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A hollow smooth uniform sphere A of mass 'm' roll without sliding on a smooth horizontal surface. It collides elastically and head on with another stationary solid sphere B of the same mass m and same radius. The ratio of kinetic energy of 'B' to that of 'A' just after the collision is

\[\begin {array} {1 1} (a)\;5:2 & \quad (b)\;1:1 \\ (c)\;2:3 & \quad  (d)\;3:2 \end {array}\]

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Since the collision in elastic, and head on between the two identical masses, they interchange their linear velocities.
Since no tangential forces are presents, the angular velocity of ball A is not affected.
hence body A has pure rotational with angular velocity w and body B has pure translation with linear velocity V.
$\large\frac{KE_A}{KE_B}$ after collision = $\large\frac{1/2 I w^2}{1/2 I wV^2}$
$\qquad= \large\frac{\Large\frac{1}{2} \frac{2}{3}mR^2w^2}{\frac{1}{2} m V_0^2}$
$\qquad= \large\frac{2}{3}$$\qquad $$[Since\; V_0=Rw]$
Hence d is the correct answer
answered Dec 5, 2013 by meena.p
edited Jun 22, 2014 by lmohan717

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