P-dog's blog: boring but important

Astronomy and physics education research and comments, field-tested think-pair-share (peer instruction) clicker questions, flashcard questions, in-class activities (lecture-tutorials), current events questions, backwards faded scaffolding laboratories, Hake gains, field-tested multiple-choice and essay exam questions, indices of discrimination, presentation slides, photos, ephemerae, astronomy in the marketplace, unrelated random sketches and minutiae.

20111031

Astronomy quiz question: star cluster with white dwarfs

Astronomy 210 Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

A star cluster containing white dwarfs would also have:
(A) medium-mass main sequence stars.
(B) massive protostars.
(C) red dwarfs.
(D) supergiants.

Correct answer: (C).

White dwarfs are the end-stage of medium-mass stars, so medium-mass main sequence stars cannot be in the same star cluster as white dwarfs. Supergiants are the end-stages of the massive stars; while massive protostars are the start of massive stars, and as they evolve much more quickly than medium-mass stars, they cannot be in the same star cluster as white dwarfs. Red dwarfs are the main sequence stage of low-mass stars, and as they evolved much slower from protostar to main sequence stars, they are the most plausible of the given choices to be found in the same star cluster as white dwarfs.

Section 70158
Exam code: quiz05SpRn
(A) : 15 students
(B) : 5 students
(C) : 6 students
(D) : 6 students
(No response: 1 student)

Success level: 25% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

Section 70160
Exam code: quiz05nUc7
(A) : 9 students
(B) : 3 students
(C) : 11 students
(D) : 4 students

Success level: 45% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

20111030

Astronomy quiz question: giant energy source

Astronomy 210 Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

The energy source in the core of a giant is:
(A) hydrogen fusion.
(B) gravitational contraction.
(C) fusion of elements heavier than hydrogen.
(D) radioactive decay.
(E) (Does not produce energy.)

Correct answer: (C).

A giant is a medium-mass star that has converted all of the hydrogen in its core to helium at the end of its main sequence lifetime. After a brief intermediate stage where energy is produced by hydrogen fusion in a shell surrounding its core, energy is then produced in the core by the fusion of helium into carbon.

Section 70158
Exam code: quiz05SpRn
(A) : 18 students
(B) : 0 students
(C) : 13 students
(D) : 0 students
(E) : 2 students

Success level: 45% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.22

20111029

Astronomy quiz archive: stellar evolution

Astronomy 210 Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

Section 70158, version 1
Exam code: quiz05SpRn

Section 70158

Quiz 5 results (max score = 40):
   0- 8.0 : 
 8.5-16.0 : **********  [low = 9.5]
16.5-24.0 : ********  [mean = 21.3 +/- 7.5]
24.5-32.0 : ************
32.5-40.0 : ***  [high = 33.0]

20111028

Astronomy quiz question: red dwarf energy source

Astronomy 210 Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

The energy source in the core of a red dwarf is:
(A) hydrogen fusion.
(B) gravitational contraction.
(C) fusion of elements heavier than hydrogen.
(D) radioactive decay.
(E) (Does not produce energy.)

Correct answer: (A).

A red dwarf is a low-mass main sequence star, which produces energy from fusing hydrogen into helium during its (extremely long) main sequence lifetime.

Section 70160
Exam code: quiz05nUc7
(A) : 15 students
(B) : 1 student
(C) : 3 students
(D) : 3 students
(E) : 5 students

Success level: 61% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73

Astronomy quiz archive: stellar evolution

Astronomy 210 Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

Section 70160, version 1
Exam code: quiz05nUc7

Section 70160

Quiz 5 results (max score = 40):
   0- 8.0 : **  [low = 7.0]
 8.5-16.0 : **
16.5-24.0 : *****  
24.5-32.0 : *********  [mean = 26.9 +/- 9.2]
32.5-40.0 : *********  [high = 40.0]

20111027

Online reading assignment question: Milky Way tags

Astronomy 210 Reading Assignment 9, fall semester 2011
Cuesta College, San Luis Obispo, CA

111027-milkywaywordle
http://www.flickr.com/photos/waiferx/6287121530/
Originally uploaded by Waifer X

Wordle.net tag cloud for "Milky Way," generated by responses from Astronomy 210 students at Cuesta College, San Luis Obispo, CA (http://www.wordle.net/show/wrdl/4312349/Untitled).

Students have a weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

Write down five words that describe or are associated with the "Milky Way." Keep word phrases together with no spaces between them (e.g., "forexample, likethis, nospaces"). (Graded for completion.)

