GR8677 #78



Alternate Solutions 
Saint_Oliver 20130913 20:08:00 
since
Therefore the total energy is greater than 0, which leaves only choice (C).  

Comments 
Saint_Oliver 20130913 20:08:00 
since
Therefore the total energy is greater than 0, which leaves only choice (C).   thebigshow500 20081014 01:41:19  So what will happen if radiation involves? I can't get a concise answer from this case. Anyone care to explain?
Poop Loops 20081105 21:21:34 
Makes things a lot harder. Charged particles radiate when they accelerate or decelerate. Things with mass accelerate or decelerate under a force. So they'd be pushing each other away and radiating, etc. So it's weird.

  grae313 20071007 18:34:19  Yosun, you assume the particle with a velocity is moving towards the stationary particle, but what if it was moving away from the particle? The answer must be valid for both cases (moving towards the particle, or away), and in the case where it moves away, potential energy will always be zero and the total energy will just be given by the positive kinetic energy. Thus, C is clearly the answer.
511mev 20091105 21:57:33 
I know this is an old comment, but i'd still like to clarify: the potential energy is only zero at time zero, but since the coulomb potential is conservative, then T + V = a constant (of unspecified sign, since we don't know the signs of the charges).

llama 20131016 18:19:15 
T+V is a positive constant, since initially V=0 (as stated in the question) and T>0 (since .5*mv^2>0)

  dbiggerstaff 20051110 00:13:03  There\'s a simpler way to look at it: the problem says that each particle initially has zero potential energy, but that one particle has a velocity relative to the other, which implies that it will have some kinetic energy in this reference frame. This means the total initial energy is T+U = (something)+0 = (something positive.) Since there\'s no radiation, energy will be conserved, and the total energy will remain (something positive.) Thus, the correct answer is (c).
nitin 20061116 13:36:49 
Thumbs up mate!

abc111 20091002 02:54:55 
can anyone explain... the sign cannot be determined unless intial velocity is known.....i am thinking...T is only thing that contains velocity here...but it will absorb minus sign because 1/2 m v^2 has a square term...

kaic 20131009 13:37:53 
It's highly unlikely that the particles would have negative energy, don't you think? While negative energy is not forbidden, and has been demonstrated to be possible, it's unlikely to show up on the GRE unless it is explicitly included as part of a question. As you said, the sign of the initial velocity does not matter because it is squared in the kinetic energy, so T is positive, and U is 0, so total energy E = T + V is positive.

 




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