Conservation of Marbles

Transcription

1 Conservation of Marbles Pre- Lab: Introduction to Conservation Laws A Bit of History Richard Feynman was one of the great physics minds of the 20 th century. As proof, one needs to look no further than the Nobel Prize in Physics that he was awarded in 1965 for his work in quantum electrodynamics. However, that Nobel Prize does not necessarily indicate that Feynman was also one of the great personalities of the 20 th century. As proof of that, one needs only to read his autobiography Surely You re Joking, Mr. Feynman! In it, he details his antics cracking safes in Los Alamos as he worked on the Manhattan project, his forays into erotic art, and his adventures drumming in Brazil. As the subtitle of the book points out, Feynman was simply a very curious individual. In fact, his Nobel Prize- winning work grew out of a problem involving a Frisbee that he was tackling purely for the joy of it. What is a Conservation Law and Where Do They Come From? A conservation law states that, under the right circumstances, a quantity describing a system is constant. In this lab, you will look at a system where the number of marbles in the system is constant. In your first semester of Intro Physics, you will spend more time considering whether or not the total energy, momentum, or angular momentum of certain systems is constant. If you continue on to the second semester of Intro Physics, you ll consider many systems where the total electric charge is conserved. The details describing when one can and when one cannot expect a certain quantity to be conserved (i.e. remain constant) depends on which quantity is being considered. We won t concern ourselves with those details right now. In general, a physicist can expect some quantity to be conserved when there is a symmetry associated with a system. The word symmetry is used a little differently here than it is in most of our lives. As Moore puts it, a symmetry can be stated in the form the laws of physics are unaffected if you do such and such. For example, Conservation of Energy results from a symmetry in time. The laws of physics are unaffected if you wait until tomorrow. Conservation of Momentum results from a symmetry in space. The laws of physics are unaffected if you take a step to your left. Conservation of Angular Momentum results from a symmetry of rotation. The laws of physics are unaffected if you turn around. It s that simple! Actually, it was quite groundbreaking when Emmy Noether discovered the connection between symmetry and conservation laws in the early 1900s. Feynman s Introduction Richard Feynman famously taught an introductory physics course to students at the California Institute of Technology in the mid 1960s. These lectures were later collected into a text aptly titled The Feynman Lectures on Physics. The way in which Feynman introduces energy is quite charming and is well worth reading, which you are now asked to do. 1

2 Do This: Read Feynman s introduction to energy found on the In- Lab Links page. You just need to read the first section with the heading What is Energy? PL1. How does the window in Feynman s story relate to the notion of a system being isolated? PL2. Feynman gives a list of many different forms of energy. How many forms of energy appear on this list? The Neutrino: The Child of Conservation Laws In beta decay, an atomic nucleus spits out an electron and becomes a different element in the process. For example, 14 C (the parent nucleus) can emit an electron through a type of radioactive decay called beta decay and become 14 N (the daughter nucleus). In 1927, experimental physicists Ellis and Wooster showed that beta decay seemed to violate several well established conservation laws. Most famously, experiments consistently showed that the energy of the system before the decay (i.e. the energy of the parent nucleus) was greater than the energy of the system after the decay (i.e. the energy of the daughter nucleus plus the energy of the emitted electron). Furthermore, conservation of momentum and conservation of angular momentum seemed to be broken in the process. There are several ways a scientist could react to reading such a report. The first thing to do would be to make sure that the final energy of the system really was less than the initial energy of the system. Remember that all measurements have uncertainty. Perhaps the uncertainty in the energy measurements was greater than the missing energy. In that case, there would be no cause for alarm. Well, Ellis and Wooster were careful assigning the correct uncertainty to all of their measurements and calculations, and the missing energy was definitively missing. There was no question about it. So if the energy really is missing, what to do next? One approach is to say that this experiment proves that our law of energy conservation wrong. The notable physicist Neils Bohr actually took this viewpoint. He viewed this as proof that the conservation laws simply weren t valid when working on very small scales with very light particles. A second approach is to have confidence in our law of energy conservation and try to figure out what we don t understand about the experiment. It is this approach that physicists Wolfgang Pauli and Enrico Fermi took. Through their combined effort, they advanced the theory of a new particle called the neutrino. This is the approach that was ultimately proven correct when neutrinos were detected in 1956 by Reines and Cowan. Reines was awarded the Nobel Prize in Physics in 1995 for this work. 2

