Millikan Experiment Based Marble Mass Homework

[Charles Link]

I think that 3 gram minimum is inaccurate, and is resulting in extra confusion. I estimate the container to be closer to 2 grams.

 

[orangegalaxies]

I think that 3 gram minimum is inaccurate, and is resulting in extra confusion. I estimate the container to be closer to 2 grams.

hmmmm. these are the values my teacher gave me...

 

[hutchphd]

Apologies. You almost got it but you need to choose best values from noisy data:

Largest mass: 31.9 g
Approx weight of one marble: 4 g
Mass of empty container: minimum of 3 g

(31.9 - 3)/4 = 7 marbles in largest container

This is wrong, I don't follow what you are saying, can you please explain?

Can't be 7 marbles by definition. Mass of marble is more like 4.5, Try another guess. It won't be perfect.

Now make guess and plot mass vs # marbles for all data points.

 

[Charles Link]

If you look at the data, at first glance, (I'm sure you observed it as well), there appears to be increments of between 4.5 and 5 grams. There is a jump from 11 to 21, but with a small sample, it's not surprising that the 16 is absent.

 

[orangegalaxies]

Apologies. You almost got it but you need to choose best values from noisy data:
Can't be 7 marbles by definition. Mass of marble is more like 4.5, Try another guess. It won't be perfect.

Now make guess and plot mass vs # marbles for all data points.

Can you explain how you got 4.5 g?

Assuming each marble is 4.5, and the empty container is 3:

Mass : Number of marbles
11.0 g = ~2
11.1 g = ~2
20.6 g = ~4
21.4 g = ~4
21.5 g = ~4
25.7 g = ~5
25.8 g = ~5
25.9 g = ~5
31.6 g = ~6
31.9 g = ~6

 

[orangegalaxies]

If you look at the data, at first glance, (I'm sure you observed it as well), there appears to be increments of between 4.5 and 5 grams. There is a jump from 11 to 21, but with a small sample, it's not surprising that the 16 is absent.

Yeah you're right that is a patterm

 

[Charles Link]

I just spotted something else=they give you the weight of the container to be at least 3 grams=it might even be large enough to make the 11 g have just one marble=my previous post on this being inaccurate and possibly 2 grams was incorrect.

 

[hutchphd]

I think that 3 gram minimum is inaccurate, and is resulting in extra confusion. I estimate the container to be closer to 2 grams.

I disagree I think it is more like 6g. The problem says >3.

 

[Charles Link]

I disagree I think it is more like 6g. The problem says >3.

see my post 42 :)

 

[orangegalaxies]

I just spotted something else=they give you the weight of the container to be at least 3 grams=it might even be large enough to make the 11 g have just one marble=my previous post on this being inaccurate and possibly 2 grams was incorrect.

This is also true yeah

 

[hutchphd]

Yes Now the OP needs to graph the data, draw the best linear fit. Slope gives marble mass, intercept gives container mass

 

[orangegalaxies]

Yes Now the OP needs to graph the data, draw the best linear fit. Slope gives marble mass, intercept gives container mass

graph the differences in masses?

 

[Charles Link]

Graph y=total measured mass vs. x=number of marbles
y value for x=0 gives mass of container.

 

[hutchphd]

Your eyeball and ruler will draw the "best" line pretty well. Gives the best fit to all the data.

 

[orangegalaxies]

is there a way to solve this problem without graphing? my teacher never mentioned graphing, they only said you need to find the averages. they are a pretty bad teacher tbh but i wanted to confirm

 

[hutchphd]

Plot every point and then make the best line ...better than averages (apologies to your teacher).

 

[orangegalaxies]

Your eyeball and ruler will draw the "best" line pretty well. Gives the best fit to all the data.

in your previous post, is there any way you found the marble mass = 4.5 g? calculations i mean

 

[hutchphd]

Rough graph in my head. Do the graph. Excel for graph?
Your head does a good "least squares fit" to the straight line.

 

[orangegalaxies]

Rough graph in my head. Do the graph. Excel?

sure, but as Charles mentioned, y=total measured mass vs. x=number of marbles. To find the number of marbles I need the mass of one so I wanted to know how you can say its 4.5 g

 

[Charles Link]

in your previous post, is there any way you found the marble mass = 4.5 g? calculations i mean

The increments are about 4.5 other than the missing 16. This is a very simple graph to draw by hand=if you have some graph paper it helps. Otherwise EXCEL is often used, but that can take some work to learn.

 

[hutchphd]

You only need the integer number of marbles to plot the data. I guessed the 4.5 because it seemed to give integers within reasonable error limits. The error in the mass obtained from the best fit line (slope) from the graph should be good to a few percent. This slope in the Millikan oil drop experiment gave Millikan the charge increment for little balls of oil, electron by electron as he irradiated them). A truly beautiful experiment.
Incidentally it is possible that all the changes in the graph involved always moving 2 marbles (or3 or4) but it is less and less likely More data would make it more certain.

