Astronomy midterm question: metal-poor vs. metal-rich stars

Astronomy 210 Midterm 2, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[20 points.] Discuss what observations tell you that a star is metal-poor, and whether this is an old or young star, using the properties and evolution of stars in your explanation.

Solution and grading rubric:

• p = 20/20:
  Correct. Discusses (a) how observations of absorption spectra classify stars as metal-poor, and (b) how production of metals in cores of massive and medium-mass stars is dispersed via type II/type Ia supernovae to be incorporated into the outer layers of subsequent generation stars.
  
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. One topic is complete/correct, other other is problematic.
  
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Only one of two discussion topics is complete and correct, the other is missing or has serious errors (typically spectroscopy); or both topics have minor errors.
  
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Discusses metal-poor stars as being old, but with conceptual errors; no discussion on spectroscopy.
  
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. May discuss how metals are used to fuel a star; less metals would then indicate an older star that has used up all its metals.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 70158
p: 5 students
r: 6 students
t: 23 students
v: 14 students
x: 22 students
y: 1 student
z: 0 students

A sample "p" response (from student 0711):
A sample "t" response (from student 1990):
Another sample "t" response (from student 3259):
A sample "x" response (from student 7027):

Astronomy midterm question: apparent magnitude, absolute magnitude, and distance

Astronomy 210 Midterm 2, fall semester 2008
Cuesta College, San Luis Obispo, CA

Consider a star with an apparent magnitude that is brighter than its absolute magnitude. Discuss whether this star is farther than 10 parsecs, exactly at 10 parsecs, or closer than 10 parsecs from the Earth. Explain your reasoning.

Solution and grading rubric:

• p:
  Correct. Differentiates between apparent ("as is") magnitude, and absolute ("true") magnitude, and explains reasoning as to why a star that seems brighter than it actually is must be closer than the standard 10 parsec distance in the absolute magnitude definition.
  
• r:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
  
• t:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. May involve other factors such as the Stefan-Boltzmann law, lookback time caused by the finite speed of light, etc.
  
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
  
• y:
  Irrelevant discussion/effectively blank.
  
• z:
  Blank.
  

Grading distribution:
Section 70158
p: 47 students
r: 6 students
t: 6 students
v: 11 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response, with a graphical representation of the relationship between apparent magnitude, absolute magnitude, and distance (from student 0725):
Another "p" response, with a slightly different graphical representation (from student 4492):
A "p" response based on common-sense principles, from a student who claims to have no recollection that this topic was ever covered in class (from student 6789):
A "p" response based on the relating this phenomenon to the movie Jurassic Park (from student 9933):

Physics midterm problem: inelastic rotational collision

Physics 205A Midterm 2, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.76

[20 points.] A 0.500 kg point mass is dropped onto a disk rotating at 4.00 rad/s that has a mass of 0.700 kg and a radius of 0.150 m. The point mass eventually rotates with the disk at a distance of 0.100 m from the axis. What is the final kinetic energy of the disk and point mass system? Show your work and explain your reasoning.

Solution and grading rubric:

• p = 20/20:
  Correct. Applies conservation of angular momentum to find the final angular velocity of the disk and point mass system; and then calculates the change in initial and final rotational kinetic energies in this inelastic rotational collision.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Some attempt at applying conservation of angular momentum before evaluating the final kinetic energy.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Applies kinetic energy conservation to find final rotational kinetic energy, even though this is an inelastic rotational collision (as seen in lab), and if it were elastic, the final rotational kinetic energy would be exactly equal to the initial kinetic energy.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 70854, 70855
p: 1 student
r: 3 students
t: 3 students
v: 31 students
x: 3 students
y: 0 students
z: 2 students

A sample of the sole "p" response, applying angular momentum conservation to find the final angular velocity of the point mass and disk system, and then calculating the final (rotational) kinetic energy of this system (from student 0215):
A sample "t" response (from student 0420), with an attmempt to calculate the final kinetic energy of the system, but with a realization that the final angular velocity is not yet determined:

Physics midterm problem: elastic collision carts

Physics 205A Midterm 2, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 7.42, 7.49

[20 points.] A 0.200 kg cart collides elastically with a 0.700 kg cart that is initially at rest. If the 0.200 kg cart was traveling in the +x direction at 1.40 m/s before the collision, what is the velocity (speed and direction) of the 0.200 kg cart after the collision? Show your work and explain your reasoning.


