m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

efficiency (m_1 v_{1i}² + m_2 v_{2i}²) = m_1 v_{1f}² + m_2 v_{2f}²

The intersection(s) of the 
momentum line and the 
KE ellipse represent 
possible collisions.  
If there are two collisions, 
one involves the masses 
passing through each other 
or never colliding (moving 
in opp. dir.).

\text{momentum: }m_1 v_{1i}+m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \cr
\text{KE: }(\text{efficiency})\left(\frac{1}{2}m_1 v_{1i}^2+\frac{1}{2}m_2 v_{2i}^2\right) = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2

m_1 = 1 (slider from 0 to 10, step 0.1)
m_2 = 2.3 (slider from 0 to 10, step 0.1)
v_{1i} = 1 (slider from -5 to 5, step 0.1)
v_{2i} = -1 (slider from -5 to 5, step 0.1)
efficiency = 0.6950000000000001 (slider from 0 to 1, step 1.0E-5)
momentum: (m_1 * v_{1i}) + (m_2 * v_{2i}) = (m_1 * x) + (m_2 * y)
KE: (efficiency * ((m_1 * v_{1i}^(2)) + (m_2 * v_{2i}^(2)))) = (m_1 * x^(2)) + (m_2 * y^(2))
A, B = Intersect(KE, momentum)
f: y = x