[Responses have been edited to consolidate related common subjects.]

Student responses
Sections 70158, 70160

galaxy spiral allaroundus halo nuclearbulge
diskcomponent, spiralarms, halo, nuclearbulge, darkmatter.
greatstarsystem, Cepheids, propermotions, halo, disk
spiralgalaxy, candybar, 9byold, SagittariusA, Sun
milkyroad, foggy, diameter100000ly, 13.2byold, spiral
clusters, stars, Sun, spiralgalaxy, oursunislocatedonthesideofit.
notquitesurehowtoanswerthis
spiralgalaxy, candybar, 9byold, SagittariusA, nuclearbulge
spiralgalaxy, diameter100000ly, spiralarms, oldastheuniverse, 200billionstars
nightskyglowingbandstars
dust, supernova, hydrogen, helium, iron.
halo metals clusters stars galaxy
milky vast habitated beautiful home
galaxy, solarsystem, spiralgalaxy, celestialsphere, stars.
galaxy, solarsystem, candybar stars
havenoidea
white, stars, cluttered, band, line
milkyroad, nuclearbulge, spiralarms, sphericalcomponent, rotationcurve.
beautiful, enormous, stars, spiral, solarsystem
BigDipper, blackhole, celestialobject, clusters, comets
spiralgalaxy, 13.2byold, SagittariusA, disk, stars dust
stars galaxy spiral dust cloudy
clusterofstars particles spiral.
caramel chocolate milky quick cheap
planets stars comets asteroids
greatrift, spiralarms, halo, disk, darkmatter
barredspiralgalaxy, darkmatter, supermassiveblackhole, gammaraybubbles, binarysystem
billionsandbillions viagalactica supermassiveblackhole spiralarms home
candybar, galaxy, stars, bright
clockwise galaxy dusty spiral large
delicious, chocolate, caramel, bitesized, candy
birdspath, roadtoSantiago, silverriver, rivertoheaven, starryway
lotsofdifferentvarietieofstars
hazyband, solarsystem, galaxy, barredspiral, darknebulae
galaxy solarsystem dimglow
thisisreallyweirdwow
vast candy Earth galaxy pretty
200billionstars, localgroup, Galileodiscovered, supermassiveblackhole SagittariusA, alwaysonthemove
peanuts, caramel, chocolate, darkchocolate MilkyWayMidnights, nougat
supermassiveblackhole spiral beautiful incomprehensible immense
cloudy, big, chocolate, bright, preferSnickers
ourcluster spiral candy disk
clustersofstars, swirlsoflightinthesky, galaxy, candybar, MilkyWay

20111025

Astronomy current events question: warmer/wetter Mars evidence

Astronomy 210L, fall semester 2011
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)

Astronomy.com editors, "Caltech Researchers Take the Temperature of Mars' Past," October 17, 2011
http://astronomy.com/en/News-Observing/News/2011/10/Caltech%20researchers%20take%20the%20temperature%20of%20Mars%20past.aspx

Researchers at the California Institute of Technology analyzed __________ from Mars to provide evidence that Mars was warmer and wetter in its past.
(A) look-back time photons.
(B) polar ice reflections.
(C) a Russian probe that returned.
(D) a meteorite that originated.
(E) surface gamma-ray emissions.

Correct answer: (D)

Student responses
Sections 70178, 70186, 70200
(A) : 2 students
(B) : 6 students
(C) : 2 students
(D) : 31 students
(E) : 5 students

Astronomy current events question: dark matter survey

Astronomy.com editors, "Hubble Survey Carries Out a Dark Matter Census," October 13, 2011
http://astronomy.com/en/News-Observing/News/2011/10/Hubble%20survey%20carries%20out%20a%20dark%20matter%20census.aspx

The NASA/ESA Hubble Space is surveying the presence of dark matter by looking for:
(A) matter-antimatter annihilations.
(B) cosmic background radiation distortions.
(C) distant galaxies images distorted by gravity.
(D) concentrations of dark energy.
(E) pockets of empty, but energetic space.

Correct answer: (C)

Student responses
Sections 70178, 70186, 70200
(A) : 2 students
(B) : 3 students
(C) : 28 students
(D) : 5 students
(E) : 6 students

Astronomy current events question: past impact on Iapetus?

