Two particles, each having a mass m are placed at a separation d in a uniform magnetic field B as shown in they have opposite charges of equal magnitude q. at time t =0, the particles are projectd towards each other,each with a speed v. suppose the coulomb force between the charges is switched off. (a) Find the maximum value `v_m` of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if `v = v_m /2?` (c) At what instant will a collision occur between the particles if `v = 2v_m` ? (d) Suppose `v =2v_m` and the collision between the particles is completely inelastic. Describe the motion after the collision.

Explanation

The particles will not collide if
`d = r_(1) + r_(2)`
`rArr d = ((mV)/(qB) + (mV)/(qB) + (2mV)/(qB))`
`v= (qBd)/(2m)`
`d_(1) = r_(1) +r_(2) = 2r`
`rArr = (d)/(2)` (minimum distance)
Max distance ,
`d_(2) = + d + 2r = d + (d)/(2) = (3d)/(2)`
`( c ) V= 2V _(2) m`
`r_(1) = r_(2) = d ` The are is `(1)/(6)`
`(d ) V = 2V_(m)`
The particle will collide at point P . At point P , both the particle will have motion `m` in upward direction . Since the particle collide inelastically they stick together distance `i` between centers `= d`,
`sin sintheta = (i)/(2r)` velocity upward
`= v cos (90 - sin theta ) = (vi)/(2r)`
`(mv^(2) )/(r) = qvB`
` rArr r = (mV)/(qB)`
Using `(mu^(2) )/(r) = (qvB)/(2m) = V_(m)`
Hence the combine mass will move with the velocity `V_(m)`