Linear Momentum Conservation Theorem

up:: Linear Momentum

Let $\{{\vec{r}}_i{\}}_{i=1}^n$ be the positions of particles $i$, of mass $\{m_i{\}}_{i=1}^n$, with Center of Mass $\vec{R}$ and external forces $\left\{{\vec{F}}_i^{(e)}\right\}$. Then

$$M\frac{d^2\vec{R}}{d{t}^2}=\sum _{i=1}^n{\vec{F}}_i^{(e)}$$

That is, the sum of the external forces over a system of particles induces Newton’s Second Law to the equivalent particle $(M,\vec{R})$, located at the center of mass and with mass equal to the system’s total mass.

Proof

By Newton’s Second Law, we have that, for each particle $i$¹,

$$\frac{d\overrightarrow{p_i}}{dt}=\sum _{\begin{array}{c}j=1\\ j\ne i\end{array}}^n{\vec{F}}_{j\to i}+{\vec{F}}_i^{(e)}$$

Summing for all particles $j$, we have that

$$\sum _i\frac{d\overrightarrow{p_i}}{dt}=\sum _{\begin{array}{c}i,j=1\\ j\ne i\end{array}}^n{\vec{F}}_{j\to i}+\sum _{i=1}^n{\vec{F}}_i^{(e)}$$

The sum of all reciprocal forces in a system of particles is always zero $\sum _{\begin{array}{c}i,j=1\\ j\ne i\end{array}}^n{\vec{F}}_{j\to i}=\vec{0}$². Thus,

$$\sum _i{m}_i\frac{d\overrightarrow{p_i}}{dt}=\sum _{i=1}^n{\vec{F}}_i^{(e)}$$

Using the Center of Mass definition, and supposing that $\frac{d{m}_i}{dt}=0$, one can write the above as

$$\sum _{i=1}^n{\vec{F}}_i^{(e)}=M\frac{d^2\vec{R}}{d{t}^2}$$

Corollaries

A system’s total momentum is conserved if it has no net external forces: If the external forces sum to $\vec{0}$, then the system’s total momentum

$$\vec{P}=\sum _{i=1}^n{m}_i{v}_i=\sum _{i=1}^n{m}_i\frac{d{\vec{r}}_i}{dt}=M\frac{d\vec{R}}{dt}$$

is conserved.

Examples

Let a system of two particles with no external forces acting upon them. Then

$$m_1\overrightarrow{v_1}=m_2\overrightarrow{v_2}$$

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References

• LEMOS, Nivaldo A. Mecânica analítica. Editora Livraria da Física, 2007.

Footnotes

1. Denoting $F_{j\to i}$ as the force the particle $j$ exerts upon particle $i$. ↩

2. Due to Newton’s Third Law. One can think about it as summing all values of an antisymmetric matrix. ↩