20081031

Physics clicker question: pressure underwater

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

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 9.18

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

P_atm = 101.3 kPa
Density of water rho_water = 1,000 kg/m^3

The pressure at point [2] is:
(A) 6.5 x 10^3 Pa.
(B) 1.1 x 10^5 Pa.
(C) 2.1 x 10^5 Pa.
(D) (I'm lost, and don't know how to answer this.)

Sections 70854, 70855
(A) : 1 student
(B) : 16 students
(C) : 20 students
(D) : 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.

Sections 70854, 70855
(A) : 0 students
(B) : 7 students
(C) : 28 students
(D) : 0 students

Correct answer: (C)

The pressure at point [1] is P_1 = P_atm = 1.013 x 10^5 Pa. Due to the depth underwater of point [2], the pressure there is:

P_2 = P_1 + rho_water*g*d,

where g = 9.80 m/s^2, and d = 11.0 m. Response (B) is the rho_water*g*d term only, while response (A) is P_1 - rho_water*g*d.

Pre- to post- peer-interaction gains:
pre-interaction correct = 53%
post-interaction correct = 80%
Hake, or normalized gain = 58%

20081030

Astronomy clicker question: helium fusion 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:

__________ stars will be able to fuse helium.
(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 70160
(A) : 3 students
(B) : 8 students
(C) : 4 students
(D) : 10 students
(E) : 1 student
(F) : 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 70160
(A) : 1 student
(B) : 2 students
(C) : 0 students
(D) : 25 students
(E) : 0 students
(F) : 0 students

Correct answer: (D)

Low-mass stars (red dwarfs) will never get hot enough to fuse helium. Both medium mass and massive stars will eventually fuse helium in their giant and supergiant stages, respectively.

Clarification from one student who wanted all (A), (B), and (C) choices incorporated in his (D) reply: "I thought the question meant which stars make helium." (I.e., does "bake a cake" refer to making a cake from ingredients, or taking a finished cake and putting it into the oven?)

Pre- to post- peer-interaction gains:
pre-interaction correct = 37%
post-interaction correct = 89%
Hake (normalized) gain <g> = 83%

20081029

Overheard: accretion vs. secretion

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

(Overheard in class during an in-class activity on mass transfer in close binary star systems.)

Student: "What does 'accretion' stand for?"

Instructor: "As in an 'accretion disk?' ...It's 'gathering' stuff from the companion star."

Student: "So, it's like the opposite of 'secretion?' Like in giving out stuff?"

Instructor: "Yeah--like exuding or oozing stuff out."

Esprit d'escalier:
Instructor: "'Accrete' is an obscure term, but it's similar to 'acquire' and 'accumulate.'"

20081028

Astronomy clicker question: isolated white dwarf--nova?

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:

An isolated white dwarf cannot explode as a nova because it:
(A) has no companion star to heat up.
(B) expended all of its extra energy during the planetary nebula phase.
(C) has no new source of hydrogen.
(D) does not have enough degeneracy pressure.
(E) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 15 students
(B) : 2 students
(C) : 10 students
(D) : 0 students
(E) : 0 students
(F) : 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 70160
(A) : 9 students
(B) : 0 students
(C) : 19 students
(D) : 0 students
(E) : 0 students
(F) : 0 students

Correct answer: (C)

A white dwarf in a close binary system will take hydrogen from its companion star, and if this mass transfer is sufficiently slow, will undergo a nova explosion. (A rapid mass transfer would result in a type Ia supernova explosion.)

Pre- to post- peer-interaction gains:
pre-interaction correct = 37%
post-interaction correct = 68%
Hake (normalized) gain <g> = 49%

20081027

Erasing slate: Happy Day!

"Happy Day! [Heart]" by Anonymous
October 20, 2008
Cuesta College, San Luis Obispo, CA

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

20081026

Physics quiz question: cart on track

Physics 205A Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 6.10, 6.26

A 2.20 kg cart moving to the right has a speed of 12.0 m/s on the lower horizontal part of this track, and travels to the upper horizontal part of this track. Neglect friction and drag.

[3.0 points.] What is the kinetic energy of the cart on the lower horizontal part of this track?
(A) 16.2 J.
(B) 26.4 J.
(C) 142 J.
(D) 158 J.

Correct answer: (D)

This is a simple calculation of K_i = (1/2)*m*(v_i)^2. Response (A) is m*g*y_f; response (B) is m*v_i; response (C) is the final kinetic energy of the cart at the top of the ramp.

Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 10 students
(C) : 3 students
(D) : 26 students

"Difficulty level": 61%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47

[3.0 points.] What is the speed of the cart on the upper horizontal part of this track?
(A) 3.83 m/s.
(B) 7.35 m/s.
(C) 11.4 m/s.
(D) 12.6 m/s.

Correct answer: (C)

From before, K_i = 158 J. The increase in gravitational potential energy as it climbs up to the upper horizontal part of the track is delta(U) = m*g*delta(y) = 16.2 J. Thus the final kinetic energy will be 158 J - 16.2 J = 142 J; and the final speed will be v_f = sqrt(2*K_f/m) = 11.4 m/s.

Response (A) is sqrt(2*g*delta(y)), which is the speed that the cart would have at the lower part of the track, if it had started at rest from the upper part of the track. Response (B) is the maximum height that the cart can reach uphill (with the wrong units). Response (D) is sqrt((v_i)^2 + 2*g*delta(y)), which has a sign error inside of the square root, and results in a final speed greater than the initial speed.

Student responses
Sections 70854, 70855
(A) : 12 students
(B) : 8 students
(C) : 17 students
(D) : 4 students

"Difficulty level": 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.57

20081025

Astronomy clicker question: very young star cluster

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, during a review session before a midterm:

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

Section 70160
(A) : 17 students
(B) : 2 students
(C) : 1 student
(D) : 6 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results.

