Answer
The final answer is :
$m_{\rho}c^2$ = $2E_{\pi}$ = $769 MeV$
Work Step by Step
Using Table 44-4, the rest energy of each pion is $139.6 MeV$.
The momentum magnitude of each pion is $P_{\pi}=(358.3 MeV)/c$
Then we use the relativistic relationship between energy and momentum using Eq 37-54 to find the total energy of each pion:
$E_{\pi}= \sqrt (P_{\pi}c)^2+(m_{\pi}c^2)^2$
= $\sqrt (358.3 MeV)^2 + (139.6 MeV)^2= 384.5 MeV$
Finally, using the conservation of energy, we have:
$m_{\rho}c^2$ = $2E_{\pi}$ = $769 MeV$
