Conservation of energy in decay

A Λ0 particle of rest mass 1115.7 MeV and kinetic energy 172.6 MeV decays into proton of mass 938.8 MeV/c^2 and momentum 428.8 MeV/c and a "mystery" particle - calculate mass of mystery particle and hence deduce what type of particle it is.

I have done the calculation and get a total energy of 215.8 MeV and momentum of 129.05 MeV/c for the "mystery" particle. Hence the rest mass I get is 172.96 MeV/c^2. I know the particle must be a negative pion rest mass 140 MeV/c^2. What have I done wrong? How can I get to the correct answer? thanks

• You can use this equation $E=mc^{2}\gamma$ and $\gamma= \dfrac{1}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$ – Eli Oct 19 '19 at 13:14
• @Eli While you could involve the Lorentz factor in a successful solution that would be longer, more complicated, less clear, and easier to mess up than need be. – dmckee --- ex-moderator kitten Oct 19 '19 at 14:45
• Can you show your work? It's not easy to see where you went wrong from your final numbers. If you gave an algebraic result, there's a better chance... but still may be not so easy. (Using rapidities and the Law of [Hyperbolic-]Cosines, I get 138.8.) – robphy Oct 20 '19 at 19:27
• @robphy I found my error. I calculated the total energy of the neutral lambda seemingly in a non relativistic manner when dealing with the total energy equation (E^2=(mc^2)^2+(pc)^2). I used its kinetic energy given to find momentum via T=p^2/2m - then put it into the equation with the rest energy - the error stemmed from there. Once I found total energy via E=(gamma)mc^2, the calculation came out correct to about 139.8 Mev/c^2, 139.57MeV/c^2 being the rest energy of a negative pion. I used the kinetic energy in both cases, to get momentum and gamma so I am not sure exactly what went wrong. – Σ baryon Oct 21 '19 at 19:49
• The relativistic kinetic energy is not related to the relativistic momentum like you wrote. That is, $K_{rel} \neq \frac{p_{rel}^2}{2m_0}$. In terms of rapidities (the Minkowski angle between worldlines, so $(v/c)=\tanh\theta$ and $\gamma=\cosh\theta$) we have $p_{rel}=m_0c\sinh\theta$ and $K_{rel}=m_0c^2(\cosh\theta -1)$. With some trig identities, you can write $K_{rel}$ in terms of $p_{rel}$ as follows: $K_{rel}=\frac{p_{rel}^2}{(\cosh\theta +1)m_0}$. – robphy Oct 22 '19 at 19:36