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Two particles hv equal mass 5g each & opp. charges 4 X 10-5 .They r released frm rest with separation of 1 m b\w them.Find speeds of the particle when the seperation is reduce to 50cm.
applying conservation of momentam bw initial and final stage
Pi=0
Pf=mv1 +mv2
Pf=Pi
v1=-v2..........1
now applying energy conservation bw two points
KEi + (P.E)i = KEf + (P.E)f
(mv1^2 -mv2^2)/2 =kq^2(2/r -1/r )
mv1^2=154/10
v1=53.66m/s=-v2
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