When stationary, the upthrust must balance the tension in the rubber band plus the weight of the ping pong ball.
ρVg = T + mg
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When the bucket is accelerated upward at a, the apparent g of the water becomes g+a. The water thus exerts a larger upthrust on the ball than before. Applying N2L on the ball,
ρV(g+a) – T – mg = ma’
ρVa = ma’
a’=ρV/m a
Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket accelerates at a, the ping pong ball accelerates at >a, and the rubber becomes more stretched.
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On the other hand, when the bucket has a downward acceleration of a, the apparent g of the water becomes g-a. The water exert a smaller upthrust on the ball than before. Applying N2L on the ball,
T + mg – ρV(g–a) – = ma’
ρVa = ma’
a’=ρV/m a
Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket has a downward acceleration of a, the ping pong ball has a downward acceleration >a, and the rubber becomes less stretched.
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The simplest case to consider is if the bucket is released from rest. As the bucket free falls, the upthrust acting on the ball disappears. While the bucket and the water free fall at g, the ball accelerates at larger than g, because it would be pulled down by gravitational force plus the tension in the rubber band.
xmDemo 