Section 70160
(A) : 15 students
(B) : 0 students
(C) : 0 students
(D) : 10 students
(E) : 0 students

Correct answer: (A)

Low-mass protostars take a very long time to become a main sequence star; while massive protostars will take a very short time to reach a main sequence, live out its main sequence lifetime, and then become a supergiant. In-between are medium mass main sequence stars, which become giants, planetary nebulae, and then white dwarfs.

During the instructor-facilitated discussion following the second round of clickers, students who selected (D) said that a very young star cluster would have both supergiants and white dwarfs--which was then immediately shouted down by their fellow students.

Pre- to post- peer-interaction gains:
pre-interaction correct = 65%
post-interaction correct = 60%
Hake (normalized) gain = -16%

20081024

Astronomy clicker question: young or old star cluster?

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, during a review session before a midterm:

The H-R diagram of a star cluster shown at right shows:
(A) stars that are all old.
(B) stars that are all young.
(C) stars that are a mixture of young and old stars.
(D) (This H-R diagram is not possible for a star cluster.)
(E) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 12 students
(B) : 1 student
(C) : 11 students
(D) : 1 student
(E) : 0 students

This question was not asked again after displaying the tallied results, but was discussed with the whole class with the instructor facilitating students who gave opinions on either (A) or (C).

Correct answer: (A)

Low-mass protostars take a very long time to reach the main sequence; while massive stars take a very short time to reach the main sequence. Low-mass stars also never leave the main sequence (this is yet to happen, and would require about three times the current age of the universe).

Some students had thought that the massive main sequence stars signaled a young star cluster, while the low-mass stars had left the main sequence, indicating a very old star cluster.

Student: "All stars in a star cluster have to be the same age!"

Pre- to post- peer-interaction gains: (N/A)

20081023

Collision flowchart

"Collision Flowchart," Patrick M. Len
October 17, 2008

Flowchart detailing the decision tree in determining which conservation laws apply to a particular type of collision.

20081022

Astronomy quiz question: blue main sequence star vs. blue supergiant

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] A blue main sequence star is known to be smaller than a blue supergiant because it is:
(A) same temperature as, but more luminous than a blue supergiant.
(B) same temperature as, but less luminous than a blue supergiant.
(C) same luminosity as, but hotter than a blue supergiant.
(D) same luminosity as, but cooler than a blue supergiant.

Section 70158
(A) : 9 students
(B) : 50 students
(C) : 8 students
(D) : 2 students

Correct answer: (B)

From Wien's law, the main sequence star and the supergiant must have the same temperature, because they have the same color. From the Stefan-Boltzmann law, for two stars of the same temperature, the less luminous star must be smaller in size.

"Difficulty level": 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.38

20081021

Astronomy quiz question: stellar distances

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] Sirius A is a star that has an apparent magnitude m = –1.5 and an absolute magnitude M_v = +1.41. Approximately how far away is Sirius A from Earth?
(A) Much farther away than 10 parsecs.
(B) A little farther away than 10 parsecs.
(C) Exactly 10 parsecs away.
(D) Closer than 10 parsecs away.

Section 70158
(A) : 9 students
(B) : 14 students
(C) : 0 students
(D) : 45 students

Correct answer: (D)

When brought to the "fair distance" of 10 parsecs (thus changing its apparent magnitude to absolute visual magnitude), Sirius A now seems to be dimmer, indicating that it must have moved away to the Earth during this process. Thus Sirius A must be closer than this fair distance of 10 parsecs.

"Difficulty level": 67%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.64

20081020

Astronomy quiz question: continuous spectrum source

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] A continuous spectrum is caused by:
(A) blackbody radiation passing through diffuse, cool gas atoms.
(B) electrons moving to lower energy orbitals.
(C) hot, agitated electrons and atoms.
(D) the Doppler effect.

Correct answer: (C)

The electrons and atoms in a hot, dense, opaque object (a "blackbody") will emit a continuous spectrum.

Student responses
Section 70158
(A) : 10 students
(B) : 14 students
(C) : 41 students
(D) : 4 students

"Difficulty level": 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.55

20081019

Astronomy quiz question: red supergiant vs. red giant

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] A red supergiant is known to be larger in size than a red giant because it is:
(A) same temperature as, but more luminous than a red giant.
(B) same temperature as, but less luminous than a red giant.
(C) same luminosity as, but hotter than a red giant.
(D) same luminosity as, but cooler than a red giant.

Section 70160
(A) : 29 students
(B) : 0 students
(C) : 1 student
(D) : 1 student

Correct answer: (A)

From Wien's law, the supergiant and the giant must be the same temperature, because they have the same color. From the Stefan-Boltzmann law, for two stars of the same temperature, the more luminous star must be larger in size.

"Difficulty level": 93%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25

20081018

Astronomy quiz question: stellar distances

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] Procyon A is a star that has an apparent magnitude m = +0.3 and an absolute magnitude M_v = +2.64. Approximately how far away is Procyon A from the Earth?
(A) Much farther away than 10 parsecs.
(B) A little farther away than 10 parsecs.
(C) Exactly 10 parsecs away.
(D) Closer than 10 parsecs away.

Section 70160
(A) : 3 students
(B) : 3 students
(C) : 0 students
(D) : 25 students

Correct answer: (D)

When brought to the "fair distance" of 10 parsecs (thus changing its apparent magnitude to absolute visual magnitude), Procyon A now seems to be dimmer, indicating that it must have moved away to the Earth during this process. Thus Procyon A must be closer than this fair distance of 10 parsecs.

"Difficulty level": 81%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50

20081017

Astronomy quiz question: absorption spectrum source

Astronomy 210 Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[4.0 points.] An absorption spectrum is caused by:
(A) the Doppler effect.
(B) hot, agitated electrons and atoms.
(C) electrons moving to lower energy orbitals.
(D) blackbody radiation passing through diffuse, cool gas atoms.

Correct answer: (D)

The electrons in the atoms in a cool, diffuse gas will absorb select wavelengths from a continuous spectrum source.

Student responses
Section 70160
(A) : 1 student
(B) : 11 students
(C) : 6 students
(D) : 13 students

"Difficulty level": 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.49

20081016

Overheard: midterm studying

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

(Overheard in class before their first midterm.)

