There is a topic called variation of mass with velocity. In which to get the equation of mass in relativistic case we consider collision of two particles. We take steady frame S1 and moving frame S2. When we consider moving frame S2 we assume collision of two masses, each of having same value m moving in the opposite direction to each other with the same speed. using such information we get value of velocities of the particles for corresponding steady frame S1 using velocity transformation formula. When we write equation of conservation of the momentum for the masses in steady frame S1 we consider masses to be different i.e. m1, m2. In moving frame masses are considered identical but for steady frame they are considered different. Why?
[Physics] Special theory of relativity: Variation of relativistic mass with velocity
inertial-framesmassspecial-relativityvelocity
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"Suppose two objects collide and combine into a single object, will the total relativistic momentum and relativistic mass stay the same?"
The answer is "yes", or rather their sums over the system of bodies will stay the same, but I would counsel you to stop using the term relativistic mass. It's going out of use for a number of good reasons that I won't bore you with now.
$m \gamma c^2$ in which m is the body's mass (formerly called 'rest mass') represents the sum of the body's internal energy, $mc^2$ and its kinetic energy ($\gamma m - m)c^2$. So for a closed system its sum over the bodies of the system is conserved in collisions, elastic or inelastic. Think of $\gamma m$ as the body's total energy, expressed in mass units.
The beauty of it is that a body's total energy (divided by the mere constant, c) and the three components of its momentum $(m \gamma u_x, m \gamma u_y, m \gamma u_z)$ make up a 4-component vector (or 4-vector): $(m \gamma c, m \gamma u_x, m \gamma u_y, m \gamma u_z)$. So for a closed system, despite collisions elastic or inelastic, the vector sum of these vectors is conserved, that is the sums of each component separately. One conserved 4-vector deals with conservation of energy and conservation of momentum.
Note also that the modulus of the 4-vector, defined as $\sqrt{(m \gamma c)^2 - (m \gamma u_x)^2 - (m \gamma u_y)^2 - (m \gamma u_z)^2}$, is simply $mc$, the body's mass multiplied by the mere constant, c.
$m$ is a constant for the body (provided we don't tamper with the body, e.g. by changing its internal energy!) and doesn't vary from frame to frame. It is a Lorentz invariant. [Beware: the sum of the masses (rest masses) of bodies in a system has no obvious significance; it is certainly not the mass (rest mass) of the system!]
I've gone on longer than I should have done. It's all so wonderful. A classic introduction to Special Relativity, first rate on concepts, is Spacetime Physics by Taylor and Wheeler.
The easiest way is to use four-vectors. If an object has three-velocity $\mathbf{u} =\begin{bmatrix} u_x \\ u_y \\ u_z \end{bmatrix}$ in one frame, its four-velocity is $$\gamma_u\begin{bmatrix} c \\ u_x \\ u_y \\ u_z\end{bmatrix}$$
The general form of a Lorentz transformation from frame $S$ to frame $S'$ is $$\begin{bmatrix} \gamma&-\gamma\beta_x&-\gamma\beta_y&-\gamma\beta_z\\ -\gamma\beta_x&1+(\gamma-1)\frac{\beta_x^2}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}\\ -\gamma\beta_y&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&1+(\gamma-1)\frac{\beta_y^2}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}\\ -\gamma\beta_z&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}&1+(\gamma-1)\frac{\beta_z^2}{\beta^2}\\ \end{bmatrix}$$ where $\begin{bmatrix} \beta_x \\ \beta_y \\ \beta_z \end{bmatrix}$ is the velocity of $S'$ relative to $S$. Since the four-vector transforms by matrix multiplication, and both four-vectors are given, it is then a straightforward task to solve for the relative velocity in the matrix.
Since the velocity of the particle in the y direction is the same with respect to both frames, can we say that the relative velocity of the frames in the y direction at least is 0?
Yes, this implies that all relative velocity must either be zero or be in the perpendicular direction.
Best Answer
The simple answer to your question is that since (presumably) we're considering mass to be a function of speed, since the bodies have the same speed in S2 they will have the same mass, assuming that they are identical bodies.
I feel bound to point out, though, that it's becoming less and less common among physicists to talk about mass varying with speed. The argument concerns how to interpret the formula for the momentum, $\vec{p}$, of a body moving at velocity $\vec{u}$, speed u, namely$$\vec{p}=\gamma m_0 \vec{u}\ \ \ \text{in which}\ \ \ \gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}.$$ Old interpretation
$m_0$, a constant for the body independent of its motion, was called the rest mass of the body. $γm_0$, which was sometimes denoted by m, was called the body's relativistic mass, and is speed-dependent. The idea of calling $γm_0$ 'relativistic mass' was that you could continue to use the Newtonian formula, $\vec{p}=m \vec{u}$, provided that you used so-called 'relativistic mass' instead of rest mass.
One of the reasons why this interpretation has fallen out of favour is that putting $γm_0 $ instead of $m_0$ doesn't make other Newtonian formulas into relativistic ones. For example, it won't make $KE=\frac{1}{2}m_0 u^2$ into the relativistic formula $$KE=(\gamma-1)m_0 c^2.$$ Modern interpretation
Instead of regarding $γm_0$ as a speed-dependent mass that, when multiplied by $\vec{u}$, gives us $\vec{p}$, it's at least as logical to regard $\vec{p}=\gamma m_0 \vec{u}\ $ as $\ \vec{p}= m_0 (\gamma \vec{u})$, that is as $m_0$, a constant for the body, multiplied by $\gamma \vec{u}$, a kinematic quantity (called proper velocity) that replaces $\vec{u}$ in the ordinary Newtonian expression.
This means that we don't call $γm_0$ 'relativistic mass'. If we want to call it anything, it's the mass equivalent of the body's total energy (KE + rest energy), $\gamma m c^2$. Remember that $c^2$ is merely a constant!
In that case there's no need to call $m_0$ 'rest mass' – it's the only sort of mass we talk about. Nor is there any need for the zero subscript! So the two formulas we quoted earlier are usually written$$\vec{p}=\gamma m \vec{u}$$and$$KE=(\gamma-1)m c^2.$$