

And, as above, you could argue that no matter how rigid the balls are, the delay in the sequence of shock waves is still finite, so the same behaviour would persist in the limit.
 
  • #35
osilmag said:
Conservation of momentum and energy are why two balls are knocked. If the final was only one, it would go higher than the initial two. The PE is transferred to KE to maintain velocity.
I don't think energy is conserved in a collision. I hear a click.
 
<h2>1. Why do two balls launch from the right in a Newton's cradle instead of just one?</h2><p>Two balls launch from the right in a Newton's cradle because of the principle of conservation of momentum. When one ball is pulled back and released, it transfers its momentum to the next ball, causing it to move. This process continues until the last ball on the right side receives the momentum and launches off the cradle.</p><h2>2. How does the Newton's cradle demonstrate Newton's third law of motion?</h2><p>The Newton's cradle demonstrates Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When one ball is pulled back and released, it exerts a force on the next ball, causing it to move in the opposite direction. This reaction continues throughout the cradle, demonstrating the law in action.</p><h2>3. Can a Newton's cradle work with different sized balls?</h2><p>Yes, a Newton's cradle can work with different sized balls as long as the balls have the same mass. This is because the principle of conservation of momentum depends on the mass of the objects involved, not their size.</p><h2>4. Why do the balls eventually stop moving in a Newton's cradle?</h2><p>The balls eventually stop moving in a Newton's cradle due to the presence of friction. As the balls collide with each other and the cradle, some of their energy is converted to heat due to friction. This loss of energy causes the balls to eventually come to a stop.</p><h2>5. Can a Newton's cradle be used to generate energy?</h2><p>No, a Newton's cradle cannot be used to generate energy. While the balls may continue to move for a while, the energy they lose due to friction is greater than the energy they gain from the initial pull, so there is a net loss of energy. In order to generate energy, there must be a source of energy input, such as a motor or a person pulling the balls back and releasing them repeatedly.</p>

1. Why do two balls launch from the right in a Newton's cradle instead of just one?

Two balls launch from the right in a Newton's cradle because of the principle of conservation of momentum. When one ball is pulled back and released, it transfers its momentum to the next ball, causing it to move. This process continues until the last ball on the right side receives the momentum and launches off the cradle.

2. How does the Newton's cradle demonstrate Newton's third law of motion?

The Newton's cradle demonstrates Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When one ball is pulled back and released, it exerts a force on the next ball, causing it to move in the opposite direction. This reaction continues throughout the cradle, demonstrating the law in action.

3. Can a Newton's cradle work with different sized balls?

Yes, a Newton's cradle can work with different sized balls as long as the balls have the same mass. This is because the principle of conservation of momentum depends on the mass of the objects involved, not their size.

4. Why do the balls eventually stop moving in a Newton's cradle?

The balls eventually stop moving in a Newton's cradle due to the presence of friction. As the balls collide with each other and the cradle, some of their energy is converted to heat due to friction. This loss of energy causes the balls to eventually come to a stop.

5. Can a Newton's cradle be used to generate energy?

No, a Newton's cradle cannot be used to generate energy. While the balls may continue to move for a while, the energy they lose due to friction is greater than the energy they gain from the initial pull, so there is a net loss of energy. In order to generate energy, there must be a source of energy input, such as a motor or a person pulling the balls back and releasing them repeatedly.

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