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Q.

A ball strikes a horizontal floor at an angle θ=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

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a

38

b

5/8

c

18

d

78

answer is A.

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Detailed Solution

v=usin2θ+e2cos2θ

=u12×14+12=u58

v=58u

Loss of K.E (fraction) =u2−v2u2

=u2−58u2u2=38

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