## 20071231

### Astronomy final exam question: lookback time

Astronomy 10 Final Exam, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q12.1

[15 points.] Sir Martin Rees, Astronomer Royal of Great Britain remarked that, "we in astronomy have an advantage in studying the universe, in that we can actually see the past." Explain what makes this is possible.

Solution and grading rubric:
• p = 15/15:
Correct. Due to the finite speed of light, it takes one year of time for light to travel one light year, and thus observing distant objects means that they appear as they did in the past.
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Grading distributions:
Section 0135
p: 24 students
r: 4 students
t: 0 students
v: 0 students
x: 0 students
y: 1 student
z: 1 student

Students were for the most part successful in answering this question, particularly expressed in a number of different ways. A sample of a "p" response (from student 1929) is shown below:A "p" response (from student 1524) is somewhat more poetic: A whimsical "p" response (from student 1985):
One more "p" response (from student 9389) with an illustration of how the finite speed of light affects observations:

## 20071228

### Astronomy final exam question: spiral arm mapping

Astronomy 10 Final Exam, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q11.2

[15 points.] Consider the following observation:
"The central reason astronomers have been slow to understand the Milky Way is simply because we are deep in the thick of things: The other stars, the gas and especially all the dust in the disk prevent us from seeing the full extent of the galaxy's structure."
--Henry Freudenreich, Am. Sci. vol 87 no. 5, p. 418 (1999)
Explain how it is still possible to map the spiral arm structure of the Milky Way.

Solution and grading rubric:
• p = 15/15:
Correct. O- and B-type stars, due to their extremely short lifetimes, are born and die only in the spiral arms; and are bright enough to be seen through some of the obscuring gas and dust. Similarly for HII (emission) nebulae. 21 cm wavelength radio waves emitted from cold hydrogen gas are not obscured by gas and dust. Observing the locations of all three result in a spiral arm map of the Milky Way.
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. Discusses locating nearby stars without being very specific about the type of stars (typically O- and B-type).
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Describes direct or indirect evidence for the thin disk structure of the Milky Way, but not the spiral arm structure.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Grading distributions:
Section 1073
p: 14 students
r: 3 students
t: 6 students
v: 6 students
x: 1 student
y: 0 students
z: 1 student

A sample of a "p" response (from student 1000) is shown below:Another "p" response (from student 8556), with an illustration of density wave star formation:

## 20071227

### Astronomy final exam question: close-pair mass transfer

Astronomy 10 Final Exam, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q10.3

[15 points.] The stars in a binary star system are a 4 M_Sun main sequence star and a 1 M_Sun giant, where no mass transfer is taking place right now. Discuss which star had originally first transferred mass to the other star, and explain your reasoning.

Solution and grading rubric:
• p = 15/15:
Correct. Massive stars have shorter main sequence lifetimes than more massive stars. Star "B," the 1 M_Sun giant, had to have been the larger star in order to have reached the giant phase first. Since it is now less massive than star "A," the 4 M_Sun main sequence star, star "B" must have transferred mass to star "A."
• r = 12/15:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t = 9/15:
Contains right ideas, but discussion is unclear/incomplete or contains major errors.
• v = 6/15:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x = 3/15:
Implementation/application of ideas, but credit given for effort rather than merit.
• y = 1.5/15:
Irrelevant discussion/effectively blank.
• z = 0/15:
Blank.
Grading distributions:
Section 1073
p: 10 students
r: 13 students
t: 4 students
v: 4 students
x: 0 students
y: 0 students
z: 0 students

A sample of a "p" response (from student 1000) is shown below:
Another "p" response (from student 1256):

## 20071226

### Astronomy quiz question: dark energy evidence

Astronomy 10 Quiz 12, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q12.3

[3.0 points.] Which one of the following choices best describes the evidence for the existence of dark energy?
(A) Stars in the inner and outer parts of the disk orbit the center of the Milky Way with the same speed.
(B) Each and every galaxy sees all other galaxies moving away from its location.
(C) Nearly equal amounts of cosmic background radiation are detected from all directions.
(D) Distant galaxies appear to be much younger than nearby galaxies.
(E) Type Ia supernovae in distant galaxies appear to be dimmer than expected from their redshifts.

