Collision between spheres: general result

Deriving general vector expressions for the velocities of two smooth spheres following a collision. We include both elastic and inelastic collisions, by allowing for an arbitrary coefficient of restitution. Using the zero-momentum frame allows us to simplify the maths a little!
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About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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[@mikip3242]

It's interesting that conservation of energy was not involved in the calculation. I guess that a coeficient of restitution of e = 1 would have been the subrepticious way of introducing it, but this is not immediately apparent. Good video as always.

[@DrBenYelverton]

Thanks - yes, using the coefficient of restitution makes the algebra easier and has the advantage that the result works for both elastic and inelastic collisions!

[@Nxck2440]

If we wanted to allow for rough spheres so that we can have spinning balls, what extra things would we need to know? Can we still use a zero momentum frame since now we will have angular momentum?

[@DrBenYelverton]

The coefficient of friction is enough extra information in principle, but the complication is that the frictional force is governed by an inequality (F≤μR) rather than an equation. That makes it hard to deal with analytically, but I've implemented it in a simulation before. The idea is to first apply a normal impulse given by the equation from the video, then compute the tangential impulse that would reduce the relative velocity of the points of contact to zero, clamp this tangential impulse to μ × (normal impulse) if it's too big, then apply it. I suppose you could follow the same logic analytically and come up with some conditions and corresponding equations for changes in tangential and angular velocity, but it would be a little complicated! The ZMF is still very useful in the sense that the calculation of the normal impulse is the same as in the video - the linear momentum will still be zero after the collision even if the balls have angular momentum about their centres of mass.

[@darwinvironomy3538]

@@DrBenYelverton As long the roughness is small or somehow ball are rotating so fast we can assume F = μR right?. does this equation you derived work for 3D? because , following your derivations, there's no assumption regarding the dimension of the problem.

[@DrBenYelverton]

@@darwinvironomy3538 The result does indeed work in 3D. If they are rough enough and/or rotating fast enough then F will be equal to μR, but the difficult part is determining what exactly "enough" means here!