A glass ball collides with a smooth horizontal surface (`xz` plane) with a velocity `V = ai- bj`. If the coefficient of restitution of collision be `e`, the velocity of the ball just after the collision will be

Explanation

Collision takes place along the normal. Therefore the magnitude normal component `(v_(y))` of the velocity of the glass ball is changed to` v_(y')="e v"_(y)` just after the collision whereas the horizontal component `(v_(x))` of its velocity remains constant due to the absence of any horisontal force.
`rArr` The velocity of the ball just after the impact
`=vecv=vecv_(x)+vecv_(y)`
`rArr" "vecv'=v'_(x)hati+v'_(y)hatj`
where, `v'_(x)=a` & `v'_(y)=eb`
`rArr" "vecv'=ahati+ebhatj`
Therefore the magnitud of the velocity `vecv'=|vecv'|=sqrt(a^(2)+e^(2)b^(2))` and the direction is given as
`theta=tan^(-1)((v_(x'))/(v_(y')))=tan^(-1)((a)/(eb))` to the normal (vertical)