A proton \(\mathrm{p}_{1}\) moving in the Earth reference frame collides with a proton \(p_{2}\) that is at rest in that reference frame. The collision creates a particle that has mass \(m\) and an internal energy 40 times the internal energy of a proton: \(m c_{0}^{2}=40 m_{\mathrm{p}} c_{0}^{2}\). The velocity of this new particle in the Earth reference frame is not given. What is the ratio of the energy of \(p_{1}\) to its internal energy in the Earth reference frame? (First analyze the collision in the zeromomentum reference frame in which \(\mathrm{p}_{1}\) and \(\mathrm{p}_{2}\) are both in motion, moving in opposite directions at the same initial speed.)