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Perfectly elastic collision of spheres 
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#1
Nov2812, 11:48 AM

P: 4

Hey guys!
I am currently developing a simulation that involves sphere (or if you like particle) collision in 3D space. And I want it to be accurate (on the level of classic mechanics). The algorithm to do the job would take in the velocities, masses and relative position (aka line of impact) of two colliding particles and compute their new velocities short after the collision. So I tried to develop this algorithm and ran into a problem. I set up the equations for the conservation of momentum and conservation of kinetic energy. That gave me four equations, one for the kinetic energy (scalar) and three for the momentum (one for each of the three dimensions). Since the result of this calculation has six variables (the two new velocity vectors) it needs six equations to be solved. Obviously two more are necessary. I dont know how to set them up but they certainly have to take the relative position into account. Well, I just hope that someone here can help me solve this. In case I have left something unclear, just ask. Regards, janismac 


#2
Nov2812, 12:18 PM

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#3
Nov2812, 12:20 PM

P: 70

You can simplify by using spherical coordinates based on the centerofmass.
Conserve angular momentum. 


#4
Nov2812, 03:25 PM

P: 4

Perfectly elastic collision of spheres
my equations are: momentum kinetic energy (equation multiplied by 2) 


#5
Nov2812, 04:36 PM

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P: 3,941

In physical terms, imagine the two balls approaching each other at equal speeds. The could bounce away at any angle, and as long as they're traveling at the same speed and in exactly opposite directions after the collision, you have a solution that conserves energy and momentum. Your two extra unknowns correspond to the the two angles the balls actually take. Tadchem's answer (use conservation of angular momentum  that will give you the two extra equations you need to solve for those last two unknowns) is good. 


#6
Nov2912, 07:33 AM

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#7
Nov2912, 01:10 PM

P: 4

Vector x is the absolute position of a particle in my coordiante system. This equation would set the centerofrotation to the origin. May I just do that? This would give me three new equations instead of two... so it's most likely wrong. But I dont know any better. Couldnt someone just give me the equation? 


#8
Nov2912, 01:23 PM

P: 4

I have another idea that may lead to a solution.
The old velocity vector, the new velocity vector and the force that acts on a particle have to lie in one plane (dont they?) so I could express this in two equations, one for each particle: does that make sense? 


#9
Nov2912, 11:10 PM

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I assume you start off with known trajectories for the particles before impact. From those you can compute the point of impact on each particle. I'll further assume the particles are smooth, so you don't have to worry about spin.
You can arbitrarily fix one particle as the reference frame by subtracting its velocity from the other. Further, you can rotate the frame so that the mass centres, point of impact and velocity of approaching particle are all in the XY plane. The direction of departure of the previously 'stationary' particle is obvious, so you only have three scalar unknowns to determine and still 3 scalar equations to make use of. 


#10
Nov3012, 06:22 AM

P: 70

Back in 1993 I wrote up the details of this in a nonpeerreviewed paper: "A simple and accurate method for calculating viscosity of gaseous mixtures", US Bureau of Mines Report of Investigations 9456, Govt. Doc Number I 28.23:9456. In the appendix I analyzed in detail the mechanics of the transfer of momentum in an elastic collision between two perfect spheres of arbitrary masses.
The choice of centerofmass coordinates allows one to disregard the net velocity of the system (conserved before and after the collision), and aligning the zaxis of a spherical coordinate system with the angular momentum pseudovector removes two degrees of freedom. There is no individual momentum of the spheres parallel to the zaxis, leaving r and theta. All momentum *transfer* occurs along the r coordinate, and (angular) momentum remains unchanged along the theta coordinate. The problem is thus reduced to the onedimensional problem along the line connecting the centers of the spheres. The physics revealed is that the efficiency of the transfer of momentum is a maximum when the masses are equal. Dissimilar masses result in the smaller mass recoiling from the impact while the larger mass only transfers some of its momentum to the smaller mass, yielding a lower efficiency of momentum transfer The importance is that the efficiency of the transfer of momentum from one portion of a fluid to another is called fluidity, which is better known through its reciprocal  viscosity. This helps explain the viscosity anomaly  the fact that a mixture of fluids of different molecular weights always has a higher viscosity than the corresponding linear combination of the separate viscosities of the individual fluids. 


#11
Nov3012, 06:27 AM

P: 70

NB: The impact parameter may be useful in other contexts, but as long as the overall collision is elastic, i.e. there is no conversion of kinetic energy to thermal energy or other forms, then the details of the interaction such as the impact parameters are insignificant. What matters is the momenta of the individual particles before and after the interaction. This is true whether the interaction is mechanical, gravitational, or coulombic.



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