Physics quiz archive: energy conservation, momentum conservation

Physics 205A Quiz 4, fall semester 2012
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1
Exam code: quiz04Bmw3



Sections 70854, 70855 results

0- 6 :
7-12 :	******** [low = 9]
13-18 :	**********
19-24 :	****************** [mean = 21.3 +/- 6.0]
25-30 :	**************** [high = 30]

Astronomy quiz archive: sun/spectra/star properties

Astronomy 210 Quiz 4, fall semester 2012
Cuesta College, San Luis Obispo, CA

Section 70158, version 1
Exam code: quiz04Spr6


Section 70158

0- 8.0 :	*** [low = 4.0]
8.5-16.0 :	****
16.5-24.0 :	***********
24.5-32.0 :	******* [mean = 24.5 +/- 9.8]
32.5-40.0 :	********** [high = 40.0]

Section 70160, version 1
Exam code: quiz04nNm5


Section 70160

0- 8.0 :
8.5-16.0 :	** [low = 12.0]
16.5-24.0 :	*********
24.5-32.0 :	*********** [mean = 25.4 +/- 7.2]
32.5-40.0 :	**** [high = 40.0]

Twitter: watching the two moons rise

"Incidental Comics: Steampunk Summer" (excerpt)
by Grant Snider
http://www.incidentalcomics.com/2012/06/steampunk-summer.html

Astronomy 210 Midterm 1, student 2210 response
Cuesta College, San Luis Obispo, CA
http://waiferx.blogspot.com/2012/10/astronomy-essay-question-watching-two.html

@patrick_m_len This is awesome! And apparently I need to enroll in some astronomy classes. :)

— Grant Snider (@grantdraws) October 15, 2012

Twitter: work habits of the moon and sun

"Doogie Horner's Flowchart: Work Habits of the Moon and Sun" (excerpt)
by Doogie Horner
http://boingboing.net/2012/03/16/doogie-horners-flowchart-wo.html

@doogiehorner infographic: work habits of the moon and sunboingboing.net/2012/03/16/doo… #astro101

— Patrick M. Len (@Patrick_M_Len) March 17, 2012

"hides behind clouds"? #astro101 #fail RT: @doogiehorner infographic: work habits of moon and sunow.ly/1IE6ME (via @patrick_m_len)

— Peter Newbury (@polarisdotca) March 17, 2012

@doogiehorner To me, your infographic suggest clouds cause Moon phases, a common misconception that #astro101 teachers strive to remedy.

— Peter Newbury (@polarisdotca) March 17, 2012

@doogiehorner Glad thats not what you meant. My spidey-senses tingle whenever I see Moon phases and clouds in the same sentence :)

— Peter Newbury (@polarisdotca) March 18, 2012

Astronomy midterm question: watching two moons rising?

Astronomy 210 Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

[20 points.] Shown below is an excerpt from an online comic strip(*) posted on June 6, 2012. Assuming that Earth can have two moons, discuss why this view of these two moons is not possible for an observer in San Luis Obispo, CA, and how you know this. Support your answer using a diagram showing the positions of the sun, the moon, Earth, and an observer on Earth.


*Source: "Steampunk Summer," Incidental Comics, Grant Snider, http://www.incidentalcomics.com/2012/06/steampunk-summer.html.

Solution and grading rubric:

• p = 20/20:
  Correct. The full moon rises at 6 PM, while the smaller waning crescent moon would rise at 3 AM, so it would not be possible to watch both moons rising at the same time. May instead demonstrate it is impossible for an observer to see these moons with two different phases simultaneously in the same part of the sky, as two moons in the same part of the sky must have the same (or nearly identical) phases. Complete diagram and reasoning.
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Depicts how these moons with two different phases are in different locations in their orbits, but does not explicitly discuss how these moons cannot be seen rising at the same time, or seen together in the same part of the sky, as these moons could still be seen in (different parts of) the same sky between 12 AM to 3 AM.
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problems with either diagram or discussion.
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Diagram and discussion problematic.
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. Misconceptions or non-relevant concepts: moon phases created by Earth's shadow, diagrams with the moon orbiting the sun, etc.
• y = 2/20:
  Irrelevant discussion/effectively blank.
• z = 0/20:
  Blank.

