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A uniform rod AB of mass M is hinged at end A. It is released from rest in the horizontal position and swings down to the vertical, where it strikes a stationary sphere of mass m resting on a smooth horizontal surface. The collision is along the horizontal direction and perfectly elastic. After the collision, the rod swings up to a maximum angle of 60 degrees with the vertical. Given that the rod length L = sqrt(2) * r where r = (6 * sqrt(2)) / 10 m, select the correct statement(s).

1. The ratio M/m equals 3/2
2. The ratio M/m equals 2/3
3. The speed of the sphere just after collision is 6 m/s
4. The speed of the sphere just after collision is 3 m/s

Correct answer: The ratio M/m equals 3/2

Solution

L = sqrt(2) * (6*sqrt(2)/10) = 1.2 m. Before collision: omega = sqrt(3g/L) = sqrt(3*10/1.2) = 5 rad/s. Tip speed u = omega*L = 6 m/s. After collision, rod rises to 60 deg with vertical: height = L/2*(1-cos60) = L/4. Energy: (1/2)(ML²/3)omega'² = Mg*L/4 => omega'² = 3g/(2L) => omega'=sqrt(12.5)=5/sqrt(2) rad/s (rod may reverse). Tip speed after: u' = omega'*L. Angular momentum: (ML²/3)omega = (ML²/3)omega' + m*V*L. Elastic: V - omega'*L = omega*L. These equations give M/m = 3/2 (with rod bouncing back) and V = 6 m/s.

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