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A solid sphere falls at terminal velocity 20 m/s through air where g = 9.8 m/s². The same sphere is then taken into a gravity-free chamber containing air at the same pressure and is given a downward push at 20 m/s. (Treat the density of air as negligible.) Which statement is correct?
- Its initial acceleration will be 9.8 m/s² downward.
- Its initial acceleration will be 9.8 m/s² upward.
- The magnitude of acceleration will decrease as time passes.
- It will eventually stop.
Correct answer: Its initial acceleration will be 9.8 m/s² upward.
Solution
On Earth, at terminal velocity the upward viscous drag balances gravity: F_drag = m*g. In the gravity-free chamber gravity is absent, so the only force at 20 m/s is the same-magnitude viscous drag, now directed opposite to the velocity (upward, since motion is downward). Thus initial acceleration = F_drag/m = g = 9.8 m/s², directed upward (decelerating the sphere). As speed drops, drag drops, so deceleration decreases, and the sphere asymptotically slows but never abruptly stops; the correct listed statement is that the initial acceleration is 9.8 m/s² upward.
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