1.6 KiB
3.
Firstly, it is essential to comprehend how the velocity of the particles after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian).
3.1
Consider two particles, designated as particles 1 and 2, with respective masses m_1 and m_2. Before the collision, the velocities of particles 1 and 2 are v_1 and v_2, respectively. After the collision, the velocities of particles 1 and 2 are v_1' and v_2', respectively.
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.
$$
m_{1}\vec{v}{1} + m{2}\vec{v}{2} = m{1}\vec{v'}{1} + m{2}\vec{v'}_{2}
$$
\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}
$$
\begin{cases}
m_{1}({v_{1}} - {v'{1}}) = m{2}({v'{2}} - {v{2}}) \
m_{1}({v_{1}}^2 - {v'{1}}^2) = m{2}({v'{2}}^2 - {v{2}}^2)
\end{cases}
$$
\frac{v_{1} - v'{1}}{{v{1}^2} - {v'{1}}^2} = \frac{v'{2} - v_{2}}{{{v'{2}}^2} - {v{2}}^2}
$$
\frac{v_{1} - v'_{1}}{(v_1 + v'1)(v_1 - v'1)} = \frac{v'{2} - v{2}}{(v'_2 - v_2)(v'_2 + v_2)}
$$
\frac{1}{v_1 + v'_1} = \frac{1}{v'_2 + v_2}
$$
v_1 + v'_1 = v'_2 + v_2
$$
\begin{cases}
v'_1 = v_2 + v'_2 - v_1 \
v'_2 = v_1 + v'_1 - v_2
\end{cases}
$$
\begin{cases}
m_1v_1 + m_2v_2 = m_1{v'}_1 + m_2(v_1 + {v'}_1 - v_2) \
m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
\end{cases}
$$
\begin{cases}
v'_1 = \frac{2m_2v_2 + v_1(m_1-m_2)}{m_1+m_2} \
v'_2 = \frac{2m_1v_1 + v_2(m_2-m_1)}{m_1+m_2}
\end{cases}