Grading distribution:
Section 70158
Exam code: midterm01sLa6
p: 7 students
r: 10 students
t: 9 students
v: 7 students
x: 6 students
y: 0 students
z: 0 students

A sample "p" response (from student 2210):
Another sample "p" response (from student 8669):

Astronomy midterm question: Mercury higher than Venus at sunrise?

Astronomy 210 Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

An astronomy question on an online discussion board[*] was asked and answered:

anon: If Mercury and Venus are both morning stars, can Mercury be higher in sky than Venus at dawn?
"?": Yes, Mercury can have a higher altitude [in the sky] than Venus. It depends on exactly where Earth, Mercury and Venus are in their orbits in relationship to each other and your location on Earth.

Discuss why this answer is correct for an observer in San Luis Obispo, CA, and how you know this. Support your answer using a diagram showing the positions of the sun, Mercury, Venus, Earth, and an observer on Earth.

[*] answers.yahoo.com/question/index?qid=20120922122209AAFhP8t.

Solution and grading rubric:

• p:
  Correct. Diagram with observer at sunrise Earth (outer heliocentric orbit) and Mercury and Venus (inner heliocentric orbits); shows and discusses how Mercury at or near western elongation would be higher in the sky above the horizon at sunrise if Venus is just after inferior conjunction, or just before superior conjuction.
• r:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. May show that Venus in a heliocentric orbit inside of Mercury's, and/or how Mercury would be higher than Venus for an observer at sunset instead of sunrise.
• t:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Shows Mercury higher than the sun in the morning sky after sunrise, with Venus below the sun, such that Mercury would be higher than Venus in the morning sky after sunrise, even though both planets would no longer be visible in the sky with the sun above the horizon.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Diagram and discussion problematic. Shows planets orbiting Earth, or in heliocentric orbits outside of Earth's, but still shows how Mercury could be higher in the sky (morning or evening) than Venus.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Misconceptions or non-relevant concepts.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Section 70158
Exam code: midterm01sLa6
p: 7 students
r: 10 students
t: 9 students
v: 7 students
x: 6 students
y: 0 students
z: 0 students

A sample "p" response (from student 0640):

Astronomy midterm question: why no full moon in afternoon?

Astronomy 210 Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

An astronomy question on an online discussion board(*) was asked and answered:

rraghunathan_in: In some afternoons we see both the sun and the moon in the sky... It is clear that the moon is directly exposed to the sun in the daytime sky, but only three-fourths or less...of the moon is illuminated. Why is it not a full moon at that time?
TicToc...: The moon is not full because it's not in opposition, which means the sun will be setting and the moon is rising at the same time.

Discuss why this answer is correct for an observer in San Luis Obispo, CA, and how you know this. Support your answer using a diagram showing the positions of the sun, the moon, Earth, and an observer on Earth.

*Source: http://answers.yahoo.com/question/index?qid=20120511191408AARz0Af.

Solution and grading rubric:

• p = 20/20:
  Correct. The moon cannot be full in the afternoon because the full moon rises at sunset, thus the opposition argument is correct. Complete diagram and reasoning.
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Depicts a non-full moon visible during the afternoon, but does not directly address why the moon cannot be full in the afternoon.
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problems with either diagram or discussion.
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Diagram and discussion problematic.
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. Misconceptions or non-relevant concepts: moon phases created by Earth's shadow, diagrams with the moon orbiting the sun, etc.
• y = 2/20:
  Irrelevant discussion/effectively blank.
• z = 0/20:
  Blank.

Grading distribution:
Section 70160
Exam code: midterm01n4rN
p: 6 students
r: 10 students
t: 4 students
v: 3 students
x: 6 students
y: 0 students
z: 1 student

A sample "p" response (from student 2513):
A sample "p" response (from student 1105):
A sample "x" response (from student 1964):

Astronomy midterm question: Venus as morning and evening star in same day?