Solution and grading rubric:

• p = 20/20:
  Correct. Applies both momentum conservation and kinetic energy conservation for this elastic colllision, and finds both the magnitude and direction of the final velocity of the 0.200 kg cart.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least applies both momentum and kinetic energy conservation.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically only attempts momentum conservation.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 70854, 70855
p: 16 students
r: 1 student
t: 16 students
v: 8 students
x: 1 student
y: 1 student
z: 0 students

A sample of a "p" response, applying both momentum conservation and (the vector form of) kinetic energy conservation, and then methodically feeding the answers back in to verify that the two conservation laws were satisfied (from student 7503):

Cuesta College North County campus star party

081120-1060659
http://www.flickr.com/photos/waiferx/3059302272/
Originally uploaded by Waifer X

Cuesta College Astronomy 210L students and guests from both North County and San Luis Obispo campuses during a North County campus star party. The SLO campus students made the inevitable In-n-Out Burger stop on the way back down the Cuesta grade.

Sights seen:
M57 (Ring Nebula)
M31 (Andromeda Galaxy)
M45 (Pleiades)
Epsilon Lyrae (The "Double-Double")
Aldebaran
Vega

Astronomy clicker question: Moon geology

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the start of their learning cycle, after being shown pictures of the Moon (note that mare are referred to "lava plains"):

Which feature on the Moon is the oldest?
(A) Craters partially filled in with flat lava plains.
(B) Craters on top of flat lava plains.
(C) Flat lava plains.
(D) (More than one of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Section 70158 (pre-)
(A) : 12 students
(B) : 7 students
(C) : 9 students
(D) : 2 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.

Section 70158 (post-)
(A) : 17 students
(B) : 3 students
(C) : 10 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The principle of superposition from geology holds that the bottommost layer must be the oldest geological feature.

Pre- to post- peer-interaction gains:
pre-interaction correct = 40%
post-interaction correct = 57%
Hake (normalized) gain <g> = 28%

Which feature on the Moon is the youngest?
(A) Craters partially filled in with flat lava plains.
(B) Craters on top of flat lava plains.
(C) Flat lava plains.
(D) (More than one of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Section 70160 (pre- only)
(A) : 10 students
(B) : 2 students
(C) : 0 students
(D) : 4 students
(E) : 1 student
(F) : 9 students
(G) : 0 students

Correct answer: (A)

Correct = 54%

Astronomy clicker question: reionization

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

__________ is/are evidence of reionization in the early universe.

(A) Galaxy redshifts are proportional to galaxy distances.
(B) The near zero curvature of space-time.
(C) The amounts of helium and lithium in extremely old stars.
(D) Extremely distant galaxies surrounded by neutral hydrogen gas.
(E) The cosmic microwave background.
(F) (None of the above choices.)
(G) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 4 students
(B) : 1 student
(C) : 3 students
(D) : 10 students
(E) : 8 students
(F) : 0 students
(G) : 0 students

Correct = 38%

Correct answer: (D)

Reionization occurred when the first generation of stars formed the first galaxies, at the end of the dark ages. The light from these stars ionized hydrogen around them, creating emission nebulae.

Astronomy clicker question: main sequence to neutron star evolution

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

A __________ main sequence star will eventually become a neutron star.
(A) massive.
(B) medium-mass.
(C) low-mass.
(D) (More than one of the above choices.)
(E) (None of the above choices.)
(F) (I'm lost, and don't know how to answer this.)