Correct answer: (E)

Since type Ia supernovae in distant galaxies appear to be dimmer than expected from their redshifts, this shifts the slope of the Hubble plot for the long-ago (distant) universe, compared to the recent (nearby) universe. Thus the expansion rate of the universe has been accelerating recently, and this is the evidence for dark energy.

Response (A) describes the evidence for dark matter in the halo of the Milky Way. Response (B) is the consequence of an expanding universe. Response (C) describes the "horizon problem" of the cosmic microwave background, and is the evidence for the inflation era of expansion. Response (D) is the consequence of the finite speed of light.

Student responses
Section 0135
(A) : 4 students
(B) : 3 students
(C) : 13 students
(D) : 2 students
(E) : 7 students

## 20071224

### Erasing slate: cool tablet

"Cool Tablet :)" by Anonymous
Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Latest scribbling on the lift-and-erase slate in the hallway, outside the office door.

## 20071221

### Astronomy quiz question: dark energy destiny

Astronomy 10 Quiz 12, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q12.3

[3.0 points.] Which one of the following choices best describes what dark energy may eventually do to the universe in the distant future?
(A) The expansion of space will continue forever, until matter fades away, or is torn apart.
(B) The expansion of space will eventually slow down and stop, and begin contracting as galaxies and stars collide in a "Big Crunch."
(C) Eventually produce enough antimatter to annihilate all matter in the universe.
(D) Stabilize the expansion caused by inflation, such that universe will remain the same size.
(E) Stabilize the contraction caused by dark matter, such that the universe will remain the same size.

Correct answer: (A)

Dark energy is responsible for the recent accelerating expansion of space in the universe. As dark energy is apparently created by space itself, the expansion of space creates more dark energy, which would in turn accelerate the expansion of space, etc. Thus the expansion would steadily increase such that universe would be subject to a "Big Rip" or "Big Tear," as distant galaxies and nearer galaxies recede faster than the speed of light, disappearing from our view, and then locally the expansion of space would rip apart matter itself, overcoming attractive gravitational, electromagnetic, and then nuclear forces.

The "Big Crunch" response (B) had previously been the prevailing prediction for the fate of the universe, before the discovery of accelerating expansion caused by dark energy. For a closed, matter-dominated universe, gravitational forces could eventually overcoming the expansion of space, resulting in slowing down, stopping, and possibly reversing the expansion of space.

Student responses
Section 1073
(A) : 16 students
(B) : 20 students
(C) : 2 students
(D) : 5 students
(E) : 2 students

## 20071220

### Physics quiz question: moles of ideal gas

Physics 5A Quiz 7, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.70

[3.0 points.] There are two identical volume tanks of gas at 15.0° C and at atmospheric pressure, one containing H2 molecules and the other He atoms. The atomic masses of hydrogen and of helium are 1.00 u and 4.00 u, respectively. Which tank contains more moles of gas?
(A) The H2 tank.
(B) The He tank.
(C) (The H2 tank and the He tank each contain the same moles of gas.)
(D) (Not enough information is given to determine this.)

Correct answer: (C)

From the macroscopic form of the ideal gas law:

P*V = n*R*T,

and as each tank is at the same pressure, has the same volume, and is at the same temperature, and thus must contain the same number of moles of gas.

Student responses
Sections 0906, 0907
(A) : 17 students
(B) : 12 students
(C) : 5 students
(D) : 0 students

## 20071219

### Physics quiz question: moles of dry ice

Physics 5A Quiz 7, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.36

[Version 1]

[3.0 points.] How many moles of CO2 are there in 60.0 g of CO2? The atomic masses of carbon and of oxygen are 12.0 u and 16.0 u, respectively.
(A) 0.467 moles.
(B) 0.733 moles.
(C) 1.36 moles.
(D) 2.14 moles.

