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

Conservation of linear momentum requires that

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
and consequently
m₁(u₁−v₁) = −m₂(u₂−v₂).
 

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·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(u₁ − v₁) = 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

D² = (x₁ − x₂)·(x₁ − x₂)
 

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