**Two-dimensional Collision of Hard Balls** is a classic physical task, explained in this article.

Demonstration of the two dimensional collision. In the top diagram, both balls move in the opposite directions, illustrating that the centre of mass always stays at the same place. In the bottom diagram, the moving ball hits the standing ball. Use the slider in the middle to navigate through the all history of the collision.

We begin with two balls colliding, one moving, one at rest.

Immediately move to center of mass coordinates where

In these coordinates, the total linear momentum is zero.

The primes denote the velocities after the collision. The effect of this momentum conservation on the kinetic energy is that the particle velocities are related by

Also, the speeds before and after the collision are unchanged

Because and are so well connected, we

can complete the calculation by studying alone. Our main physical consideration is the moment of impact of the balls. If we define an impact parameter, *B*, as the distance between the centers of the two balls, we can define a single angle to describe the collision,

where *r* is the ball radius. For balls of differing radii, just use

During a head-on collison, . A glancing blow of above is at and below is at .

At the moment of impact, the impulse of the impact is along the point of contact of the two balls. For the first ball, we can write the impulse as

The and components simplify to

Making these a fraction gives us the needed angle,

In the last step, we replaced and . The angle at which leaves the collision is now measured from the axis. We now solve this equation.

Squaring both sides gives

Using the quadratic formula, we find

Simplifying shows

Clearly we take the negative root to find

With this simple solution, let's turn to technicalities.

## Acknowledegments

- This article has been originally written by Drew Dolgert for Michael Fowler.

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