 

[orangegalaxies]

You only need the integer number of marbles to plot the data. I guessed the 4.5 because it seemed to give integers within reasonable error limits. The error in the mass obtained from the best fit line (slope) from the graph should be good to a few percent. This slope in the Millikan oil drop experiment gave Millikan the charge increment for little balls of oil, electron by electron as he irradiated them). A truly beautiful experiment.
Incidentally it is possible that all the changes in the graph involved always moving 2 marbles (or3 or4) but it is less and less likely More data would make it more certain.

Hello, after graphing, I got a slope of 4.49992 and a y-intercept of 3.00000061. Is this reasonable?

 

[Charles Link]

Hello, after graphing, I got a slope of 4.49992 and a y-intercept of 3.00000061. Is this reasonable?

Please draw the graph. I don't think you graphed it, because I just graphed it, and I got round numbers, but not real close to 4.5, and certainly not 3 for the other number. The accuracy is only one decimal place or thereabouts.

If you want a shortcut to the graph, you can take the n=1 point the n=5 point, (n=number of marbles), and compute the slope between these two points. Once you know the mass of the marble, it is easy to compute the mass of the container without a y-intercept.

The graph is the preferred method=I used paper with 1/4" squares.

 

[hutchphd]

And I think the first data cluster is for n=1 marble so the mass of the container is ~6g. Good that you are on it!

 

[Charles Link]

And I think the first data cluster is for n=1 marble so the mass of the container is ~6g

The graph shows this answer very clearly, and I'm hoping the OP will draw the graph like I did to see this for himself.

 

[Charles Link]

One note about this problem for the OP is that it was important that they stated that the container weighed at least 3 grams, because if the container were only about one gram or thereabouts, the mass readings of about 11 grams could have had two marbles each.
This would not affect the slope of the graph, but would simply shift the x-axis, so that the y-axis intercept (at x=0) would be at 1g instead of 6g. (I encourage you to draw the graph=this should then all make sense).
With a container of at least 3 grams, it became clear that there was only one marble for the 11 gram data points.
Looking at the graph (at x=0) is then the simplest way to then get this 6g answer for the mass of the container.
I originally thought that n=2 for this case of 11 grams, but that was my mistake.

 

[orangegalaxies]

thanks guys, i don't have any graph paper on me so i tried to graph it digitally which didn't work apparently. i will get some today to finish it off. would you be able to help me answer these questions?
- Would having more containers make this easier?: i think so since it would give more data right?
- Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?
- How does this relate to Millikan's experiment?
- What are some sources of error?: damage to either the container/marbles which would alter their mass?

 

[orangegalaxies]

also can you please explain the average method if i can't find any graph paper?

 

[PeroK]

thanks guys, i don't have any graph paper on me so i tried to graph it digitally which didn't work apparently. i will get some today to finish it off. would you be able to help me answer these questions?
- Would having more containers make this easier?: i think so since it would give more data right?
- Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?
- How does this relate to Millikan's experiment?
- What are some sources of error?: damage to either the container/marbles which would alter their mass?

To be honest, in the 63 posts so far, there have not been many contributions from you. You need to show a better grasp of the problem yourself.

 

[orangegalaxies]

To be honest, in the 63 posts so far, there have not been many contributions from you. You need to show a better grasp of the problem yourself.

okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute. i can only grasp the problem if someone explains which is not the case since it seems more like everyone is asking me why I'm not doing this a certain way instead of telling me WHY AND HOW i am supposed to do it that way.

 

[PeroK]

okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute

You need to take responsibility for your own academic progress. I'm not convinced that it's your teacher's fault - or the people who've posted ideas or suggestions here.

 

[Charles Link]

okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute. i can only grasp the problem if someone explains which is not the case since it seems more like everyone is asking me why I'm not doing this a certain way instead of telling me WHY AND HOW i am supposed to do it that way.

Please read my post 58. There is a shortcut presented there that you can do without graph paper that is reasonably accurate.

 

[orangegalaxies]

You need to take responsibility for your own academic progress. I'm not convinced that it's your teacher's fault - or the people who've posted ideas or suggestions here.

i'm not saying it's their fault. my teacher doesn't offer any support to their students, i can't teach myself everything through youtube videos and tutorials. i need someone to explain things to me which is why i came here. i am new to these concepts, if someone tells me something once, i won't just take it to be correct, i need to understand why that method was used which takes time.

 

[PeroK]

i'm not saying it's their fault. my teacher doesn't offer any support to their students, i can't teach myself everything through youtube videos and tutorials. i need someone to explain things to me which is why i came here. i am new to these concepts, if someone tells me something once, i won't just take it to be correct, i need to understand why that method was used which takes time.

Here's the first thing I said on this thread:

This is perhaps a tough question, as you may actually have to think about what to do.

We can't think for you: you have to do that yourself.

 

[orangegalaxies]

Here's the first thing I said on this thread:
We can't think for you: you have to do that yourself.

yes i don't need you to, thanks.