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» <html><body><p></p>Solution :Let u be the velocity of ball before <a href="https://interviewquestions.tuteehub.com/tag/collision-922060" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISION">COLLISION</a>. Speed of the ball after collision will become <br/> `v= <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(u^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) sin^(2) <a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a> + e^(2)u^(2) cos^(2) theta)= sqrt(((u)/(sqrt2))^(2) + ((u)/(2 sqrt2))^(2))= sqrt((5)/(8))u` <br/> `therefore` Praction of KE lost in collision <br/> `=((<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2) m u^(2) - (1)/(2) mv^(2))/((1)/(2) m u^(2))= 1- ((v)/(u))^(2)=1- (5)/(8)= (3)/(8)`</body></html>