3 The big idea was that the neutrino was a difficult- to- detect particle that was emitted along with the electron in beta decay. The neutrino must be a neutral particle (to conserve charge) with a spin of ½ (to conserve angular momentum). In any given beta decay, its velocity would be such that the energy and momentum of the system would also be conserved. With this elegant addition to the world of subatomic particles, all of the conservation laws were again being followed. Problem solved! Read This: Answer the following questions about the discovery of the neutrino based on the above reading. PL3. Let E! be the measured energy of the parent nucleus before beta decay. Let E! be the measured total energy of the daughter nucleus and the electron emitted by beta decay. Finally, let δe be the uncertainty in the measurements of E! and E!. If E! = 100 units and E! = 94 units which of these values of δe can allow us to conclude that there is missing energy? PL4. Which scientists performed the 1927 experiment that definitively showed that there was missing energy when atoms underwent beta decay? PL5. Which famous physicist was inclined to believe that the experiments on beta decay were definitive proof that the Law of Conservation of Energy was simply not valid on very small scales? End of Pre- Lab 3

4 Part I: Preliminary Measurements The Story In the Pre- Lab, you read a story about a boy and his blocks as an introduction to energy. Here you will play a game with marbles that is inspired by that story. Equipment Bowl of 12 marbles Graduated cylinder Toy box Digital balance The Rules There are three places that marbles are allowed to go: the bowl, the box, and the graduated cylinder. If you want to know how many marbles are in the bowl, all you have to do is look and count. If you want to know how many marbles are in the box, you can t open it; you have to weigh it like the mother from Feynman s story. If you want to know how many marbles are in the graduated cylinder you have to look at the level of the water. 1. The Volume of a Marble Before playing the game, we need to have a good value for the volume of one of the marbles. You will investigate three different ways to determine the volume of a marble after which you will decide which method is the best. Read This: The first method is to simply use the equation for the volume of a sphere. Checkpoint 1.1: What s the equation for the volume of a sphere? Read This: The company that sold us these marbles claims that they have a diameter of inch. Checkpoint 1.2: Calculate the volume of one of these marbles. Read This: That was easy! But there s a little bit more to the problem than that. In today s lab, we will be paying close attention to uncertainties in measured values and in calculated values. The volume you just wrote down in your notebook is a calculated value. You used a previously measured value (the diameter) in your calculation. The uncertainty in your calculated volume depends on the uncertainty in the diameter that was measured by the company that sold us the marbles. The company that sold us the marbles assures us that the uncertainty in the diameter is inches. 4

5 Read This: It can be shown (Appendix A) that the uncertainty (δv) in a spherical volume (V) that is calculated using a diameter (d) whose uncertainty is δd is given by δv = 1 2 πd! δd Checkpoint 1.3: Calculate the uncertainty in the volume that you calculated in Checkpoint 1.2. Checkpoint 1.4: Write the volume of the marble in the form V!"#$%& ± δv!"#$%&. Read This: Now for the second method. In this method, you will use the water displaced by a single marble as the volume of the marble. Checkpoint 1.5: How much water does a single marble displace? Read This: Let s consider the uncertainty of the volume that you found in Checkpoint 1.5. You made two measurements in order to determine the volume of the marble. Thus the uncertainty in the volume must depend on the uncertainty in your two measurements. Let s call the initial water level V! and the final water level V!. Checkpoint 1.6: What uncertainty would you assign to V! and V!? Take into account the smallest increment on the graduated cylinder as well as the ease with which you can identify the meniscus. We ll refer to this uncertainty as δv throughout the rest of the lab. Read This: It can be shown (Appendix B) that the uncertainty in the volume of the single marble as calculated using V! and V! is given by δv!"#$%& = 2δV Checkpoint 1.7: Calculate the uncertainty in the volume that you found in Checkpoint 1.5. Checkpoint 1.8: Write this second value for the volume of the marble in the form V!"#$%& ± δv!"#$%&. Read This: Now, for the final method. Checkpoint 1.9: Measure the displacement of multiple marbles at the same time and use that displacement to calculate the volume of one marble. Read This: In your calculation you divided the change in water level by the number of marbles you dropped in the cylinder. It can be shown (Appendix C) that this clever technique reduces the uncertainty in your calculated volume by a factor of the number of marbles. For example, placing two marbles in the cylinder cuts in half the uncertainty in the volume of a single marble. Checkpoint 1.10: What is the uncertainty in your third value for the volume of a marble? 5

