Two particles A and B of masses m and 2m and carrying charges +Q and +2Q respectively are initially at a very large separation. Both particles are free to move if particle A is projected towards particle B with velocity v, The velocity of both the particles at minimum separation is [Ignore gravitational force of attraction]
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a
v
b
v/2
c
v/3
d
zero
answer is C.
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Detailed Solution
When separation between A and B is minimum, velocity of the particles will be equal. Conserving momentum, m+2mv'=mv⇒v'=v3.