# Detailed simulation or SlowMo of elastic collision?

I was thinking about Newton's Cradle the other day, and I wondered how those collisions actually look like in detail. Which then got me thinking that my understanding of even a basic elastic collision of two macroscopic objects is weak to say the least.

Simplified, the two objects could be two rods of a certain length but negligible width, and they simply meet head on.

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I would imagine when they meet, they will compress each other's ends, which will cause a pressure shockwave to propagate through both of them at the specific speed of sound. These shockwaves will probably get reflected at the respective ends etc.

It will probably get rather complex pretty quickly, but are there simulations of or SlowMo videos of this, and particularly how it results in such a clean end result of both objects traveling away from each other? I could imagine the two could just as well just "ring" at their respective resonance frequency, wasting a lot of energy that way (and thus no longer being elastic).

sophiecentaur

Chestermiller
Mentor
Your analysis is, in my judgment, spot on. The compression zone will grow at the speed of sound.

To quantify things, the contact force would be $$F=\rho c A v$$ where c is the speed of sound ##\sqrt{E/\rho}## and v is the approach velocity. Upstream of the compression zone, the velocity will be v, and inside the compression zone, the velocity will be zero. The compressive axial strain within the compression zone will be ##\epsilon=v/c##.

Once the compression zone reaches the far upstream end of the bar, the entire bar will be stationary, but compressed. After this, the compression begins getting released, starting at the far end. In the released portion, the velocity will be -v, while the remainder of the bar is stationary and compressed (and the contact force remains constant at F).. The release zone travels at the speed of sound until the compression is released over the entire bar, and the entire bar is moving at velocity -v. At this point, the bars separate.

sophiecentaur and rumborak
Actually, when I think about it, this reflection at the end of the rods is probably the classic "closed end" wave reflection where the phase reverses. Meaning, the compression turns into expansion and travels backwards through the rod, as you mention.

I'm still a bit surprised that no energy remains in these waves after separation, which would result in that "ringing" of the rods. Is this one of those cases where the effects exactly cancel out, and all energy results in motion?

Chestermiller
Mentor
Actually, when I think about it, this reflection at the end of the rods is probably the classic "closed end" wave reflection where the phase reverses. Meaning, the compression turns into expansion and travels backwards through the rod, as you mention.

I'm still a bit surprised that no energy remains in these waves after separation, which would result in that "ringing" of the rods. Is this one of those cases where the effects exactly cancel out, and all energy results in motion?
Hmmm. Interesting. It seems like all the initial kinetic energy is accounted for, and momentum is conserved.

ehild
Homework Helper
I was thinking about Newton's Cradle the other day, and I wondered how those collisions actually look like in detail. Which then got me thinking that my understanding of even a basic elastic collision of two macroscopic objects is weak to say the least.
This is an amazing Slo-Mo video of colliding a golf ball with steel.

nasu and jerromyjon
That's interesting though, because the golf ball exhibits exactly that "ringing" I was talking about (it keeps deforming after it separated from the wall). Clearly this wasn't a purely elastic collision because part of the energy is now used in that ringing.

Interesting question: usually energy losses like this are marked down to plastic deformation. But, is the idea of a macroscopic object exhibiting elastic collision, even under perfect conditions, unreasonable, because the object will always have a certain part of its initial energy siphoned into that kind of "ringing"?

Chestermiller
Mentor
That's interesting though, because the golf ball exhibits exactly that "ringing" I was talking about (it keeps deforming after it separated from the wall). Clearly this wasn't a purely elastic collision because part of the energy is now used in that ringing.

Interesting question: usually energy losses like this are marked down to plastic deformation. But, is the idea of a macroscopic object exhibiting elastic collision, even under perfect conditions, unreasonable, because the object will always have a certain part of its initial energy siphoned into that kind of "ringing"?
There are big difference between this and the rod collision. First, and most important, the golf ball deformation is huge, compared to the small deformations required for Hooke's law to apply (as in the rod case). Secondly the geometry is very different (spherical, vs head-on cylindrical). Thirdly, if the two rods are not of exactly the same length, there will be unsymmetry, and the waves will reflect from the ends at different times. This should almost certainly result in ringing. So, it would see that the rod collision is a very special case.

That's all true. Somehow I still have a hard time imagining even a theoretical setup in which *any* object, entirely within Hooke's Law's bounds, would come out with zero ringing. Negligible amount, maybe. Entirely? Somehow I doubt that.

Chestermiller
Mentor
That's all true. Somehow I still have a hard time imagining even a theoretical setup in which *any* object, entirely within Hooke's Law's bounds, would come out with zero ringing. Negligible amount, maybe. Entirely? Somehow I doubt that.
I hear ya. But, as far as I can tell, the solution I gave to this linear problem satisfies the differential wave equation (in terms of the axial strain and/or the displacement), the boundary conditions, and the initial conditions. So since the equations are linear, it must be the unique solution. Here is the wave equation in terms of displacement u: $$\frac{1}{c^2}\frac{\partial^2 u}{dt^2}=\frac{d^2 u}{dx^2}$$ where $$c^2=\frac{E}{\rho}$$. The axial strain in terms of the displacement is given kinematically by: $$\epsilon=\frac{\partial u}{\partial x}$$So the strain also satisfies the wave equation also. The solution for the strain as a function of time and position I described is a d'alembert form of the solution. If you can spot a logical or mathematical error in the solution I described, I'm anxious to hear about it.

Dagnabbit, I'm in no position to make an educated criticism of what you posted :) I easily understand the parts of what you posted, but I can't say whether the conclusion you draw is logically binding.
At the very least it will give me something to do on an upcoming 3-hour flight of mine. I should be reasonably able to simulate the scenario as a collection of connected ideal springs, and see how they behave after separation.

Chestermiller
Mentor
Dagnabbit, I'm in no position to make an educated criticism of what you posted :) I easily understand the parts of what you posted, but I can't say whether the conclusion you draw is logically binding.
At the very least it will give me something to do on an upcoming 3-hour flight of mine. I should be reasonably able to simulate the scenario as a collection of connected ideal springs, and see how they behave after separation.
Cool. Springs and masses in series. Very nice.

JRMichler
If you simulate the collisions of a Newton's Cradle with springs, start with two rigid balls with the right mass for steel and an ideal spring at the contact. Pick a spring constant such that the first ball is in contact with the second ball for about 1 millisecond, or maybe slightly less, when it swings in.

I've done a lot of hammer testing to find natural frequencies, and 1 msec is about the right contact time for steel balls of the size used in a typical Newton's Cradle. I would expect ringing of the balls to be minimal because the force vs time curve of the impact will be a half sine of 1 msec duration, so will not put much energy into ringing the much higher ball natural frequency.