6 Checkpoint 1.11: Write the third value for the volume of the marble in the form V!"#$%& ± δv!"#$%&. S1 Synthesis Question 1 (30 Points): Discuss your three values for the volume of a marble and choose the value you will use in the rest of this experiment. In your discussion make sure you include the following: Each value written in the form V!"#$%& ± δv!"#$%& A quick explanation of how you found each value (you do NOT need to explain the uncertainty calculations) A graphical representation of the volume measurements. You can find a pre- made scale on the In- Lab Links page that might speed up the process. A brief discussion of the discrepancies between your values. For starters, are any of the discrepancies significant? The value you have chosen to use for the rest of the lab, along with support for your choice 2. The Mass of a Marble The game also requires a good value for the mass of a marble. The company that sold us the marbles does not give us detailed information regarding their masses. That leaves us with the digital balance as our best bet for determining the mass of a marble. Read This: The best way to assign an uncertainty to a value that you measure using a digital instrument is to look at the manual. The fine people at Ohaus tell us that these scales have an uncertainty of 0.1 grams. For the rest of the lab, the uncertainty in the scale will be referred to as δm. Thus, δm = 0.1 g. Read This: As is often the case, the uncertainty is equal to the last digit of precision. However, this is NOT always the case. S2 Synthesis Question 2 (15 Points): Find a value for the mass of a marble that has an uncertainty of 0.01 g. Explain your method, show some data, and show some math. 6

7 Part II: Feynman s Marbles 3. Almost Time to Play At this point we are getting very close to playing the game. But first you must develop the equation that you will use when testing the Law of Conservation of Marbles. Read This: As you play the game, we will refer to the following values: N! The number of marbles you start with. It should be 12. N The total number of marbles at some later time. N!"#$ The number of marbles seen in the bowl. N!"# The number of marbles in the box (inferred, not seen). N!"# The number of marbles in the graduated cylinder (inferred, not seen). m! The mass of the empty box. m! The mass of the full box (mass of the box and the marbles inside the box). m m! m! m!"#$%& The mass of the marble from Synthesis Question 2. V! The initial water level in the graduated cylinder (no marbles). V! The final water level in the graduated cylinder (after marbles are dropped in). V V! V! V!"#$%& The volume of the marble from Synthesis Question 1. Read This: If you look in your bowl you should see that N! = 12. When the game starts, marbles will be placed in the box and/or in the cylinder. Others may be left in the bowl. Thus, at any time, the total number of marbles N is given by N = N!"#$ + N!"# + N!"# = N! The Law of Conservation of Marbles states that N is always equal to N!. S3 Synthesis Question 3 (10 Points): Write an algebraic expression for N in terms of N!"#$, m, m!"#$%&, V, and V!"#$%&. Your expression should be completely symbolic. It should contain no numbers. Make sure you give a brief explanation of the steps you take on the way to your answer. 4. Finally! Preliminary Games As we begin playing the game, we ll focus on whether or not we can detect violations of the Law of Conservation of Marbles. Detecting the violation of a conservation law is something that requires a close look at uncertainty. Read This: It can be shown (Appendix D) that the uncertainty in the number of marbles in the graduated cylinder is approximately given by 7

8 δn!"# = 2 δv V!"#$%& where δv is the uncertainty in the graduated cylinder and V!"#$%& is the volume of the marble. Note that the uncertainty in the volume of a marble does not enter into the equation. This is because the uncertainty in the volume of a marble is much, much smaller than the uncertainty in a reading of the water level. Read This: We begin by seeing if any marbles disappear when we drop them in the graduated cylinder. Do This: Make sure there are no marbles in the graduated cylinder. Fill the graduated cylinder to somewhere between 150 ml and 170 ml. Checkpoint 4.1: Record the level of water in the graduated cylinder. This is V!. Do This: Drop some number of marbles into the graduated cylinder. Checkpoint 4.2: Record the number of marbles that you dropped into the cylinder. Checkpoint 4.3: Record the new water level. This is V!. Checkpoint 4.4: Calculate the number of marbles in the graduated cylinder using your measurements of V! and V! along with your previously calculated value for V!"#$%& (Synthesis Question 1). Checkpoint 4.5: Calculate the uncertainty in your response to Checkpoint 4.4. Checkpoint 4.6: What is the discrepancy between your responses to Checkpoint 4.2 and Checkpoint 4.4. How does the discrepancy compare to the uncertainty you found as Checkpoint 4.5? Checkpoint 4.7: Do your results support the hypothesis that marbles are conserved when you drop them in the graduated cylinder? Discuss, making sure to refer to your responses to Checkpoint 4.2, Checkpoint 4.5 and Checkpoint 4.6. Read This: Now let s consider what happens when you place marbles in the box and close it. It can be shown (Appendix E) that the uncertainty in the number of marbles in the closed box is approximately given by δn!"# = 2 δm m!"#$%& where δm is the uncertainty of the balance (0.1 g) and m!"#$%& is the mass of the marble. 8

