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Applied Mechanics - Engineering

Q1:

A ball is dropped from a height of 2.25 m on a smooth floor and rises to a height of 1.00 m after the bounce. The coefficient of restitution between the ball and the floor is

A 0.33

B 0.44

C 0.57

D 0.67

ANS:D - 0.67

Here, V(velocity after seperation)=eU(velocity before impact).
(2gh)^1/2=(2gH)^1/2 *coefficient of restitution.
c.o.r(e)=(2g*1/2g*2.25)^1/2.
c.o.r(e)=0.44.
h=velocity with which ball impinges on the floor.
H=velocity with which ball rebounds.