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

- [[Math:c| \begin{eqnarray} \vec{v_0} = v_1\hat{x} \\ \vec{v}_2 = 0 \end{eqnarray} ]]

Immediately move to center of mass coordinates where

- [[Math:c| x_{\mbox{cm}}=\frac{m_1\vec{x}_1\plusm_2\vec{x}_2}{m_1\plusm_2} ]]

In these coordinates, the total linear momentum is zero.

- [[Math:c| \begin{eqnarray} m_1\vec{v}_1\plusm_2\vec{v}_2 = 0 \\ m_1\vec{v}'_1\plusm_2\vec{v}'_2 = 0 \end{eqnarray} ]]

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

- [[Math:c| \begin{eqnarray} \vec{v}_2 = -\frac{m_1}{m_2} \vec{v}_1,\\ \vec{v}_2'=-\frac{m_1}{m_2}\vec{v}_1'. \end{eqnarray} ]]

Also, the speeds before and after the collision are unchanged

- [[Math:c| \begin{eqnarray} v_1' = v_1, \\ v_2' = v_2. \end{eqnarray} ]]

Because [[Math:c|\vec{v}_1]] and [[Math:c|\vec{v}_2]] are so well connected, we

can complete the calculation by studying [[Math:c|\vec{v}_1]] alone. Our main physical consideration is the moment of impact of the balls. If we define an impact parameter, *B*, as the [[Math:c|\hat{y}]] distance between the centers of the two balls, we can define a single angle to describe the collision,

- [[Math:c|\sin\theta = \frac{B}{2r}]]

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

- [[Math:c|B \over {(r_1\plusr_2)}]]

During a head-on collison, [[Math:c|\theta=0]]. A glancing blow of [[Math:c|m_1]] above [[Math:c|m_2]] is at [[Math:c|\theta=\pi/2]] and below is at [[Math:c|\theta=-\pi/2]].

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

- [[Math:c| m_1(\vec{v}_1'-\vec{v}_1) = I (-\cos\theta \hat{x}\plus\sin\theta \hat{y}). ]]

The [[Math:c|\hat{x}]] and [[Math:c|\hat{y}]] components simplify to

- [[Math:c| \begin{eqnarray} m_1v_{1x}'-m_1v_1 = -I\cos\theta \\ m_1v_{1y}' = I\sin\theta. \end{eqnarray} ]]

Making these a fraction gives us the needed angle,

- [[Math:c| \tan\theta = \frac{m_1v_{1y}'}{m_1v_1-m_1v_{1x}'}= \frac{\sin{\phi}}{1-\cos{\phi}}. ]]

In the last step, we replaced [[Math:c|v_{1x}=v_1\cos{\phi}]] and [[Math:c|v_{1y}=v_1\sin{\phi}]]. The angle at which [[Math:c|m_1]] leaves the collision is now [[Math:c|\phi]] measured from the [[Math:c|\hat{x}]] axis. We now solve this equation.

- [[Math:c| \tan\theta-\cos\phi\tan\theta=\sin\phi ]]

Squaring both sides gives

- [[Math:c| \tan^2\theta(\cos^2\phi-2\cos\phi\plus1)=\sin^2\phi=1-\cos^2\phi. \\ (\tan^2\theta\plus1)\cos^2\phi-2\tan^2\theta\cos\phi\plus\tan^2\theta-1 = 0 ]]

Using the quadratic formula, we find

- [[Math:c| \cos\phi=\frac{2\tan^2\theta\pm\sqrt{4\tan^4\theta-4(\tan^2\theta\plus1) \\ (\tan^2\theta-1)}}{2(\tan^2\theta\plus1)}. ]]

Simplifying shows

- [[Math:c| \cos\phi = \frac{\tan^2\theta\pm1}{\tan^2\theta\plus1}. ]]

Clearly we take the negative root to find

- [[Math:c| \cos\phi = 2\sin^2\theta - 1 = 1-2\cos^2\theta. ]]

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

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

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