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two dimensional collision

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

Run Review after Drew Dolgert for Michael Fowler * Alone

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.

$\begin{eqnarray} \vec{v_0} = v_1\hat{x} \\ \vec{v}_2 = 0 \end{eqnarray}$

Immediately move to center of mass coordinates where

$x_{\mbox{cm}}=\frac{m_1\vec{x}_1\plusm_2\vec{x}_2}{m_1\plusm_2}$

In these coordinates, the total linear momentum is zero.

$\begin{eqnarray} m_1\vec{v}_1\plusm_2\vec{v}_2 = 0 \\ m_1\vec{v}'_1\plusm_2\vec{v}'_2 = 0 \end{eqnarray}$

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

$\begin{eqnarray} \vec{v}_2 = -\frac{m_1}{m_2} \vec{v}_1,\\ \vec{v}_2'=-\frac{m_1}{m_2}\vec{v}_1'. \end{eqnarray}$

Also, the speeds before and after the collision are unchanged

$\begin{eqnarray} v_1' = v_1, \\ v_2' = v_2. \end{eqnarray}$

Because $\vec{v}_1$ and $\vec{v}_2$ are so well connected, we

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

$\sin\theta = \frac{B}{2r}$

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

$B \over {(r_1\plusr_2)}$

During a head-on collison, $\theta=0$ . A glancing blow of above is at $\theta=\pi/2$ and below is at $\theta=-\pi/2$ .

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

$m_1(\vec{v}_1'-\vec{v}_1) = I (-\cos\theta \hat{x}\plus\sin\theta \hat{y}).$

The $\hat{x}$ and $\hat{y}$ components simplify to

$\begin{eqnarray} m_1v_{1x}'-m_1v_1 = -I\cos\theta \\ m_1v_{1y}' = I\sin\theta. \end{eqnarray}$

Making these a fraction gives us the needed angle,

$\tan\theta = \frac{m_1v_{1y}'}{m_1v_1-m_1v_{1x}'}= \frac{\sin{\phi}}{1-\cos{\phi}}.$

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