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Thayer Watkins
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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 + m2v2and consequently m1(u1−v1) = −m2(u2−v2).

For elastic spheres kinetic energy is also conserved and hence

This reduces to

m1(u1·u1−v1·v1) = −m2(u2·u2−v2·v2) and hence m1(u1−v1)·(u1+v1) = −m2(u2−v2)·(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 = (x1 − x2)/|x1 − x2|.

Then the change in momentum is given by

m1(u1 − v1) = −m2(u2 − v2) = 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·(u1 − u2) = a[1/m1 + 1/m2] so a = 2k·(u1 − u2)/[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