Section 70158
(A) : 33 students
(B) : 9 students
(C) : 1 students
(D) : 0 students
(E) : 0 students

Correct = 74%

Correct answer: (A)

Massive stars main sequence stars will become supergiants, undergo type II supernovae explosions, and then can either become black holes or neutron stars, depending on the mass of the remaining core.

Section 70160
(A) : 22 students
(B) : 3 students
(C) : 1 student
(D) : 0 students
(E) : 0 students

Correct = 85%

Astronomy clicker question: extremely young cluster stars

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

__________ will be found in an extremely young star cluster.
(A) Supergiants.
(B) Red dwarfs.
(C) White dwarfs.
(D) Blue-hot main sequence stars.
(E) Yellow-hot main sequence stars.
(F) (More than one of the above choices.)
(G) (I'm lost, and don't know how to answer this.)

Section 70158 (pre-)
(A) : 20 students
(B) : 6 students
(C) : 6 students
(D) : 5 students
(E) : 2 students
(F) : 4 students
(G) : 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.

Section 70158 (post-)
(A) : 0 students
(B) : 6 students
(C) : 6 students
(D) : 22 students
(E) : 5 students
(F) : 4 students
(G) : 0 students

Correct answer: (D)

Massive stars (blue-hot main sequence stars, which become supergiants) evolve faster than medium mass stars (yellow-hot main sequence stars, which become giants, then white dwarfs), which evolve faster than low mass stars (red dwarfs). Thus blue-hot main sequence stars will be the youngest type of stars, and be found in extremely young stars clusters.

Pre- to post- peer-interaction gains:
pre-interaction correct = 12%
post-interaction correct = 51%
Hake (normalized) gain <g> = 45%

Section 70160 (pre-)
(A) : 10 students
(B) : 2 students
(C) : 0 students
(D) : 4 students
(E) : 1 student
(F) : 9 students
(G) : 0 students

Section 70160 (post-, after students were directed to not pick (F))
(A) : 20 students
(B) : 1 student
(C) : 5 students
(D) : 5 students
(E) : 0 students
(F) : 0 students
(G) : 0 students

Pre- to post- peer-interaction gains:
pre-interaction correct = 15%
post-interaction correct = 16%
Hake (normalized) gain <g> = 1%

Astronomy clicker question: same luminosity main sequence vs. giant star

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

A main sequence star will be ___________ compared to a giant star that has the same luminosity.
(A) cooler and smaller.
(B) cooler and larger.
(C) hotter and smaller.
(D) hotter and larger.
(E) (I'm lost, and don't know how to answer this.)

Section 70158 (pre-)
(A) : 11 students
(B) : 2 students
(C) : 31 students
(D) : 2 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.

Section 70158 (post-)
(A) : 3 students
(B) : 0 students
(C) : 38 students
(D) : 3 students
(E) : 0 students

Correct answer: (C)

From an H-R diagram, main sequence stars are hotter than giants that have the same luminosity. From the Stefan-Boltzmann law (luminosity proportional to size and temperature^4), since the main sequence star has the same luminosity as, but is hotter than the giant, then the main sequence star must be smaller in size.

Pre- to post- peer-interaction gains:
pre-interaction correct = 67%
post-interaction correct = 86%
Hake (normalized) gain <g> = 58%

Section 70160 (pre- only)
(A) : 3 students
(B) : 0 students
(C) : 22 students
(D) : 1 student
(E) : 0 students

Correct = 85%

Physics quiz question: comparing wire vs. string wave speeds

Physics 205A Quiz 6, fall semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 11.6, Multiple-Choice Question 11.5

A string and a wire are each 10.0 m long, and are stretched by the same tension of 30.0 N. The string has a mass of 2.40×10^–3 kg, and the wire has a mass of 1.90×10^–3 kg.

