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.

Explanation

Again in this question we have to do the conservation of mechanical energy, Initially, there is only potential energy but finally there is both kinetic and potential energy.
initial state energy :
Bodies are at rest so kinetic energy = 0
Potential energy = `(Q_(1)Q_(2))/(4 pi epsilon_(0)R) = (9 xx 10^(9) (4 xx 10^(-5))(-4 xx 10^(-5)))/(1 ) = 14.4` joule
Final state energy :
Let us suppose both the charges approach each other with individual speed v. (both will have same speed because masses are same and momentum will be conserved . since initial momentum is zero, they will have final momentum also equal to zero, since momentum is vector, they will have equal and opposite momentum )
Final kinetic energy of boty = `2 xx (1)/(2) mv^(2) = mv^(2)`
Final potential energy = `(Q_(1)Q_(2))/(4 pi epsilon_(0)R) = (9 xx 10^(9) (4 xx 10^(-5))(-4 xx 10^(-5)))/(0.5) = -28.8` joule
Now, initial total energy = final total energy
= 14.4 `mv^(2) - 28.8`
therfore, `mv^(2) = 14.4 "or" v = sqrt((14.4)/(5 xx 10^(-3)))` = 54 m/s