Correct answer: (C)

The mass of one mole of CO2 is 12.0 u + 2*(16.0 u) = 44.0 u. Then the number of moles of CO2 is (60.0 u)/(44.0 u) = 1.36 moles. Choice (A) is (12.0 u + 16.0 u)/(60.0 u). Choice (B) is (44.0 u)/(60.0 u). Choice (D) is (60.0 u)/(12.0 u + 16.0 u).

Student responses
Sections 0906, 0907
(A) : 3 students
(B) : 1 student
(C) : 11 students
(D) : 2 students

[Version 2]

[3.0 points.] How many moles of CO2 are there in 15.0 g of CO2? The atomic masses of carbon and of oxygen are 12.0 u and 16.0 u, respectively.
(A) 0.341 moles.
(B) 0.536 moles.
(C) 1.87 moles.
(D) 2.93 moles.

Correct answer: (A)

The mass of one mole of CO2 is 12.0 u + 2*(16.0 u) = 44.0 u. Then the number of moles of CO2 is (15.0 u)/(44.0 u) = 0.341 moles. Choice (B) is (15.0 u)/(12.0 u + 16.0 u). Choice (C) is (12.0 u + 16.0 u)/(15.0 u). Choice (C) is (44.0 u)/(15.0 u).

Student responses
Sections 0906, 0907
(A) : 7 students
(B) : 3 students
(C) : 4 students
(D) : 3 students

## 20071218

### FCI post-test comparison: Cuesta College versus UC-Davis

Students at both Cuesta College (San Luis Obispo, CA) and the University of California at Davis were administered the Force Concept Inventory (Doug Hestenes, et al.) during the last week of instruction, in order to follow up on the pre-test results from the first week of instruction (which showed no statistical difference between pre-test scores).
`     Cuesta College    UC-Davis     Physics 5A        Physics 7B     Fall Semester     Summer Session II     2007              2002N    31 students       76 studentslow   4                 3mean 13.4 +/- 5.0      12.7 +/- 5.4high 22                26`

A "Student" t-test of the null hypothesis results in p = 0.51, thus there is no significant difference between Cuesta College and UC-Davis FCI post-test scores.

The pre- to post-test gain for this semester at Cuesta College is:
`Physics 5A Fall Semester 2007 sections 0906, 0907         N = 31<initial%> = 30% +/- 14%  <final%> = 46% +/- 17%       <g> = 0.23`

This is higher than the gains for algebra-based introductory physics at UC-Davis (0.16), and for calculus-based introductory physics at Cuesta College (0.14-0.16), as discussed in previous postings on this blog.

Notable about this Physics 5A class at Cuesta College is the implementation of electronic response system "clickers" (Classroom Performance System, einstruction.com), compared to the traditional lecture of Physics 8A at Cuesta College, and the reformed peer-instruction centered approach at UC-Davis. More analysis on the impact of using clickers on this introductory physics class will be forthcoming.

## 20071217

### Physics clicker question: heat capacity during phase change

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 13.11

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:

[0.6 participation points.] The heat capacity of a material undergoing a phase change (such as melting/freezing, or boiling/condensing) is typically:
(A) zero.
(B) infinity.
(C) a finite number.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 10 students
(B) : 12 students
(C) : 11 students
(D) : 0 students
(E) : 0 students

Correct answer: (B)

The heat capacity is defined as the ratio of the heat Q (put in) to the resulting temperature increase delta(T):

C = Q/delta(T).

But when heat is put into, or extracted from a system, its temperature remains constant, making delta(T) = 0, and thus its heat capacity will be undefined as it would approach infinity.