Waves will travel faster along the:
(A) string.
(B) wire.
(C) (There is a tie.)
(D) (Not enough information is given to determine this.)

Correct answer (highlight to unhide): (B)

Transverse waves on a string or wire is given by v = √(F/(m/L)), where both strings have the same tension F, but have different linear mass densities ("thicknesses") (m/L). Since the wire has a smaller mass per length, waves will travel faster along it.

Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 34 students
(C) : 2 students
(D) : 0 students

Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.21

Physics quiz question: load-bearing wire

Physics 205A Quiz 6, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 10.4

[3.0 points.] A wire of length 3.00 m with a cross-sectional area of 1.80e-5 m^2 stretches by 0.00250 m when a load of 2.00e+3 N is hung from it. What is the Young's modulus for this wire?
(A) 9.26e+4 N/m^2.
(B) 1.11e+8 N/m^2.
(C) 3.33e+8 N/m^2.
(D) 1.33e+11 N/m^2.

Correct answer: (D)

Hooke's law is (F/A) = Y*(delta(L)/L), so Y = (F*L)/(A*delta(L)). Response (A) is (F*delta(L))/(A*L); response (B) is (F/A), and response (C) is (F*L)/A.

Student responses
Sections 70854, 70855
(A) : 4 students
(B) : 3 students
(C) : 0 students
(D) : 34 students

"Difficulty level": 82%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50

Kepler's third law: Galilean satellites

081116-1060598-invert
http://www.flickr.com/photos/waiferx/3035594981/
Originally uploaded by Waifer X

Tracker chart for the Galilean satellites of Jupiter, adapted from the "Jupiter Moon Tracker Results" data from the NASA Planetary Rings Node website (http://pds-rings.seti.org/tools/tracker2_jup.html), used to determine the mass of Jupiter using Kepler's third law.

Astronomy quiz question: observing the early universe

Astronomy 210 Quiz 6, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] What makes it possible to observe the very early universe, just after the start of the big bang?
(A) Dark energy.
(B) Curvature of space-time in the universe.
(C) Doppler effect.
(D) Finite speed of light.

Section 70160
(A) : 0 students
(B) : 6 students
(C) : 5 students
(D) : 10 students

Correct answer: (C)

The finite speed of light means that it takes time for light from distant objects to reach the Earth. Thus looking at an object a given number of light years away means that it appears as it did that same number of years in the past. This is used to great advantage when observing extremely distant objects nearly as far out in light years as the universe is old.

Apparently students successfully discriminated against the dark energy distractor.

"Difficulty level": 65% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.40

Folded mountain range analog

081115-1060592
http://www.flickr.com/photos/waiferx/3032576847/
Originally uploaded by Waifer X

Carpet wrinkles, simulating a folded mountain range due to motion caused by plate tectonics.

Astronomy quiz question: Milky Way shape

Astronomy 210 Quiz 6, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

How do we know that the Milky Way is shaped like a flat disk?
(A) Statistically, most other galaxies are shaped like flat disks.
(B) Distant stars appear to be dimmer than nearby stars.
(C) Naked-eye observations at dark sky locations.
(D) The precession of the axis of the Earth's rotation.

Section 70158
(A) : 10 students
(B) : 15 students
(C) : 23 students
(D) : 8 students

Correct answer: (C)

Since the stars of the Milky Way form a band across the night sky, as seen from Earth, its overall shape must be thin and flat, as opposed to spherical.

"Success level": 40% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.51

Section 70160
(A) : 7 students
(B) : 9 students
(C) : 11 students
(D) : 2 students

"Success level": 44% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43

Astronomy quiz question: metal rich vs. metal poor stars

Astronomy 210 Quiz 6, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Older stars are metal poor, while newer stars are metal rich because:
(A) metals produced by older stars were released by supernova explosions, and became part of newer stars.
(B) dark matter gradually converts metal rich stars into metal poor stars.
(C) older stars concentrate more of their metals into their cores, leaving their outer layers metal poor.
(D) older stars have had more time to break down heavy elements into lighter elements.

