M=P^2/2KE: How is this derived?

by JHCreighton
Tags: derivation, kinetic energy, mass, momentum, physics
 PF Gold P: 7,120 It's fairly simple, no trickery involved. What does $$P^2$$ equal? Now, that almost looks like something you have with your kinetic energy equation. Can you convince yourself that $$KE = \frac{m^2 v^2}{2m}$$ is the same as your original equation? If so, simply plug in $$P^2$$. From there, simply solve for m.
 P: 3,014 By solving the system: $$p = m \, v$$ $$K = \frac{1}{2} \, m \, v^{2}$$ with respect to $m$ and eliminating $v$.