Two balls of charges `q_(1)` and `q_(2)` initially have a velocity of the same magnitude and direction. After a uniform electric field has been applied for a certain time. The direction of the first ball changes by `60^(@)` and the velocity magnitude is reduced by half. The direction of velocity of the second ball changes by `90^(@)`. In what ratio will the velocity of the second ball change? Determine, the magnitude of the charge to mass ratio of the second ball if it is equal to `alpha_(1)` for the first ball. Ignore the electrostatic interaction between the balls.
,

Answer

`v/sqrt(3)`

Explanation

Momentum of 1st ball

`P_(1)=mvi, P_(f)=mv/2 (cos 60^(@) hat(i)+sin 60^(@) hat(j))`
change in momentum
`DeltaP=P_(f)-P_(i)=-(3v)/4hat(i)+(sqrt(3)v)/4 hat(j)`
Momentum of Iind ball :
`P_(i)=mvhat(i) " " P_(f)=mvhat(j)`
Change in momentum `=mv[hat(j)-hat(i)]`