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@ -9,15 +9,20 @@ Consider two particles, designated as particles `1` and `2`, with respective mas
In any collision, momentum is conserved. An elastic collision is an collision, as a result of which the total kinetic energy of the colliding particles is conserved.
The conservation of momentum before and after the collision is expressed by the following equation:
$$
m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2} = m_{1}\vec{v'}_{1} + m_{2}\vec{v'}_{2}
$$
And the conservation of the total kinetic energy before and after the collision is expressed by the following equation:
$$
\frac{m_{1}{v_{1}}^2}{2} + \frac{m_{2}{v_{2}}^2}{2} = \frac{m_{1}{v'_{1}}^2}{2} + \frac{m_{2}{v'_{2}}^2}{2}
$$
The aforementioned equations can be transformed into the following system:
$$
\begin{cases}
m_{1}({v_{1}} - {v'_{1}}) = m_{2}({v'_{2}} - {v_{2}}) \\
@ -25,22 +30,32 @@ m_{1}({v_{1}}^2 - {v'_{1}}^2) = m_{2}({v'_{2}}^2 - {v_{2}}^2)
\end{cases}
$$
The division of each side of the top equation by each side of the bottom equation yields:
$$
\frac{v_{1} - v'_{1}}{{v_{1}^2} - {v'_{1}}^2} = \frac{v'_{2} - v_{2}}{{{v'_{2}}^2} - {v_{2}}^2}
$$
The preceding equation may be transformed using the difference of two squares formula to obtain the following:
$$
\frac{v_{1} - v'_{1}}{(v_1 + v'_1)(v_1 - v'_1)} = \frac{v'_{2} - v_{2}}{(v'_2 - v_2)(v'_2 + v_2)}
$$
This expression can be transformed as follows:
$$
\frac{1}{v_1 + v'_1} = \frac{1}{v'_2 + v_2}
$$
From this expression follows:
$$
v_1 + v'_1 = v'_2 + v_2
$$
From this equation, the following system can be derived:
$$
\begin{cases}
v'_1 = v_2 + v'_2 - v_1 \\
@ -48,6 +63,8 @@ v'_2 = v_1 + v'_1 - v_2
\end{cases}
$$
The law of conservation of momentum then yields the following:
$$
\begin{cases}
m_1v_1 + m_2v_2 = m_1{v'}_1 + m_2(v_1 + {v'}_1 - v_2) \\
@ -55,6 +72,8 @@ m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
\end{cases}
$$
From this, we can derive the velocity of the particles following the collision as follows:
$$
\begin{cases}
v'_1 = \frac{2m_2v_2 + v_1(m_1-m_2)}{m_1+m_2} \\