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A ball strickes a horizontal floor at an...

A ball strickes 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 function of its kinetic energy lost in the collision is.

Text Solution

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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)/(sqrt(2)))^(2)+ ((u)/(2 sqrt(2)))^(2)) = sqrt((5)/(8)) u`
`:.` Fraction 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)`.
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