## 20071214

### Hot coffee

"Any Tool (That's Some Powerful Hot Coffee!)," by MissChief
Worth1000.com
More Power 7 contest

"Caution: the coffee you are about to drink may be very hot."

## 20071213

### Physics clicker question, movie: Space Shuttle tile

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 13.9 (extended)

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:

[0.6 participation points.] A 50 g Space Shuttle heat tile, and a 50 g iron sample each absorb an equal amount of heat from a bunsen burner. Which sample will have a greater increase in temperature?
(A) Space Shuttle tile.
(B) 50 g iron sample.
(C) (Both samples will have approximately the same increase in temperature.)
(D) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 0 students
(B) : 26 students
(C) : 3 students
(D) : 0 students

Correct answer: (B)

The heat capacity is defined as the ratio of the heat Q (put in) to the resulting temperature increase delta(T):

C = Q/delta(T).

The Space Shuttle TPS (thermal protection tile) has a larger heat capacity, as it experiences a small increase in temperature for a given amount of heat, while the iron sample has a smaller heat capacity, experiencing a large increase in temperature with the same amount of heat.

Space Shuttle Thermal Protection System (TPS) Tile
Posted by ntschke
http://www.youtube.com/watch?v=XuSlkEob-3w

Do not touch the tile immediately after heating!

## 20071212

### Physics clicker question: molecular rms speeds

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 13.1 (extended)

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:

[0.6 participation points.] Which type of air molecule has the fastest rms speed at room temperature?
(A) CO2.
(B) H2O.
(C) N2.
(D) O2.
(E) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 5 students
(B) : 9 students
(C) : 14 students
(D) : 4 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 0906, 0907
(A) : 5 students
(B) : 14 students
(C) : 9 students
(D) : 2 students
(E) : 0 students

Correct answer: (B)

The rms speed of a particle is given by:

v_rms = sqrt(3*k*T/m),

where k is Boltzmann's constant, T is the absolute temperature (in Kelvin), and m is the particle mass (in kg). In unified atomic mass units, the masses of these molecules are 44 u, 18 u, 28 u, and 32 u, respectively, making H2O (the lightest molecular mass in this list) the molecule with the fastest rms speed.

## 20071211

### Physics clicker question: molar mass

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 13.11

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:

[0.6 participation points.] The mass of a single aluminum atom is 27.0 u. What is the mass of one mole of aluminum atoms?
(A) (27.0/N_A) g = 4.48*10^-23 g.
(B) 27.0 g.
(C) 27.0 kg.
(D) (N_A/27.0) g = 2.23*10^22 g.
(E) (27.0*N_A) g = 1.63*10^25 g.
(F) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 11 students
(B) : 5 students
(C) : 2 students
(D) : 4 students
(E) : 9 students
(F) : 2 students

Correct answer: (B)

The unified atomic mass unit, "u" is defined to be exactly 1/12th the mass of one mole of carbon-12 atoms (as opposed to the defunct "amu" atomic mass unit, which is either defined to be exactly 1/16th the mass of one mole of oxygen-16 atoms, or 1/16th the mass of one mole of all isotopes of oxygen as found in their naturally occurring proportions). Thus the mass of any particle, in unified atomic mass units, is exactly the mass of one mole of these particles.

## 20071210

### Physics clicker question: opening in heated metal plate

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 13.4 (extended)

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:

[0.6 participation points.] A square opening is made in a metal plate. If the metal plate is heated up from room temperature by 100 degrees C, what happens to the size of the square opening?
(A) The square opening becomes smaller.
(B) The square opening becomes larger.
(C) The square opening remains the same size as before.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 19 students
(B) : 12 students
(C) : 2 students
(D) : 0 students
(E) : 0 students

Correct answer: (B)

All dimensions, external as well as internal in the metal plate expand when it is heated, and this includes the dimensions of the square opening. However, the hole in the middle of a ring of donut batter will get smaller as the donut cooks in oil, but this is an entirely different phenonomenon--no "fluffy donuts!"

