Regarding light as a particle, is there something like momentum from light when it shines on objects? Regarding light as a particle, is there something like momentum from light when it shines on objects?
We can see light as a particle or a wave. Regarding it as a particle, is there some momentum given to objects which are struck by light?
It should not be, as light has no mass - or does it have mass due to movement?
 A: Photons do have momentum, and the equation for the momentum of a photon is given by:
$$p = \frac{E}{c}$$
where E is the energy of the photon.
These equations can be derived from Einstein's Equation:
$$E^2 = p^2c^2 + m^2c^4$$
For a photon with mass = 0, the equation is reduced to:
$$E = pc$$
This means that mass is not required in order to have momentum, so long the particle has energy (which a photon surely has - as shown through the photelectric effect)
Thus, it is a common misconception that mass is required for momentum.
Edit: momentum of a photon can also be written as $$p = \frac{h}{\lambda}$$
What this λ represents in the context of a particle is simply the EM wave associated with that particular photon since particles themselves do not have wavelengths.
Hope this helps!
A: In addition to the other correct answers, it's worth noting that even classically (i.e., even in the old days when people thought light was just a wave), the electromagnetic field carries energy and momentum. The momentum of a classical electromagnetic wave leads to radiation pressure. Of course we have a deeper understanding of this now this now in terms of quantum mechanics and photons, but the effect is there even in classical physics.
A: Yes the electromagnetic field carries a momentum density proportional to $\mathbf{E}\times\mathbf{B}$, when a charged particle is accelerated by said field some of this momentum is transferred. The same is true for energy and angular momentum.
A: Yes it does transfer momentum. Although it has no mass, from the energy momentum relation you get the following
$ E = \sqrt{p^2c^2+m^2c^4} = h\nu$
And hence,
$ |p| = \frac{h\nu}{c}$
As Kurt G. pointed out, that was the idea for Crooke's radiometer, but eventually not the correct explanantion.
Other examples I think would be the trail of a coment pointing away from the sun.
Or the project "Breakthrough Starshot", where they want to accelarate a nano-spacecraft by laserlight.
