How Is m=P^2/2KE Derived?

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  • Thread starter Thread starter JHCreighton
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Discussion Overview

The discussion revolves around the derivation of the formula m = P^2 / 2KE, exploring how momentum and kinetic energy equations can be manipulated to arrive at this expression. The scope includes algebraic manipulation and the relationships between physical quantities in classical mechanics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the algebraic process needed to derive m = P^2 / 2KE from the definitions of momentum (P = mv) and kinetic energy (KE = 1/2 mv^2).
  • Another participant suggests that recognizing the relationship between P^2 and the kinetic energy equation can simplify the derivation, proposing to substitute P^2 into the kinetic energy formula.
  • A third participant expresses appreciation for the simplicity of the approach suggested, indicating a realization of the method's clarity.
  • Another participant mentions alternative methods for deriving the formula using the definitions of force, distance, and work.
  • One participant proposes solving the system of equations for momentum and kinetic energy with respect to mass and eliminating velocity as a step in the derivation process.

Areas of Agreement / Disagreement

The discussion does not present a consensus on a single method for the derivation, as multiple approaches are suggested, and no definitive resolution is reached regarding the best method.

Contextual Notes

Participants do not clarify specific assumptions or limitations in their approaches, and the discussion includes various methods without resolving potential discrepancies in the algebraic steps involved.

Who May Find This Useful

This discussion may be useful for students or individuals interested in classical mechanics, particularly those looking to understand the relationships between momentum, kinetic energy, and mass through algebraic manipulation.

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

<br /> p = m \, v<br />

<br /> K = \frac{1}{2} \, m \, v^{2}<br />

with respect to m and eliminating v.
 

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