I believed I could use $0.5\times M_{ej} V_{ej}^2$, with $M_{ej}$ being the ejected mass and $V_{ej}$ being the velocity of the ejected mass. But I noticed in this and this that the the mean velocity is somehow calculated as following : $V_{mean}=(3/5)^{0.5}\times V_{ej}$. And so the kinetic energy was calculated as : $(3/5)(0.5)\times M_{ej} V_{mean}^2 =(3/10)\times M_{ej} V_{mean}^2$. So is there any explanation for how this factor of $0.3$ or the mean velocity was calculated ?
No comments:
Post a Comment