Two-dimensional Collision of Hard Balls is a classic physical task, explained in this article.
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
Clearly we take the negative root to find
With this simple solution, let's turn to technicalities.
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