GTOM said:
A : If a relativistic particle hit something, does its kinetic energy derived from base mass, or relativistic mass?
B : Do we have any data, how it affects a gravity based interaction between two relativistic objects?
C : Let's suppose a relativistic isotope emits two particles, one in the direction of acceleration (where we accelerated it) one in the opposite direction. Will be speed difference between them? (Compared to emitter isotope.)
A: There's no need for relativistic mass. It's an old-fasioned idea, which is obsolete since 1908, when Minkowski found the fully developed math behind special relativity. The energy and momentum of a classical particle in terms of its invariant mass ##m## and the usual three-velocity ##\vec{v}=c \vec{\beta}##(which is what you seem to call "base mass", an expression I've never heard before) are
$$E=\frac{m c^2}{\sqrt{1-\beta^2}}, \quad \vec{p}=\frac{m c}{\sqrt{1-\beta^2}}.$$
You can also rewrite the energy in terms of the momentum
$$E=c \sqrt{m^2 c^2+\vec{p}^2},$$
as you can easily check from the above conditions. This leads to the definition of the invariant mass in terms of four-vectors, showing that it is a scalar, i.e., independent of the frame of reference
$$(p^{\mu}):=(E/c,\vec{p}) \; \Rightarrow \; p_{\mu} p^{\mu} = (E/c)^2-\vec{p}^2=m^2 c^2.$$
This is also called the "on-shell condition".
B: Gravity is understood in terms of General Relativity. There idea of a "relativistic mass" here becomes totally unusable! Just last week we had a breakthrough with the first ever direct detection of a gravitational wave by the LIGO and VIRGO collaborations. The accurate analysis of this signal lead to the conclusion that its source are two supermassive black holes orbiting around each other and then merging into one even larger black hole. The comparison between the measured signal and the expectation for the (numerical) solution of the corresponding general-relativistic Einstein field equations lead to a very accurate confirmation of General Relativity even under stuch extreme conditions.
Even more accurate tests of General Relativity are possible using "pulsar timing", i.e., one measures the radio signals of pulsars very accurately. Sometimes you also have a pulsar orbiting around another object, and sometimes you are even as lucky to have such a double-star system where both stars are pulsars whose signals are observable on Earth. With these very accurate measurments of the electromagnetic signals from such systems you get also very accurate (in fact the most accurate) confirmations of General Relativity.
C: The most simple way to analyse such reactions is to use the covariant four-vector formalism and analyze the energy-momentum conservation together with the on-shell conditions. Then you can use the most simple reference frame for the process and write it down in a way that it applies in any frame of reference. For your decay example, the most simple reference frame is where the decaying particle is at rest. Let's call the four-momenta of the mother particle ##p_1## and that of the daughter particles ##p_2## and ##p_3##. Further let's use "natural units", setting the speed of light ##c=1##. Then you have
##p_1=(m_1,0,0,0)## and ##p_2=(E_2,\vec{p})##, ##p_3=(E_3,-\vec{p})##. The latter follows from the conservation of energy and momentum
$$p_1=p_2+p_3 \; \Rightarrow \vec{p}_2=\vec{p}=-\vec{p}_3.$$
The energies are determined by the on-shell conditions:
$$E_2=\sqrt{m_2^2+\vec{p}^2}, \quad E_3=\sqrt{m_3^2+\vec{p}^2}.$$
Now you can determine the momentum using the invariant
$$p_1^2=m_1^2=(p_2+p_3)^2=p_2^2+p_3^2+2p_2 \cdot p_3=m_2^2+m_3^2 + 2 (E_2 E_3+\vec{p}^2).$$
Solving for the magnitude of the momentum gives
$$|\vec{p}|=\frac{\sqrt{[m_1^2-(m_2+m_3)^2][m_1^2-(m_2-m_3)^2]}}{2m_1}.$$
Since the momenta in this reference frame (the center-momentum frame of the decay) are collinear, you can calculate the relative velocity in the naive way,
$$|\vec{v}_{\text{rel}}|=\left |\frac{\vec{p}_2}{E_2}-\frac{\vec{p}_3}{E_3} \right|.$$
One can show (try it!) that this can be expressed in a manifest covariant way by
$$|\vec{v}_{\text{rel}}|=\frac{\sqrt{(p_2 \cdot p_3)^2-m_2^2m_3^2}}{E_2 E_3}.$$