Section 70158
(A) : 37 students
(B) : 0 students
(C) : 10 students
(D) : 17 students

Correct answer: (A)

Stars produce metals (elements heavier than hydrogen and helium) in their cores during their giant/supergiant phases, up through type Ia/II supernovae explosions. Along with their unused hydrogen, these metals are then scattered into the interstellar medium, which are then incorporated into later generations of stars. An old, early generation star will have metals only in its core, while a young, later generation star will have metals sprinkled in its outer layers.

Apparently all students successfully discriminated against the dark matter distractor.

"Difficulty level": 60% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63

The "Cooper Cooler(TM) Effect"

081106-1060446
http://www.flickr.com/photos/waiferx/3018649612/
Originally uploaded by Waifer X

Cooper Cooler(TM) apparatus in action for testing the hypothesis that an off-center impact on Uranus, which may have tilted its axis, would have also cooled it adversely due to stirring its interior, compared to Neptune, which would have remained relatively static, and cooled off at a slower rate. Two bottles containing equal amounts of warm water at the same initial temperature are either put into a Cooler Cooler(TM) beverage chiller, and spun while deluged by a cascade of ice water, or placed in a static ice bath for an equal amount of time. Afterwards, the final temperatures of the two bottles are compared.

Carbonaceous chondrite meteorite analog

081109-1060476
http://www.flickr.com/photos/waiferx/3016613907/
Originally uploaded by Waifer X

Raw spinach leaves versus cooked spinach leaves, as an analogy for meteorites containing carbonaceous chondrites (volatile materials which would not survive extensive heating, and thus may represent a sampling of the early solar nebula), versus achondrite meteorites, which lack volatile materials (and thus experienced extensive heating in its past).

Found physics: "Turkey/Cornish Hen Effect"

081108-1060468
http://www.flickr.com/photos/waiferx/3016610525/
Originally uploaded by Waifer X

The "Turkey/Cornish Hen Effect," illustrating the effect of mass on thermal capacity. Low-mass objects cool off much quicker than massive objects, which cool off at a slower rate.

Cuesta College North County campus star party

081108-1060468
http://www.flickr.com/photos/waiferx/3021862660/
Originally uploaded by Waifer X

Cuesta College Astronomy 210L students Nadia and Mary use digital cameras to take pictures through the eyepiece of one of the North County campus 8" reflectors during a North County campus star party.


081108-1060460
http://www.flickr.com/photos/waiferx/3021031647/
Originally uploaded by Waifer X

First quarter Moon (inverted), November 6, 2008, photographed with a hand-held Panasonic Lumix LZ-8 through a Meade LX200 8" reflector at Cuesta College North County campus, Paso Robles, CA.


081108-1060461
http://www.flickr.com/photos/waiferx/3021032599/
Originally uploaded by Waifer X

Cuesta College Astronomy 210L students and guests (aka "Starry Night Gangstas") during a North County campus star party.

Sights seen:
M57 (Ring Nebula)
M31 (Andromeda Galaxy)
M45 (Pleiades)
Epsilon Lyrae (The "Double-Double")
Alberio (Beta Cygni)
Vega
First quarter Moon

Physics quiz question: manometer

Physics 205A Quiz 5, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 9.24

A water manometer is attached to a flask containing a gas, where the other end is open to the atmosphere.

[3.0 points.] How does the pressure of the gas inside the flask compare to atmospheric pressure?
(A) P_flask = P_atm.
(B) P_flask > P_atm.
(C) P_flask < P_atm.
(D) (Not enough information is given to determine this.)

Correct answer: (C)

The manometer water level will be lower on the side that is at lower pressure, thus P_flask < P_atm. If the water levels were the same in either side of the U-tube, then P_flask = P_atm. If the water level was higher on the side of the U-tube open to the atmosphere, then P_flask > P_atm.

