Relativistic Elastic Collision

I am having trouble getting my head around the transfer of energy in a relativistic elastic collision. My understanding of a relativistic elastic collision is one in which the total rest mass on each side of the equation is unchanged, i.e:

$$\sum_{\text{before}} m_{i}=\sum_{\text{after}} m_{j}$$

If we have a particle of rest mass $M$ and relativistic energy $E$ colliding with a particle at rest with rest mass $m$, then we have by conservation of 4-momentum:

$$\mathsf{P}_{1}\cdot\mathsf{P}_{2}=\mathsf{P}_{1}'\cdot\mathsf{P}_{2}'$$

Initially we have: $\mathsf{P}_{1}=(\frac{E}{c},p,0,0)$ and $\mathsf{P}_{2}=(mc,0,0,0)$, and therefore:

$$\mathsf{P}_{1}'\cdot\mathsf{P}_{2}'=\frac{E'E_{2}}{c^{2}}-p_{1}'p_{2}=mE$$

However, this leaves us with a lot of unknowns $E'$, $E_{2}$ and $p_{2}$, so I'm not sure how I'd go about reducing this to one unknown $E'$ (assuming it's possible)?

Energy-momentum conservation is a stronger statement than the statement* that the inner product $p_\mu p'^\mu$ is conserved. It states that the sums are conserved individually/coordinatewise - $P_1+P_2=P_1'+P_2'$.
So then, we have (in the 1D case), separately: $$E_1+E_2=E_1'+E_2'$$ $$p_1+p_2=p_1'+p_2'$$