Two balls of charges `q_(1)` and `q_(2)` initially have a velocity of the same nagnitude and direction. After a unform 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 ofvelocity of the second ball changes by `90^(@)`. In whata ratio will hte velociyt of the second ball change? Determine, the magnitude of the charge to mass ratio of the ball if it is equal to `alpha_(1)` for the firt ball. Ignore the electrostatic interactin between the balls.
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Explanation

Let the electric field on each ball be given by
`E=E_(x)hat i+E_(y)hat j`
From the impulse-momentum equation, we have
Impules=Change in momentum
Let the final velocities of the ball be `v_(1)` and `v_(2)`. Nothing that `v_(1)=v//2`, we have
`q_(1)(E_(x)hat i +E_(y)hat j)Delta t=m_(1)((v)/(2) cos60^(@)hat i+(v)/(2) cos60^(@)hat i)-m_(1)vhat i`...(i)
`q_(2)(E_(x)hat i +E_(y)hat j)Delta t=m_(2)(v_(2) cos60^(@)hat i+v_(2) cos60^(@)hat i)-(m_2)vhat i`...(ii)
On comparing the x and y-components on both sides of eq(i) we get
`(q_(1))/(m_(2))(E_(x))Delta t=-(3)/(4)upsilon` and `(q_(1))/(m_(2))E_(y)Delta t=-(sqrt(3))/(4)upsilon` ...(iii)
Similarly for Eq. (ii), we get
`(q_(2))/(m_(2))E_(x)Delta t=-v` and `(q_(2))/(m_(2))E_(y)Delta t=-v_(2)` ...(iv)
From Eqs. (iii) and (iv), by dividing the equation for x-components, we get
`(q_(1)//m_(1))/(q_(2)//m_(2))=(3)/(4)`
or `(q_(2))/(m_(2))=(4)/(3)(q_(1))/(m_(1))=(4)/(3)alpha_(1)`
Also `(q_(1)//m_(1))/(q_(2)//m_(2))= (sqrt(3)v)/(4v_(2))` or `(sqrt(3)v)/(4v_(2)) =(3)/(4)` or `v_(2)=(v)/(sqrt(3))`.