A uniform horizontal rod of length l falls vertically from height h on two identical blocks placed symmetrically below the rod as shown in figure. The coefficients of restitution are e1 and e2. The maximum height through which the centre of mass of the rod will rise after bouncing off the blocks is
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a
he1+e2
b
e1+e224h
c
e1+e22h
d
(e1+e2)2h2
answer is D.
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Detailed Solution
velocity of the two end after collision is e12gh and e22ghVcm=e1+e222ghmaximum height of center of mass=(Vcm)22g