A ball strikes a horizontal floor at an angle theta= 45^(@) with the normal to floor. The coefficient of restitution between the ball and the floor is e=1/2. The fraction of its kinetic energy lost in collision is

A ball strikes a horizontal floor at an angle theta= 45^(@) with the n

1.	A ball strikes a horizontal floor at an angle theta= 45^(@) with the normal to floor. The coefficient of restitution between the ball and the floor is e=1/2. The fraction of its kinetic energy lost in collision is
Answer» Solution :Let u be the velocity of ball before COLLISION. Speed of the ball after collision will become `v= SQRT(u^(2) sin^(2) THETA + e^(2)u^(2) cos^(2) theta)= sqrt(((u)/(sqrt2))^(2) + ((u)/(2 sqrt2))^(2))= sqrt((5)/(8))u` `therefore` Praction of KE lost in collision `=((1)/(2) m u^(2) - (1)/(2) mv^(2))/((1)/(2) m u^(2))= 1- ((v)/(u))^(2)=1- (5)/(8)= (3)/(8)`