Ping-Pong Ball Submerged in Water and then Released

In summary, the problem involves a ping-pong ball being pushed down to a depth of 1m below the surface of a swimming pool and then released. The question asks how high above the surface the ball will shoot as it emerges from the water, assuming a density of 0.25 that of water and neglecting drag force and energy transfer to splashing and waves. The solution involves using Archimedes' Principle and Bernoulli's Equation to determine the buoyant force, mass, weight, and force balance on the ball in terms of its volume (which cancels out in the end). The final result is an acceleration of 0.75g, but it is unclear if the ball will actually travel above the water due
  • #1
ColtonCM
33
2

Homework Statement



Direct problem statement: Imagine you are playing with a Ping-Pong ball in a swimming pool. You push the ball down to a depth of 1 m below the surface, then release it. How high above the water surface will the ball shoot as it emerges from the water? Assume the ball has a density of 0.25 that of water. You can neglect any drag force as well as any transfer of energy to the splashing and waves produced by the emerging ball.

Given variables: 1m below the surface. The density of the ball is 0.25(density of water).

Homework Equations



Total pressure = pressure above surface + (density)(g)(h)

Maybe Archimedes' Principle? Buoyant force = (mass of water displaced)(g)

Maybe Bernoulli's Equation? p + 1/2(density)(v^2) + (density)(g)(h)

The Attempt at a Solution



I'm struggling with setting this problem up. I'm thinking that if I can calculate the buoyant force, that will go a long way to figuring out how high above the water the ball will travel, but I'm not sure how to determine that force when I"m not given the mass of the ping-pong ball. I can't calculate its mass from its density because I'm not given the volume that the ball occupies or the volume of water in the pool. I suppose Bernoulli's equation could give me a constant, but I'm not sure how that would factor into this question. Any help with getting me started on this problem would be appreciated. If I can get my foot in the door I'm pretty sure the rest will follow easily.
 
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  • #2
Let V be the volume of the ping pong ball. In terms of V, what is the buoyant force acting on it? In terms of V, what is the mass of the ping pong ball? In terms of V, what is the weight of the ping pong ball? In terms of V, what is the force balance on the ping pong ball?

Chet
 
  • #3
Chestermiller said:
Let V be the volume of the ping pong ball. In terms of V, what is the buoyant force acting on it? In terms of V, what is the mass of the ping pong ball? In terms of V, what is the weight of the ping pong ball? In terms of V, what is the force balance on the ping pong ball?

Chet

Forgive me, but I'm unsure what you mean. Are you saying to take Density = (Mass)/(V) where V is the volume of the ping-pong ball, solve for V so that it is in terms of mass and density, then substitute that into something else to eliminate the V variable entirely? That would still leave mass of the ball which is unknown which is needed for buoyant force equation.

Thanks,

Colton
 
  • #4
ColtonCM said:
Forgive me, but I'm unsure what you mean. Are you saying to take Density = (Mass)/(V) where V is the volume of the ping-pong ball, solve for V so that it is in terms of mass and density, then substitute that into something else to eliminate the V variable entirely? That would still leave mass of the ball which is unknown which is needed for buoyant force equation.

Thanks,

Colton
I don't mean this at all. I'm going to answer my own first question, and then leave it up to you to answer my other questions. But, in the end, the volume of the ball is going to cancel out.

Buoyant force on ball: F = ##ρ_wgV##, where ρw is the density of water, g is the acceleration of gravity, and V is the volume of the ball.

Now, you need to supply equations (in terms of V) for:

Mass of ball: M = ?

Weight of ball: W = ?

Force Balance: F - W = ?

Chet
 
  • #5
Chestermiller said:
I don't mean this at all. I'm going to answer my own first question, and then leave it up to you to answer my other questions. But, in the end, the volume of the ball is going to cancel out.

Buoyant force on ball: F = ##ρ_wgV##, where ρw is the density of water, g is the acceleration of gravity, and V is the volume of the ball.

Now, you need to supply equations (in terms of V) for:

Mass of ball: M = ?

Weight of ball: W = ?

Force Balance: F - W = ?

Chet

Oh okay, let me attempt to complete your initial statements then. Buoyant force =

F = ρwgV

Mass of ball: m = 0.25ρV

W = 0.25ρgV (from W = mg).

Force Balance: ρwgV - 0.25ρgV = 0.75pgV

Is this how you meant?

Thanks,

Colton
 
  • #6
ColtonCM said:
Oh okay, let me attempt to complete your initial statements then. Buoyant force =

F = ρwgV

Mass of ball: m = 0.25ρV

W = 0.25ρgV (from W = mg).

