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  1. A small ball of density $\rho$ is immersed in a liquid of density $\sigma$, ($\sigma>\rho$) to the depth $h$ and released. How can we find the height above surface of the liquid up to which the ball will jump?

  2. And what if we change the question so that the ball is dropped from a height of $h$ how deep in the liquid the ball will go? Will the this will be solved with the same method as the first question?

  3. Will the ball continue what is called a simple harmonic motion if surface tension and Drag is ignored?


My work

1.Net upward force acting on the body; $F$=$\sigma vg-\rho vg$=$ma$

$a=g(\frac{\sigma}{\rho}-1)$

velocity after height $h$; $v=\sqrt{2g(\frac{\sigma}{\rho}-1)h}$

conserving mechanical energy at max height $h'$ and at surface,

$mgh'$=$\frac{1}{2}m(2g(\frac{\sigma}{\rho}-1)h)$

$\fbox{$h'=(\frac{\sigma}{\rho}-1)h$}$

  1. velocity just before touching the liquid= $\sqrt{2gh}$

negative acceleration= $g(\frac{\sigma}{\rho}-1)$ from part 1.

using, $v^2=u^2-2g(\frac{\sigma}{\rho}-1)h'$

$\fbox{$h'=\frac{h}{(\frac{\sigma}{\rho}-1)}$} $

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closed as off-topic by Danu, Neuneck, ACuriousMind, Kyle Kanos, user10851 Dec 12 '14 at 18:33

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  • $\begingroup$ Hi Dheeraj Kumar. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Dec 12 '14 at 12:35
  • $\begingroup$ BTW @Qmechanic can you answer Part 3 please $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 12:40
  • $\begingroup$ It won't be simple harmonic, because one of the restoring forces (buoyancy) changes at the water-air interface, while gravity remains constant. It will be some sort of harmonic behavior so long as friction, surface tension, etc. are ignored. $\endgroup$ – Carl Witthoft Dec 12 '14 at 12:53
  • $\begingroup$ @CarlWitthoft I tried and found that it will do a SHM for $\sigma$=2, $\rho$=1, and initial depth=1 unit, you can check it by putting different sets of values in part 1 and 2 $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 13:03
  • $\begingroup$ What's your density for air? Are you using zero, aka vacuum? $\endgroup$ – Carl Witthoft Dec 12 '14 at 13:08
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First you must find the speed when the ball comes out of water, the same way you calculate the speed of free fall of an item in the air. $$m\vec{a}=\Sigma{\vec{F_i}}$$ where the forces are : friction, weight, Archimedes force.

Then again in the air where the Archimedes force may be neglected. If the air friction is neglected, it's easier using the potential energy the kinetic energy of the ball as it comes out water is equal its potential energy when its speed becomes 0.

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  • $\begingroup$ will the answer depend on the volume of the ball $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 9:09
  • $\begingroup$ I think you're missing an important factor here: surface tension of the water-air interface. This will result in an opposing force at the moment the ball tries to break the interface which will reduce the height. Additionally, the water-tail that you often see for balls that jump out of the water will increase the effective weight of the ball and reduce its height further $\endgroup$ – Michiel Dec 12 '14 at 9:46
  • $\begingroup$ @Michiel That's right. This force is hard to evaluate. In this document experimental measures have been done (pages 137 and so forth). $\endgroup$ – TZDZ Dec 12 '14 at 10:06

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