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

Options

A. `sqrt(e^(2)a^(2)+b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical
B. `sqrt(a^(2)+e^(2)b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical
C. `sqrt(a^2+b^2/e^2)` at angle `tan^-1((ea)/b)` to the vertical
D. `sqrt(a^(2)/e^(2)+b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical

Answer

B

Explanation

Collision takes place along the normal. Therefore the masgnitude normal component `(V_(y))` of the velocity of the glass ball is changed to `V_(y)^(')=-eV_(y)^(`) just asfter the collision where as the horizotnal component `(V_(x))` of its velocity remains constant due to the absence of any horizontal force.
`=V'=vecV_(x)^(')+vecV_(y)^(')impliesvecv'=V_(x)^(')hati+V+y'hatj`
where `V_(x)^(')=a` and `V_(y)^(')=eb` (`:'vecV=ahati-bhatj`)
`impliesvecV'=hati+ebhatj`
Therefore the magnitude 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.