Force Balance: ρwgV - 0.25ρgV = 0.75pgV

Is this how you meant?

Thanks,

Colton
Yes. Now set that equal to Ma, and solve for a. You will see that V cancels out.

Chet
 
  • #7
Setting it equal gives 0.75g after both density and V cancel out, which equals 7.35m/s2. I am good so far. How about relating this acceleration to how high above the surface the ball will travel once released? Or is it a trick question (my concepts here might be wrong but bear with me) and the ball will not travel above the water at all because the acceleration upwards is not greater than the acceleration downwards due to gravity once above the water where there is no buoyant force (or rather it is negligible from the atmosphere). Is there an equation I am missing involving acceleration? Try to be as vague as possible, I realize I'm getting walked through this problem pretty heavily.

Thanks,

Colton
 
  • #8
ColtonCM said:
Setting it equal gives 0.75g after both density and V cancel out, which equals 7.35m/s2. I am good so far. How about relating this acceleration to how high above the surface the ball will travel once released? Or is it a trick question (my concepts here might be wrong but bear with me) and the ball will not travel above the water at all because the acceleration upwards is not greater than the acceleration downwards due to gravity once above the water where there is no buoyant force (or rather it is negligible from the atmosphere). Is there an equation I am missing involving acceleration? Try to be as vague as possible, I realize I'm getting walked through this problem pretty heavily.

Thanks,

Colton
Something does not seem quote right regarding your 0.75g value for acceleration. Try that again. This template might help (similar to Chestermiller's):

What is the mass of the ball in terms of V and ρw?

What is the ball's weight in terms of V, ρw and g?

What is the weight of the displaced water in terms of V, ρw and g?

What is the net force acting on the ball in terms of V, ρw and g?

What is the acceleration of the ball? (use Newton's second law for this)

Can you think of a kinematics equation that relates velocity, acceleration and displacement?

That should get you out of the water at least. Then use another kinematics equation once out of the water.

[Edit: alternatively, you could use the work-energy theorem and conservation of energy instead of the kinematics equations if you choose. It's up to you. For what it's worth though, this work-energy theorem/conservation of energy approach leads to a much simpler and more elegant solution -- no need to muck around with kinematics or Newton's second law.]
 
Last edited:
  • #9
collinsmark said:
Something does not seem quote right regarding your 0.75g value for acceleration. Try that again. This template might help (similar to Chestermiller's):

What is the mass of the ball in terms of V and ρw?

What is the ball's weight in terms of V, ρw and g?

What is the weight of the displaced water in terms of V, ρw and g?

What is the net force acting on the ball in terms of V, ρw and g?

What is the acceleration of the ball? (use Newton's second law for this)

Can you think of a kinematics equation that relates velocity, acceleration and displacement?

That should get you out of the water at least. Then use another kinematics equation once out of the water.

[Edit: alternatively, you could use the work-energy theorem and conservation of energy instead of the kinematics equations if you choose. It's up to you. For what it's worth though, this work-energy theorem/conservation of energy approach leads to a much simpler and more elegant solution -- no need to muck around with kinematics or Newton's second law.]

I followed Chet's template exactly and he okay'd my results. I did locate the error in the acceleration though.

Copy and pasted from above:

Mass = 0.25ρV
Weight = 0.25ρgV
Buoyant force = ρwatergV
Weight of water = ρgV
The net force is the buoyant force, going upwards.
So Newton's second law: F = ma where m is expressed in terms of the ball. ρwatergV - 0.25ρwatergV = 0.75ρgV
Now 0.75ρgV = ma from Newton's second law. So 0.75ρgV = 0.25ρV = 3g = 29.4m/s2.

I'm familiar with the equations of kinematic motion but have no clue which to use, or if I have to solve for certain quantities in some and then substitute into others. Here are the equations:

v=v0+at
x-x0=v0t + 1/2(at2)
v2=v02+2a(x-x0)
x-x0=1/2(v0+v)t
x-x0=vt-1/2(at2)

All I have is acceleration. And maybe displacement of 1m before leaving the water? But no velocity or time, so should I solve for velocity and time in terms of acceleration and displacement then sub that into whatever equation works? Also, is x0simply = 0?

Thanks,

Colton
 
  • #10
ColtonCM said:
I followed Chet's template exactly and he okay'd my results. I did locate the error in the acceleration though.

