Two steel ball of equal diameter are connected by a rigid bar of negligible weight as shown and are dropped in the horizontal position from height `h` above the heavy steel and brass base plates. If the coefficient of restitition between the ball and steel base is `0.6` and the between the other ball and the brass base is `0.4`. The angular velocity of the bar immediately after rebound is. (Assume the two impacts are simultaneous).
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Options

A. `(2)/(5) rad//sec`
B. `(1)/(5) rad//sec`
C. `(3)/(5) rad//sec`
D. `(1)/(4) rad//sec`

Answer

A

Explanation

(a) If `v` is the velocity by which balls strike the bases the their velocities after collision are `0.6 v` and `0.4 v` as shown 'u' is velocity of `COM` just after collision and `omega` in angular velocity acquired by the rod.
For end `A` : `0.6 v = u + (l)/(2). omega`
For end `B` : `0.4 v = u - (l)/(2). omega`....(ii)
Solving (i) and (ii), we get
`0.2 v = omega l rArr 3 omega = (v)/(5 l)`
Now using `[v^2 = u^2 + 2a S]`
`v^2 = 0 + 2 xx 10 xx 0.2 = 4 rArr v = 2 m//s`
`omega = (2)/(5 xx 1) = (2)/(5) rad//sec`.
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