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Question

Two particles have equal masses of 5.0 g each and opposite charges of +4×105C and 4×105C. They are released with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

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Solution

Applying conservation of momentum by initial and final stage
Pi=0Pf=mv1+mv2Pf=Piv1=v2...(1)
Now applying energy conservation by two points
KEi+(P.E)i=KEf+(P.E)f
mv21mv212=kq2(2r1r)
mv21=15410
v1=53.66m/s=v2.


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