# Derivation of the Elastic Collision Formula

In the scenario of an one-dimensional elastic collision between two objects, $1$ and $2$, their final velocities, $v_1$ and $v_2$ can be found with the following formula knowing their individual masses, $m_1$ and $m_2$, and their initial velocities, $u_1$ and $u_2$.

Derivation:

In an elastic collision, momentum is conserved:

\begin{aligned} \sum p_f&=\sum p_i\\ m_1v_1+m_2v_2&=m_1u_1+m_2u_2\\ m_1v_1-m_1u_1&=m_2u_2-m_2v_2\\ \textcolor{#3D99F6}{m_1(v_1-u_1)}&=\textcolor{#3D99F6}{m_2(u_2-v_2)} \end{aligned}

In an elastic collision, kinetic energy is conserved:

\begin{aligned} \sum K_f&=\sum K_i\\ \frac12m_1v_1^2+\frac12m_2v_2^2&=\frac12m_1u_1^2+\frac12m_2u_2^2\\ m_1v_1^2+m_2v_2^2&=m_1u_1^2+m_2u_2^2\\ m_1v_1^2-m_1u_1^2&=m_2u_2^2-m_2v_2^2\\ m_1(v_1^2-u_1^2)&=m_2(u_2^2-v_2^2)\\ \textcolor{#3D99F6}{m_1(v_1-u_1)}(v_1+u_1)&=m_2(u_2-v_2)(u_2+v_2)\\ \textcolor{#3D99F6}{m_2(u_2-v_2)}(v_1+u_1)&=m_2(u_2-v_2)(u_2+v_2)\\ v_1+u_1&=v_2+u_2\\ \textcolor{#20A900}{v_2}&=\textcolor{#20A900}{u_1+v_1-u_2} \end{aligned}

\begin{aligned} \sum p_f&=\sum p_i\\ m_1v_1+m_2\textcolor{#20A900}{v_2}&=m_1u_1+m_2u_2\\ m_1v_1+m_2(\textcolor{#20A900}{u_1+v_1-u_2})&=m_1u_1+m_2u_2\\ m_1v_1+m_2u_1+m_2v_1-m_2u_2&=m_1u_1+m_2u_2\\ m_1v_1+m_2v_1&=m_1u_1+2m_2u_2-m_2u_1\\ (m_1+m_2)v_1&=(m_1-m_2)u_1+2m_2u_2\\ \textcolor{#D61F06}{v_1}&=\textcolor{#D61F06}{\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2} \end{aligned}

\begin{aligned} v_2&=u_1+\textcolor{#D61F06}{v_1}-u_2\\ v_2&=u_1+\textcolor{#D61F06}{\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2}-u_2\\ v_2&=\left(1+\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}-1\right)u_2\\ v_2&=\left(\frac{m_1+m_2}{m_1+m_2}+\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}-\frac{m_1+m_2}{m_1+m_2}\right)u_2\\ v_2&=\left(\frac{2m_1}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2 \end{aligned}

\therefore \boxed{ \begin{aligned} v_1&=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2\\ v_2&=\left(\frac{2m_1}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2 \end{aligned} }

or, equivalently:

$\boxed{ \def\arraystretch{1.5} \begin{bmatrix} v_1\\v_2 \end{bmatrix}= \frac1{m_1+m_2}\begin{bmatrix} m_1-m_2&2m_2\\ 2m_1&m_2-m_1 \end{bmatrix} \begin{bmatrix} u_1\\u_2 \end{bmatrix} }$ Note by Gandoff Tan
1 year, 7 months ago

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