Not a good assumption. That leads to infinite forces.2. The spheres are PERFECTLY rigid.
Which is why I said "for example".The configuration in post 7 also satisfies those constraints.
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.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.
Any displacement of the angle of the string during the period of interest can be made arbitrarily small.The balls are on strings and are therefore, pendula
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.The initial and final motion should be symmetric about that axis.
"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.But you do have constraints, and those constraints will provide additioonal equations (only one thing happens).
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.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 don't think energy is conserved in a collision. I hear a click.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.
There's more to it! To model it with point particles connected by strings you have to adjust the spring constants and masses such as to guarantee a dispersion-free wave (see the AJP papers cited in #26). Newton's craddle is by far not as simple as it looks!In another current thread on the subject I modelled 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.
Dimensional analysis suggests adjusting only the spring constants and masses, but keeping the balls identical, is not going to change the qualitative behaviour. Changing the initial conditions, e.g. to construct a phonon, sounds more helpful.There's more to it! To model it with point particles connected by strings you have to adjust the spring constants and masses such as to guarantee a dispersion-free wave (see the AJP papers cited in #26). Newton's craddle is by far not as simple as it looks!
The click is sound which is particles in the air transferring energy to your ear drum. That energy is lost with each impact as is the friction caused between the balls and the air when they swing back and forth.I don't think energy is conserved in a collision. I hear a click.
Two balls mass m in at speed v have momentum 2mv and KE mv^{2}. If only one ball comes out and momentum is conserved it has speed 2v and KE 2mv^{2}.I cannot follow the arguments why two balls produce two balls, not one ball higher.
If I pushed one ball much harder could I eventually produce two?
P=mv so increasing the velocity is essentially the same as increasing the mass?
Yes, read the article. It's really surprising that you need this condition of ##k \propto \omega##. It's obviously what's realized to good approximation with the real cradle. I wonder, how one can understand this properly from the theory of bouncing metallic spheres!Dimensional analysis suggests adjusting only the spring constants and masses, but keeping the balls identical, is not going to change the qualitative behaviour. Changing the initial conditions, e.g. to construct a phonon, sounds more helpful.
I'll try to get access to the article.
Say you have the initial ball mass 2m hits 4 balls all mass m will 2 balls be produced?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.
The simplest model is what I posted in post #34. If you imagine small gaps between the balls then all four balls will move:Say you have the initial ball mass 2m hits 4 balls all mass m will 2 balls be produced?
How do you know that all the rising mass has the same velocity?Let the total falling mass be m and the velocity of impact be v. Let the total rising mass be M and its initial velocity be V.
You don't know and my answer is an approximation only based on the assumption that the collision is perfectly elastic. It's the same sort of analysis you could apply to mono atomic gases where the collisions can be perfectly elastic.How do you know that all the rising mass has the same velocity?
And how do you know that the falling mass doesn't rebound?
It's not an approximation. If you drop a ball onto the Earth it rebounds. And, in fact, in any elastic collision where a smaller mass impacts a larger mass at rest, the smaller mass will rebound.You don't know and my answer is an approximation only based on the assumption that the collision is perfectly elastic. It's the same sort of analysis you could apply to mono atomic gas molecules which can collide perfectly elastically.
The Newton's cradle is not the earth or a solid lump but is made out of separate parts which can be put into contact with each other. I think the experimental results speak for themselves. If you haven't already done so look at the you tube videos on the cradle. There is no noticeable rebound.It's not an approximation. If you drop a ball onto the Earth it rebounds. And, in fact, in any elastic collision where a smaller mass impacts a larger mass at rest, the smaller mass will rebound.
That's another possible solution in the case of Newton's cradle. The falling ball rebounds and the others all move off.
In fact, the impact ball only stops in the special where the object is of the same mass. Which is why we have analysed the problem as though there are small gaps; or, in any case, as a sequence of simple collisions.