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Physics LibreTexts

3.6: Force and Rate of Change of Momentum

  • Page ID
    6944
  • [ "article:topic", "authorname:tatumj" ]

    Theorem:

    The rate of change of the total momentum of a system of particles is equal to the sum of the external forces on the system.

    Thus, consider a single particle. By Newton’s second law of motion, the rate of change of momentum of the particle is equal to the sum of the forces acting upon it:

    \[ \dot{\textbf{P}}_{i} = \textbf{F}_{i} + \sum_i \textbf{F}_{ij} \qquad (j \neq i ) \label{eq:3.6.1} \]

    Now sum over all the particles: 

    \[\dot{\textbf{P}}_{i} =\sum_i \textbf{F}_{i} + \sum_i\sum_j \textbf{F}_{ij} \qquad (j \neq i )\]

    \[\textbf{F} + \frac{1}{2}\sum_i\sum_j \textbf{F}_{ij} + \frac{1}{2}\sum_j\sum_i \textbf{F}_{ij}\]

    \[ \textbf{F} + \frac{1}{2}\sum_i\sum_j \textbf{F}_{ji}+ \textbf{F}_{ij} \label{eq:3.6.2} \]

    But, by Newton’s third law of motion, \(\textbf{F}_{ji}+ \textbf{F}_{ij} = 0\), so the theorem is proved.

    Corollary:

    If the sum of the external forces on a system is zero, the linear momentum is constant. (Law of Conservation of Linear Momentum.)

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