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.
,

Options

A. `(4alpha)/(3)`
B. `(2alpha)/(3)`
C. `(alpha)/(3)`
D. `alpha`

Answer

A

Explanation

Let `vec(E)=E_(x)hat(i)+e_(y)hat(j)`
By impuse-momentum therorem
Impulse =change in momentum
`rArr` Let the final velocity of balls be `v_(1)&v_(2)=(v)/(2)`
`q_(1)[E_(x)hat(i)+E_(y)hat(j)]Deltat=m_(1)((v)/(2)cos60hat(i)+(v)/(2)sin60hat(j))-m_(1)vhat(j)`
`q_(2)[E_(x)hat(i)+E_(y)hat(j)]Deltat`
` =m_(2)(v_(2)cos90hat(i)+v_(2)sin90hat(j))=m_(2)vhat(i)`
`q_(1)E_(x)(Deltat)=-(m_(1)3v)/(4),q_(1)E_(y)Deltat=(m_(1)sqrt(3)v)/(4)`
`q_(2)E_(x)(Deltat)-m_(2)v,q_(2)E_(y)(Deltat)=m_(2)v_(2)`
`rArr (q_(1)ExDeltat)/(q_(2)E_(x)Deltat)=(m_(1))/(m_(2))(3)/(4) rArr (q_(2))/(m_(2))=(4alpha)/(3)`