Collision Between Two Spheres
Let us consider the collision between two rigid spheres of masses m1 and m2. We assume that during collision, a large force F of (mutual) interaction acts between the spheres for a short interval of time Δ t; and any external force acting at the same time is negligible. The force F acts normal to the surface of contact, towards left on m1 and towards right on m2.
The force F applies a normal impulse, I = F Δ t, on both the spheres. Consequently, change in normal components of momenta of the two spheres is given by,
–F Δ t = m1 (v1n – u1n)
And, F Δ t = m2 (v2n – u2n)
Where normal components are taken positive towards right.
This gives,
m1 v1n + m2 v2n = m1 u1n + m2 u2n
which is the statement of conservation of momentum of the combined system along normal direction.
Since there is no impulse on the spheres in the tangential direction, the tangential components of momenta remain unchanged:
m1 v1t = m1 u1t ; v1t = u1t
and, m2 v2t = m2 u2t ; v2r = u2t
That is, m1 v1t + m2 v2t = m1 u1t + m2 u2t
Thus, both the normal and tangential components of momentum of the system remains conserved in the process of collision.
Combining normal and tangential components, we get the conservation of total momentum for the system:
m1 v1 + m2 v2 = m1 u1 + m2 u2
In order to find the values of v1n and v2n, we need one more relation between them, besides eq. This is done by defining the coefficient of restitution e, as the negative ratio of the component of the relative velocity normal to the surfaces of the colliding spheres, after and before collision:
v2n – v1n = eu1n – eu2n
Solving eqns, we get,
If e = 0, v2n = v1n; the normal velocities of both the spheres after collision become equal. Such a collision is said to be completely inelastic. If e = 1 the normal component of relative velocities of component of approach and separation have the same magnitude:
v2n – v1n = u1n – u2n
Such a collision is said to be perfectly elastic.
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