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

by JHCreighton
Tags: derivation, kinetic energy, mass, momentum, physics
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JHCreighton
#1
Jul8-10, 07:44 PM
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
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Pengwuino
#2
Jul8-10, 07:54 PM
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It's fairly simple, no trickery involved. What does [tex]P^2[/tex] equal? Now, that almost looks like something you have with your kinetic energy equation. Can you convince yourself that [tex]KE = \frac{m^2 v^2}{2m}[/tex] is the same as your original equation? If so, simply plug in [tex]P^2[/tex]. From there, simply solve for m.
JHCreighton
#3
Jul8-10, 07:57 PM
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

russ_watters
#4
Jul8-10, 07:58 PM
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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.
Dickfore
#5
Jul8-10, 08:00 PM
P: 3,014
By solving the system:

[tex]
p = m \, v
[/tex]

[tex]
K = \frac{1}{2} \, m \, v^{2}
[/tex]

with respect to [itex]m[/itex] and eliminating [itex]v[/itex].


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