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

[JHCreighton]

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

 

[Pengwuino]

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.

 

[JHCreighton]

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

 

[russ_watters]

You can also easily derive it using f=ma, d=st and the definition of work, w=fd.

 

[Dickfore]

By solving the system:

$$p = m \, v$$

$$K = \frac{1}{2} \, m \, v^{2}$$

with respect to $m$ and eliminating $v$.