how was E=P^2/2m derived Watch

KCcrusty
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i have he equation E= p^2 / 2m. i figured it was somehow derived by Ek=1/2mv^2 and p=mv? but i dont understand how. can anybody explain it to me please?
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alveolus
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havent you already done it?

Once you know KE= m(v^2)/2,

and p=mv, so p^2 = (m^2)(v^2),

so m(v^2) is clearly (p^2)/m, right?

hence m(v^2)/2 = (p^2)/2m

Is that what you meant?
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KCcrusty
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aha thankyou. yea thats what i meant. i just couldnt get my head around it.
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G. Lee
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(Original post by KCcrusty)
i have he equation E= p^2 / 2m. i figured it was somehow derived by Ek=1/2mv^2 and p=mv? but i dont understand how. can anybody explain it to me please?
Would you like to see the original derivation?

We tend to imagine it as a linear progression. A uniformly accelerating object is usually given by three bits of information, given as:

\frac{v+v*}{2}=\hat{v}
\Delta x= \hat{v} t
a=\frac{v*-v}{t}

Combining these, we find:

\Delta v=\hat{v} t
thus
\Delta x=\hat{v} (\frac{v*-v}{a})
\Delta x=(\frac{v+v*}{2})(\frac{v*-v}{a})=(\frac{v*^{2}-v^{2}}{2a})

The task from these linear sets of equations is to find a relationship between force and mass. Rearranging the final equation gives:

a \Delta x= (\frac{v*^{2}-v^{2}}{2})

and then multiply both sides by mass gives:

Ma \Delta x=\frac{Mv*^2-Mv^2}{2}

Noting that Force = M times Acceleration, then converting gives:

F \Delta x=\frac{Mv*^2}{2}-\frac{Mv^2}{2}

which leaves the quantity of KE=\frac{1}{2} Mv^2

So this is the linear kinetic energy and is the energy related to momentum. Your work done can be given as:

\Delta KE=W_k
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GSM
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well let me simplify how the equation was derived: you know that P=mv and Ek=1/2 m(v^2)
Therefore, make v the subject in the momentum equation you will have v=P/m substitute this in to the kinetic energy equation for value of v...you will now have Ek=1/2m(p/m)^2 if you solve that you get: Ek= 1/2m(p^2)/(m^2) you can now see that one m cancel out with the other and you are left with Ek= (p^2)/2m
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