A particle of mass `m` carrying charge `q` is projected with velocity `v` from point `P` toward aninfinite line of charge from a distance `a` . Its speed reduces to zero momentarily at point `Q`, which is at a distance `a//2` from the line of charge. If another particle with mass `m` and charge `-q` is projected with the same velosity `v` from `P` toward the line of charge, what will be its speed at `Q` ?

Options

A. `v_1 = sqrt(2) v`
B. `v_1 = 2 v`
C. `v_1 = sqrtv`
D. `v_1 = sqrt(2)`

Answer

a

Explanation

A `(0 - (1)/(2) mv^2) = q (V_P - V_Q)` ….(i)
`(1)/(2) mv_1^2 - (1)/(2) mv^2 = -q(V_P - V_Q)` ….(ii)
From Eqs. (i) and (ii), `v_1 = sqrt(2) v`.