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Hey so I recently started learning physics, and came upon this wonderful site that taught me how to calculate 2D collisions between two circles.

The only part I'm confused about is how the velocity vectors are translated back to the original coordinate system from the cm frame. The equation can be seen in the self-explanitory code in the picture below:

Question

(phi is the angle of the line that goes through both circle's centers)

**Why is this equation different from the first on used to rotate the points to the new 1D coordinate system? And why is the +180/2 added into the reverse rotation equation?

How is it that force only travels along the phi line? Is there a law that explains this?

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2 Answers

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How is it that force only travels along the phi line?

In the idealization being employed here, friction between the disks is being ignored. Therefore one of them cannot apply a force to the other except perpendicular the to shared tangent plane where the two surfaces touch. (And of course the force has to be exchanged at the point where the two bodies come into contact, so these two conditions together determine the line along which the force works).

The real world does not actually work that way, so it's not any deep law -- just a convenient assumption that allows you to ignore the fact that the balls could be spinning in addition to just moving about rectilinearly.

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The $\pi/2$ in the angles is to get the transverse terms in the transformation. You may have seen it as $x'=x\cos \phi -y\sin \phi$ etc. What were the equations in the original transform? Usually the transform to the CM frame and the transform back are written more symmetrically than on that site. You might look at Wikipedia on the rotation matrix.

At the moment of collision, the center of mass is the point of tangency of the circles, as can be seen by symmetry. The velocity of each circle just before the collision is directed at the center of the other. Then the force acts from that point through the center of the circles. This shows the force acts along the phi line.

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:Is ''−ysinϕ'' the same as ''+cos(phi+pi()/2)*y'' ? and does this mean collision is totally different with friction =(, or can i apply this equation? –  Griffin Sep 24 '11 at 3:34
    
Yes, $\sin \phi=\cos(\phi + \pi/2)$, which can be proved using the angle addition formula $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. As $\cos \pi/2=0, \sin \pi/2=1$ you have it. –  Ross Millikan Sep 24 '11 at 3:37
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