Millikan Experiment Based Marble Mass Homework

[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.

 

[hutchphd]

I agree with @PeroK . It was my intention to give you a push in the correct direction and I believe that mission is accomplished. Much of the rest you need to investigate and synthesize on your own. We will be very glad to answer specific questions but "how do I do part 3 ?" is not what I mean.
Milliken's work is legend and part of every physics curriculum. As a Sophomore undergrad lab I did his entire experiment (oil, atomizer spray, beta source etc) and it was extraordinary for me. Much literature exists

 

[orangegalaxies]

i figured out how to do it the average way and i think it's correct. i will stick to this method since i couldn't get any graph paper and it's more numerical in any case so it makes more sense. thanks for all your help. i have to answer the questions still. i read my textbook but to no avail. i will probably end up making a guess for each of them. it's cool that you actually did the lab. i am still in high school and learning online so i don't have access to the kind of resources you guys did which is frustrating since these concepts are fundamental for college.

 

[Charles Link]

What did you get for an average/estimated mass of the marble? The graph shows our "guess/first look" of 4.5 to be somewhat on the low side. Using the $n=1$ and $n=5$ points as mentioned in post 58 will give a much more accurate number. This method then computes the mass of 4 marbles using these two points.

You could also similarly compute the mass of two marbles or three marbles using the other points.

The graph is perhaps one of the best ways to compute this, but you need graph paper. I'd be glad to compare your answer to what I got from the graph.

 

[orangegalaxies]

What did you get for an average/estimated mass of the marble? The graph shows our "guess/first look" of 4.5 to be somewhat on the low side. Using the $n=1$ and $n=5$ points as mentioned in post 58 will give a much more accurate number. This method then computes the mass of 4 marbles using these two points.

You could also similarly compute the mass of two marbles or three marbles using the other points.

The graph is perhaps one of the best ways to compute this, but you need graph paper. I'd be glad to compare your answer to what I got from the graph.

i got the marble to be around 5.17 g and used this to find the empty container which was 5.83 g

 

[Charles Link]

i got the marble to be around 5.17 g and used this to find the empty container which was 5.83 g

Sounds good. Looks like you took ((31.6+31.9)/2-(11.0+11.1)/2)/4 =5.17, and 11.0-5.17=5.83, but in the future please try to show us a little work.

The graph gives very nearly 5.0 grams for the marble mass and 6.0 grams for the container, but your answer is perfectly valid.

You might want to try the other points and see what they give as well, e.g. n=1 vs. n=3 for the mass of two marbles.

 

[orangegalaxies]

Sounds good. Looks like you took ((31.6+31.9)/2-(11.0+11.1)/2)/4 =5.17, and 11.0-5.17=5.83, but in the future please try to show us a little work.

The graph gives very nearly 5.0 grams for the marble mass and 6.0 grams for the container, but your answer is perfectly valid.

You might want to try the other points and see what they give as well, e.g. n=1 vs. n=3 for the mass of two marbles.

i actually found the average mass for the containers that might have had the same number of marbles (11,11.1 and 21.4, 21.5 and 25.7, 25.8, 25.9 and 31.6, 31.9), found their differences, divided the jump from 11 to 21 by 2 and then averaged the three differences. this probably makes no sense lol but yeah your method does work too.

 

[orangegalaxies]

is it okay if i run what my guesses for the additional questions are by you for feedback?

 

[Charles Link]

is it okay if i run what my guesses for the additional questions are by you for feedback?

Yes, but as others have also said, we really need more effort from you. e.g. You describe a method, but we need to see the calculations.
If you averaged all of the other points as well, I would expect it to be closer to the graphical result.

This is where you need to make an effort, and not have me or the other Homework Helpers writing out the steps that you should be coming up with.

 

[orangegalaxies]

Yes, but as others have also said, we really need more effort from you. e.g. You describe a method, but we need to see the calculations.
If you averaged all of the other points as well, I would expect it to be closer to the graphical result.

This is where you need to make an effort, and not have me or the other Homework Helpers writing out the steps that you should be coming up with.

i ran my method by a classmate and they had an answer a decimal point off by mine. i did my work by myself in a way that made sense to me, I'm sorry, i should have explained it for everyone too. as for the questions:
Would having more containers make this easier?: i believe it will make it easier since more data means you can find a pattern in the mass change more easily
Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?: I'm actually not sure on this question... would it be harder because it might be tougher to find the mass of the empty container?
How does this relate to Millikan's experiment?: multiples of a smallest value
What are some sources of error?: human error, marbles may be wet/chipped

 

[Charles Link]

We need you to show calculations. Your teacher should also make you show your work, and not simply give credit for supplying a number.
For sources of error, the biggest one is probably inaccuracy of the scale.
Here's where like the other Homework Helpers have stated, we need more effort from you.
For starters, I still think you would do well to do what you need to do to obtain a couple pieces of graph paper, and graph the data.

 

[orangegalaxies]

We need you to show calculations. Your teacher should also make you show your work, and not simply give credit for supplying a number.
For sources of error, the biggest one is probably inaccuracy of the scale.
Here's where like the other Homework Helpers have stated, we need more effort from you.
For starters, I still think you would do well to do what you need to do to obtain a couple pieces of graph paper, and graph the data.

yes I've shown my work in the assignment, i just didn't type it all out here, sorry for that. i can't go to the shop now because it's night and my paper is due in a few minutes. i went this morning and there was no graph paper there. i made the best of the situation and used the average method. the only question I'm having trouble with now is "Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?" my guess is that it would be harder but i don't really know why. thanks anyway for your help. i will make something up and hope it is right