Robert T. Gonzalez, "Astronomers Say Saturn’s Moon Iapetus May Have Once Had a Mini-moon All Its Own," October 17, 2011
http://io9.com/5848485/astronomers-say-saturns-moon-iapetus-may-have-once-had-a-mini+moon-all-its-own

Researchers at the Southwest Research Institute have hypothesized that a past impact on Saturn's moon, Iapetus, may have created:
(A) Iapetus' equatorial "walnut" ridge.
(B) Saturn's outermost ring.
(C) the nucleus of comet Elenin (C/2010 X1).
(D) minute amounts of dark matter.
(E) Iapetus' thick methane atmosphere.

Correct answer: (A)

Student responses
Sections 70178, 70186, 70200
(A) : 39 students
(B) : 4 students
(C) : 1 student
(D) : 1 student
(E) : 0 students

20111023

Physics presentation: torque and rotations

Whatever that "Nutcracker" gadget is, looks like this mechanic is going to need one to remove that stuck lug nut. We'll consider why he is probably not going to succeed, just doing what he's doing here. (Video link: "The Nutcracker - Undo Truck Wheel Nuts Easily.")

Also we'll analyze why this crane boom came crashing down in this industrial accident. (Video link: "Crane Accident - Australia....")

Here we will introduce torques and static equilibrium, much like how we developed ideas of forces and translational equilibrium, but applied here to rotations.

First, let's define torque, and some examples of how to calculate torque.

We'll emphasize a certain method of finding the torque ("the twisting") that a force exerts on an object, by using a perpendicular lever arm ℓ (sometimes called the "moment arm"). The resulting product of the force and lever arm results in the torque, in units of N·m (which in this context should not be thought of as joules), and can either have a counterclockwise or clockwise direction.

Let's revisit our hapless mechanic, who is trying two different ways to exert as much torque as possible on a lug nut. Consider the force exerted by the mechanic on the wrench (or spanner, since this shop is in Australia) be his entire weight, downwards. We've identified the lug nut, our pivot point, with a dot.

Extend a line along this force vector--this is the line of action, which is just a guide to find the perpendicular lever arm, which starts at the pivot point, and at the other end must intersect the line of action perpendicularly. In this case (but not always) the perpendicular lever arm corresponds to the physical object. The magnitude of the torque is the product of the lever arm distance (in m), and the magnitude of the force (in N). Since this force would rotate the object counterclockwise, then this is a counterclockwise torque.

Now for the second case, again extend the line of action along the force vector, and draw in the perpendicular lever arm, which starts at the pivot point, and must intersect the line of action perpendicularly. Note that in this case (as does sometimes happen) the perpendicular lever arm does not correspond to a physical object. Since this force would also rotate the object counterclockwise, this would also be a counterclockwise torque.

Compare both cases side-by-side--because the force magnitudes (the weight of the mechanic) and the perpendicular lever arm distances are approximately the same, then the torque exerted in either case is approximately the same! (How should the mechanic use the spanner to exert a greater amount of torque on the nut?)

Second, let's apply Newton's first law and second law to rotational motion.

Recall Newton's first law for vertical motion, where all of the up forces on an object would be balanced by all of the down forces on an object, such that the object's vertical motion would either be stationary, or have a constant vertical motion. (Similarly for Newton's first law for forces and motion along the horizontal direction.)

Similarly there is also a Newton's first law for rotational motion, where all of the counterclockwise torques on an object would be balanced by all of the clockwise torques on an object, such that the object would either be stationary, or have a constant rotational motion. This is the equivalent statement of saying that all the torques acting on that object add up to zero.

If you are looking forward to a career in structural engineering, then welcome to the rest of your life--keeping buildings and other structures from collapsing is all about getting the torques (and forces) to balance out (or sum to zero). Good times.

Returning to the falling crane example, let's take a look at each of the two forces that exert torques on the crane. The black-and-white checkered circle represents the center of gravity, the single point on the object at which the weight force can be considered to be acting. Then extending the line of action for the weight force allows its perpendicular lever arm to be identified, which in this case does not coincide with the physical crane. The magnitude of the weight torque is the product of the perpendicular lever arm (in m), and the magnitude of the weight (in N). Since the weight force would rotate the object clockwise, then this torque is clockwise.

Next, the cable at the end of the crane as the other force exerting a torque on it. Extending the line of action for the tension force at the end of the crane allows its perpendicular lever arm to be identified, which also in this case does not coincide with the physical crane. The magnitude of the tension torque is the product of the perpendicular lever arm (in m), and the magnitude of the tension force (in N). Since the tension force would rotate the crane counterclockwise, then this torque would be counterclockwise.