9 S4 Synthesis Question 4 (25 Points): In Checkpoints you hopefully found evidence that marbles are conserved when they are placed in the graduated cylinder. Now perform a similar experiment to determine whether or not marbles are conserved when they are sealed inside the toy box. Report on your experiment and your findings. 5. The Full Shebang Now that you have a deep understanding of marble conservation, you are going to solve a couple of problems that are very similar to problems you will look at this semester involving energy conservation. Here are two problems side- by- side for comparison. For simplicity, none of the marbles will be in the bowl: Marbles There are 12 marbles. Some of them are in a box, and some of them are in a graduated cylinder filled with water. If you measure the level of the water to be x milliliters, what is the mass of the box? Energy A projectile has 12 joules of mechanical energy. Some of that energy is kinetic energy, and some of that energy is gravitational potential energy. If you see that the projectile is x meters above the ground, what is the speed of the projectile? Checkpoint 5.1: Solve the equation you found in Synthesis Question 3 for m. This should be entirely symbolic. No numbers. Do This: Reset your station. That is, put all the marbles in the bowl and fill the graduated cylinder with 150 ml to 170 ml of water. Place the empty box on the balance. Checkpoint 5.2: Make a note of V! and m!. Do This: Take the box off of the balance. Then turn around and have a neighboring group distribute your 12 marbles among the bowl, box, and cylinder. Checkpoint 5.3: Use measurements of V!, V!, N!"#$, and N! to calculate a theoretical value for m. Checkpoint 5.4: Place the full box on the balance and record m!. What is the experimental value of m? Checkpoint 5.5: Compare the experimental value of m to your calculated value of m. Discuss what may account for any discrepancy between the values. Read This: In all likelihood, your theoretical and experimental values had a discrepancy that was considerably greater than 0.1 g. The reason for this is that you used volume measurements when you were calculating your theoretical value for m. Since these volume measurements have a relatively large uncertainty, any value that you calculate with them will have a similarly large uncertainty. 9

10 Read This: If you re interested in some numbers, your measurement of V likely had an uncertainty of 2 ml or 3 ml which would be maybe 5% or 10% of the measured value. (An uncertainty expressed as a percentage is called the fractional uncertainty.) Very roughly, any theoretical value that we calculate with this measured value of V will have an uncertainty of about 5% to 10%. Do This: Reset your station. Read This: In Checkpoints , you used volume measurements to predict the result of a mass measurement. Then you assessed the accuracy of your prediction. In this final Synthesis Question, you will use mass measurements to predict the outcome of a volume measurement. S5 Synthesis Question 5 (20 Points): Measure V! and m!. Then turn around and have a neighboring group distribute your 12 marbles among the bowl, box, and cylinder. Next, use measurements of m!, m!, N!"#$, and N! to calculate a theoretical value for V. After that, compare your calculated value to the experimental value. Finally, discuss specific sources of any discrepancy between your experimental and calculated values of V!. Make this very easy to follow. 10

11 Appendix A: Uncertainty in the Volume of a Sphere This appendix shows how to find the uncertainty in the volume of a sphere that has been calculated using a measured diameter. Since there is some uncertainty in the diameter, the calculated volume will necessarily be uncertain as well. The volume of a sphere as a function of the diameter is given by V!"#$%& = 1 6 πd! Notice that the diameter is the only variable (or in laboratory terms, measured quantity). Consider a quantity q that is a function of a single measured value x. We could write this in a very general way as q = f(x) When we have such a relationship (a function of a single variable), the uncertainty in the calculated value q is given by δq = df dx δx So to find the uncertainty of the calculated volume, all we have to do is differentiate the right side of the equation by d and make sure to tack on the δd term. We find the following expression: δv!"#$%& = 1 2 πd! δd 11

12 Appendix B: Uncertainty in the Displaced Water The measurement of the water displaced by some number of marbles relies on two separate measurements: the initial water level (V! ) and the final water level (V! ). Each of these two measurements has an uncertainty, so we know that the displaced water has some uncertainty as well. Let s begin by writing a simple equation for the displaced water, which we will call ΔV. ΔV = V! V! A quick estimate of the uncertainty in the displaced water would be simple to add the uncertainty in V! and the uncertainty in V!. We could then write δδv 2δV! = 2δV! = 2δV A more rigorous way to go about this is to consider a general quantity q that is calculated as a function of two measured values, x and y. The uncertainty in q is given by q = f(x, y) δq =! f x δx + f! y δy Let s now go back to the problem at hand. The analogy that we need to make is that ΔV is like q, V! is like x, V! is like y, and V! V! is like f. Taking the necessary partial derivatives, we find δδv = 1 δv!! + 1 δv!! If we realize that δv! and δv! are equal and if we call this quantity δv, then our expression simplifies very nicely to δδv = 2δV 12