Student: "This is like the hardest midterm I've ever had to study for."

20081015

Overheard: "ex-non"

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

(Overheard in astronomy laboratory, where students have to identify an unknown gas from observing their spectral lines.)

Student: "Ex-non." (I.e., xenon.)

20081014

Astronomy midterm question: telling "Moon-time"

Astronomy 210 Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[20 points.] Shown at right is an excerpt from a comic strip* printed on March 23, 2008. Discuss what hour it is if the Moon is highest overhead with this phase, using a diagram showing the positions of the Sun, Moon, and Earth.

*The Family Circus, Bill Keane, ©King Features Syndicate, Inc.

Solution and grading rubric:
  • p = 20/20:
    Correct. Phase as shown is waning crescent, which is highest overhead at 9 AM (rises at 3 AM, and sets at 3 PM). Explanation includes a complete Sun-Moon-Earth diagram.
  • r = 16/20:
    Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Diagram is correct, but time is wrong.
  • t = 12/20:
    Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least has a serious attempt at a Sun-Moon-Earth diagram.
  • v = 8/20:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. May involve lunar eclipses.
  • x = 4/20:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y = 2/20:
    Irrelevant discussion/effectively blank.
  • z = 0/20:
    Blank.
Grading distribution:
Section 70158
p: 36 students
r: 7 students
t: 21 students
v: 13 student
x: 0 students
y: 1 student
z: 0 students

A sample "p" response (from student 1616):

Astronomy midterm question: (non-)improvement in telescope powers?

Astronomy 210 Midterm 1, fall semester 2008
Cuesta College, San Luis Obispo, CA

Suppose that there is no turbulence in the upper atmosphere above San Luis Obispo, CA. If a telescope is used that evening, discuss which powers (light-gathering, resolving, magnifying) would not be improved because of this weather phenomenon, and explain why.

Solution and grading rubric:
  • p:
    Correct. Light-gathering power depends on the area of the primary lens or mirror, limited by light pollution; resolving power depends on the width of the primary lens or mirror, limited by atmospheric seeing conditions (turbulence); magnifying power depends on the focal lengths of the primary lens or mirror and eyepiece lens. Resolving power will be improved. Discusses light-gathering power and magnifying power as not being improved by no turbulence. (May debate magnifying power improving in that the maximum useful magnification is improved, although the factor that images are magnified is unchanged).
  • r:
    Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Resolving power improved as in (p), but only one of light-gathering power or magnifying power as not improving.
  • t:
    Contains right ideas, but discussion is unclear/incomplete or contains major errors. Claims resolving power (in addition to either light-gathering power or magnifying power) would not be improved by lack of turbulence.
  • 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:
Section 70160
p: 8 students
r: 17 students
t: 7 students
v: 1 student
x: 0 students
y: 1 student
z: 0 students

A sample "p" response (from student 8187):

Another "p" response (from student 9387):

20081013

Astronomy midterm question: Ptolemy/Copernicus as scientists?

Astronomy 210 Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

[20 points.] Explain whether Ptolemy or Copernicus (or both, or neither) could be considered scientists (basing explanations on observations and evidence, instead of first principles), by discussing their accomplishments and their theories.

Solution and grading rubric:
  • p = 20/20:
    Correct. Although Ptolemy and Copernicus both developed theories that explained observations of retrograde motion of planets, both based their theories on first principles (perfect circles) rather than strictly on observations, wich resulted in a lack of accuracy in predicting planetary motion. Discusses at least Ptolemy's or Copernicus' first principles in detail.
  • r = 16/20:
    Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Discusses Ptolemy and Copernicus as scientists because of the theories they developed to explain observations of planetary motion, even though both their theories were also based on first principles.
  • t = 12/20:
    Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least discusses the theories of Ptolemy and Copernicus.
  • v = 8/20:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Does not offer specific details on contributions of Ptolemy or Copernicus.
  • x = 4/20:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y = 2/20:
    Irrelevant discussion/effectively blank.
  • z = 0/20:
    Blank.
Grading distribution:
Section 70160
p: 11 students
r: 9 students
t: 3 students
v: 9 students
x: 0 students
y: 2 students
z: 0 students

Section 70158
p: 18 students
r: 16 students
t: 20 students
v: 22 students
x: 0 students
y: 2 students
z: 0 students

A sample "p" response (from student 4652):A shot-in-the-dark "v" response (from student 8484):

Astronomy midterm question: the "West Star"

Astronomy 210 Midterm 1, fall semester 2008
Cuesta College, San Luis Obispo, CA

[20 points.] Shown at right is an excerpt from a comic strip* printed on June 24, 1970. Discuss whether or not there is such a thing as a "West Star" that can be used for navigation. Defend your answer by clearly explaining how you used a starwheel to determine this.

*Peanuts, Charles M. Schulz, ©United Features Syndicate, Inc.

Solution and grading rubric:
  • p = 20/20:
    Correct. Discusses why there would be no "West Star," as over the course of an evening, all non-circumpolar stars rise from the east, move across the sky, and set in the west, such that a given star that was near the west horizon would set below the horizon. May instead argue for a temporary "West Star" that can be used for navigation, given a starwheel and the date/time.
  • r = 16/20:
    Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
  • t = 12/20:
    Contains right ideas, but discussion is unclear/incomplete or contains major errors.
  • v = 8/20:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
  • x = 4/20:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y = 2/20:
    Irrelevant discussion/effectively blank.
  • z = 0/20:
    Blank.
Grading distribution:
Section 70160
p: 25 students
r: 3 students
t: 4 students
v: 0 students
x: 1 students
y: 0 student
z: 1 students

A sample "p" response (from student 2639):A more elaborate "p" response (from student 4731):
A "p" response (from student) arguing how a "West Star" (albeit, a temporary one) can be used for navigation:

20081012

Physics midterm question: train engine-pulled caboose

Physics 205A Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 4.10(b)

[10 points.] A freight train consists of an engine and a caboose on level ground. If the train (engine and caboose) is moving at constant speed, discuss whether or not the engine's pull on the caboose exceeds the caboose's backward pull on the engine. Explain your reasoning using the properties of forces and of Newton's laws.