Astronomy 210 Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

[20 points.] An astronomy question on an online discussion board(*) was asked and answered:

P-d...: ...If Venus is above the east horizon at sunrise (so it is a morning star), can it be above the west horizon at sunset (so it can be an evening star) later on that day?
aladdinwa: No, it cannot be both the morning and evening star on the same day... When Venus rises before the sun, you can see it before the sun rises and it is the morning star and it disappears below the horizon while the sun is still in the sky...

Discuss why this answer is correct for an observer in San Luis Obispo, CA, and how you know this. Support your answer using a diagram showing the positions of the sun, Venus, Earth, and an observer on Earth.

*Source: http://answers.yahoo.com/question/index?qid=20120511191408AARz0Af.

Solution and grading rubric:

• p = 20/20:
  Correct. Correct and complete diagram, with observers at sunrise/sunset on Earth (outer heliocentric orbit) and Venus (inner heliocentric orbit); shows and discusses how Venus above east horizon at sunrise (as a morning star) would not be visible anywhere in sky at sunset later that day (as an evening star).
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Diagram and/or explanation has minor errors. May show that Venus visible as an evening star would not then be visible as a morning star.
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problems with either diagram or discussion.
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Diagram and discussion problematic. May have Venus orbiting Earth, or in a heliocentric orbit outside of Earth's, but still discusses how Venus could not be visible at both sunrise and sunset on the same day.
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. Misconceptions or non-relevant concepts.
• y = 2/20:
  Irrelevant discussion/effectively blank.
• z = 0/20:
  Blank.

Grading distribution:
Section 70160
Exam code: midterm01n4rN
p: 12 students
r: 5 students
t: 3 students
v: 7 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 0716):

Astronomy midterm question: better telescope choice

Astronomy 210 Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

[20 points.] An astronomy question on an online discussion board(*) was asked and answered:

Mr. Dingo: Which [telescope] has [the brightest image], best detail, [and] magnification?

     FunScope™ [Tabletop] Reflector

          [76 mm diameter mirror, 300 mm long tube, f = 10 mm eyepiece]
     Orion™ SkyScanner® Reflector

          [100 mm diameter mirror, 400 mm long tube, f = 20 mm eyepiece]
Mark H: ...the answer is the [SkyScanner®].

Discuss whether you agree or disagree with this answer, and the criteria used in your decision. Support your answer using the properties of telescopes and telescope powers.

*Source: http://answers.yahoo.com/question/index?qid=20100516131920AAPZR2i.

Solution and grading rubric:

• p = 20/20:
  Correct. Discusses how diameter of primary mirror determines the light-gathering power and resolving power, such that the larger diameter mirror of the Orion™ SkyScanner® would result in brighter, more finely resolved images. Also discusses how the primary focal length and eyepiece supplied with this telescope would result in lower magnification than the primary focal length and eyepiece supplied with the FunScope™, although this is a less important consideration in purchasing a telescope than the light-gathering and resolving power.
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. As (p), but two of three telescope powers is correct, while discussion of third telescope power is problematic.
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problematic discussion of two telescope powers, while third is complete/correct; or complete/correct discussion of two telescope powers, while discussion of third telescope power is omitted.
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least discusses some connection between telescope parameters and telescope power.
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. Does not discuss connection between telescope parameters and telescope powers.
• y = 2/20:
  Irrelevant discussion/effectively blank.
• z = 0/20:
  Blank.

Grading distribution:
Section 70158
Exam code: midterm01sLa6
p: 6 students
r: 13 students
t: 16 students
v: 2 students
x: 2 students
y: 0 students
z: 0 students

Grading distribution:
Section 70160
Exam code: midterm01n4rN
p: 9 students
r: 7 students
t: 0 students
v: 11 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 8669):

Physics midterm problem: St. Lawrence River rocket car jump

Physics 205A Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 3.53

"Rocket Lincoln jump"
elpmurc
youtu.be/tpUMSarCSQw

In 1979, a rocket-powered Lincoln Continental attempted to jump 1,600 m (approximately 1.0 mile) across the St. Lawrence River by traveling 120 m/s (approximately 270 mph) off the edge of a ramp angled 30° above the horizontal at a height of 30 m (approximately 100 ft) above the ground.[*] Three years earlier during the ramp construction process, stuntman Evel Knievel had reported for ABC's Wide World of Sports that this would not be feasible as designed. Determine whether Knievel's assessment was correct or incorrect. Neglect friction and drag. Show your work and explain your reasoning using properties of projectile motion.