If the counterclockwise torque from the tension force balanced out the clockwise torque from the weight force, then the crane would remain stationary. However, since the crane starts to rotate clockwise, these torques do not balance, as the clockwise torque from the weight force is greater than the torque from the tension force. There really should have been more tension force pulling up on the end of the crane. Oh well...

What would Newton's second law for rotations look like? If all the counterclockwise torques and clockwise torques on an object don't balance out, then the net torque can be calculated by adding the counterclockwise torques (which are positive) with the clockwise torques (which are negative, because those correspond to rotations in the opposite direction). Right now, the emphasis is on recognizing whether Newton's first law for rotations (zero net torque) or Newton's second law (non-zero net torque) for rotations apply to an object.

Because sometimes a non-zero net torque is needed, in order to make a object begin to rotate...

Or sometimes you want a zero net torque on an object, to keep it from rotating--whoops!

20111022

Physics quiz question: work done on diagonally-pulled crate

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.1

A rope that makes an angle of 35° above the horizontal drags a crate at constant speed across a floor that is not frictionless. The magnitude of the work done by friction on the crate is __________ the magnitude of the work done by the rope on the crate.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (B)

Starting with the the energy balance equation:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where ∆KE_tr = 0 (crate is being dragged at constant speed), ∆PE_grav = 0 (crate moves across a horizontal floor), and ∆PE_elas = 0 (no springs are involved), then:

W_nc = 0.

The vertical forces that act on the crate are weight (downwards) and the normal force of the floor (upwards), but since these are perpendicular to the displacement of the crate, each of these forces do zero work on the crate.

The horizontal kinetic friction force does negative work on the crate (as the angle between that force and displacement is 180°, where cos(180°) = –1), while the rope does positive work on the crate (as the angle between that force and displacement is 35°, where cos(35°) is a positive quantity).

However, since the net external work done on the crate is zero, then the (negative) work done by friction on the crate must have the same magnitude as the (positive) work done by the rope on the crate:

W_nc = 0,

W_friction + W_rope = 0.

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 31 students
(B) : 16 students
(C) : 4 students
(D) : 0 students

Success level: 31%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.36

Physics quiz question: spring-launched box, sliding up ramp

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.63

A 0.25 kg box is held against a spring that is compressed by 0.10 m. The box is released, and slides up a ramp to a height of 0.70 m above the base of the ramp. Neglect friction and drag. Before the box was released, the elastic potential energy stored in the spring was:
(A) 0.0013 J.
(B) 0.088 J.
(C) 0.25 J.
(D) 1.7 J.

Correct answer (highlight to unhide): (D)

Starting with the the energy balance equation:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where W_nc = 0 (no external gains/losses of mechanical energy) and ∆KE_tr = 0 (as the box is initially held stationary, and again momentarily stationary at its highest point on the ramp) then:

0 = ∆PE_grav + ∆PE_elas,

0 = m·g·∆(y) + (1/2)·k·∆(x²),

0 = m·g·(y_f – y₀) + (1/2)·k·(x_f² – x₀²).

With initial parameters of y₀ = 0 (initial elevation) and x₀ = –0.10 m (compressed from equilibrium), and final parameters y_f = +0.70 m (higher final elevation) and x_f = 0 (spring relaxed after launching the box), then:

0 = m·g·(y_f – 0) + (1/2)·k·(0² – x₀²),

0 = m·g·y_f – (1/2)·k·x₀².

Since the spring strength k is not given, but the mass m of the box is given, the initial elastic potential energy (the second term) cannot be directly calculated, but can be indirectly solved for, as it must be equal to the final gravitational potential energy (the first term):

(1/2)·k·x₀² = m·g·y_f,

(1/2)·k·x₀² = (0.25 kg)·(9.80 m/s²)·(+0.70 m) = 1.715 J,

or to two significant figures, the elastic potential energy that was stored in the spring was 1.7 J.

(Response (A) is (1/2)·m·(x_i)², response (B) is (1/2)·m·y_f, response (C) is m·g·x₀.)

