For the specific case of linear momentum, you can look at Newton's second law, which states that ##\dot{\boldsymbol p} = {\boldsymbol F}##, where ##\dot{\boldsymbol p}## is the continuous time derivative of the linear momentum ##{\boldsymbol p}## and ##{\boldsymbol F}(t)## is a continuous and generally time-variant forcing function.
This means that, for a given time ##t##, the system has a momentum ##{\boldsymbol p}(t)## at that time instance. Indeed, you can go further and use the momentum as a state of the system, e.g. you work with linear and angular momenta (and their time-derivatives) as your variables instead of traditional velocity and acceleration kinematic variables.
Momenta are defined quantities; for linear momentum we have the familiar{\boldsymbol p} \equiv m{\boldsymbol v}and, in one case, we have the angular momentum about a fixed point ##o##{\boldsymbol h}_o \equiv {\boldsymbol r} \times m{\boldsymbol v}.We can also define absolute and relative angular momentums about an arbitrary point. (Note that in physics ##{\boldsymbol L}## is generally used as the symbol for angular momentum.) These quantities all give instantaneous momentums, e.g. the momentum has a certain value for a given system configuration at that time instance.
For the second part of your question, let's look at the example of air resistance with a point mass model. Suppose, for simplicity, that the only force on the system is air resistance, and let it be modeled as ##{\boldsymbol f} = -cv{\boldsymbol v}##, where ##c## is assumed constant (generally it is a function of atmospheric density, projected area, and a dimensionless drag coefficient), and ##v = \|{\boldsymbol v}\|_2##. This then gives us a differential equation\dot{\boldsymbol p} = -cv{\boldsymbol v}which, if we know the initial state of the system, we can integrate numerically and observe time histories of the system states. (If we use the momentum as our state, we can obtain ##{\boldsymbol v}## at each time step by inverting the momentum-velocity relationship.)
