Impulse and Conversation of Linear Momentum

Impulse

A constant net force ${\vec F}$ is applied on a body during a time interval $\Delta t = {t_2}-{t_1}$ and the body undergoes a displacement in the direction of force. The work done is the product of the force and the displacement but the impulse is the product of the force and the time interval. Therefore impulse is the measure of how long the force is applied. The impulse $J$ is

\[\vec J = {\vec F}\Delta t \tag{1} \label{1}\]

Note that impulse is a vector quantity. Further investigation of Eq. \eqref{1} gives,

\[\begin{align*} \vec J &= {\vec F}\Delta t\\ &= m\vec a \Delta t\\ &= m\frac{{\Delta \vec v }}{{\Delta t}}\Delta t\\ &= m\Delta \vec v \\ &= m({\vec v _f} - {\vec v _i})\\ {\rm{or,}}\quad \vec J &= m{\vec v _f} - m{\vec v _i} \tag{2} \label{2} \end{align*}\]

Linear momentum is the product of mass and linear velocity of a body denoted by $p$. So $\vec p = m{\vec v}$ and in Eq. \eqref{2}, the initial linear momentum is ${\vec p _i}= m{\vec v_i}$ and final linear momentum is ${\vec p _f}= m{\vec v_f}$. So Eq. \eqref{2} can be rewritten as,

\[\vec J = {\vec p _f} - {\vec p _i} \tag{3} \label{3}\]

We can also express Newton's Second law as the rate of change of linear momentum. Newton's second law gives

\[\begin{align*} {\vec F} &= m\vec a \\ &= m\frac{{{{\vec v }_f} - {{\vec v }_i}}}{{\Delta t}}\\ &= \frac{{m{{\vec v }_f} - m{{\vec v }_i}}}{{\Delta t}}\\ {\rm{or,}}\quad {\vec F} &= \frac{{{{\vec p }_f} - {{\vec p }_i}}}{{\Delta t}} \tag{4} \label{4} \end{align*}\]

In terms of varying force we can write the force in a very small interval of time $dt$ in which $\Delta t$ in Eq. \eqref{4} approaches zero and $d\vec p = {\vec p _f} - {\vec p _i}$.

\[{\vec F} = \frac{{d\vec p }}{{dt}} \tag{5} \label{5}\]

We can show that Eq. \eqref{3} is true even if the force is not constant. We first find the impulse in infinitesimally small time interval $dt$ and integrate it from the initial linear momentum ${p _i}$ to the final linear momentum ${p _f}$ as

\[J = \int\limits_{{{p }_i}}^{{{ p }_f}} {\frac{{dp }}{{dt}}dt} = {p _f} - {p _i}\]

Which is the same as Eq. \eqref{3} without vector sign. Of course, you can put the vector arrow to indicate it as a vector quantity.

Conservation of Linear 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,

\[{{\vec{F}}_{A\text{ on }B}}+{{\vec{F}}_{B\text{ on }A}}=0 \tag{6} \label{6}\]

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 Eq. \eqref{6} becomes,

\[\begin{align*} \frac{{d{{\vec p }_1}}}{{dt}} + \frac{{d{{\vec p }_2}}}{{dt}} &= 0\\ {\rm{or,}}\quad \frac{{d{{\vec p }_1} + d{{\vec p }_2}}}{{dt}} &= 0\\ {\rm{or,}}\quad \frac{{d\vec P }}{{dt}} &= 0 \tag{7} \label{7} \end{align*}\]

In Eq. \eqref{7} the sum of the total linear momentum is $d\overrightarrow P = d{\overrightarrow p _1} + d{\overrightarrow p _2}$ which is constant. Note in Eq. \eqref{7} 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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