Solution and grading rubric:
  • p = 10/10:
    Correct. The force of the engine pulling on the caboose must be equal and opposite to the force of the caboose pulling on the engine by Newton's third law, as they are an interaction pair. May also discuss Newton's first law, which applies in this situation, but this does not by itself demand that these two forces be equal (which would also be true even if the engine and caboose were accelerating!).
  • r = 8/10:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
  • t = 6/10:
    Nearly correct, but argument has conceptual errors, or is incomplete. Has some understanding that Newton's third law is used to analyze these two forces.
  • v = 4/10:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Uses Newton's first law (which cannot be used on two forces that act on two different objects), or Newton's second law (which is only applied to accelerating objects).
  • x = 2/10:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y = 1/10:
    Irrelevant discussion/effectively blank.
  • z = 0/10:
    Blank.

Grading distribution:
Sections 70854, 70855
p: 8 students
r: 4 students
t: 3 students
v: 27 students
x: 2 students
y: 0 students
z: 1 student

A sample of a "p" response involving Newton's third law (from student 5880) is shown below (although discussion takes a tangent into Newton's first law):

Physics midterm question: constant velocity, straight-line trajectory?

Physics 205A Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 3.8

[10 points.] If an object is traveling at constant velocity, discuss whether or not it is necessarily traveling in a straight line. Explain your reasoning using the properties of velocity and acceleration vectors.

Solution and grading rubric:
  • p = 10/10:
    Correct. Recognizes that constant velocity has constant speed and direction; or that acceleration must be zero; or displacement must be changing at a constant rate; or that Newton's first law must apply in order that the object must be traveling in a straight line (at constant speed). May instead argue that a stationary object has a constant velocity (of zero!), but does not move in a straight line.
  • r = 8/10:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May interpret problem as asking about average velocity, in which case uniform circular motion would have a constant average velocity, if the time interval is the same as the period of revolution.
  • t = 6/10:
    Nearly correct, but argument has conceptual errors, or is incomplete.
  • v = 4/10:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
  • x = 2/10:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y = 1/10:
    Irrelevant discussion/effectively blank.
  • z = 0/10:
    Blank.

Grading distribution:
Sections 70854, 70855
p: 20 students
r: 6 students
t: 2 students
v: 13 students
x: 4 students
y: 0 students
z: 0 students

A sample of a "p" response (from student 5880) where the stationary case is argued:
An "r" response (from student 7220) who reinterprets the question as referring to average velocity:

Physics midterm question: highest point in trajectory

Physics 205A Midterm 1, fall semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 3.37

From the edge of the rooftop of a building, at t = 0 a Physics 205A student throws a ball with an initial speed of 15.5 m/s at angle of 40.0° above the horizontal. Find (a) the maximum height of the ball as measured from the point from which it was thrown, and (b) the time that the ball reaches its maximum height. Neglect air resistance. Show your work and explain your reasoning.

Solution and grading rubric:
  • p:
    Correct. Finds x- and y-components of initial velocity: v0x = +11.9 m/s, v0y = +9.96m/s. First solves for either the t or y of the highest point, and uses this to solve for the remaining unknown.
  • r:
    Nearly correct, but includes minor math errors.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has t or y correct, has problems finding the remaining unknown.
  • v:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  • x:
    Implementation of ideas, but credit given for effort rather than merit.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.

Grading distribution:
Sections 70854, 70855
p: 19 students
r: 3 students
t: 5 students
v: 17 students
x: 1 student
y: 0 students
z: 0 students

A sample of a "p" response, solving for the time that the vertical velocity would be zero, and then using that time to solve for the vertical displacement (from student 0818):Another "p" response (from student 1336), solving for the displacement when the vertical velocity is zero, and the going back and solving for the time:

Yet another "p" response (from student 7175), where the time when the displacement is zero is solved for, and halved to find the time of the highest point, which is then used to solve for the maximum vertical displacement:

Physics midterm problem: penny on turntable

Physics 205A Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problem 5.80(b)

[20 points.] A Physics 205A student places a penny (mass 2.50 g) on an old phonograph turntable, at a distance of 10.0 cm from the center. The coefficient of static friction between the penny and the turntable is 0.19. The turntable starts with an initial angular velocity of 33.3 rpm, and speeds up to 45.0 rpm. Will the penny stay on the turntable? Neglect air resistance. Show your work and explain your reasoning.

Solution and grading rubric:
  • p = 20/20:
    Correct. Argues that the penny will slide off the turntable before reaching its final speed of 45 rpm, by either solving for the required radial (centripetal) net force and showing that it is greater than the maximum static friction force; or by solving for the rotational speed that the penny would still remain on the turntable (when its static friction force is maxed out), and showing that it is less than 45 rpm = 4.7 rad/s. Free-body diagram, identification of forces, application of Newton's laws (first law, and second law for uniform circular motion), and interpretation of results is clearly shown.
  • 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 has free-body diagram with magnitudes of forces calculated, and/or rational attempt at applying Newton's second law for uniform circular motion to these force magnitudes.
  • v = 8/20:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically calculates w_i, w_f, f_s, and/or m*r*w^2, but does not implement Newton's second law for uniform circular motion.
  • 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: 8 students
r: 1 students
t: 9 students
v: 20 students
x: 6 students
y: 1 student
z: 0 students

A sample of a "p" response, demonstrating that the angular speed that the penny would begin to slide off the turntable is less that the final angular speed (from student 1977):
Another "p" response (from student ), demonstrating that the maximum static friction force is less than the centripetal net force required to keep the penny on the turntable at its final angular speed:

20081011

Astronomy in-class activity: OBAFGKM poetry slam

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

Astronomy 210 In-class activity 13: OBAFGKM Poetry Slam

Students were instructed to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.

Oh Be A Friend Give Kelsey Money--Rent Needs Some Contributions
--K. B.