Copy and pasted from above:

Mass = 0.25ρV
Weight = 0.25ρgV
Buoyant force = ρwatergV
Weight of water = ρgV
The net force is the buoyant force, going upwards.
So Newton's second law: F = ma where m is expressed in terms of the ball. ρwatergV - 0.25ρwatergV = 0.75ρgV
Now 0.75ρgV = ma from Newton's second law. So 0.75ρgV = 0.25ρV = 3g = 29.4m/s2.
That's more like it. :smile: Good job.

I'm familiar with the equations of kinematic motion but have no clue which to use, or if I have to solve for certain quantities in some and then substitute into others. Here are the equations:

v=v0+at
x-x0=v0t + 1/2(at2)
v2=v02+2a(x-x0)
x-x0=1/2(v0+v)t
x-x0=vt-1/2(at2)

All I have is acceleration. And maybe displacement of 1m before leaving the water?
Yes, you have acceleration and displacement (the displacement over which the acceleration applies).

But no velocity or time, so should I solve for velocity and time in terms of acceleration and displacement then sub that into whatever equation works?
You could! That's one way to do it. (There's not much reason to solve for time, but you could pick an appropriate equation to solve for velocity.)

Also, is x0simply = 0?

That decision is up to you. It depends on how you define your variables and how you set up the problem.

----------------

There's another way to solve the problem using the Work-energy Theorem (work is equal to "force times distance", sort-of-thing) and conservation of energy (specifically invoking gravitational potential energy, P.E. = mgh) that you can also use to solve for the final answer. You can do it this way, or the method above using the kinematics equations. Either way gives the same answer.
 
  • #11
collinsmark said:
That's more like it. :smile: Good job.Yes, you have acceleration and displacement (the displacement over which the acceleration applies).You could! That's one way to do it. (There's not much reason to solve for time, but you could pick an appropriate equation to solve for velocity.)
That decision is up to you. It depends on how you define your variables and how you set up the problem.

----------------

There's another way to solve the problem using the Work-energy Theorem (work is equal to "force times distance", sort-of-thing) and conservation of energy (specifically invoking gravitational potential energy, P.E. = mgh) that you can also use to solve for the final answer. You can do it this way, or the method above using the kinematics equations. Either way gives the same answer.

Hmm no matter what I solve for, there is always a v, v0, or t in the equations, making it unsolvable.
 
  • #12
ColtonCM said:
Hmm no matter what I solve for, there is always a v, v0, or t in the equations, making it unsolvable.

You already know what the acceleration is, so there shouldn't be an a left over. You've already calculated what a is.

As far as v0 is concerned, what is the ball's initial velocity the moment it is released from 1 m below the surface?
 
  • #13
The way you outlined doing it will work. You know the acceleration and the initial velocity (zero), so you can calculate the time required to travel 1 meter. Knowing the time, you can calculate the velocity of the ball when it launches from the water surface. Once the ball leaves the water surface, you have a second problem to solve. This is the same problem you would be doing if you threw the ball up into the air at the launch velocity. I'm sure you solved that problem a zillion times.

Chet
 

Related to Ping-Pong Ball Submerged in Water and then Released

What is the purpose of submerging a ping-pong ball in water and then releasing it?

The purpose is to observe the behavior of the ping-pong ball as it floats to the surface after being released from a submerged state. This can help us understand the properties of air and water, as well as the forces acting on the ball.

How does the ping-pong ball behave when submerged in water?

When the ping-pong ball is submerged in water, it initially sinks to the bottom due to the force of gravity. However, as it sinks, it displaces water and experiences an upward buoyant force. This causes the ball to slow down and eventually reach a state of neutral buoyancy where the upward and downward forces are balanced.

What happens when the ping-pong ball is released from a submerged state?

Once the ping-pong ball is released, the upward buoyant force becomes greater than the force of gravity and the ball begins to rise to the surface. As it rises, it accelerates due to the decrease in water pressure and eventually reaches the surface with a constant velocity.

How does the density of the ping-pong ball affect its behavior when released from a submerged state?

The density of the ping-pong ball affects its behavior because it determines how much water it displaces and therefore the strength of the buoyant force acting on it. A denser ball will displace more water and experience a stronger upward force, causing it to rise faster than a less dense ball.

Can the behavior of a ping-pong ball submerged in water and then released be used to study other objects?

Yes, the principles observed in this experiment can be applied to other objects submerged in fluids. This can help us understand the behavior of objects such as ships, submarines, and even marine animals in water. It can also be used to study the behavior of objects in other fluids, such as air or oil.

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