how was E=P^2/2m derivedWatch this thread
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?
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?
We tend to imagine it as a linear progression. A uniformly accelerating object is usually given by three bits of information, given as:
Combining these, we find:
The task from these linear sets of equations is to find a relationship between force and mass. Rearranging the final equation gives:
and then multiply both sides by mass gives:
Noting that Force = M times Acceleration, then converting gives:
which leaves the quantity of
So this is the linear kinetic energy and is the energy related to momentum. Your work done can be given as:
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