If a relativistic proton e.g. $\gamma=5$ hits a stationary proton will this collision take all its momentum? In classical physics when a billiard ball hits a stationary one of same mass it transfers all its momentum to it so becoming itself stationary while the other gathers their whole momentum.
But what about relativistic particles? Will a relativistic proton transfer immediately its momentum to a stationary one or the collision will go on until the second acquires relativistic mass and so the whole momentum is transferred?
 A: There is a totalitarian principle in quantum mechanics: any process not forbidden by a conservation law is compulsory (will happen with some rate). The elastic scattering process you describe is one in which the energy-momentum vectors are swapped, so it satisfies conservation of energy-momentum, regardless of the details of how energy and momentum depend on velocity in relativity.
However, you will also have inelastic processes that compete with the elastic one.
A: The collision target proton can, in principle, take all of the projectile's momentum, elastically, but this is freakishly unlikely, given the energies involved. It can also let the target proton go through it, without being affected, also comparably unlikely.
Most likely (QM!), the collision will produce a hadronic fireball, a few fermis in size (the characteristic distance of the strong interactions, the Compton wavelength of a pion) consisting of quarks and gluons, and the fireball will disintegrate into two protons and several pions, (as many as 10!) in a few $10^{-24}$ s, the characteristic time of the strong interactions: the time it takes to traverse fermis with the speed of light.
So the γ you chose still pumps so much kinetic energy in the system that you might as well think of the two protons as very loose marshmallows full of quarks and gluons colliding, not billiard balls. (Accelerators these days choose γs in the 10000s.)
A quick summary of the kinematics you chose, in HEP units, where speed is measured in units of c. (So, effectively, $\hbar$ and c are set equal to 1.) The Lorentz contraction almost always fakes you out, so you'll transition to the center of mass from the lab frame. Call the pion mass is crudely 0.14 m, where m is the proton mass.
In the lab frame, the projectile and target (energy, momentum) vectors are
$$(E,p)=(\gamma m, \sqrt{\gamma^2-1} ~ m), \qquad (m,0), $$ respectively.
Υou know that the 4-vector invariant of the total energy and momentum
$$
s=2(\gamma+1)m^2
$$ is the energy squared in the center of momentum frame, where there is no total momentum.
So the total energy in the c.m. frame is
$$
E_t=\sqrt{2(1+\gamma)} ~m.
$$
What is the maximum number n of pions this energy can convert to, beyond the outgoing protons, necessary by baryon number conservation? For your $\gamma=5$, as many as 10,
$$ E_t= \sqrt{12} m= 2m+ n~0.14 m,$$
also highly unlikely, given the minute amounts of kinetic energy left over. The most likely number is a few pions. Check on your own that the one pion production threshold is just γ = 1.29 !
Perusing the PDG will give you a better quantitative grasp of the collision, still a low energy one.
