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 restitution between the ball and steel base is 0.6 and that 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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$\frac{2}{5} rad / s$

$\frac{1}{4} rad / s$

$\frac{1}{5} rad / s$

$\frac{3}{5} rad / s$

answer is A.

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Detailed Solution

If v is the velocity by which balls strike the bases then their velocities after collision are 0.6 v and 0.4 v as shown 'u' is velocity of COM just after collision and $ω$ in angular velocity acquired by the rod.

For end A: $0.6 v = u + \frac{l}{2} . ω$ ---------(i)

For end B: $0.4v = u- \frac{l}{2} . ω$ --------(ii)

Solving (i) and (ii), we get

$0.2 v = ωl \Rightarrow 3 ω = \frac{v}{5 l}$

Now using [ $v^{2} - u^{2} + 2 aS$ ]

$v^{2} = 0 + 2 \times 10 \times 0.2 = 4 \Rightarrow v = 2 m / s$

$ω = \frac{2}{5 \times 1} = \frac{2}{5} rad / \sec$

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