13 Appendix C: Reducing Uncertainty by Measuring Many Marbles at Once Consider a displaced volume of water, ΔV, that represents the total volume of some number of marbles N. Mathematically, the volume of a single marble, V!"#$%&, can be given by V!"#$%& = ΔV N Here the volume of the marble is a function of only one uncertain variable measurement (ΔV). Therefore, we proceed just like we did in Appendix A. To find the uncertainty in V!"#$%& we take the derivative of the right side with respect to ΔV and then remember to multiply that side by δδv. δv!"#$%& = 1 N δ V Since δ V does not change as V increases or decreases, this equation shows that increasing N will reduce δv!"#$%&. 13

14 Appendix D: Uncertainty in Number of Marbles in the Cylinder The number of marbles in the graduated cylinder can be calculated using the initial water level, the final water level, and the volume of a single marble. or more simply N!"# = V! V! V!"#$%& N!"# = ΔV V!"#$%& Since N!"# is calculated using two measured values (ΔV and V!"#$%& ), its uncertainty will depend on the uncertainty in those two measured values. But before we dive into any calculus, we should think about these uncertainties. It will save us some work! We know that the uncertainty in ΔV is given by 2δV! which you hopefully estimated to be something like 1.4 ml to 3 ml. A typical value of ΔV would be between 10 ml and 100 ml. That means that the uncertainty in ΔV is at least 1.4% of the measurement itself, perhaps reaching 30%. (When we quote an uncertainty as a percentage of the measured value, this is known as the fractional uncertainty.) Now consider the uncertainty in the volume of the marble. Since we were able to calculate the volume of the marble using very precise measurements made by a fancy company, it has a very small uncertainty. You hopefully found δv!"#$%& to be about 0.03 ml. When compared to the volume of the marble, this uncertainty is only about 0.3%. That s much smaller than the uncertainty in the displaced water. Thus, the uncertainty in the displaced water is the dominant source of uncertainty. We can save some time by ignoring the uncertainty in the volume of the marble without losing much accuracy. We can essentially act like the volume of the marble is a well- known constant. We can then treat N!"# as a function of a single variable. To find its uncertainty, we simply differentiate each side of the equation by ΔV and make sure to tack on a δδv. which can be written as δn!"# = δn!"# = 2 1 V!"#$%& δδv δv V!"#$%& 14

15 Appendix E: Uncertainty in Number of Marbles in the Sealed Box The number of marbles in the toy box can be calculated using the initial mass, the final mass, and the mass of a single marble. or more simply N!"# = m! m! m!"#$%& N!"# = Δm m!"!"#$ Since N!"# is calculated using two measured values (Δm and m!"#$%& ), its uncertainty will depend on the uncertainty in those two measured values. But before we dive into any calculus, we should think about these uncertainties. It will save us some work! We know that the uncertainty in Δm is given by 2δm! which you hopefully estimated to be 0.14 grams. A typical value of Δm would be between 10 g and 100 g. That means that the uncertainty in Δm is likely between 0.1% and 1% of the measurement itself. (When we quote an uncertainty as a percentage of the measured value, this is known as the fractional uncertainty.) Now consider the uncertainty in the mass of the marble. You were asked to come up with a value for the mass of the marble that has an uncertainty of 0.01 g. When compared to the mass of the marble, this uncertainty is only about 0.1%. This analysis is not as cut and dry as with the graduated cylinder. We can see that when Δm is relatively small (i.e. when there are just a few marbles in the box) then the uncertainty in Δm dominates. However, when there are close to ten marbles in the box and Δm is close to 100 g, the fractional uncertainty in the mass of the marble is just as large as the fractional uncertainty in Δm. When Δm is small and we can ignore the uncertainty in the mass of the marble, we find an equation that looks very much like what we has for the number of marbles in the cylinder: δn!"# = 2 δm m!"#$%& [small Δm] However, as Δm gets closer to 100 grams, it would be more appropriate to treat N!"# as a function of two variables, like we did with ΔV in Appendix B. Doing so leads us to the equation: δn!"# = 2 δm m!"#$%& [Δm~100 g] where we have made the good approximation that the fractional uncertainty in Δm is nearly equal to the fractional uncertainty in m!"#$%&. For simplicity, we can use this expression for all Δm. 15