Impulse is the measure of how long a force is applied. The relationship between impulse and momentum allows us to understand collisions in the system of particles. But here we consider the force that has been applied in terms of time, and you'll soon see that the result will be the change in momentum. In work-kinetic energy theorem you do the similar thing, you applied force in terms of displacement, and the result you get is the change in kinetic energy.

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,

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

The above equation is the relationship between impulse and momentum also called impulse-momentum theorem, similar to work-kinetic energy theorem. And this relationship basically helps us understand the collisions more like conceptually than practically.The original form of Newton's second law was in terms of momentum; that is the rate of change of momentum is net force.

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

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.

The work kinetic energy theorem was the real gem on the relationship between work and energy. Here in the momentum thing, similar to you integrated the force for infinitesimal displacements and the result was the change in kinetic energy, you integrated the force \(dp/dt\) for infinitesimal times. The force as a function of time is not approachable in practice as it is approachable in force as a function of displacement. Because we see in nature that the force depends on the position and the force depending on time is not naturally abundant or it is rare.