A ball strikes a horizontal floor at an angle theta = 45^(@) with the normal to floor. The coefficient of restitution between the bal 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

1.	A ball strikes a horizontal floor at an angle theta = 45^(@) with the normal to floor. The coefficient of restitution between the bal 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 ballbefore 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)/(sqrt(2)))^(2)+((u)/(2sqrt(2)))^(2))=sqrt((5)/(8)).`u `therefore` Fraction of KE lost in collision `((1)/(2)mu^(2)-(1)/(2)MV^(2))/((1)/(2)m u^(2))=1-((v)/(u))^(2)=1-(5)/(8)=(3)/(8)`