A particle of mass m strikes a horizontal smooth floor with a velocity u making an angle 'theta' with the floor and rebound with velocity v making an angle 'phi' with the floor. The coefficient of restitution between the particle and the floor is e. Then

A particle of mass m strikes a horizontal smooth floor with a velocity

1.	A particle of mass m strikes a horizontal smooth floor with a velocity u making an angle 'theta' with the floor and rebound with velocity v making an angle 'phi' with the floor. The coefficient of restitution between the particle and the floor is e. Then
Answer» the IMPULSE delivered by the floor to the body is mu(1+e) sin `THETA` `tan phi = e tan theta` `v = usqrt(1-(1-e^(2))sin^(2) theta)` the RATIO of the FINAL kinetic energy to the initial kinetic energy is `cos^(2) theta = e^(2) sin^(2) theta` Answer :A::B::D