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My Epic Fail at Deriving an Equation with Lagrange
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[QUOTE="Arm, post: 6853687, member: 730000"] I see where I made the mistake sign now [S]Ignoring variable of integration and +C for the umpteenth time has gotten me the wrong answer yet again; when will I learn my lesson?[/S] Here's the actual correct answer $$v^2 = \frac{-2k q_1 q_2}{mr} + C$$ The particle is released from rest ##x## meters away from the other particle. Velocity is 0 at this moment $$0^2 = \frac{-2k q_1 q_2}{mx} + C$$ $$ C = \frac{2k q_1 q_2}{mx}$$ $$v^2 = \frac{-2k q_1 q_2}{mr} + \frac{2k q_1 q_2}{mx}$$ $$v^2 = \frac{2k q_1 q_2}{mx} - \frac{2k q_1 q_2}{mr}$$ General solution: $$v = \sqrt{ \frac{2 k q_1 q_2}{m}( \frac{1}{x} - \frac{1}{r} ) }$$ If x = 1 then r also = 1 so $$v = \sqrt{ \frac{2 k q_1 q_2}{m}( \frac{1}{1} - \frac{1}{1} ) }$$ $$v = \sqrt{ \frac{2 k q_1 q_2}{m}( 1 - 1 ) }$$ $$v = \sqrt{ \frac{2 k q_1 q_2}{m}( 0 ) }$$ $$v = 0$$ Thanks for all the help! [/QUOTE]
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My Epic Fail at Deriving an Equation with Lagrange
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