tiny-tim said:
Hi Ookke!
nooo … the mass of X
increases when it absorbs the light,
and since the horizontal momentum stays the same, the horizontal velocity must
decrease 
Ookke said:
That's right. I didn't see the mass increase at all. Thanks for the solution!
tiny-tim is correct that the mass increases, but that's not the whole story.
In relativity, momentum is defined by
\vec p = \gamma m \vec v
where \gamma is
\gamma \equiv \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
When a particle is at rest, it's \gamma is equal to 1. As the particle approaches the speed of light, its \gamma becomes greater than 1.
In your thought experiment, not only does the mass of the particle increase, but \gamma increases too. The final speed of the particle is greater than its initial speed (even though its velocity in the x-direction decreases). So there are two things going on.
Let's talk about the mass for a moment. The mass does increase, and by this I mean the particle's rest mass (invariant mass). That's because in your thought experiment, the collision is inelastic. The particle has gained internal energy (such as exciting an electron, maybe placing the nucleolus in an excited state, or whatever). This internal energy increases the particle's rest mass by E = mc^2.
But the particle also gains kinetic energy because its speed increases. So not all of the photon's energy goes into the particle's rest mass; some of it ends up as the particle's kinetic energy.
The
total energy of a particle is given by \gamma m c^2. That includes both its rest mass energy and its kinetic energy.
The energy of a photon is given by \frac{hc}{\lambda} where h is Planck's constant and \lambda is the photon's wavelength.
Taking into account the particle's increase in rest mass, both energy and momentum are conserved in this inelastic collision.
Energy final = Energy initial
Energy final = Initial energy of particle + Photon energy
\gamma_f m_f c^2 = \gamma_0 m_0 c^2 + \frac{hc}{\lambda}
Dividing by c^2,
\gamma_f m_f = \gamma_0 m_0 + \frac{h}{\lambda c}
Keep that above equation handy. We're going to use it a couple times below.
Let's move onto momentum. Not only is momentum conserved, it is conserved individually in each x- and y-component.
Let's start with the x-component, since that is what your original thought experiment was about. Consider the particle initially moving on the x-axis, and the photon on the y-axis.
\gamma_f m_f v_x = \gamma_0 m_0 v_{0x}
Now we substitute \gamma_f m_f into that equation and solve for the final v_x
v_x \left( \gamma_0 m_0 + \frac{h}{\lambda c} \right)= \gamma_0 m_0 v_{0x}
v_x = v_{0x} \frac{\gamma_0 m_0 }{\gamma_0 m_0 + \frac{h}{\lambda c}}
Multiplying both the numerator and denominator by c^2 we get
v_x = v_{0x} \frac{\gamma_0 m_0 c^2}{\gamma_0 m_0 c^2 + \frac{hc}{\lambda}}
And interestingly, that is
v_x = v_{0x} \left( \frac{\mathrm{Initial, \ total \ energy \ of \ particle}}{\mathrm{Initial, \ total \ energy \ of \ particle \ + \ Photon \ energy}} \right)
So as you can see, the x-component of the particle's velocity is reduced after the collision.
------------------
Although not part of your original thought experiment, we can do the same thing for the y-component of momentum. The momentum of a photon is \frac{h}{\lambda}.
\gamma_f m_f v_y = \frac{h}{\lambda}
Making our substitution,
v_y \left( \gamma_0 m_0 + \frac{h}{\lambda c} \right)= \frac{h}{\lambda}
v_y = \frac{\frac{h}{\lambda}}{\gamma_0 m_0 + \frac{h}{\lambda c}}
Multiply numerator and denominator by c^2
v_y = c \frac{\frac{hc}{\lambda}}{\gamma_0 m_0 c^2 + \frac{hc}{\lambda}}
And interestingly, that is
v_y = c \left( \frac{\mathrm{\ Photon \ energy}}{\mathrm{Initial, \ total \ energy \ of \ particle \ + \ Photon \ energy}} \right)
