Two circles with positions \( \mathbf{p_1} = (x_1, y_1) \), \( \mathbf{p_2} = (x_2, y_2) \),
velocities \( \mathbf{v_1}, \mathbf{v_2} \), and masses \( m_1, m_2 \):
\[
\mathbf{v_1'} = \mathbf{v_1} - \frac{2 m_2}{m_1 + m_2} \frac{ \langle \mathbf{v_1} - \mathbf{v_2},
\mathbf{p_1} - \mathbf{p_2} \rangle }{ \|\mathbf{p_1} - \mathbf{p_2}\|^2 } (\mathbf{p_1} -
\mathbf{p_2})
\]
\[
\mathbf{v_2'} = \mathbf{v_2} - \frac{2 m_1}{m_1 + m_2} \frac{ \langle \mathbf{v_2} - \mathbf{v_1},
\mathbf{p_2} - \mathbf{p_1} \rangle }{ \|\mathbf{p_2} - \mathbf{p_1}\|^2 } (\mathbf{p_2} -
\mathbf{p_1})
\]
Simplified Version (Equal Masses) which is what I use!
\[
\mathbf{v_1'} = \mathbf{v_1} - \frac{ \langle \mathbf{v_1} - \mathbf{v_2}, \mathbf{p_1} -
\mathbf{p_2} \rangle }{ \|\mathbf{p_1} - \mathbf{p_2}\|^2 } (\mathbf{p_1} - \mathbf{p_2})
\]
\[
\mathbf{v_2'} = \mathbf{v_2} - \frac{ \langle \mathbf{v_2} - \mathbf{v_1}, \mathbf{p_2} -
\mathbf{p_1} \rangle }{ \|\mathbf{p_2} - \mathbf{p_1}\|^2 } (\mathbf{p_2} - \mathbf{p_1})
\]
Angular Momentum
When a circle spins, it has angular momentum L.
\[
L = I \omega
\]
and
- \( I \) is the moment of inertia.
- \( \omega \) is the angular velocity (rate of spin).
Now what is a moment of inertia:
\[
I = \frac{1}{2} m r^2
\]
When two circles hit each other off-center, they create a torque \( \tau \), which changes the spin:
\[
\tau = \frac{dL}{dt}
\]
This torque comes from how their edges slide past each other (relative tangential velocity), causing them to spin more or less.
