A particle strikes a horizontal frictionless floor with a speed u at an angle `theta` with the vertical, and rebounds with a speed v at an angle `phi` with vertical. The coefficient of restitution between the particle and floor is e. The magnitude of v is

Options

A. `eu`
B. `(1 - e)u`
C. `usqrt(e^2 sin^2 theta + cos^2 theta)`
D. `usqrt(sin^2 theta + e^2 cos^2 theta)`

Answer

D

Explanation

As the floor is frictionless and there is no horizontal force, therefire, momentum must be conserved in the horizontal direction.
i.e., `m u sin theta = mv sin phi " or " u sin theta = v sin phi " " ..(i)`
And in vertical direction, `(v cos phi)/(u cos theta) = 3`
or `v cos phi = eu cos theta " " ........(ii)`
Squaring and adding (i) and (ii), we get
`v^2 (sin^2 phi + cos^2 phi) = u^2 sin^2 phi + e^2 u^2 cos^2 theta`
or `v = u sqrt(sin^2 phi + e^2 cos^2theta)`.