Calculating Maximum Depth for a Sphere Dropped into Water

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A sphere with a density of 0.75 times that of water is dropped from a height of 11 meters, and the goal is to determine how deep it will sink. Initial calculations using kinematics yield a velocity of 14.68 m/s upon impact, leading to an incorrect maximum depth of 8.25 meters. The discussion highlights the importance of considering buoyant force as the only non-conservative force in the system and emphasizes the need to account for total height when calculating depth. A revised approach suggests using the relationship between the buoyant force and the weight of the sphere, ultimately leading to confusion about the correct depth due to assumptions made in the problem. The conversation concludes with a realization that neglecting energy transfer during impact significantly affects the results.
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Homework Statement



A small sphere 0.75 times as dense as water is dropped from a height of 11 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.

for clarity's sake, I'm letting v = volume, and v = velocity

Homework Equations



vf^2 = Vi^2 + 2ad, Fb = pgv, K = 1/2mv^2, Work(non conservative) = change in mechanical energy


The Attempt at a Solution



the first thing i did was basic kinematics to get a velocity of 14.68 m/s upon hitting the water.
next i set Fb equal to nonconservative work and i got pgv(d) = Kf -Ki...
simplified down i got it to pvg(d) = 1/2mv^2
= pvg(d) = 1/2pvv^2

at this point i cancled out the volumes and got a final equation of
1000*9.8*(d) = .5*750*(14.68^2)

this got me a final answer of 8.25 meters which seems to be too large, and in fact it was wrong.
can anyone find where I'm going wrong?
 
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Gravity still acts on the object when it's under water.
 
oh i see, but how do i take that into account? the buoyant force still is non conservative right?
 
Yes, buoyant force is the only non-conservative force acting (in this simplified problem).

You want the super-duper easy way to solve this? Measured from the final position of the object (distance d under the water), what's the total mechanical energy of the object just before it's dropped? After it reaches the final position? So what must the non-conservative work equal?
 
or can i make an Fnet equation

like, mg - Fb = ma

p(ball)vg - p(water)vg = p(ball)*v*a

volumes cancle, so => 750(9.8) -1000(9.8) = 750*a
this yielding an acceleration of 3.267 m/s^2

would that be a correct approach, or is what i just laid out above incorrect?
 
Nothing wrong with that approach. Go for it.
 
Doc Al said:
Yes, buoyant force is the only non-conservative force acting (in this simplified problem).

You want the super-duper easy way to solve this? Measured from the final position of the object (distance d under the water), what's the total mechanical energy of the object just before it's dropped? After it reaches the final position? So what must the non-conservative work equal?

ok so since there's potential at the top and nothing(?) at the end, can i say

p(water)vg(d) = p(ball)vgh

1000(d) = 750(11)

d = .12 meters?

i really appreciate the help
 
islanderfan said:
ok so since there's potential at the top and nothing(?) at the end, can i say

p(water)vg(d) = p(ball)vgh
That "h" has to be the total change in height, not just the height above the water surface.
 
Doc Al said:
That "h" has to be the total change in height, not just the height above the water surface.

ok that makes sense, but won't that leave me with two unknowns?
 
  • #10
or can i make h = 11+d?

but when i did that i got a final d of 33 meters. this seems way too large to me
 
  • #11
Looks OK to me. Solve it the other way and compare.
 
  • #12
i just don't understand how it can sink that far...

btw thanks for all the help, i really understand how to solve this type of question now, i appreciate it, thanks!
 
  • #13
islanderfan said:
i just don't understand how it can sink that far...
That's because you have some common sense. :smile:
The problem is the bogus assumption: "Neglect any energy transferred to the water during impact and sinking." That's a huge effect to ignore!
 
  • #14
im still confused, my height above water was 12 instead of 11, and i tried what he did, with 1000(d)=750(12+d) because h=12+d... i got my h to be 48 and d to be 36... i tried both those answers and neither worked. his was the correct answer and I am pretty sure what i did is what he did... how did he get 33 for his final answer?

nvmmmm i used the wrong coefficients!
 
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