August 29, 2007 - Department of Physics
November 16th, 2009
PHY2048 Discussion-Fall ‘09
Quiz 10
Name: UFID:
A small solid ball is released from rest while fully submerged in a liquid and its kinetic energy is measured when it has moved 5.00 cm in the liquid. The figure below gives the results after many liquids are used: The kinetic energy is plotted versus the liquid density ρliq, and Ks = 4.00 J sets the scale on the vertical axis.
a) Express the acceleration of the ball in terms of the liquid density ρliq and the density of the ball ρb.
Gravitational force and buoyant force are exerted on the ball. Applying the Newton’s 2nd law, we have
ma = B-W ⇒ ρBVBa = ρliqVBg-ρBVBg ⇒ a = (ρliq-ρB)g/ρB
b) What is the density of the ball?
The measured kinetic energy is zero only if the ball is in equilibrium (a = 0). When the acceleration is zero, we have
(ρliq-ρB)g/ρB = 0 ⇒ρB = ρliq = 1.5 g/cm3
c) What is the volume of the ball?
When the density of the liquid is zero, the buoyancy on the ball is zero and the ball falls with the gravitational acceleration. The only force that does work on the ball is gravity. Therefore, the mechanical energy of the ball is conserved. Applying the energy conservation equation, we have
K = U = mgh = ρBVBgh ⇒VB = K/(ρBgh) = 4/(1.5×103×9.8×0.05) = 5.44×10-3 m3
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