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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?

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    $\begingroup$ A quadratic is ok, because typically there are two answers which conserve both momentum and energy: one answer where no collision happened and the two bodies just keep their original motion, and one answer where there is a collision and you get some other final momenta. The algebra is not too bad; your approach was sound. $\endgroup$ Apr 22, 2021 at 18:24
  • $\begingroup$ You get 2 answers: one for forward scattering and one for backward scattering. $\endgroup$
    – JEB
    Apr 22, 2021 at 18:39
  • $\begingroup$ @AndrewSteane Is there a way for me to mathematically determine which quadratic solution was the state before the collision and which is the state after the collision? I'm trying to figure out a way to pick which quadratic solution I need programmatically. $\endgroup$
    – Sage King
    Apr 22, 2021 at 20:15
  • $\begingroup$ I found a solution. sciencecalculators.org/mechanics/collisions I don't know how they derived this equation, so if somebody wants to take the time to explain this to me I would appreciate it. I programmed the equation and it works though so I'm going to stick with this. $\endgroup$
    – Sage King
    Apr 22, 2021 at 21:02

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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: $({m_1} + {m_2})v_{cm} = {m_1}{u_1} + {m_2}{u_2}$.

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  • $\begingroup$ I don't know what this means. Your explanation of each of these is unfollowable for me. $\endgroup$
    – Sage King
    Apr 22, 2021 at 21:03
  • $\begingroup$ @SageKing, research "center of mass reference frame". $\endgroup$ Apr 22, 2021 at 21:26
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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:

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


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.

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