A small sphere 0.60 times as dense as water is dropped from a height 8.2 m above the surface of a smooth lake?

Question

A small sphere 0.60 times as dense as water is dropped from a height 8.2 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Disregard any energy transferred to the water during impact and sinking.

Answer 1

This is a very interesting, and fun (believe it not), problem and utilizes many import concepts. Likewise it can be solved using a number of different methods.
I will go about solving the problem using conservation of energy.


The conversion between Kinetic Energy (KE) and Potential Energy (PE) is critical.

KE = 1/2 mv^2
Where m is the mass of the object and v is the speed of the object.

Gravitational Potential Energy = mgh
Where m is again the mass of the object, g is the gravitational acceleration experienced by the object and h is the height of the object above the ground.
This is a very nice formula and I think it is so nice that I will generalize it a bit more so that I can use it more than once in this problem.
PE = m * a * d
Where a is just some acceleration the object is experiencing, not necessarily due to gravity (you will see where I am going with this), and d is some distance above the “zero” potential point.

In fact, now what we have this more generalized form of Potential Energy, we don’t even need to use Kinetic Energy any more if we don’t feel like it as it is just an unnecessary middle step.


In the initial conditions of the problem the small sphere is some height, h, above the water, so the sphere has some amount of gravitational Potential Energy.
As the sphere falls, this Potential Energy is converted into Kinetic Energy. Once the sphere hits the water it will still continue to travel downward, but since it is less dense than water, there will be a net force acting upward on the sphere. Eventually, this net upward, buoyant, force will cause the sphere to reach some maximum depth and start traveling upward again toward the surface.
When the sphere is at its maximum depth, you might (and I will) say that the sphere has a different type of Potential Energy…we will call it the buoyant Potential Energy for lack of a better term. The deeper the sphere descends into the water, the more and more buoyant PE it gains at the expense of its original Kinetic Energy as it enters the water.

All of the initial Potential Energy the sphere has when it is still hovering in the air above the water is converted into ‘Buoyant’ Potential Energy when the sphere is at its lowest point underwater.

We already know how to find the sphere’s initial [gravitational] Potential Energy,
PE = mgh
Where g = 9.81 m/s^2 downward due to gravity.
We are given a value for h,
h = 8.2 meters
But we do not know the sphere’s mass…but this is OK, we will find that it will cancel out in the end anyway.
For now we will just leave the value of PE as PE = mgh.

Now to find the value for the sphere’s PE when it is at its lowest point in the water using our more general PE equation,
PE = m * a * d
Where a is the net acceleration the sphere feels while in the water and d is the maximum depth underwater (and m is the mass).
We are trying to solve for h and we do not know a.

To find a we need to find the differences in the accelerations (caused by the two forces acting on the sphere) which are acting on the sphere. Gravity is of course acting downward on the sphere to accelerate it downward at what we will here call “negative” 9.81 m/s^2. But the water is pushing the ball upward due to the buoyant force at some other rate. The buoyant force acting on the sphere is equal to the mass of the displaced water by the sphere which is related to the density of the sphere.
We know the sphere has a density which is equal to 6/10 that of water, so for a given volume, the sphere has a mass of 0.60 that of the water’s mass (water’s density = 1 kg/L). This means that the buoyant force acting upward on the object will cause an acceleration of g/.6 upward on the object…or 16.35 m/s^2.
So you have a force causing a 9.81 m/s^2 acceleration downward and a force causing a 16.35 m/s^2 acceleration upward…making for an overall acceleration of 6.54 m/s^2 upward. This is our “a” value for the buoyant Potential Energy which we can write as PE = m * a * d.


So since the initial gravitational PE gets converted into Buoyant PE we can sets these two energy values equal to each other,
m * g * h = m * a * d
we can see that the mass term cancels our on both sides so we get,
g * h = a * d, and we can re-arrange to solve for d,
d = g * h / a
we know the values for g, h, and a, so now we just plug in and solve.

d = (9.81 m/s^2) * (8.2 m) / (6.54 m/s^2)
d = 12.3 meters
So the object descends to a maximum depth of 12.3 meters beneath the water’s surface.



EDIT:
“Mihir Durve”,

When the sphere is in the air it is surrounded by air pressure. The air is exerting a force on the sphere from all sides. In the top the sphere is being ‘pushed’ down by the air and on the bottom the sphere is being ‘pushed’ up (we need not concern ourselves with the horizontal component of any push). If we assume the sphere is incompressible…the only way for the sphere to respond to these pushes would be to accelerate. However, there is a push up from the bottom and a push down from the top, how do we know what the net push is? The net push is determined by the difference in pressure between the top and bottom of the sphere. Pressure decreases with altitude so the air pressure on the top would be less than the air pressure on the bottom. So the sphere would be getting a slight push upward due to the air pressure. However, we were told that this is a small sphere, and being a small sphere that means that the difference in altitude between the top and bottom of the sphere would be very small and thus the difference in pressure would be very small…so small that I would so we could neglect it.

A similar argument could be made for when the sphere is in the water and experiences a difference in water pressures between the top and bottom. But again, if the sphere is small (and we are told that it is), then this force would be negligible compared to that of the sphere’s weight and buoyant force.

Answer 2

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Answer 3

Very good mrjeffy321 but as that ball will go down and down the hydrostatic pressure will also push the ball downward as we know that pressure is h*d*g where h is changing. So we are required to take this under consideration. Also atm pressure will first push the ball in to the water and then will oppose the ball to come up. so what should be done??