Position, velocity, and speed | Physics | Khan Academy

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[@bharath__100]

Hello mahesh sir,.... Here's a small doubt.. maybe out of context... If 2 electrons are approaching each other with velocity v then their distance of closest approach is given by using the relation total K. E = potential energy of system of 2 electrons i. e. ½mv²+½mv²= ke²/r...from which r= ke²/mv² But if I try to solve the question using the perspective of first electron , then the relative velocity of second electron should be 2v , therefore, the relation will be ½m(2v) ²= ke²/r....from which r= ke²/2mv²...why the both results are different....? Could you please clarify.. where am I wrong in the analysis?

[@christinejackson6630]

I think the issue is that you can only say (total initial KE) = (U at closest approach) if both particles indeed come to rest at the moment of closest approach. However, in your second case of a reference frame with same velocity of one of the initial electrons, this isn't true. The reference frame may start out moving with the electron. However, for it to be an inertial reference frame, it must continue with that velocity, whereas the electron will accelerate (slow down) as it begins interacting with the other electron. So in that frame of reference the first electron will appear to start moving backwards. The incoming electron will also appear to slow down. But since the electron we started with is already moving backwards, the two electrons cannot possibly come to rest simultaneously in this frame. So, even at the moment of closest approach there will be some KE that must be accounted for in the right side of the calculation. Just my thought - I'm not positive!

[@bharath__100]

@@christinejackson6630 thanks a lot for responding on it... I think the force on the second electron will be 2*ke²/r²..because the first electron is not inertial... it feels a force pushing it away so there arises a pseudo force on the second electron which pushes it away (whose magnitude is ke²/r²) and also the second electron faces another ke²/r² .. Which makes the total force 2ke²/r² and hence the total regarding work = 2ke²/r??

[@christinejackson6630]

hmmm, I'm not sure. Both electrons accelerate in both frames of reference. And by Newton's third law they must experience equal and opposite forces in both cases. I think the magnitude of the force on each electron would always be ke²/r². For the second frame I get this: Initially, the first electron appears to have v=0, and the second electron v=2v, so total initial energy of the system is ½m(2v) ² as you have above. But considering the closest approach from the first frame of reference, it occurs when both electrons' velocities have changed by -v opposite the direction they're moving. In the second frame this looks like the first electron is now moving backwards at speed v, and the second (incoming) electron's speed has decreased from 2v to v. I believe this is the instant of closest approach in the second frame. Since both electrons have speed v at the moment of closest approach, the COE of energy calculation becomes: ½m(2v) ² = ke²/r + ½mv² + ½mv² which gives the same r= ke²/mv² that you got for the first frame.

[@bharath__100]

​@@christinejackson6630 sounds correct... let me know if you are sure about the answer...

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