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

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><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :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= sqrt(u^(2)sin^(2)theta+e^(2)u^(2)cos^(2)theta)`<br/>`v = sqrt(((u)/(sqrt(2)))^(2)+((u)/(2sqrt(2)))^(2))=sqrt((5)/(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)).u`<br/>`therefore` Fraction of KE lost in collision<br/>`= ((1)/(2)m u^(2)-(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a>^(2))/((1)/(2)m u^(2))=1-((v)/(u))^(2)=1-(5)/(8)=(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)/(8)`</body></html>