  The real beauty of momentum is seen when talking about collisions. The complex forces involved in the deformation and energy used in heat and sound can be overlooked and conservation of momentum can give us valuable information about the collision, such as initial and final masses and velocities.

There are basically three types of collisions:

• Inelastic: Some kinetic energy is lost and momentum is conserved
• Completely Inelastic: The maximum amount of kinetic energy is lost.
• Elastic: Kinetic energy and momentum are conserved

We shall look at each case separately.

Inelastic

Most collisions are of this type. If two cars smash into each other and then bounce back a bit, it's an example of an inelastic collision. There are too many variables in these types of collisions to make a simple analysis possible. Mathematically, we wind up with more unknowns than equations, making the problem intractable. We should know what these collisions are, but we won't do much with them.

Completely Inelastic

Some special cases of the above collisions are called completely inelastic collisions. After the collision, the maximum amount of kinetic energy has been dissipated. In the above car crash example, if the two cars had been completely stuck together after the crash, it would have been an example of a completely inelastic collision. Other examples of this type are:

• Balls of clay or mashed potatoes being smashed together
• Train Cars interlocking with each other
• Projectile weapons being lodged into their targets

In essence these problems are of the same type as that of two ice skaters pushing off from each other, just done in reverse. If two skaters approach each other at certain velocities and upon meeting grasp the other so that they remain together, they have had an inelastic collision.

The equation for these types of collisions is the same conservation of momentum equation used before:

m1fv1f + m2fv2f = m1iv1i + m2iv2i

For example, if you know the masses (m1 and m2) and velocities (v 1 and v2) of two interlocking train cars before a collision, you also know their mass after collision (m 1 + m2) and can thus find the velocity of the interlocked car after collision.

Elastic Collisions

These collisions occur when all kinetic energy is conserved in a collision. Because of our universe's tendency toward disorder (Entropy, baby!), this kind of collision doesn't often occur in the "real world." If a ball bounced on the floor came back up to the point from which it was dropped, that would be an elastic collision. If two cars had a collision and each bounced backward without any energy being converted to heat or sound, it would be an elastic collision.

Even though perfectly elastic collisions are rare, we can often come close to simulating them. An example of this is the air tracks we use in the classroom. The rubber bands allow for the two gliders to collide without actually touching and converting energy. Some other examples of pretty good near elastic collisions are:

• Billiard Balls on a Pool Table
• The Newton's Cradle Desk Toy (with several steel balls hanging from a stick)
• Some low speed car crashes with good bumpers. (Sorry, no SUV's!)

The analysis of Elastic Collisions is fairly straight forward, although slightly more complex than completely inelastic collisions. Whereas with completely inelastic collisions we know that the final mass is just the mass of the two together, we have no such luxury with elastic collisions. That means we have one equation and two unknowns. We work around this problem by taking conservation of energy into consideration as well. Thus we have the two equations:

m1fv1f + m2fv2f = m1iv1i + m2iv2i

(1/2)m1fv1f2 + (1/2)m2fv2f2 = (1/2)m1iv1i2 + (1/2)m2iv2i2

Sometimes, solving this problems lead to difficult algebra, but they can be solved.

Let's assume that we have two different mass objects that collide elastically, and further that the first object is moving and the second is stationary. Using the definition of relative velocity:

v1f - v2f = - (v1i - v2i)

we can find that:

v1f = (((m1 - m2)v1i) / (m1 + m2)) + ((2m2v2i) / (m1 + m2))

and that:

v2f = (((m2 - m1)v2i) / (m1 + m2)) + ((2m1v1i) / (m1 + m2))

Which leads us to the following general conclusions:

1. When m1 is less than m2, v1f is opposite to v1i. The lighter object bounces off the heavier one.
2. When m1 = m 2, v1f = 0 and v2f = v1i. The colliding object stops, and the target moves off with the same velocity.
3. When m1 is greater than m 2, the colliding object continues to move, but with a slower velocity, while the target moves off with a much greater velocity.

Conservation of Momentum and Collisions in 2-D and 3-D

The process is the same as above except now the momentum must be broken up into its vector components, so we have conservation of momentum in the y-direction, etc.

Remember:  Energy is just a scalar number, not a vector, and so it doesn't have to be broken up into vector components.