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

by JHCreighton
Tags: derivation, kinetic energy, mass, momentum, physics
 P: 6 I am just curious as to how this fits in. If momentum P=mv, and kinetic energy KE=1/2mv^2, how would one combine, derive, switch and swap (whatever the process is called), these two equations to end up with the formula m=P^2/2KE. It seems like a no-brainer, but I can't seem to make sense of the algebra. Thanks, JHCreighton
 PF Gold P: 7,125 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: 6 Hey, that's great! You're right, it is pretty simple. I almost feel foolish for not thinking to solve like that. Thanks for the speedy response. JHCreighton
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P: 22,008

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

You can also easily derive it using f=ma, d=st and the definition of work, w=fd.
 P: 3,015 By solving the system: $$p = m \, v$$ $$K = \frac{1}{2} \, m \, v^{2}$$ with respect to $m$ and eliminating $v$.

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