Accounting for Air Resistance in Conservation of Energy Lab with Bouncing Ball

[nbrady]

Homework Statement

Ball mass = 0.0027 kg
PE (0.5 meters) = 0.01323 J
height of ball after three trials = 0.19 meters
KE before bounce = 0.01323 J
PE at new max height = 0.00503
Total energy = KE + PE
Ke after bounce = ?

Homework Equations

mgh

The Attempt at a Solution

I did a lab where I bounced a ball at a certain height. I found the potential energy with the equation mgh (m = 0.0027 g, g=9.8, h=0.5m), so it is 0.01323J. I am supposed to find the kinetic energy before the ball bounced (I think it's equal to the PE that I calculated), and I'm supposed to find the PE at the maximum height the ball hit (h = 0.19 m), so when I plug it into the mgh equation, I get 0.00503J. I know the total energy would be KE before the bounce + PE at the maximum height, which was 0.00503J, so I just add those up. How do I find the KE after the ball bounced?

 

[gneill]

Hi nbrady, Welcome to Physics Forums.

You should describe the lab procedure in your problem statement so that the information you've listed has a context that helpers can understand. Remember, no one here (most likely) has performed the lab with you, so we don't know what you've done, what the instructions were, or how/what data was collected. For example, you write, "height of ball after three trials = 0.19 meters". What is a trial? Is it an important distinction for the given data?

What is the question that you need to answer or value you need to find? Is it the kinetic energy after one bounce? Or 3 bounces (again, what is a "trial"?).

Under relevant equations you've listed one expression but no equations. mgh is not an equation. What does it represent? Are there other equations that might be of interest for this problem? How about conservation of energy relating kinetic energy to potential energy?

I did a lab where I bounced a ball at a certain height. I found the potential energy with the equation mgh (m = 0.0027 g, g=9.8, h=0.5m), so it is 0.01323J. I am supposed to find the kinetic energy before the ball bounced (I think it's equal to the PE that I calculated),

You gave the mass of the ball to be 0.0027 g, which is 2.7 x 10^-6 kg. If the height in the potential energy equation is just 0.5 m, the resulting potential energy would be much smaller than the value you've calculated -- something like 1.3 x 10^-5 J. You might want to verify your specification for the mass of the ball. Your value seems awfully small for a typical lab situation. Perhaps the units for the mass should have been kg rather than grams?

and I'm supposed to find the PE at the maximum height the ball hit (h = 0.19 m), so when I plug it into the mgh equation, I get 0.00503J. I know the total energy would be KE before the bounce + PE at the maximum height, which was 0.00503J, so I just add those up. How do I find the KE after the ball bounced?

Again, I think your mass units are suspect so your energy values are likely off by several orders of magnitude. Even so, you should rethink your energy computations. When a ball with real elastic properties bounces it tends to lose some energy during the time it is in contact with the floor (heat dissipation from the flexing of the material mostly). So you wouldn't expect energy conservation (KE + PE) to hold over that period of contact. But before and after the ball is being compressed, energy conservation would apply (so while it is falling and while it is rising, not in while it's in contact with the floor). So most likely they are wanting you to find the KE immediately before and after the ball is in contact with the floor. Conservation of energy will apply for the trip the ball makes from the initial drop height to first contact with the floor, and the trip it takes leaving the floor to its new maximum height (some energy being lost during the bounce itself).

 

[Simon Bridge]

I did a lab where I bounced a ball at a certain height. I found the potential energy...

... of what? Where?
When you are trying to understand what to do, it is often useful to be specific about what you say.
What do you mean when you say you "bounced a ball at a certain height"?
Do you mean you bounced a ball at the top of the bulding as opposed to at ground level?
Were you required to bounce the ball so it's max height it traveled up to was 0.5 meter or something like that?

... with the equation mgh (m = 0.0027 g, g=9.8, h=0.5m), so it is 0.01323J.

If you took the floor to be PE=0, then you have calculated the change in potential energy after the ball has traveled from the floor to a height 0.5m up from the floor.
Note: 0.0027g is a very small mass for this sort of experiment.
Are you sure that's not 2.7g (which would be 0.0027kg - a ping-pong ball).

I am supposed to find the kinetic energy before the ball bounced (I think it's equal to the PE that I calculated),

Is that a good assumption?
If you air resistance is small, then the kinetic energy of the ball just after the bounce would be very close to the PE of the ball at the top of the bounce.
You have to think in terms of where the energy goes and what sort of collision the ball makes with the ground: ie. is it an elastic collision?

... and I'm supposed to find the PE at the maximum height the ball hit (h = 0.19 m), so when I plug it into the mgh equation, I get 0.00503J. I know the total energy would be KE before the bounce + PE at the maximum height,

No it isn't. The total energy is the KE+PE at all times - and this is a constant for all closed systems. That is what conservation of energy means.

