A charged particle having charge `q` and mass `m` is projected into a region of uniform electric field of strength `vecE_(0)`, with velocity `vecV_(0)` perpendicular to `vecE_(0)`. Throughout the motion apart from electric force particle also experiences a dissipatice force of constant magnitude `qE_(0)` and directed opposite to its velocity. If `|vecV_(0)|=6m//s`, then find its speed when it has turned through an angle of `90^(@)`.

Explanation

Consider the situation when particle has turned through an angle of `phi`
`(dv)/(dt)=-(qE_(0))/(m)[1-sinphi]`
`(dv_(x))/(dt)=(qE_(0))/(m)[1-sinphi]`
`rArr(dv)/(dt)=-(dv_(x))/(dt)`
`rArr v=-v_(x)+C`
`rArrv=v_(0)-v_(x)=v_(0)-vsinphi`
`rArr v=(v_(0))/(1+sinphi)=(6)/(1+1)=3m//s`