Special Relativity problem using conservation of energy and momentum

A particle of mass $M$ at rest decays to 2 particles of mass $M_1$ and $M_2$

Question: Write $v_1$ in terms of $M_1$ and $M_2$.

Conservation of momentum

$0=γ(v_1)M_1v_1-γ(v_2)M_2v_2$ , $γ(v_1)$ means gamma as a function of $v_1$

$γ(v_1)M_1v_1=γ(v_2)M_2v_2$

$\frac{(M_1v_1)^2}{c^2-v_1^2}= \frac{(M_2v_2)^2}{c^2-v_2^2}$ (1)

Conservation of Energy

$Mc^2 =γ(v_1)M_1c^2+γ(v_2)M_2c^2$

$M =γ(v_1)M_1+γ(v_2)M_2$

$M=\frac{cM_1}{c^2-v_1^2}+\frac{cM_2}{c^2-v_2^2}$ (2)

now having both of these equations from conservation of energy and conservation of momentum, i have 5 unknowns, if i combine both equation by eliminating $v_2$, i will still have $M$, is there an equation or relationship that can allow me to eliminate $M$?

• The velocity depends on the initial mass, because it affects the amount of energy available. So you can't eliminate it from your equations unless you have more data. – Javier Sep 13 '16 at 17:11

It's easiest to solve this sort of problem using 4-vectors and writing conservation of 4-momenta in such a way that squaring gets rid of things you don't care about. $$\vec{P_2}=\vec{P}-\vec{P_1}$$ Square both sides and write the components of $\vec{P}$ and $\vec{P_1}$
$$M_2^2=M^2-2(M,\vec{0})\cdot(E_1,\vec{p_1})+M_1^2$$ $$M_2^2=M^2-2ME_1+M_1^2$$ $$M_2^2=M^2-2M(\gamma_1M_1)+M_1^2$$ You can easily rearrange this last expression to get $\gamma_1$ and then extract $v_1$ from the $\gamma_1$.