How Deep Will a Less Dense Ball Sink in Water?

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    Bouyancy Motion
AI Thread Summary
A ball with a density of 0.4*10^3 kg/m3 is dropped into water from a height of 9cm, and the discussion revolves around calculating how deep it will sink based on buoyancy, ignoring viscous retardation. The initial approach involved calculating the final velocity upon entering the water and the buoyant force acting on the ball. The key realization was that the retardation due to buoyancy must account for the ball's weight, leading to a correction in the calculations. After addressing this oversight, the participants confirmed that the correct approach would yield a depth consistent with the provided answer choices. The final conclusion emphasizes the importance of considering both buoyant force and weight in such calculations.
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Homework Statement



A ball whose density is 0.4*103 kg/m3 falls into water from a height of 9cm. To what depth does the ball sink? Only consider bouyancy and ignore retardation due to viscocity.

(a) 9cm (b) 6cm (c) 4.5cm (d) 2.25cm

Homework Equations



Bouyant force = Volume of Body*Density of Liquid*Gravitational Acceleration

v2 = u2 + 2as

F = ma

Density = Mass / Volume


The Attempt at a Solution



Well, first I figured out the final velocity with which the body entered into the water using the third equation of motion (second equation stated above).

Then I figured out the retarding force acting on the body because of the bouyancy offered by the water (using the second, the first, and then the third equation stated above).

Finally I plugged in the retardation offered by bouyancy into the third equation of motion, and figured out the final answer. My answer was 3.6cm, but that is nowere in the answer choices.

Please help.
 
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In the retardation equation, did you include the fact that the ball's weight is still acting downwards on it, and that the retardation is caused by a resultant upwards force equal to buoyancy minus weight?
 
Umm... no I didn't. I'm stuipid, right?

Yup, well that solved the problem.

Thanks a lot.
 
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