Orchids Bloom After Furry Giants Kill Mammoths, Rendering New Sweet Coats
--C. B.

Only Brahe Accepted Friendly German Kepler's Mind
--M. D.

One Bat And Four Gorillas Killed Mice!
--T. D.

Obsessive Boyfriends Are From Girlfriends Kissing Me
--R. G.

Oh Bartender, Another Full Glass--Keeps Me Sitting Next To Rachel
--S. G.

Observe, But Actually Find--Great Knowledge [comes from] Mind
--N. G.

On Birthdays All Friends Go Kayaking Merrily
--T. H.

Open Bets Are For Getting Kindly Manipulated Losers' Trust
--C. H.

Older Brother Asked For Good Knitted Mittens
--P. K.

Organic Bananas Always Feed Good Kids' Minds
--A. M.

Orion Built And Forged God-King's Monoceros [and] Rounded New Space Continuums
--D. M.

Oxygen Buys A Few Good Kind Minutes
--L. M.

Only Big Alligators Fight Giant Killing Monsters
--A. M.

Oh Boy! Apples From Grandma Keep Me Light [and] Trim
--H. R.

Oregan Beavers Are Freakin' Geeks; Kill Me Right Now, Southern California
--S. R.

Opossom Bones Aptly Fill Glad K9s Mouths, Really Nicely Saving Cats
--M. S.

Other Babes Are Fully Grouchy Kissing Michael
--K S.

Only Burp After Finishing Good Killer Munchies
--T. S.

Outrageously Blunt Axes Fight Gigantic Killer Mice
--P. S.

Only Best And Friendly Galileo Killed Mayhem
--K. U.

Ocean Bears Are Furry Generous Kind Mammalia; Reptiles Not So Cuddly
--B. U.

Ordinarily Big Apes Falter Giving Kinky Massages
--E. V.

Oh Behave! Another Former Girlfriend Keeps Me Reminded: Never Stop Calling
--R. W.

On Bank Appointment Frank Gave Kim Money
--L. Y.

Oh Baby, Another Fun Guy Kisses Me Late Tonight
--E. A.

Oh Boys! All Freakin' Girls Know More!
--B. D.

Obnoxious Bad Asses [Who] Find Good Karma Meaningless Really Need Survival Clues
--C. G.

Open Books And Finals Get Kids Mad Remember Never Study Chemistry
--J. G.

Older Baby Anteaters Feel Great Killing Mosquitoes Not Locusts
--M. G.

Only Besides Angels Furious Girls Kill Men
--J. K.

Oh Bummer Ass Farmer Genghis Khan Killed Me Remember No Such Character
--D. L.

Oooh Baked Applecrisp From Grandma's Kitchen Mmmm
--L. M.

Oh Boy A Free Green Kitchen Mug
--J. P.

Obviously Big Asteroids Foreshadow God Killing Mankind
--J. R.

[Detail of previous OBAFGKM poem]


Only Buy Acid From Grey Killer Monkeys
--D. U.

Previous post: Astronomy in-class activity: OBAFGKM poetry slam instructions.

Astronomy in-class activity: OBAFGKM poetry slam (NCC)

Astronomy 210, Fall Semester 2008
Cuesta College, San Luis Obispo, CA
(North County Campus)

Astronomy 210 In-class activity 13: OBAFGKM Poetry Slam

Students were instructed to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.

Only Ballin' Astronomy Funny Guys Kick Moons Right Near [the] Sun's Core
--N. C.

Obama Blamed Angels For Grotesquely Killing McCain Right Near South Carolina
--R. F.

Orbital Boundaries Away From Gravity Kept Moons Running North-South Losing Time
--J. H.

Oh Biannually A Footballing Giant Kicks Martians Lightyears
--C. J.

Only Bad Asses Forgive Girls Kicking Mr. Len's Tail
--W. L.

Oftentimes Brainlessness And Failing Grades Keep Me Retaking Nasty School Classes
--M. P.

Oh Boy Another Freaking Gangmember Killing Musicians Like Tupac
--J. P.

On Birthdays And Fiestas [you] Get Kissed More
--H. R.

Oh Boy Astronomy Facts Give Killer Migraines
--M. R.

Oprah Boasted About Finding Good Komfy Mocasins
--J. S.

Occasionally Bullets And Frag Grenades Kill Marmosets
--R. S.

Onion Bagels Are For Great Kings' Meals
--J. T.

Officer Brad Always Finds Grizzled Killer Murderers Really Near Small Communitiess
--R. W.

Oh Bender Aid Friends Grooming Killer Monkeys
--A. H.

On [the] Boat A Fiery Goat Killed Me
--Z. M.

Oh Brother, Another Female Girl Kicked Me
--D. S.

Old Bitches Are Far Greater Kite Makers
--J. W.

Previous post: Astronomy in-class activity: OBAFGKM poetry slam instructions.

Education research: partial credit for multiple-choice questions (PCMC) student attitudes (Fall semester 2008)

During Fall semester 2008, Cuesta College students taking Astronomy 210 (introductory astronomy) at Cuesta College, San Luis Obispo, CA have piloted the use of partial credit for multiple-choice (PCMC) questions.

During the seventh week of instruction (after taking three 10-question quizzes), students were given the opportunity to evaluate PCMC in an online "Partial Credit for Multiple-Choice (PCMC) Survey" hosted by SurveyMonkey.com.

These are the complete survey results. Analysis will be forthcoming after more data has been compiled from future semesters. Values for the mean and standard deviations are given next to the modal response category for each question. Note that the order of questions within sections II and III were randomly scrambled for each student.
Partial Credit for Multiple-Choice Survey
Cuesta College
Astronomy 210 Fall Semester 2008 sections 70158, 70160
(N = 40)

I. In order to receive credit for completing this survey,
first enter your first and last name below:
____


II. Recall that for this semester there is PCMC = "partial
credit for multiple choice," where you can get partial
credit even after circling a wrong answer, if you have
also eliminated an incorrect answer by marking it with
an "x."
Answer the following statements which may or may not
describe your beliefs about PCMC in this class.
Because of PCMC...

