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Two identically sized balls of different masses m1 and m2 (with m2 > m1) are tied to the two ends of a light thread and dropped together from rest. Air exerts a viscous drag (depending on size and speed) and a buoyant force (equal to the weight of air displaced) on each ball. After a long time both reach a common terminal velocity. Find the tension in the connecting thread at this stage.
- 0.5(m2 - m1)g
- (m2 - m1)g
- 0.5(m2 + m1)g
- zero
Correct answer: 0.5(m2 - m1)g
Solution
Both balls have the same size and the same terminal speed, so the combined upward (drag + buoyancy) force F is identical for each. For the heavier ball the thread pulls up (tension T), for the lighter ball the thread pulls down. Writing equilibrium for each and solving gives T = 0.5(m2 - m1)g.
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