# One Dimensional Elastic Collisions

A demonstration of one dimensional elastic collisions highlighting the conservation of both momentum and energy

Imagine two billiard balls of mass $$m_1$$ and $$m_2$$, travelling at velocities $$\vec{v_{1i}}$$ and $$\vec{v_{2i}}$$ respectively (the $$_{i}$$ stands for initial). They collide and bounce off each other with velocities $$\vec{v_{1f}}$$ and $$\vec{v_{2f}}$$ ( the $$_{f}$$ stands for final).

Assuming we know the masses and initial velocities, we can calculate the final velocities by making use of two key conservation principles: the conservation of momentum, and the conservation of energy.

The conservation of momentum (ie total momentum before the collision equals total momentum after) gives us equation 1. Note that because we are dealing with one dimension we only require the magnitude of the vecotrs the so vector notation is not needed.

$$\label{eq:momentum-conservation} m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

The conservation of energy (ie the total energy before the collision equals the total energy afterwards) gives us equation $$\eqref{eq:kinetic-energy}$$. The scenario we are dealing with is perfectly elastic so no energy is lost in the collision itself allowing us to deal purely in terms of kinetic energy.

$$\label{eq:kinetic-energy} \frac{1}{2}m_1v_{1i}{}^2 + \frac{1}{2}m_2v_{2i}{}^2 = \frac{1}{2}m_1v_{1f}{}^2 + \frac{1}{2}m_2v_{2f}{}^2$$

And we can factor out the $$\frac{1}{2}$$ from both sides of this equation to give equation $$\eqref{eq:kinetic-half-factored}$$.

$$\label{eq:kinetic-half-factored} m_1v_{1i}{}^2 + m_2v_{2i}{}^2 = m_1v_{1f}{}^2 + m_2v_{2f}{}^2$$

Rearranging equation $$\eqref{eq:kinetic-half-factored}$$ gives us

\begin{align} m_1v_{1i}{}^2 - m_1v_{1f}{}^2 &= m_2v_{2f}{}^2 - m_2v_{2i}{}^2 \label{eq:kinetic-rearranged} \\ m_1 (v_{1i}{}^2 - v_{1f}{}^2) &= m_2(v_{2f}{}^2 - v_{2i}{}^2) \end{align}

The difference of two squares on each side (the term in the bracket) can now be factored out.

$$\label{eq:kinetic-factored} m_1 (v_{1i} + v_{1f})(v_{1i} - v_{1f}) = m_2(v_{2f} + v_{2i})(v_{2f} - v_{2i})$$

Now, going back to equation $$\eqref{eq:momentum-conservation}$$, we group all terms involving the same mass.

\begin{align} m_1v_{1i} - m_1v_{1f} &= m_2v_{2f} - m_2v_{2i} \\ m_1(v_{1i} - v_{1f}) &= m_2(v_{2f} - v_{2i}) \label{eq:momentum-rearranged} \end{align}

We can now divide equation $$\eqref{eq:kinetic-factored}$$ by equation $$\eqref{eq:momentum-rearranged}$$ to give us

\begin{align} \label{eq:divided} v_{1i} + v_{1f} &= v_{2f} + v_{2i} \\ v_{1f} &= v_{2f} + v_{2i} - v_{1i} \label{eq:velocities} \end{align}

From here, plug equation $$\eqref{eq:velocities}$$ into $$\eqref{eq:momentum-conservation}$$

\begin{align} m_1v_{1i} + m_2v_{2i} &= m_1(v_{2f} + v_{2i} - v_{1i}) + m_2v_{2f} \\ m_1v_{1i} + m_2v_{2i} &= m_1v_{2f} + m_1v_{2i} - m_1v_{1i} + m_2v_{2f} \\ m_1v_{1i} + m_2v_{2i} &= (m_1 + m_2)v_{2f} + m_1v_{2i} - m_1v_{1i} \\ \end{align} \begin{gather} 2m_1v_{1i} + m_2v_{2i} - m_1v_{2i} = (m_1 + m_2)v_{2f} \\ 2m_1v_{1i} + v_{2i}(m_2 - m_1) = (m_1 + m_2)v_{2f} \\ v_{2f} = \frac{2m_1v_{1i} + v_{2i}(m_2 - m_1)}{m_1 + m_2} \end{gather}

This is the solution for $$v_{2f}$$, and we can now plug the result into $$\eqref{eq:velocities}$$ to get our final solution for $$v_{1f}$$

\begin{align} v_{1f} &= \frac{2m_1v_{1i} + v_{2i}(m_2 - m_1)}{m_1 + m_2} + v_{2i} - v_{1i} \\ v_{1f} &= \frac{2m_1v_{1i} + v_{2i}(m_2 - m_1) + v_{2i}(m_1 + m_2) - v_{1i}(m_1 + m_2)}{m_1 + m_2} \\ v_{1f} &= \frac{2m_2v_{2i} + v_{1i}(m_1 - m_2)}{m_1 + m_2} \end{align}