Student responses
Sections 70854, 70855
(A) : 2 students
(B) : 7 students
(C) : 31 student
(D) : 1 student

"Difficulty level": 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.15

[3.0 points.] How does the pressure at point [1] compare to the pressure at point [2]?
(A) P_1 = P_2.
(B) P_1 > P_2.
(C) P_1 < P_2.
(D) (Not enough information is given to determine this.)

Correct answer: (A)

According to Pascal's principle, "same fluid, same level, same pressure." In fact, P_1 = P_2 = P_atm.

Student responses
Sections 70854, 70855
(A) : 30 students
(B) : 6 students
(C) : 5 students
(D) : 0 students

"Difficulty level": 73%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.20

Physics quiz question: cable-held boom

Physics 205A Quiz 5, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.14

[3.0 points.] A uniform boom of length 1.20 m is held at an angle of 30.0° above the horizontal. A cable, at angle of 30.0° below the horizontal, is attached to the middle of the boom, and fastened to a wall. The boom is fastened to a hinge attached to the wall. Which is the lesser magnitude force acting on the boom?
(A) Tension.
(B) Weight.
(C) (Tension and weight have the same magnitude.)
(D) (Not enough information is given to determine this.)

Correct answer: (C)

The magnitude of the counterclockwise torque of the weight on the boom is:

tau_weight = r*F_perp = (0.60 m)*w*sin(60.0°).

This must be equal to the magnitude of the clockwise torque of the cable tension on the boom:

tau_tension = r*F_perp = (0.60 m)*T*sin(60.0°).

From inspection, the magnitudes of the weight and tension forces must be equal.

Student responses
Sections 70854, 70855
(A) : 11 students
(B) : 14 students
(C) : 15 students
(D) : 1 student

"Difficulty level": 36%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.12

Erasing slate: I love this!

"I love this!" by Anonymous
November 5, 2008
Cuesta College, San Luis Obispo, CA

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

Open, flat, and closed universes

081104-1060413
http://www.flickr.com/photos/waiferx/3003896971/
Originally uploaded by Waifer X

Models representing a negative curvature "open" universe, a zero curvature "flat" universe, and a positive curvature "closed" universe. The closed universe is constructed by removing two triangles from a flat sheet, and taping the resulting cutout edges together. These triangles are then taped into slits cut into another flat sheet, producing the open universe. The black-and-white inverted version of the same photo is shown below.

081104-1060413-inverted
http://www.flickr.com/photos/waiferx/3004952710/
Originally uploaded by Waifer X

Astronomy clicker question: spiral arm with supergiants and red dwarfs

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

It is possible for red dwarfs and supergiants to be found within the same spiral arm if:
(A) they began their main sequence lives in different spiral arms.
(B) the supergiants have already exploded as type II supernovae, but this light has not yet reached the Earth.
(C) they contain different concentrations of metals (elements heavier than hydrogen and helium).
(D) they are both remnants of medium mass main sequence stars.
(E) (I'm lost, and don't know how to answer this.)

Section 70158
(A) : 25 students
(B) : 7 students
(C) : 6 students
(D) : 13 students
(E) : 1 student

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.

Section 70158
(A) : 38 students
(B) : 5 students
(C) : 3 students
(D) : 8 students
(E) : 0 students

Correct answer: (A)

Massive stars live for a brief amount of time, evolving rapidly from protostars to main sequence stars to supergiants to type II supernovae. At the other extreme, low mass stars live for an inordinately long time, evolving from protostars to the main sequence, where they all still exist to this day as red dwarfs. So in order for a supergiant to be located in the same spiral arm as a red dwarf, they must have been born in separate locations--the massive star is born, lives, and dies within the same spiral arm, while the red dwarf is on its nth round trip, passing through the spiral arm. Responses (B) and (C) are not an important factors, while response (D) is obviously false.

