What is a body's momentum really equal to? I know a body's momentum is equal to the product of its velocity and mass.  But what kind of mass?  Relativistic mass or rest mass?
 A: The modern answer

I know a body's momentum is equal to the product of its velocity and mass.

No, it's not. At least, not in general. $p=mv$ is an approximate formula that works well for massive particles or particle-like objects traveling at slow speeds. But for fast-moving objects, or massless objects, or things that don't even count as objects (I'm thinking of waves), momentum will have a different formula. For example, the momentum of an object moving at relativistic speed is
$$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
or the momentum of a photon is
$$p = \frac{h}{\lambda}= \frac{hf}{c}=\frac{E}{c}$$
Underlying this answer is the idea that mass is just mass. It's an intrinsic property of an object, more or less; in particular, an object's mass does not depend on how fast it's moving.
The outdated answer
In the old days, one would say that momentum could be expressed as the product of relativistic mass and velocity, since
$$m_\text{rel} = \frac{m}{\sqrt{1 - v^2/c^2}}$$
A: The equation by David Z can only be applied for particles having mass not for photons or other massless particles.
The general equation is 
$$E^2-p^2=m^2,$$
where $E$, $p$ and $m$ are energy, momentum and mass of the particle resp.
And this mass is rest mass, there is nothing like relativistic mass in modern books.
A: Relativistic momentum  is given by calculating Newtonian momentum, but instead of using mass, we use relativistic mass. Therefore, starting with
$p = m_0v$
To become correct, we use relativistic mass
$p = \dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}v$
Or, in a more attractive form
$p = \dfrac{m_0v}{\sqrt{1-\dfrac{v^2}{c^2}}}$