II.1 ...I read questions more carefully.
1. Strongly disagree 1 : *
2. Disagree 1 : *
3. Neutral 3 : ***
4. Agree 19 : ******************* [4.2 +/- 0.9]
5. Strongly agree 16 : ****************

II.2 ...I spend more time taking a test.
1. Strongly disagree 1 : *
2. Disagree 4 : ****
3. Neutral 9 : *********
4. Agree 16 : **************** [3.7 +/- 1.0]
5. Strongly agree 9 : *********

II.3 ...I feel more confident about circling a correct choice.
1. Strongly disagree 2 : **
2. Disagree 5 : *****
3. Neutral 9 : *********
4. Agree 19 : ******************* [3.5 +/- 1.0]
5. Strongly agree 5 : *****

II.4 ...I look for incorrect choices before circling a correct choice.
1. Strongly disagree 2 : **
2. Disagree 7 : *******
3. Neutral 6 : ******
4. Agree 11 : *********** [3.7 +/- 1.3]
5. Strongly agree 14 : **************

III. Answer the following statements which may or may not describe
your beliefs about PCMC in this class.

III.1 PCMC helped me answer test questions better.
1. Strongly disagree 0 :
2. Disagree 3 : ***
3. Neutral 10 : **********
4. Agree 17 : ***************** [3.9 +/- 0.9]
5. Strongly agree 10 : **********

III.2 I would recommend using PCMC in future semesters of this class.
1. Strongly disagree 0 :
2. Disagree 0 :
3. Neutral 0 :
4. Agree 22: ********************** [4.5 +/- 0.5]
5. Strongly agree 18 : ******************

III.3 Having PCMC a positive experience.
1. Strongly disagree 0 :
2. Disagree 0 :
3. Neutral 1 : *
4. Agree 29 : ***************************** [4.2 +/- 0.5]
5. Strongly agree 10 : **********

III.4 Using PCMC was difficult.
1. Strongly disagree 22 : **********************
2. Disagree 15 : *************** [1.5 +/- 3.5]
3. Neutral 3 : ***
4. Agree 0 :
5. Strongly agree 0 :

III.5 I would identify incorrect choices even without PCMC.
1. Strongly disagree 0 :
2. Disagree 5 : *****
3. Neutral 14 : **************
4. Agree 16 : **************** [3.5 +/- 0.9]
5. Strongly agree 5 : *****

IV. (Optional.) Please type in any comments you may have regarding
the use of PCMC in this class.
The following are all of the student responses to this question, verbatim and unedited.
"P-Pre-ordained C-Cows M-Maul C-Children Mwaaahahaaahaa"

"I think it makes you read the questions more carefully so you can make sure that if you don't know the answer you can at least get some credit for knowing what's NOT the answer, which I think somewhat reinforces knowledge."

"I try my hardest in this class adn I still suck!....Ah"

"I think that using PCMC is another way we can earn points on out tests. Knowing the wrong answer is like know what the right answer is. I believe it is helpful and that every class should use this system."

"In my opinion, PCMC is a really great idea. I don't know how anyone could not like the ability to receive extra credit. It doesn't change the way I read or answer questions on a quiz, but I do appreciate the partial credit if I'm not sure on a question. Every point counts on these killer Astronomy quizzes, right? :)"

"As far as the top questions I'm not sure I understood the exact meaning, I wasn't trying to contradict myself if i inadvertently did..."
It helps out.

"I like using PCMC because if I get a question wrong on the test but I "X" out the wrong answer, I still get partial creadit and it still gives me a chance to do well on the test even if i don't answer correctly."

"Partial credit for multiple choice is a great way to get points even if you miss the question. Sometimes i know wrong answers but i can not make up my mind on the correct answer. In this case PCMC is a great way to reward students for at least being able to identify wrong answers."

"At least you get credit if get the wrong answer and some credit is better than none"

"I dig it."

"My tactic is to find the right answer first, when I achieve this, I cross out the one I know is incorrect. Other times, if I am unsure with a correct answer, the thought of PCMC gives me a glimpse of hope. Otherwise, PCMC is not a system I employ to answer a question correctly, but it does make me consider each answer, and this perhaps leads to better scores in the end."

"I feel like I am always sure which answer could be incorrect on quiz's and i use PCMC everytime whether i am sure on a question or not. I don't always feel like my points get counted for crossing out an answer. I think it really would be helpful to use PCMC if their were good results for it. It shows we know the answer, but the question was confusing."

"I think it is beneficial overall. I have never crossed another instructor who has done it, so I hope you inspire others. The only complaint would be it seems it would be harder to grade with precision."

"PCMC is helpful because it directs my thoughts to eliminate the incorrect answers and that helps with choosing the correct one."

"I like PCMC, because sometimes I am not sure about the answers, but I know some answers are not correct. Because of PCMC, I still have chances to get points. This is a good system."

"I don't know a lot about astronomy, so the partial credit really helps"

20081010

Education research: partial credit for multiple-choice questions (PCMC) bibliography

Fall semester 2008 has partial credit for multiple-choice questions (PCMC) being piloted for Astronomy 210 (introductory one-semester astronomy) at Cuesta College (San Luis Obispo, CA). The intent is to get students to spend more time critically selecting their answers, especially with regards to eliminating incorrect answers. Observations of students in previous semesters is that students spend approximately one minute per question before turning in their completed quizzes, which could be considered much too little time, given that there are two minutes per question allotted.

Quizzes have 10 multiple-choice questions, each with four (A)-(D) choices.

Sample quiz questions from Fall semester 2008:
http://tinyurl.com/3tbhnz

From the course policy:
"Partial credit (0.5 points) will be given for an incorrectly circled answer to a multiple-choice question if a single "x" is also made on one incorrect response to that question. No other circumstance other than that previously described will result in partial credit."
A briefer statement on the quiz itself:
"Closed book, closed notes. Clearly circle ("O") the one choice that you think is most definitely correct. Cross out ("×") only one choice that you think is definitely incorrect."
A correct circled answer is 4.0 points, and 0.5 points is awarded for a successfully identified incorrect answer only if the circled answer is incorrect. Thus the partial-credit option is 1/8th of the full-credit possible.

