A fixed point charge Q is at origin At `t=0` a charge q with m is at `x=a` with leftward velocity `V_(0)` which satisfies `(kQq)/a=3mV_(0)^(2)` The particle turns around and starts to move right ward at the position b lt a

Options

A. the ratio `b/a` is `6/7`
B. the ratio `b/a` is `3/7`
C. the velocity of the particle at a large distance from the origin is `sqrt(7)V_(0)`
D. the velocity of the particle at a large distance from the origin is zero

Answer

A, C

Explanation

Using law of conservation of energy ,`E_(A)=E_(B)`
`:.1/2mV_(0)^(2)+(kQq)/a=0+(kQq)/a`
[:. At B,V=0and zr=b]
`:.(kQq)/(6a)+(kQq)/a=(kQq)/b`
`:.7/(6a)=1/b Rightarrowb/a=6/7`
At infinity `E_(A)=E_(infty)`

:.`1/2mV_(0)^(2)+(kQq)/a=1/2mV_(infty)^(2)+0 [:.(kQq)/rrightarrow0as r rightarrowinfty]`
`:.7/2mV_(0)^(2)=1/2mV_(infty)^(2)`
Rightarrow`V_(infty)=sqrt(7)V_(0)`