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A particle of mass 2m is attached by an inextensible string of length 1.2 m to a ring of mass m that can slide freely on a horizontal smooth rod. Initially the ring and the particle are at the same height with the string taut and horizontal. Both are released from rest simultaneously. How far (in metres) does the ring move by the time the string becomes vertical?

1. 0
2. 0.4
3. 0.8
4. 1.2

Correct answer: 0.8

Solution

Since horizontal momentum (and hence the centre of mass) is conserved, the ring (mass m) and particle (mass 2m) move so that m*d_ring = 2m*d_particle; with their relative horizontal displacement equal to the string length 1.2 m, the ring moves 0.8 m.

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