# Using the Center of Mass Reference Frame

## The Problem:

A particle of mass m1 and velocity v1 collides elastically (in one dimension) with a particle of mass m2 and velocity v2. What are the velocities of m1 and m2 after the collision?

 A particle of mass m1 and velocity v1 collides elastically with a particle of mass m2, with initial velocity v2. After the collision, m1 has velocity v1', and m2 has velocity v2'. What are v1' and v2'?

An interesting fact about elastic collisions is that they are symmetric with respect to the center of mass. If you stand at the center of mass to observe an elastic collision, you see mass m1 approach with velocity V1(not the earth-frame-of-reference velocity v1 above), and mass m2 approaching with velocity V2. The masses collide at the center of mass (Ouch!). Then, you see mass m1 leaving the center of mass with velocity -V1, and mass m2 leaving the center of mass with velocity -V2.

To find the velocities of the particles after the collision, you can:

1. Find the velocity of the system center of mass:
2. Switch to the center of mass reference frame. To do this, simply subtract vcm from each particle's velocity.
3. Have the collision. The particles' velocities reverse.
4. Switch back to the original frame of reference, by adding vcm to each particle's velocity.
5. Smile, you're done.

Gee, this seems awfully easy - one simple calculation, a couple of additions and subtractions - does it really work, always? Well, yes, it does. Here is a proof.

Actually, the solution for a one-dimensional elastic collision is even easier than that! It can be shown (quite easily) that

v1_after = 2vcm - v1, and v2_after = 2vcm - v2

## Example:

A 2 kg meatball with a velocity of 5 m/s collides head-on, elastically, with a 3 kg meatball with a velocity -5 m/s. If the collision is confined to one dimension, what will be their velocities after the collision?

The center of mass of the two-meatball system is easy to find:

The velocity of the 2 kg. meatball is 2vcm - v1 = 2(-1 m/s) - 5 m/s = -7 m/s

The velocity of the 3 kg. meatball is 2vcm - v2 = 2(-1 m/s) - (-5 m/s) = 3 m/s

Can it really be that easy? Yes, it can.

## Some Interesting Special Cases:

### A. m1 >> m2:

If m1 is much more massive that m2, then:

Therefore,

v1_after = 2vcm - v1 = 2v1 - v1 = v1 (approximately)

The equations predict that the velocity of m1 will be essentially unchanged, which corresponds to experience, and:

v2_after = 2vcm - v2 = 2v1 - v2 (approximately)

If m2 started at rest (v2 = 0), its velocity after the collision will be approximately 2v1.

### B. m1 = m2:

If the two masses are equal,

Therefore:

In other words, if the masses are equal, the two objects simply exchange velocities in an elastic collision.

### C. m1 = m2, v2 = 0:

If the masses are equal and m2 is initially at rest (a special case of part B above):

The velocities after the collision are:

So the two objects exchange velocities - m1 stops and m2 takes its velocity. Try it on an air track!

last update November 10, 2007 by JL Stanbrough