How do I find the KE after the ball bounced?

When thinking of conservation of energy problems, start out by listing the energy transformations:
What is happening is that you throw (or drop) the ball ... gravitational PE (+ any energy from your arm) gets exchanged for kinetic energy until the ball hits the ground. At which point the ball experiences a complicated set of forces that result in a bounce - the KE just before the impact gets turned into KE after the bounce, as well as elastic PE in the ball structure, oscillations, heat, and sound. ie. not all the initial energy ends up as kinetic energy after the bounce.
Having bounced, the KE the ball has left gets converted to PE in gravity, and work against air resistance. Neglecting air resistance, when gravitational PE is equal to the initial KE, the ball stops - this is the max height.
At the max height, KE=0, and PE=Emax.
Just after the bounce, KE=Emax, and PE=0
See how the total energy is a constant?

 

[nbrady]

Sorry everyone,

Thank you for trying to help me. Let me try this again. This was the procedure to my lab:

1. Find a room with a hard, flat surface that you will be able to drop a ball on.
   Hint: The harder the surface the better.
2. Using the tape measure, measure 0.50 meters above the flat surface. Use masking tape to secure the
   tape measure to the wall so you will be able to read the height of the ball.
3. Take the ping pong ball and place the bottom of the ball at the 0.50 meter mark.
4. Drop the ball and record the height the bottom of the ball reaches after one bounce in Table 2.
5. Repeat Steps 3 - 4 two more times for the ping pong ball.
6. Repeat Steps 3 - 5 for two other balls of your choice.

My data was for the three trials of 3 different kinds of balls:
For example for the ping pong ball:
Trial 1 - 18cm
Trial 2 - 19cm
Trial 3 - 20cm
Average for all 3 trials = 19 cm = 0.19 m, so I use this value for calculations.

The lab gave a chart and the weight of the ping pong ball according to the chart was 0.0027 kg.

Then, we have to fill out a chart. We have to find the potential energy at 0.5 meters, which I got to be 0.0027 kg * 9.8 m/s^2 * 0.5 meters = 0.01323 J.
Then it asks to find the KE before the ball bounced (I don't think I got that right, because I thought it was equal to PE at 0.5 meters).
From here, I have to calculate the potential energy at 0.19 m (which is the average of my three trials)
I then have to calculate the KE after the ball bounced.
Then, I have to get the total energy.

I hope I fully explained this for everyone

Hi nbrady, Welcome to Physics Forums.

You should describe the lab procedure in your problem statement so that the information you've listed has a context that helpers can understand. Remember, no one here (most likely) has performed the lab with you, so we don't know what you've done, what the instructions were, or how/what data was collected. For example, you write, "height of ball after three trials = 0.19 meters". What is a trial? Is it an important distinction for the given data?

What is the question that you need to answer or value you need to find? Is it the kinetic energy after one bounce? Or 3 bounces (again, what is a "trial"?).

Under relevant equations you've listed one expression but no equations. mgh is not an equation. What does it represent? Are there other equations that might be of interest for this problem? How about conservation of energy relating kinetic energy to potential energy? You gave the mass of the ball to be 0.0027 g, which is 2.7 x 10^-6 kg. If the height in the potential energy equation is just 0.5 m, the resulting potential energy would be much smaller than the value you've calculated -- something like 1.3 x 10^-5 J. You might want to verify your specification for the mass of the ball. Your value seems awfully small for a typical lab situation. Perhaps the units for the mass should have been kg rather than grams?

Again, I think your mass units are suspect so your energy values are likely off by several orders of magnitude. Even so, you should rethink your energy computations. When a ball with real elastic properties bounces it tends to lose some energy during the time it is in contact with the floor (heat dissipation from the flexing of the material mostly). So you wouldn't expect energy conservation (KE + PE) to hold over that period of contact. But before and after the ball is being compressed, energy conservation would apply (so while it is falling and while it is rising, not in while it's in contact with the floor). So most likely they are wanting you to find the KE immediately before and after the ball is in contact with the floor. Conservation of energy will apply for the trip the ball makes from the initial drop height to first contact with the floor, and the trip it takes leaving the floor to its new maximum height (some energy being lost during the bounce itself).

 

[Simon Bridge]

OK I got you ... see last paragraph post #3.
You drop the ball from height h above the ground - and it hits the floor.
It starts with KE=0 and PE=mgh so total energy E=KE+PE=mgh
Ignoring air resistance, all the PE gets converted to KE at the floor (because you defined it to have PE=0 at the floor level)
Therefore $\frac{1}{2}mu^2=mgh$ ... where u is the speed of the ball just before it hits the ground.
... see how this works?

If v is the speed of the ball just after the bounce, just as it leaves the ground, and it flies up to max height, after the bounce, of y, then what is the relationship between v and y?

 

[nbrady]

Would v and y be equal??

