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 angle phi is equal to

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 angle phi is equal to
Answer» `theta` `(1 + e) theta` `TAN^(-1)(e tan theta)` `tan^(-1)(1/e tan theta)` Solution :As the FLOOR is frictionless and there is no horizontal force, therefore, momentum must be conserved in the horizontal direction. i.e, `m U sin theta = mv sin phi " or " u sin theta = V sin theta "" ….(i)` And in VERTICAL direction, `(v cos phi)/(u cos theta) = e` or `v cos phi = eu cos theta "".....(ii)` Divide (i) by (ii), we get `tan phi = 1/e tan phi " or " phi = tan^(-1) [1/e tan phi]`.