8.3: Impulse and momentum of a system

To extend the principle of impulse and momentum to a system of point masses, it is convenient to simplify the expression of the total momentum of such a system, by using its centre of mass.

Concept: Momentum of a system of point masses

The total momentum of a system of point masses \(\sum_{i} \overrightarrow{\boldsymbol{p}}_{i}\) is identical to the momentum \(\overrightarrow{\boldsymbol{p}}_{G}\) of a single point mass \(m_{\mathrm{tot}}=\sum_{i} m_{i}\) that moves along with the centre of mass of the system:

\[\sum_{i} \overrightarrow{\boldsymbol{p}}_{i}=\sum_{i} m_{i} \overrightarrow{\boldsymbol{v}}_{i}=\frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} m_{i} \overrightarrow{\boldsymbol{r}}_{i}=\frac{\mathrm{d}}{\mathrm{d} t} m_{\mathrm{tot}} \overrightarrow{\boldsymbol{r}}_{G}=m_{\mathrm{tot}} \overrightarrow{\boldsymbol{v}}_{G}=\overrightarrow{\boldsymbol{p}}_{G} \tag{8.6} \label{8.6}\]

Here Equation 7.39 was used for the CoM: \(\sum_{i} m_{i} \overrightarrow{\boldsymbol{r}}_{i}=m_{\text {tot }} \overrightarrow{\boldsymbol{r}}_{G}\). Let us now analyse the forces that generate impulse on the system in more detail. We can subdivide them into two groups, as shown in Figure 8.2: the external forces \(\overrightarrow{\boldsymbol{F}}_{i, \text { ext }}\) acting on the masses in the system and the internal forces \(\overrightarrow{\boldsymbol{F}}_{i j \text { int }}\) that the point masses generate on each other. For this second group of forces, we know from Newton’s third law, that the force vector of mass \(m_{j}\) on mass \(m_{i}\) is equal and opposite to the force vector of \(m_{i}\) on mass \(m_{j}: \overrightarrow{\boldsymbol{F}}_{i j, \text { int }}=-\overrightarrow{\boldsymbol{F}}_{j i, \text { int }}\).

Figure 8.2: The internal and external forces on a system of point masses.

Derivation:

We are now going to derive that the sum of all internal force vectors that act between point masses in a system is zero, as can intuitively be understood from Figure 8.2.

To determine the total impulse on a system, we sum first over all forces \(j\) that act on a point mass to determine the total force on that point mass: \(\overrightarrow{\boldsymbol{F}}_{i, \text { tot }}=\sum_{j \neq i} \overrightarrow{\boldsymbol{F}}_{i j, \text { int }}+\overrightarrow{\boldsymbol{F}}_{i, \text { ext }}\). Then we sum over all point masses \(i\) in the system: \(\overrightarrow{\boldsymbol{F}}_{\text {sys,tot }}=\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \text { tot }}\). Finally we integrate over time and obtain the total impulse on the system:

\[\overrightarrow{\boldsymbol{J}}_{\text {sys,tot }, 12}=\int_{t_{1}}^{t_{2}} \sum_{i}\left(\overrightarrow{\boldsymbol{F}}_{i, \text { ext }}+\sum_{j \neq i} \overrightarrow{\boldsymbol{F}}_{i j, \text { int }}\right) \mathrm{d} t \tag{8.7} \label{8.7}\]

Now we note that the sum \(\sum_{i} \sum_{j \neq i} \overrightarrow{\boldsymbol{F}}_{i j \text {,int }}\) includes for each term \(i j\) also a term \(j i\) with the inverted indices. Since we know from Newton’s third law that \(\overrightarrow{\boldsymbol{F}}_{i j, \text { int }}+\overrightarrow{\boldsymbol{F}}_{j i, \text { int }}=\overrightarrow{\mathbf{0}}\), the sum of internal forces in a system is always zero:

\[\sum_{i, j \neq i} \overrightarrow{\boldsymbol{F}}_{i j, \text { int }}=\overrightarrow{\mathbf{0}} \tag{8.8} \label{8.8}\]

Therefore, only the impulse due to external forces contributes to the change of the total momentum. By summing Equation 8.3 over all particles \(i\) in a system we obtain:

Concept: Principle of impulse and momentum for a system

The change of the total momentum of a system of point masses, during the time interval between \(t_{1}\) and \(t_{2}\) is identical to the sum of the impulses \(\sum_{i} \overrightarrow{\boldsymbol{J}}_{i, \mathrm{ext}, 12}\) of all acting external forces.

\[\begin{align} \sum_{i} m_{i} \overrightarrow{\boldsymbol{v}}_{i}\left(t_{1}\right)+\sum_{i} \int_{t_{1}}^{t_{2}} \overrightarrow{\boldsymbol{F}}_{i, \text { ext }} \mathrm{d} t & =\sum_{i} m_{i} \overrightarrow{\boldsymbol{v}}_{i}\left(t_{2}\right) \tag{8.9} \label{8.9}\\[4pt] \overrightarrow{\boldsymbol{p}}_{G, 1}+\sum_{i} \overrightarrow{\boldsymbol{J}}_{i, \text { ext }, 12} & =\overrightarrow{\boldsymbol{p}}_{G, 2} \tag{8.10} \label{8.10}\end{align}\]

where we used Equation 8.1 and Equation 8.6 to obtain an expression for the effect of an impulse on the momentum of the centre of mass of a system of point masses.

We can now derive \({ }^{1}\) Euler’s first law by taking the time derivative of Equation 8.9 and using Equation 8.6:

Concept: Euler’s first law

Euler’s first law states that the total mass \(m_{\mathrm{tot}}\) times the acceleration \(\overrightarrow{\boldsymbol{a}}_{G}\) of the CoM of a system of point masses equals the sum of external forces acting on the system:

\[\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}=m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G} \tag{8.11} \label{8.11}\]

This is an important equation, since it states that the acceleration of the CoM of a system of point masses only depends on the sum of the external forces on it. Interestingly, the points of action of the forces do not matter, and also internal forces can fully be neglected, which can significantly simplify analysis.