Suppose that (Newtonian) momentum is conserved in a given frame, that is \begin{equation} \sum m_i v_i = \sum \mtwo_j \vtwo_j \end{equation} (Both of these would be zero in the center-of-mass frame.) Changing to another frame moving with respect to the first at speed $v$, we have \begin{eqnarray} v_i &=& v'_i + v \\ \vtwo_j &=& \vtwo'_j + v \end{eqnarray} so that \begin{equation} \sum m_i (v'_i+v) = \sum \mtwo_j (\vtwo'_j+v) \end{equation} We therefore see that \begin{equation} \sum m_i v'_i = \sum \mtwo_j \vtwo'_j \iff \sum m_i = \sum \mtwo_j \end{equation} that is, momentum is conserved in all inertial frames provided it is conserved on one frame and mass is conserved.

Repeating the computation for the kinetic energy, we obtain starting from \begin{equation} \frac12 \sum m_i v_i^2 = \frac12 \sum \mtwo_j \vtwo_j^2 \end{equation} that \begin{equation} \frac12 \sum m_i (v'_i+v)^2 = \frac12 \sum \mtwo_j (\vtwo'_j+v)^2 \end{equation} Expanding this out, we discover that (kinetic) energy is conserved in all frames provided it is conserved in one frame and both mass and momentum are conserved.

The situation in special relativity is quite different.

Consider first the momentum defined by the ordinary velocity, namely \begin{equation} p = mu = m \frac{dx}{dt} \end{equation} This momentum is not conserved!

We use instead the momentum defined by the 2-velocity, which is given by \begin{equation} p = m \frac{dx}{d\tau} = mc \sinh\alpha \end{equation} Suppose now that, as seen in a particular inertial frame, the total momentum of a collection of particles is the same before and after some interaction, that is \begin{equation} \sum m_i \cc \sinh\alpha_i = \sum \mtwo_j \cc \sinh \atwo_j \label{momconsI} \end{equation}

Consider now the same situation as seen by another inertial reference frame, moving with respect to the first with speed \begin{equation} v = \cc\,\tanh\beta \end{equation} We therefore have \begin{eqnarray} \alpha_i &=& \alpha'_i + \beta \\ \atwo_j &=& \atwo'_j + \beta \end{eqnarray} Inserting this into the conservation rule ($\ref{momconsI}$) leads to \begin{eqnarray} \sum m_i \cc \sinh\alpha'_i &=& \sum m_i \cc \sinh(\alpha_i-\beta) \\ &=& \left( \sum m_i \cc \sinh\alpha_i \right) \cosh\beta - \left( \sum m_i \cc \cosh\alpha_i \right) \sinh\beta \qquad\quad \end{eqnarray} and similarly \begin{equation} \sum \mtwo_j \cc \sinh\atwo'_j = \left( \sum \mtwo_j \cc \sinh\atwo_j \right) \cosh\beta - \left( \sum \mtwo_j \cc \cosh\atwo_j \right) \sinh\beta \end{equation} The coefficients of $\cosh\beta$ in these 2 expressions are equal due to the assumed conservation of momentum in the original frame. We therefore see that conservation of momentum will hold in the new frame if and only if we have in addition that the coefficients of $\sinh\beta$ agree, namely \begin{equation} \sum m_i \cc \cosh\alpha_i = \sum \mtwo_j \cc \cosh\atwo_j \end{equation} But what is this?