Question

Two particles each have a mass of 5.6 x 10-4 kg. One has a charge of +5.4 x 10-6 C, and the other has a charge of -5.4 x 10-6 C. They are initially held at rest at a distance of 0.72 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-third its initial value?

Answer 1

use the conservation of energy

when the two particles are held at a distance r apart, they have only potential energy, and the potential energy between two charges is:

PE=kq1 q1/r

when they are a distance r/3 apart, they have both PE and KE; the conservation of momentum tells us:

total energy before=total energy after

kq1 q1/r = kq1q2/(r/3) + 2(1/2 mv^2)

and we have 2(1/2mv^2) since there are two particles of mass m moving with speed v

the equation above gives us

2k q1q2/r = mv^2

for k=9x10^9
q1 x q2= -29.16x10^(-12)
m=5.6 x 10-4 kg
r=0.72m
we have

2*(9x10^9)(29.16x10^(-12)/0.72 =5.6 x 10-4 kg *v^2
v=36.08m/s