Brief bibliography, with abstracts/excerpts:

"Humanizing the multiple-choice test with partial credit"
Barry M. Cherkas, Joseph Roitberg, International Journal of Mathematical Education in Science and Technology, Volume 24, Issue 6 November 1993, pages 799-812.
"We describe an innovative method for machine-grading precalculus students: the partial-credit, multiple-choice test (PCMCT). Our scheme is to assign partial credit to any wrong multiple-choice answer for which we can ascribe partially correct reasoning. This permits us to accumulate additional information about the extent of a student's incomplete understanding. We add the assigned partial credit to arrive at our final measure of a student's total knowledge—the student's grade. In comparison with the standard multiple-choice test (MCT), the PCMCT judiciously avoids the necessity to scale scores—an educationally unsound but common practice with MCTs. In addition, the PCMCT turns out to be better at distinguishing uneducated guessing than the MCT. We conducted student surveys in our pre-calculus classes and found that most students believed the PCMCT is a more accurate measure of their knowledge than the MCT. We also uncovered the remarkable result that most students are less math-anxious when taking the PCMCT than when taking either the MCT or a standard essay test."
"Comparative Review: Partial-Credit Scoring Methods for Multiple-Choice Tests"
Robert B. Frary, Applied Measurement in Education, Volume 2, Issue 1 January 1989, pages 79-96.
"This review covers multiple-choice response and scoring methods that attempt to capture information about an examinee's degree or level of knowledge with respect to each item and use this information to produce a total test score. The period covered is mainly from the early 1970s onward; earlier reviews are summarized. It is concluded that there is little to be gained from the complex responding and scoring schemes that have been investigated. Although some of them have confirmed potential to increase internal-consistency reliability, this outcome is often obtained only at the expense of validity. Also, the extra responding time required by some methods would permit lengthening a conventional multiple-choice test sufficiently to obtain the same reliability improvement. Partial-credit response and scoring methods that continue to be used will probably earn this status due to secondary characteristics such as providing feedback to enhance learning."
"A Multiple Choice Test that Rewards Partial Knowledge"
M. Bush, Journal of Further and Higher Education, Volume 25, Number 2, 1 June 2001, pp. 157-163(7).
"A new multiple choice test format is presented that allows examinees to select more than one answer to a question if they are uncertain of the correct one. Negative marking is used to penalise incorrect selections. The aim is to explicitly reward examinees who possess partial knowledge as compared with those who are simply guessing. The result is a test method that forces examinees to think more carefully about their answers, and that yields results of a higher resolution than standard multiple choice tests. After describing the new format, the paper presents and critiques several existing methods which have the same or similar aims. The paper ends with a discussion of the feedback and experience gained to date in using the new format."
"A singular choice for multiple choice"
Gudmund S. Frandsen, Michael I. Schwartzbach, ACM SIGCSE Bulletin Volume 38, Issue 4 (December 2006), p. 34-38.
"How should multiple choice tests be scored and graded, in particular when students are allowed to check several boxes to convey partial knowledge? Many strategies may seem reasonable, but we demonstrate that five self-evident axioms are sufficient to determine completely the correct strategy. We also discuss how to measure robustness of the obtained grades. Our results have practical advantages and also suggest criteria for designing multiple choice questions."

20081009

Astronomy clicker question: speed of electromagnetic radiation

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, during a review session before a midterm:

Which type of electromagnetic radiation wave travels with the slowest speed?
(A) Gamma rays.
(B) X rays.
(C) Microwaves.
(D) TV.
(E) (All travel at same speed.)
(F) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 3 students
(B) : 5 students
(C) : 7 students
(D) : 8 students
(E) : 7 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results.

Section 70160
(A) : 3 students
(B) : 1 student
(C) : 0 students
(D) : 23 students
(E) : 3 students

Correct answer: (E)

All forms of electromagnetic radiation travel with the same speed!

Pre- to post- peer-interaction gains:
pre-interaction correct = 23%
post-interaction correct = 10%
Hake (normalized) gain = -17%

20081008

Astronomy clicker question: hours of daylight

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, during a review session before a midterm:

If the Sun rises in the southeast horizon, as seen by an observer in San Luis Obispo, CA, how many hours will there be before the Sun sets on that day?
(A) Less than 12 hours.
(B) Approximately 12 hours.
(C) More than 12 hours.
(D) (The sun will not set below the horizon on that day.)
(E) (I'm lost, and don't know how to answer this.)

Section 70160
(A) : 15 students
(B) : 7 students
(C) : 7 students
(D) : 0 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results.

Section 70160
(A) : 26 students
(B) : 0 students
(C) : 1 student
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The Sun will set in the southwest later that day, meaning that this is some date between the autumnal equinox and the vernal equinox, thus there will be less than 12 hours of daylight from sunrise to sunset.

Pre- to post- peer-interaction gains:
pre-interaction correct = 51%
post-interaction correct = 96%
Hake (normalized) gain = 92%

20081007

Astronomy clicker question: supergiants vs. giants

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

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

A supergiant will be ___________ compared to a giant star that has the same size.
(A) cooler and dimmer.
(B) cooler and brighter.
(C) hotter and dimmer.
(D) hotter and brighter.
(E) (I'm lost, and don't know how to answer this.)

(Blank comparison table provided to students for reference.)

Section 70158
(A) : 11 students
(B) : 9 students
(C) : 8 students
(D) : 30 students
(E) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results.

Section 70158
(A) : 3 students
(B) : 6 students
(C) : 7 students
(D) : 39 students
(E) : 0 students

Correct answer: (D)

Given that this supergiant is the same size as a giant, and from inspection of a Hertzsprung-Russell diagram, is more luminous than a giant, then from applying the Stefan-Boltzman law, the supergiant must be hotter than the giant.

