In an elastic collision, conservation of momentum is not sufficient, since both bodies can have different velocities after the collision.
The equation therefore contains two unknowns, one of which must be eliminated. This is achieved by considering a second condition that must be met: the conservation of kinetic energy.
Both equations together can be written as follows:
\begin{eqnarray}
m_1 v_1 + m_2v_2 &=& m_1v_1'+m_2v_2'\label{eq:inelastic_momentum}\\
\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\label{eq:inelastic_energy}
\end{eqnarray}
To calculate the quantities $v_1'$ and $v_2'$ from this system of equations, the factor 1/2 can first be canceled out of the lower equation.
Then we get
\begin{equation}
m_1 v_1^2 + m_2 v_2^2 = m_1 v_1'^2 + m_2 v_2'^2
\end{equation}
Rearranging and applying the third binomial formula yields:
\begin{equation}
m_1(v_1-v_1')(v_1+v_1') = m_2(v_2-v_2')(v_2+v_2')
\end{equation}
A similar rearrangement can be performed with the above equation, resulting in the following relationship:
\begin{equation}
m_1(v_1 - v_1') = m_2(v_2 - v_2')
\end{equation}
Now dividing the first equation by the second, we get
\begin{equation}
v_1 + v_1' = v_2 + v_2'
\end{equation}
and the following relationship between the relative velocities before and after the collision:
\begin{equation}
v_1 - v_2 = v_2' - v_1'
\end{equation}
Using this equation, one of the two final velocities can now be eliminated.
Rearranging for $v_1'$ and inserting it into the momentum conservation equation gives:
\begin{equation}
m_1 v_1 + m_2 v_2 = m_1(v_2' - v_1 - v_2) + m_2 v_2'
\end{equation}
The equation thus obtained must then be expanded and rearranged for $v_2'$. The final velocity of the second colliding partner is thus obtained:
\begin{equation}
{v_2' = \frac{m_1(2v_1 - v_2) + m_2 v_2}{m_1 + m_2}}
\end{equation}
Similarly, by rearranging for $v_2'$, the final velocity of the first colliding partner can be calculated:
\begin{equation}
{v_1' = \frac{m_1 v_1 + m_2 (2v_2 - v_1)}{m_1 + m_2}}
\end{equation}
Three important special cases will be considered below:
- Equal masses of the bodies ($m_1 = m_2 = m$) and the colliding partner at rest ($v_2 = 0$): The colliding body is at rest after the collision, since $v_1' = 0$, and the colliding body continues to move with the velocity $v_2' = v_1$ that the first body had before the collision.
- Equal masses ($m_1 = m_2 = m$) and opposite velocities ($v_2 = -v_1$): The magnitudes of the velocities do not change, but the direction reverses ($v_1' = -v_1$) and $v_2' = -v_2$
- Stationary colliding body ($v_2 = 0$) with infinite mass ($m_2\rightarrow\infty$): The colliding body remains at rest ($v_2 = 0$), while the $v_1'$ becomes $-v_1$, i.e., the direction of the colliding body reverses, but the magnitude of the velocity remains the same.
Elastic collisions can also be illustrated using two gliders that collide with each other at different speeds and masses. Another example is the up-and-down motion of a rubber ball. Here, the mass of the colliding partner (the ground) can be considered infinite, so the rubber ball reaches approximately the same height each time.