Conservation of Momentum

The collision of two bodies $A$ and $B$ in straight line motion is shown in Figure 1. According to Newton's third law the force body $A$ exerts on body $B$ must be equal and opposite to the force the body $B$ exerts on body $A$ that is, ${{\vec{F}}_{A\text{ on }B}}=-{{\vec{F}}_{B\text{ on }A}}$. So,

\[{{\vec{F}}_{A\text{ on }B}}+{{\vec{F}}_{B\text{ on }A}}=0 \tag{1} \label{1}\]

Figure 1 The total momentum before collision is equal to the total momentum after collision.

The force ${{\vec{F}}_{A\text{ on }B}}$ in terms of the rate of change of linear momentum is ${{\vec{F}}_{A\text{ on }B}}=\frac{d{{\vec{p}}_{1}}}{dt}$ and similarly ${{\vec{F}}_{B\text{ on }A}}=\frac{d{{\vec{p}}_{2}}}{dt}$. The Equation \eqref{1} becomes,

\[\begin{align*} \frac{{d{{\vec p }_1}}}{{dt}} + \frac{{d{{\vec p }_2}}}{{dt}} &= 0\\ {\rm{or,}}\quad \frac{{d{{\vec p }_1} + d{{\vec p }_2}}}{{dt}} &= 0\\ {\rm{or,}}\quad \frac{{d\vec P }}{{dt}} &= 0 \tag{2} \label{2} \end{align*}\]

In Equation \eqref{2} the sum of the total linear momentum is $d\overrightarrow P = d{\overrightarrow p _1} + d{\overrightarrow p _2}$ which is constant. Note in Equation \eqref{2} that the derivative of a constant is zero and hence the total momentum of a system of particles is constant or conserved if no net external force acts on the system which we call the conservation of linear momentum. We can conclude from the conservation of linear momentum that the total momentum before collision is equal to the total momentum after collision which is discussed in the next article in elastic and inelastic collisions in more details.

Mechanics
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