# 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. Sep 13, 2016 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 $\gamma_1$.