Recall our work energy theorem for a single particle,

or

or

If W_NC ≠ 0 then energy will not be conserved. For a two-body collision process, then an inelastic collision is one in which energy is not conserved (i.e. W_NC ≠ 0), but an elastic collision is one in which energy is conserved (W_NC = 0). Now if you think of a collision of two billiard balls on a horizontal pool table then U_f = mgyf and U_i = mgyi, but y_f = y_i and thus U_f = U_i or ΔU = 0. Thus the above work-energy theorem would be

Thus for collisions where U_i = U_f , we often say more simply that an elastic collision is when the kinetic energy alone is conserved and an inelastic collision is when it is not conserved. In this section we first will deal only with elastic collisions in 1-dimension.

Example A billiard ball of mass m₁ and initial speed v_1i hits a stationary ball of mass m₂. All the motion occurs in a straight line. Calculate the final speeds of both balls in terms of m₁, m₂, v_1i, assuming the collison is elastic (Is this a good assumption?).

Solution All the motion is in 1-dimension and so conservation of momentum (with v_2i = 0) is just

and conservation of kinetic energy is

Here we have two equations with the two unknowns v_1f and v_2f . Thus the rest of the problem is simply doing some algebra. Let's solve for v_1f in the first equation and then substitute into the second equation to get v_2f. Thus

or

Substituting this into the conservation of kinetic energy equation gives

which simplifies to

giving

which is finally

Substituting this back into the conservation of momentum equation gives

which gives

or

There are some interesting special situations to consider.

1) Equal masses (m₁ = m₂). This implies that v_1f = 0 and v_2f = v_1i. That is the projectile billiard ball stops and transfers all of its speed to the target ball. (This is also true if the target is moving.)

2) Massive target (m₂ >> m₁). In this case we get v_1f   -v_1i and  which means the projectile bounces off at the same speed and the target remains stationary.

3) Massive projectile (m₁ >> m₂). Now we get v_2f   -v_2i and v_1f   -v_1i meaning that the projectile keeps charging ahead at about the same speed and the target moves off at double the speed of the projectile.