Two charged particles of mass m and 2m is moving in gravity free space...
Problem Statement
Two charged particles of mass m and 2m are moving in gravity-free space as shown in the figure. The initial distance between the particles is d, and the velocities are perpendicular to each other with 2m having velocity along the line joining the particles. In the subsequent motion of the particles, they are found to be at a distance 3d from each other twice. Find the possible values of the charge of each particle.
Solution
Let q1 and q2 be the charges on the particles of mass m and 2m, respectively. The initial velocities are perpendicular to each other, so the initial momentum is zero.
Let v1 and v2 be the final velocities of the particles. By conservation of momentum,
m v1 + 2m v2 = 0
or
v1 = -2 v2
Let r1 and r2 be the final positions of the particles. The distance between the particles is given by
|r1 - r2| = 3d
Squaring both sides, we get
(r1 - r2)2 = 9d2
Expanding the left-hand side, we get
r12 + r22 - 2 r1 r2 = 9d2
Substituting r1 = vt and r2 = -2vt (where t is the time taken and v is the common magnitude of the velocities), we get
(v2 - 4v2) t2 = 9d2
or
t2 = 9d2/3v2 = 3d2/v2
Let q be the magnitude of the charge on each particle. The force between the particles is given by Coulomb's law as
F = k q2 / r2
where k is the Coulomb constant and r is the distance between the particles. At the final position, the force between the particles is attractive and equal in magnitude to the centripetal force required to keep them moving in a circular path.