## 20071207

### Astronomy quiz question: lavender/pink and blue nebulae

Astronomy 10 Quiz 11, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q11.1

[Version 1]

[3.0 points.] Which one of the following choices best describes what causes the color of a lavender/pink nebula?
(A) Photons emitted by electrons.
(B) Scattered light from stars.
(C) Blackbody radiation.
(D) Gravitational contraction.
(E) Dark matter.

Correct answer: (A)

The lavender/pink color is the mixture of several visible light photons of different wavelengths, emitted when an electron jumps down several hydrogen orbitals.

Student responses
Section 0135
(A) : 17 students
(B) : 9 students
(C) : 3 students
(D) : 2 students
(E) : 1 student

[Version 2]

[3.0 points.] Which one of the following choices best describes what causes the color of a blue nebula?
(A) Photons emitted by electrons.
(B) Scattered light from stars.
(C) Blackbody radiation.
(D) Gravitational contraction.
(E) Dark matter.

Correct answer: (B)

Very small dust particles will scatter short wavelengths of light more than long wavelengths. Thus red light will typically pass through this type of nebula, while blue light will be scattered in all directions.

Student responses
Section 1073
(A) : 12 students
(B) : 27 students
(C) : 1 student
(D) : 0 students
(E) : 2 students

## 20071206

### Erasing slate: have a nice day

"[flower] Have a nice day!" by Anonymous
Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Latest scribbling on the lift-and-erase slate in the hallway, outside the office door.

## 20071205

### Student scribble: sleepy time

"Sleepy Time" by Anonymous
Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Found artifact from student notebook.

## 20071204

### Physics midterm question: closed-open pipe versus string standing waves

Physics 5A (currently Physics 205A) Midterm 2, fall semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 10.4

"Mozart: Allegro - Sonata For Bassoon And Cello in B-Flat Major | Kroth, Bagratuni"
Michigan State University | College of Music
http://youtu.be/INfybEkRtgM

Discuss why the fundamental frequency of a bassoon (approximated as a closed-open pipe) is more sensitive to a change in air temperature than the fundamental frequency of a cello (a string fixed at both ends). Explain your reasoning using the properties of sound waves, string waves, temperatures, and standing waves.

Solution and grading rubric:
• p:
Correct. The standing wave frequencies for a bassoon will change, as the velocity of sound waves in air along the pipe depend on air temperature. The standing wave frequencies for the cello will not change, as the velocity of waves along the string do not depend on air temperature.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. At least recognizes the dependence of sound wave velocity on temperature.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Argument based on a factor of two difference between f1 = v/(4·L) and f1 = v/(2·L) for closed-open pipes and a string fixed at both ends, instead of on the difference of the temperature dependence of v in either case. May also argue that the bassoon is affected more by changes in air temperature because it somehow involves a larger volume of air.
• x:
Implementation/application of ideas, but credit given for effort rather than merit.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.

Grading distribution:
p: 4 students
r: 5 students
t: 13 students
v: 16 students
x: 0 students
y: 0 students
z: 0 students

A sample of a "p" response (from student 1129) is shown below:

## 20071203

### Physics midterm question: compressing short and long bars

Physics 5A Midterm 2, fall semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 10.4

A cylindrical steel bar is compressed by the application of forces of magnitude F at each end. What magnitude forces would be required to compress by the same amount a steel bar of the same cross-sectional area, but twice the length? Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:
• p:
Correct. Applies Hooke's law in a quantitative manner to argue that less force is required to compress a longer bar by a certain amount than a shorter bar compressed by the same amount.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. Applies Hooke's law in a quantitative manner, but argues that more force is required to compress the longer bar.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
Implementation/application of ideas, but credit given for effort rather than merit.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.

Grading distribution:
p: 16 students
r: 2 students
t: 13 students
v: 6 students
x: 1 student
y: 0 students
z: 0 students

A sample of a "p" response (from student 4500) is shown below: