# Remembering the Final Velocities

## The Problem:

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

 A particle of mass m1 and velocity v collides elastically with a particle of mass m2, initially at rest. After the collision, m1 has velocity v1, and m2 has velocity v2. What are v1 and v2?

One way to solve a problem of this type is to solve the problem in general, and then remember the solution. This is not usually a good strategy, since there can be a lot of variation in problems, and you can't memorize everything. In this case, however, the general solution can be worthwhile to know, and certainly enlightening to study,

First of all, note that the problem as stated above can be considered the most general case. If both particles have a velocity before the collision, it is always possible (and quick and easy) to switch to a frame of reference in which one of the particles is at rest, then find the velocities after the collision in this new frame of reference, and then switch back to the original reference frame.

Given that, it can be shown that the velocities of the particles after the collision are given by:

## Examining the Solutions:

Are these solutions reasonable? Let's look at three cases:

### Case 1: m1 is much less than m2

Think of a ping-pong ball (m1) colliding with a stationary bowling ball (m2),. First, notice that any time m1 is less than m2, v1 will be negative. Physically, this means that if the "bullet" particle is less massive than the "target" particle the "bullet" particle will bounce back in the direction from which it came. This is certainly what would happen if a ping-pong ball hit a bowling ball. In fact, if m1 is very small compared to m2, the equation for v1 becomes approximately:

(Formally, .) So the ping-pong ball would bounce back with approximately its original speed. This happens. What about the bowling ball? If m1 is approximately zero (compared to m2), then

In other words, the equations predict that the bowling ball would hardly move. This happens.

### Case 2: m1 = m2 = m

What if an air-track glider collides elastically with an identical glider initially at rest, or a billiard ball collides elastically with another billiard ball of the same mass (neglecting spin effects)?

The equations predict that the moving particle will come to rest (v1 = 0) and the initially stationary particle will "steal" its velocity (v2 = v). In effect, the two particles trade velocities. This happens (Try it!).

### Case 1: m1 is much greater than m2

What happens if a bowling ball collides with an initially-stationary ping-pong ball?

Physically, the equations predict that the bowling ball will continue with about the same velocity (hardly even noticing the collision!), and the ping-pong ball will fly away with almost twice the bowling-ball's velocity. This happens.

last update September 28, 2004 by JL Stanbrough