Momentum

When two bodies 1 and 2 collide, Newton III must be satisfied at any time $t$. However, we are only interested in the initial and final states at this time. The processes that occur during the collision will therefore not play a role in the following discussion. Rather, for further calculations, only the initial velocities $v_1$ and $v_2$, or the final velocities $v_1'$ and $v_2'$ of the two bodies, are of interest.
These can be calculated using the equation $F=ma$ mentioned in Newton II.

Inserting this into $F_{1\rightarrow 2} = -F_{2\rightarrow 1}$ and then integrating both sides over the entire time interval from $t_1$ to $t_2$ yields:
\begin{equation*}
m_1 \int_{t_1}^{t_2} \vec{a}_1\,\mathrm{d}t = m_2 \int_{t_1}^{t_2} \vec{a}_2\,\mathrm{d}t
\end{equation*}
Using the previously discussed relationship between velocity and acceleration, we get:
\begin{equation*}
m_1 \vec{v}_1' - m_1 \vec{v}_1 = m_2 \vec{v}_2' - m_2 \vec{v}_2
\end{equation*}
Transforming this equation ultimately leads to the following Form:
\begin{equation*}
m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v}_1' + m_2 \vec{v}_2'
\end{equation*}
The quantity
\begin{equation}\label{eq:momentum}
{\vec{p} = m\vec{v}}
\end{equation}
is called momentum.
With this definition, the previously formulated equation can be written as follows:
\begin{equation*}
\vec{p}_1 + \vec{p}_2 = \vec{p}_1' + \vec{p}_2'
\end{equation*}
The equation in its current form states that the total momentum in a closed system must always be conserved.
This means that the sum of the momenta before a certain point in time (e.g., a collision process) must be equal to the sum of the momenta after that point in time.
Of course, this relationship applies not only to two mass points but can easily be applied to systems with multiple mass points.
For a system with $N$ mass points, the following applies:
\begin{equation}
\sum_{i = 1}^{N} \vec{p}_i = \sum_{i = 1}^{N} \vec{p}_i'
\end{equation}

The force acting on a body is equal to the change in its momentum over time.
This results in the following relationship between force and momentum:
\begin{equation}
\vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} + \frac{\mathrm{d}m}{\mathrm{d}t}\vec{v}
\end{equation}
In the last step, the product rule of differential calculus was applied. If the mass does not change over time, the last term disappears, leaving us with
\begin{equation}
\vec{F} = m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = m\vec{a}
\end{equation}
Newton's Second Law, as already described.