Can something have momentum but not velocity? The idea of momentum is fundamental, even more fundamental than velocity or mass. But I was wondering can momentum exist without velocity, since momentum can exist without needing mass?
Thinking classically, I deduced that if such an object exists, when I apply a force to it, it shouldn't move. So another parameter should change (like in waves, the frequency changes)
I also thought about it at the Quantum level. Since momentum does not commute with position, a wave that is an eigenvalue of position has uncertainty in momentum. However, this is not what I'm searching for. I'm more interested in Classical Mechanics.
 A: Momentum does not require mass.
For example the electromagnetic  field carries momentum, the momentum density of the EM field is:
$$\vec{p} = \epsilon_{0} \vec{E} × \vec{B}$$
For light:
$$p = \frac{E}{c}$$
$$E=hf$$
It follows that
$$p = \frac{hf}{c}$$
The phase velocity of light in the context of EM  is "c".
Traditional force cannot be applied to light.
$$F=ma$$ where $$m=0$$
Yields $a=$ undefined.
Even in the realm of special relativity, taking the derivative of  the equation
$$F= \gamma m_{0} v$$
To find the force doesn't work, since $\gamma$ is undefined.
light doesn't interact electromagnetically with itself, as $q=0$,so wouldn't experience an electromagnetic force.
Light does interact with gravity however. Light experiences "gravitational redshift"
https://en.m.wikipedia.org/wiki/Gravitational_redshift
which is a change in frequency as light moves in a gravitational field
A: 
The idea of momentum is fundamental, even more fundamental than velocity or mass.

This is not correct. At best it is a matter of opinion what is "more fundamental."
In the classical Lagrangian formulation of mechanics, velocity is "fundamental." The Lagrangian $L$ is a function of the position $x$ and the velocity $v$; $L=L(x,v)$.
In moving from the Lagrangian to the Hamiltonian formalism we write the canonical momentum $p$ as:
$$
p = \frac{\partial L}{\partial v}\;.
$$
The Hamiltonian $H$ is a function of the position and momentum, $H=H(x,p)$, and in this formalism the velocity is given by:
$$
\frac{\partial H}{\partial p} = v\;.
$$


I'm more interested in Classical Mechanics.

In that case, you can write down the hamiltonian $H$ for the system you are studying and it must be a function of position of momentum $H = H(x,p)$. The velocity, by definition is:
$$
v = \frac{\partial H}{\partial p}\;.
$$
So, seemingly, you can not "have momentum" without "having" velocity.
A: thinking about photon, it has constant speed, so when a force, like gravity is applied to it, because there is no mass, It changes the direction of the photon, and also the frequency. so now about the velocity-less particle: although it's an intriguing idea to have a particle locked in space for some reason, It is only possible in Zero Kelvin. In any other condition the particle will have a wave function that determines the probability of its place and momentum. If a particle were to have a spatially locked state without zero kelvin condition, it would violate the Heisenberg uncertainty principle. you might say the uncertainty principle is for quantum mechanics. but that's not totally true. even in classic Newtonian physics, the exact location and momentum of an object are not possible to be measured together, because of the definition of velocity which needs your measuring devices to wait for small amount of time. in this small amount of time, your object will move and you will lose measurement accuracy on your location to measure a more accurate velocity. this means any object of any size has to follow the uncertainty principle. although for larger objects this uncertainty is not by much, because the size of the object is much higher than the wave length of the spatial wave function for those objects. so the uncertainty is negligible.