[*] Robert Fortier, "The Devil At Your Heels," National Film Board of Canada (1981), onf.ca/film/devil_at_your_heels. (Last-minute substitute stuntman Kenny Powers survived the failed attempt.)

Solution and grading rubric:

• p:
  Correct. Finds x- and y-components of initial velocity vector, then applies projectile motion equations in a methodical manner, and either:
  1. finds t when car reaches ground level (y = –30 m, assuming y₀ = 0), and determines that the car would horizontally be located at x = v_0x∙t, which is less than +1,600 m; or
  2. finds required t for car to travel to x = +1,600 m, and determines that the car would be below ground level at that time; and concludes that this jump is not feasible; or
  3. may also solve for time to reach highest point in trajectory, then doubles this time and finds that car would travel less than required x = +1,600 m, with implicit or explicit assumptions regarding the existence of a landing ramp of similar dimensions (y = 0, x = +1,600 from ramp edge to ramp edge), etc.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least enough steps are shown that would theoretically result in a complete answer, multiple math errors notwithstanding.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at systematic use of kinematic equations for projectile motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01sWFf
p: 14 students
r: 13 students
t: 8 students
v: 22 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1101), comparing the time for the car to travel 1,600 m horizontally to the predicted time of flight for the car until it reaches the ground. Note the alligator in the St. Lawrence River:
Another sample "p" response (from student 1408), calculating the time for the car to return to the launch height by multiplying the time to reach its highest height by two (thus supposing a landing ramp), and finding that the horizontal distance traveled would be less than 1,600 m:
Another sample "p" response (from student 1970), calculating the time for the car to travel 1,600 m horizontally, then determining that the car would need to be 238 m vertically below its starting height in order to travel that horizontal distance:

Physics midterm problem: Swiffer® Sweeper force

Physics 205A Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 4.65


"Cleans more than us"
Christopher Heschong
flic.kr/p/7G7H4n

A child pushes along the handle of a Swiffer® Sweeper at an angle of 55° with respect to the vertical. If the mass of a dry Swiffer® Sweeper pad is 0.15 kg, and the coefficient of kinetic friction between the pad and the floor[*] is 0.17, determine the amount of force applied along the handle for a Swiffer® Sweeper pad to slide along the horizontal floor with constant speed. Show your work and explain your reasoning using a free-body diagram, and the properties of forces, and Newton's laws.

[*] flic.kr/p/dftcPA.

Solution and grading rubric:

• p:
  Correct. Draws free body diagram to illustrate that the normal force upwards must have the same magnitude as the y-component of the applied force plus the weight force downwards, due to Newton's first law; and that the x-component of the applied force must have the same magnitude as the static friction force points, also due to Newton's first law. Then expresses Newton's first law in the vertical (N = w + F_applied,x) and horizontal (F_applied,y = f_k) directions, and substitutes f_k = μ_k∙N to solve for the magnitude of the applied force.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically neglects y-component of applied force, such that the upwards normal force is set equal in magnitude to the downwards weight force. Then uses N = w = m∙g result to calculate kinetic friction force f_k = μ_k∙N = 0.25 N, and equates this f_k to the x-component of the applied force, using trigonometry to determine magnitude of applied force.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Free body diagram identifies most forces and their directions, with some attempt at applying Newton's laws.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Garbled/incomplete free body diagram with little to no application of Newton's laws, etc.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01sWFf
p: 1 student
r: 3 students
t: 30 students
v: 11 students
x: 12 students
y: 0 students
z: 0 students

A sample "p" response (from student 6377):
A sample "t" response (from student 2507), where the magnitude of the normal force is set equal to the weight of the pad:

Physics quiz archive: vectors, projectile motion, forces

Physics 205A Quiz 3, fall semester 2012
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1
Exam code: quiz03sQr7



Sections 70854, 70855 results

0- 6 :	* [low = 6]
7-12 :	***
13-18 :	***********************
19-24 :	********************** [mean = 19.3 +/- 4.7]
25-30 :	****** [high = 30]