1.

Two particles have equal masses of 5.0 g each and opposite charges of `+4.0 xx10^(-5) C.` They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reducced to 50 cm.

Answer» Here, `m_(1) = m_(2) = 5.0g = 5xx10^(-3) kg`.
`r_(1) = 1.0 m, r_(2) = 50 cm = (1)/(2) m`
From symmetry , `v_(1) = v_(2) = v = ?` As increase in K.E. = loss in P.E, - initial P.E.
`:. 2xx(1)/(2) mv^(2) = (| q_(1)|| q_(2)|)/(4pi in_(0)) [(1)/(r_(2)) - (1)/(r_(1))]`
`5xx10^(-3) v^(2) = 9xx10^(9) (4xx10^(-5))^(2) [(1)/(1//2) - 1]`
`= 9xx16xx10^(-1) (2-1)`
`v^(2) = (1.44)/(5xx10^(-3)) = 2.88xx10^(3)`
`v = 53.7 m//s`


Discussion

No Comment Found

Related InterviewSolutions