OK I got you ... see last paragraph post #3.
You drop the ball from height h above the ground - and it hits the floor.
It starts with KE=0 and PE=mgh so total energy E=KE+PE=mgh
Ignoring air resistance, all the PE gets converted to KE at the floor (because you defined it to have PE=0 at the floor level)
Therefore $\frac{1}{2}mu^2=mgh$ ... where u is the speed of the ball just before it hits the ground.
... see how this works?

If v is the speed of the ball just after the bounce, just as it leaves the ground, and it flies up to max height, after the bounce, of y, then what is the relationship between v and y?

 

[nbrady]

Let me see if I understand this, based on your third paragraph. The PE at 0.5 meter (because that what I had to measure first) would be (mgh) 0.0027 kg * 9.8 m/s^2 * 0.5 meters = 0.01323 J. KE before the bounce would be 0 because it hasn't bounced yet. The potential energy at the max height (which is 0.19m), would be (mgh) 0.0027 kg * 9.8 m/s^2 * 0.19 meters = 0.00503 J. I don't think I know the KE after the bounce. I don't think I understand that part.

OK I got you ... see last paragraph post #3.
You drop the ball from height h above the ground - and it hits the floor.
It starts with KE=0 and PE=mgh so total energy E=KE+PE=mgh
Ignoring air resistance, all the PE gets converted to KE at the floor (because you defined it to have PE=0 at the floor level)
Therefore $\frac{1}{2}mu^2=mgh$ ... where u is the speed of the ball just before it hits the ground.
... see how this works?

If v is the speed of the ball just after the bounce, just as it leaves the ground, and it flies up to max height, after the bounce, of y, then what is the relationship between v and y?

 

[haruspex]

KE before the bounce would be 0 because it hasn't bounced yet

No, if the KE is zero it isn't moving, so that would only be at a maximum height, either when initially dropped or after a bounce.
At any instant, there is a total mechanical energy, the sum of its PE and KE. If we take the zero GPE (gravitational PE) as being at ground level, then the GPE is mgh.
Initially, there is GPE but no KE. As it falls, the KE increase and the GPE reduces, the total remaining almost constant.
When it hits the floor, all the GPE has been turned KE. During the bounce, the KE turns into elastic energy and back into KE, but moving upwards now, because it is not perfectly elastic, so energy is lost.
As it rises, remaining KE is turned back into GPE, the total remaining almost constant again. At max height, all that remaining mechanical energy is back to GPE.

 

[CWatters]

On the way down..

The Ball starts with some PE that is converted to KE as it falls. If we ignore air resistance then the KE just before impact with the ground is the same as the initial PE.

During the collision some unknown quantity of KE will be lost (to sound and heat).

On the way back up...

The ball leaves the ground with some unknown KE. That KE is converted to PE on the way up. If we ignore air resistance then the PE at the top will be equal to the KE just as it left the ground.

 

[nbrady]

Ok...NOW let me see if I understand this. I had to read your post a couple of times, because I suck at physics.

On the way down, the ball has PE which would be mgh = 0.0027 * 9.8 * 0.5 meters (it was dropped from this height) = 0.01323 J. The PE gets converted to KE as it falls (before it bounces), so this KE would also equal 0.01323 J.

On the way back up, the ball has KE that is converted to PE. This new PE is equal to the KE that occurred after the bounce, so in this case, PE = 0.0027 * 9.8 * 0.19 (new max height) = 0.00503J, and this value equals the KE after the bounce.

Please tell me I am understanding this correctly now.

On the way down..

The Ball starts with some PE that is converted to KE as it falls. If we ignore air resistance then the KE just before impact with the ground is the same as the initial PE.

During the collision some unknown quantity of KE will be lost (to sound and heat).

On the way back up...

The ball leaves the ground with some unknown KE. That KE is converted to PE on the way up. If we ignore air resistance then the PE at the top will be equal to the KE just as it left the ground.

 

[NascentOxygen]

Yes, that's right.

HOWEVER, 2.7 grams, if correct, would make it a featherweight ball. A 3 cm diameter "super ball" I have on my desk is 45 grams.

 

[nbrady]

Yeah, it's a ping pong ball and that mass was given

Thank you so much, everybody! I finally get it!

Yes, that's right.

HOWEVER, 2.7 grams, if correct, would make it a featherweight ball. A 3 cm diameter "super ball" I have on my desk is 45 grams.

 

[haruspex]

Yeah, it's a ping pong ball and that mass was given

Hmm... that makes our descriptions a little inaccurate, perhaps. A ping pong ball would lose quite a bit to drag on the way down and on the way up.

 

[CWatters]

+1

Your understanding in post #10 is correct but only if we ignore air resistance. That might be reasonable with a solid rubber ball but with a relatively light ball like a ping pong ball that's a bit optimistic. Did the lab instructions say anything about ignoring air resistance/drag?