A uniform rod AB of mass M is attached to a hinge at one end A, and released from the horizontal position. The rod rotates about A, and when it reaches the vertical position the rod strikes a sphere of mass m and radius r initially at rest on the smooth horizontal surface as shown in the adjacent figure. The impact is along the horizontal direction and perfectly elastic.
If at the moment of impact, the lowest end of the rod is very close to the smooth horizontal surface. After the impact, the sphere moves along the horizontal and the rod, subsequently rises to a maximum of $60^{0}$ with the vertical. Choose the correct statement(s) from the following, taking into account the information given above. The length of the rod equals $\sqrt{2} r . (r = \frac{6 \sqrt{2}}{10} m)$ (take $g = 10 m / s^{2}$ )

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The ratio M/m is 3/2

The ratio m/M is 9/2

The ratio m/M is 3/2

The ratio m/M is 2/9

answer is A.

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Detailed Solution

Using conservation of mechanical energy

$\begin{array}{l} m g \frac{L}{2} = \frac{1}{2} \frac{M L^{2}}{3} ω_{0}^{2} \\ \Rightarrow ω_{0} = \sqrt{\frac{3 g}{L}}, ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 g}{2 \sqrt{2 r}}} \\ L = \sqrt{2 r} \end{array}$

$\begin{array}{l} (L - r) ω_{0} = v - ω (L - r) \\ v = (L - r) (ω_{0} + ω) = (L - r) ω (\sqrt{2} + 1) = r ω = \sqrt{\frac{3 g r}{\sqrt{2}}} \\ r = \frac{6 \sqrt{2}}{10} m \\ v = \sqrt{\frac{3 \times 10 \times 6 \times \sqrt{2}}{2 \sqrt{2} \times 10}} = 3 m / s \end{array}$

Using COAM

$\begin{array}{l} l ω_{0} = l ω + m v (L - r) \\ l (ω_{0} - ω) = m (ω_{0} + ω) {(L - r)}^{2} \\ \frac{M L^{2}}{3} (ω_{0} - ω) = m (ω_{0} + ω) {(L - r)}^{2} \\ \frac{M}{m} = \frac{3 {(L - r)}^{2}}{L^{2} (ω_{0} - ω)} = \frac{3}{2} \end{array}$