Pre- to post- peer-interaction gains:
pre-interaction correct = 48%
post-interaction correct = 70%
Hake (normalized) gain <g> = 43%

Section 70160 (pre-)
(A) : 14 students
(B) : 7 students
(C) : 3 students
(D) : 4 students
(E) : 0 students

Section 70160 (post-)
(A) : 20 students
(B) : 8 students
(C) : 0 students
(D) : 0 students
(E) : 0 students

Pre- to post- peer-interaction gains:
pre-interaction correct = 50%
post-interaction correct = 71%
Hake (normalized) gain <g> = 43%

Astronomy clicker question: Milky Way mass

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

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

The total mass of the Milky Way can be estimated by:
(A) observing how quickly or slowly stars at different distances orbit its center.
(B) observing how quickly the supermassive black hole at its center is growing.
(C) carefully counting all the stars in the night sky.
(D) timing all of the Cepheid variable stars in the halo.
(E) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 14 students
(B) : 6 students
(C) : 0 students
(D) : 7 students
(E) : 0 students

Correct answer: (A)

Kepler's third law would be used to determine the total mass of the Milky Way. (This clicker question sets up the introduction of dark matter.) Response (D) would determine the luminosity of the Cepheid variable stars, from which the their distances could be found, resulting in the distance of the sun from the center of the Milky Way. Response (B) is a somewhat fanciful notion.

Astronomy quiz question: evolution to stellar remnants

Astronomy 210 Quiz 5, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[Version 1]

[4.0 points.] Which type of main-sequence star will eventually become a neutron star?
(A) Massive.
(B) Medium-mass.
(C) Low-mass.
(D) (None of the above choices.)

Section 70158
(A) : 34 students
(B) : 16 students
(C) : 8 students
(D) : 8 students

Correct answer: (A)

"Mass is destiny." Massive stars will become either neutron stars, or black holes; medium-mass stars will become white dwarfs, while low-mass stars become (and remain) red dwarfs, at least for now, given the current age of the universe.

"Difficulty level": 54%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.95

[Version 2]

[4.0 points.] Which type of main-sequence star will eventually become a white dwarf?
(A) Massive.
(B) Medium-mass.
(C) Low-mass.
(D) (None of the above choices.)

Section 70160
(A) : 0 students
(B) : 20 students
(C) : 5 students
(D) : 5 students

Correct answer: (B)

"Difficulty level": 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.42

Astronomy quiz question: red dwarfs as giants?

Astronomy 210 Quiz 5, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

A red dwarf will never become a giant because:
(A) it is made of degenerate matter.
(B) its core will never get hot enough to fuse helium.
(C) it will never run out of hydrogen to fuse.
(D) not a single red dwarf has died of old age anywhere in the universe.

Correct answer: (B)

A red dwarf has a main sequence lifetime much longer than the age of the universe, so none have actually "died" yet. However, when all of the hydrogen in a red dwarf has been depleted, it will gravitationally contract, but not enough to increase the temperature for helium to fuse, so it will never restart fusion to become a giant, as a medium mass main sequence star would at the end of its hydrogen fusion lifetime.

Section 70158
(A) : 2 students
(B) : 51 students
(C) : 6 students
(D) : 7 students

"Success level": 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.22

Section 70160
(A) : 1 student
(B) : 16 students
(C) : 8 students
(D) : 5 students

"Success level": 58%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.31

Overheard: large-impact hypothesis

Astronomy 210L, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

(Overheard in lab during an analysis of Moon formation theories, namely the large-impact hypothesis, where the Moon formed from the crustal debris flung off of the Earth from a Mars-sized protoplanet impactor.)

Student 1: "Wait...does that mean that the Moon is the Earth?"

Student 2: "What? The Moon is the Earth?!?"

Esprit d'escalier:
Instructor: "Dude! The Moon is the Earth. Have another brownie."