Pre- to post- peer-interaction gains:
pre-interaction correct = 51%
post-interaction correct = 71%
Hake (normalized) gain = 41%

20081006

Astronomy quiz question: telescope funding

Astronomy 210 Quiz 3, Fall Semester 2008
Cuesta College, San Luis Obispo, CA


[4.0 points.] Based on being able to detect its wavelength, and cost of location, which telescope should be funded?
(A) A gamma ray detector in Antarctica.
(B) An ultraviolet detector in the Mojave desert.
(C) A visible light telescope in the Chilean mountains.
(D) A radio telescope in space.

Correct answer: (C)

Gamma rays and ultraviolet waves cannot be detected at ground level, due to absorption of these wavelengths by the atmosphere. A visible light telescope on a mountain would be able to detect signals at much less cost than a radio telescope in space.

Student responses
Section 70158
(A) : 11 students
(B) : 15 students
(C) : 34 students
(D) : 11 students

"Difficulty level": 51%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63

20081005

Astronomy quiz question: atmospheric turbulence

Astronomy 210 Quiz 3, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

What power of an optical telescope is most affected by the turbulence in the Earth's atmosphere?
(A) Light-gathering power.
(B) Resolving power.
(C) Magnifying power.
(D) (None of the above choices.)

Correct answer: (B)

For large ground-based telescopes, resolving power is limited by atmospheric turbulence, which distorts and "twinkles" the stars.

Student responses
Section 70158
(A) : 8 students
(B) : 53 students
(C) : 7 students
(D) : 3 students

"Success level": 76%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.58

20081004

Astronomy quiz question: telescope funding

Astronomy 210 Quiz 3, Fall Semester 2008
Cuesta College, San Luis Obispo, CA


Based on being able to detect its wavelength, and cost of location, which telescope should be funded?
(A) A gamma ray detector in Antarctica.
(B) An ultraviolet detector in the Mojave desert.
(C) A near-infrared telescope on an aircraft.
(D) A radio telescope in space.

Correct answer: (C)

Gamma rays and ultraviolet waves cannot be detected at ground level, due to absorption of these wavelengths by the atmosphere. A near-infrared telescope on an aircraft would be able to detect signals at much less cost than a radio telescope in space.

Student responses
Section 70160
(A) : 6 students
(B) : 4 students
(C) : 19 students
(D) : 4 students

Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.78

20081003

Astronomy quiz question: adaptive optics

Astronomy 210 Quiz 3, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Adaptive optics is used in modern optical telescopes to improve:
(A) light-gathering power.
(B) resolving power.
(C) magnifying power.
(D) the amount of radiation transmitted through the atmosphere.

Correct answer: (B)

For large ground-based telescopes, their resolving power is limited by atmospheric turbulence, which distorts and "twinkles" the stars. Using adaptive optics to shape the mirror to compensate for these distortions will improve the resolution of images obtained by ground-based telescopes.

Student responses
Section 70160
(A) : 3 students
(B) : 25 students
(C) : 2 students
(D) : 1 student

Success level: 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.44

20081002

Astronomy clicker question: stellar distance

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:

Antares is a star that has an apparent magnitude m = +0.92 and an absolute visual magnitude M_v = –4.5. Approximately how far away is Antares from the Earth?
(A) Much farther away than 10 parsecs.
(B) A little farther away than 10 parsecs.
(C) Exactly 10 parsecs away.
(D) Closer than 10 parsecs away.
(E) (I'm lost, and don't know how to answer this.)

Section 70158
(A) : 39 students
(B) : 13 students
(C) : 2 students
(D) : 3 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. (Comment from student, as to whether the pair-share follow-up is necessary: "I think it [the response distribution, shown to the class] already looks pretty." Retort from instructor: "Then make it look prettier...or if you think it looks ugly, then fix it!") No (other) 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) : 54 students
(B) : 2 students
(C) : 0 students
(D) : 0 students
(E) : 1 student

Correct answer: (A)

When brought to the "fair distance" of 10 parsecs (thus changing its apparent magnitude to absolute visual magnitude), Antares now seems to be brighter, indicating that it must have moved closer to the Earth during this process. Since this is a substantive gain in brightness, Antares must be much farther away than this fair distance of 10 parsecs.

Pre- to post- peer-interaction gains:
pre-interaction correct = 68%
post-interaction correct = 95%
Hake (normalized) gain <g> = 83%

20081001

Physics quiz question: dragged book, with friction

Physics 205A Quiz 3, fall semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 4.77

[Version 1]
Consider a book on a horizontal table, with a tension force of 10.0 N acting on it. The magnitude of the normal force of the table on the book is 22.0 N. The coefficient of static friction is 0.42. The coefficient of kinetic friction is 0.35. Which force has the greatest magnitude?
(A) Friction force on the book.
(B) Weight of the book.
(C) Tension force on the book.
(D) (Not enough information is given to determine this.)

Correct answer: (B)

The maximum static friction force magnitude is mu_s*N = 9.2 N, where the normal force is 22.0 N. The kinetic friction force magnitude is mu_k*N = 7.7 N. Since the tension force (10.0 N) is greater than the maximum static friction force magnitude, then the book is unstuck, and is sliding and accelerating to the right. From applying Newton's first law in the vertical direction, the weight of the book is 22.0 N. Thus the weight of the book has the greatest magnitude, while the (kinetic) friction force has the least magnitude.

Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 12 students
(C) : 7 students
(D) : 0 students

[Version 2]
Consider a book on a horizontal table, with a tension force of 10.0 N acting on it. The magnitude of the normal force of the table on the book is 22.0 N. The coefficient of static friction is 0.42. The coefficient of kinetic friction is 0.35. Which force has the least magnitude?
(A) Friction force on the book.
(B) Weight of the book.
(C) Tension force on the book.
(D) (Not enough information is given to determine this.)

Correct answer: (A)

Student responses
Sections 70854, 70855
(A) : 17 students
(B) : 2 students
(C) : 4 students
(D) : 0 students

Success level: 64%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45