Elastic Collisions

For elastic scattering, the momentum conservation has to be taken into account which can be written as follows: $$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$ For the energy conservation, the following equation has to be taken into account: $$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2$$ For solving this system of equation, first the second equation can be multiplied with the factor 1/2: $$m_1 v_1^2 + m_2 v_2^2 = m_1 v_1'^2 + m_2 v_2'^2$$ Using the third binomial equation gives $$m_1(v_1-v_1')(v_1+v_1') = m_2(v_2-v_2')(v_2+v_2')$$ A similar rearranging can be done for the first equation: $$m_1(v_1 - v_1') = m_2(v_2 - v_2')$$ Dividing the first equation by the second one results in $$v_1 + v_1' = v_2 + v_2'$$ Therefore, the relative speed between both objects before and after the collision is given as $$v_1 - v_2 = v_2' - v_1'$$ This equation makes it then possible to eliminate one of the two final speeds. For \(v'_1\) it follows $$\boxed{v_1' = \frac{m_1 v_1 + m_2 (2v_2 - v_1)}{m_1 + m_2}}$$ and for $v_2$ respectively $$\boxed{v_2' = \frac{m_1(2v_1 - v_2) + m_2 v_2}{m_1 + m_2}}$$