Is it possible to find the final velocities of two objects in a perfectly elastic one dimensional collision where both objects have initial velocity?

Question

I'd like to know how to generally solve for final velocity of both objects after a perfectly elastic collision. Cannot be in a lab reference frame because I'm programming it. Must be able to solve for two final velocities from two initial velocities. I tried conservation of momentum and conservation of kinetic energy and ended up with a quadratic. Quadratics can't be in the final solution. Can't have two or no answers. Is this possible?

Answer 1

To Solve a 1D elastic collision, I recall two approaches. If you shift to a coordinate system which is moving with the center of mass, the velocities of each mass will be reversed by the collision. Then shift back. If you want to work with the conservation equations, rearrange the energy equation so that you have a difference of two squares on each side (with the masses factored out). Factor, and then divide by the momentum equation. Without the factor of (½), the energy equation can be put into this form: ${m_1}(u_1 + v_1)(u_1 - v_1) = {m_2}(v_2 + u_2)(v_2 – u_2)$, where the u's are velocity components before the collision, and the v's are after. Similarly, the momentum equation is; ${m_1}(u_1 – v_1) = {m_2}(v_2 – u_2)$. Divide the first equation by the second and you find that the sum of the two components before the collision is the same as after (some components may be negative). Combine this result with the momentum equation to solve for the v's. As for the center of mass: Remember that the momentum of the system can be expressed in terms of the velocity of the center of mass: $({m_1} + {m_2})v_{cm} = {m_1}{u_1} + {m_2}{u_2}$.

Answer 2

Solving for velocities in 1D an elastic collision for two objects $A$ and $B$ respectively boils down to solving the following system of equations (where the $'$ denotes the velocity after the collision):

$$ \left\{ m_Av_A + m_Bv_B = m_Av_A' + m_Bv_B'\quad \text{cons. of momentum} \atop \dfrac1 2m_Av_A^2 + \dfrac1 2m_Bv_B^2 = \dfrac1 2m_Av_A'^2 + \dfrac1 2m_Bv_B'^2\quad \text{cons. of KE} \right.$$ The above system yields the following solution when $v_A$, $v_B$ and the ratio of masses are known:

$$v_A' = \dfrac{m_A - m_B}{m_A+m_B} v_A + \dfrac{2 m_B}{m_A+m_B}v_B$$ $$v_{B}' = \dfrac{2m_A}{m_A+m_B} v_{A} - \dfrac{m_A-m_B}{m_A+m_B}v_{B}$$

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Now, you point out the quadratic has 2 solutions. That is true. However, the second solution is the trivial solution $v_A'=v_A$ and $v_B'=v_B$ (which is no collision). So you can discard it. Which is why the above solution is the only one usually presented.