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A uniform rod AB of mass M is pivoted at end A and released from rest in the horizontal position. When it swings down to the vertical position, it strikes a sphere of mass m (initially at rest on a frictionless surface) with a perfectly elastic, horizontal impact. After the collision, the sphere moves horizontally and the rod swings back upward to a maximum angle of 60 degrees from the vertical. Given that the length of the rod is sqrt(2)*r (where r = 6*sqrt(2)/10 m), what is the ratio M/m?

1. The ratio M/m equals 3/2
2. The ratio M/m equals 2/3
3. The ratio M/m equals 3
4. The ratio M/m equals 1/3

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

Solution

From energy conservation omega1 = sqrt(3g/L) and omega2 = sqrt(3g/(2L)). Applying angular momentum and elastic restitution gives M/m = 3(omega1 - omega2)/(omega1 + omega2) = 3(sqrt2 - 1)/(sqrt2 + 1). Evaluating with the given rod length yields M/m = 3/2.

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