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The Elastic Collision of Spheres

Consider two spheres. For i∈{1,2} let mi , xi and ui be the mass, position vector and velocity vector of the i-th sphere before the collision and vi the velocity vector after the collision. (Red symbols stand for vectors.)

Conservation of linear momentum requires that


m1u1 + m2u2 = m1v1 + m2v2
and consequently
m1(u1v1) = −m2(u2v2).
 

For elastic spheres kinetic energy is also conserved and hence


½m1(u1·u1) + ½m1(u1·u2) = ½m1(v1·v1) + ½m2(v2·v2)
 

This reduces to


m1(u1·u1v1·v1) = −m2(u2·u2v2·v2)
and hence
m1(u1v1)·(u1+v1) = −m2(u2v2)·(u2+v2)
 

In a collision the change in momentum for each sphere is in the direction of the vector between their centers at the instant of contact. Let k be the unit vector in that direction; i.e.,


k = (x1x2)/|x1x2|.
 

Then the change in momentum is given by


m1(u1v1) = −m2(u2v2) = ak.
 

This means that the conservation of kinetic energy can be expressed as


k·(u1 + v1) = k·(u2 + v2).
 

Since m1(u1v1) = ak it holds that v1 = u1 − (a/m1)k. Likewise v2 = u2 + (a/m2)k.

This means that the conservation of kinetic energy requires that


k·(2u1 − (a/m1)k) = k·(2u2 + (a/m2)k)
 

since k·k=1 the above equation reduces to


2k·(u1u2) = a[1/m1 + 1/m2]
so
a = 2k·(u1u2)/[1/m1 + 1/m2]
 

Everything on the right-hand-side (RHS) of the above equation is know; therefore a is determined. With a known v1 and v2 can be computed.

Note that the distance between the two spheres is given by


D² = (x1x2)·(x1x2)
 


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