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

A particle strikes a horizontal frictionless floor with a speed u at a

1.	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
Answer» `EU` `(1 - e)u` `usqrt(e^2 SIN^2 theta + cos^2 theta)` `usqrt(sin^2 theta + e^2 cos^2 theta)` Solution :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)`.