A particle strikes a horizontal smooth floor with a velocity a making an angle `theta` with the floor and rebounds with velocity `v` making an angle `phi` with the floor. If the coefficient of restitution between the particle and the floor is `e`, then

Options

A. The impulse delivered by the floor to the body is mu (1 + e) sin `theta`.
B. `tan phi = e tan theta`
C. `v=u sqrt(1-(1-(1-e^(2))sin^(2)theta)`
D. the ratio of final K.E. to initial K.E. is `(cos^(2)theta+e^(2)sin^(2)theta)`

Answer

A, B, C, D