Originally posted by: hiredgoons
I don't know if it has a name (it probably does); I just substituted p for mv in E = .5mv^2. It's not that useful though since any time you'd know the kinetic energy you'd almost definitely be able to calculate the momentum from the same information.
Yeah, I know, I was just curious
.
Another question: Why doesn't this work with E = mc^2?
Either I don't understand the theory or I do and it's not correct.
E = E (makes sense).
.5mv^2 = mc^2 (makes less sense, but can still be correct).
Let m = 10 kg and v = 20 m/s. c ~= 300.000.000 m/s.
.5x10x(20^2) != 10x(300.000.000^2)...
It's probably me being a dumbass, though..