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Two particles, each of mass m, are constrained to slide along a smooth circular hoop of equal mass m. They start at opposite ends of a diameter and are each given equal speeds v0 (as shown). The whole system is in gravity-free space. Find the speed of each particle just before they collide.

1. sqrt(7)/3 * v0
2. 1/sqrt(3) * v0
3. sqrt(3)/2 * v0
4. 2/sqrt(3) * v0

Correct answer: sqrt(7)/3 * v0

Solution

Applying conservation of momentum, angular momentum and energy to the free hoop and the two sliding particles gives the speed of each particle just before collision as sqrt(7)/3 times v0.

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