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M=P^2/2KE: How is this derived? 
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#1
Jul810, 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 nobrainer, but I can't seem to make sense of the algebra.
Thanks, JHCreighton 


#2
Jul810, 07:54 PM

PF Gold
P: 7,120

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.



#3
Jul810, 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 


#4
Jul810, 07:58 PM

Mentor
P: 22,286

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.



#5
Jul810, 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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