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 1 student
(B) : 8 students
(C) : 12 students
(D) : 30 students

Success level: 59%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73

Physics quiz question: Ford Mustang GT Premium kinetic energy

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.17

"First Test: 2011 Ford Mustang GT"
Motor Trend Channel
youtu.be/ydykiWKW6fE

A 2011 Ford Mustang GT Premium (mass 1,640 kg) takes 10.3 s to accelerate from rest to 45.0 m/s on a horizontal track[*]. What is the final kinetic energy of this car?
(A) 7.38×10⁴ J.
(B) 7.23×10⁵ J.
(C) 7.60×10⁵ J.
(D) 1.66×10⁶ J.

[*] motortrend.com/roadtests/coupes/112_1003_2011_ford_mustang_gt_premium_test/viewall.html.

Correct answer (highlight to unhide): (D)

The final (translational) kinetic energy of the car is given by:

KE_tr,f = (1/2)·m·(v_f)²,

KE_tr,f = (1/2)·(1,640 kg)·(45.0 m/s)² = 1,660,500 J,

or to three significant figures, 1.66×10⁶ J.

(Response (A) is m·v_f, response (B) is m·g·v_f, response (C) is m·v_f·t.)

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 2 students
(B) : 7 students
(C) : 4 students
(D) : 38 students

Success level: 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.64

Physics quiz question: rebounding truck crash

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.7

"Crash Test Ford Ranger 2012..."
MZBTZOO M
youtu.be/44UXH4AvOnY

An unmanned 2003 Ford Ranger XL Short Bed 2WD truck (mass 1,980 kg[*]) crashes into a stationary concrete wall at a speed of 11.0 m/s, and rebounds off the wall with a speed of 2.2 m/s in the opposite direction. The magnitude of the impulse exerted on the truck is:
(A) 4.4×10³ N·s.
(B) 1.7×10⁴ N·s.
(C) 2.18×10⁴ N·s.
(D) 2.61×10⁴ N·s.

[*] "Standard GVWR: 4360 lb," autos.msn.com/research/vip/spec_engines.aspx?year=2003&make=Ford&model=Ranger&trimid=95693.

Correct answer (highlight to unhide): (D)

The impulse J can be calculated as the initial-to-final change in momentum:

J = ∆p = m·∆v,

where ∆v = v_f – v₀.

The initial velocity vector is v₀ = +11.0 m/s (traveling forwards in the +x direction), and the final velocity vector is v_f = –2.2 m/s (rebounding in the reverse –x direction). Then:

J = (1,980 kg)·((–2.2 m/s) – (+11.0 m/s)) = (1,980 kg)·(–13.2 m/s) = –26,136 N·s,

or to two significant figures, the magnitude of the impulse is 2.61×10⁴ N·s (and the "–" sign indicates that it is in the backwards –x direction).

(Response (A) is the magnitude of the initial momentum, response (B) is the magnitude of the final momentum, response (C) is (1,980 kg)((–2.2 m/s) + (+11.0 m/s)).)

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 1 student
(B) : 18 students
(C) : 11 students
(D) : 20 students

Success level: 32%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.64

Physics quiz question: pellet bouncing off of block

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.49

(Ignore friction, drag, and other external forces during this brief collision.) A pellet moving to the right hits a motionless block, and the pellet bounces off to the left, while the block moves to the right. Based solely on the information given, __________ must be conserved.
(A) momentum.
(B) translational kinetic energy.
(C) (Both of the above choices.)
(D) (None of the above choices.)

Correct answer (highlight to unhide): (A)

Negligible net external force and brief time duration for this collision makes the external impulse on this system zero, such that momentum is conserved. Since the pellet bounces off of, instead of embedding itself into the block, this is not a completely inelastic collision, leaving two possibilities: a (partially) inelastic collision (such that translational kinetic energy is not conserved), or an elastic collision (such that translational kinetic energy is conserved). However, without additional information regarding the resulting permanent deformation (if any) of the pellet and/or block after the collision, only momentum can be said to be conserved.

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 33 students
(B) : 3 students
(C) : 14 students
(D) : 1 student

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.36

Physics quiz question: coupled freight car and locomotive

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.41

(Ignore friction, drag, and other external forces during this brief collision.) An empty freight car rolls along a straight level track, and collides with an initially stationary diesel locomotive, and couple (attach) together. Based solely on the information given, __________ must be conserved.
(A) momentum.
(B) translational kinetic energy.
(C) (Both of the above choices.)
(D) (None of the above choices.)

Correct answer (highlight to unhide): (A)

Negligible net external force and brief time duration for this collision makes the external impulse on this system zero, such that momentum is conserved. Since the freight car and locomotive "couple together," this is a completely inelastic collision, such that kinetic energy is not conserved.

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 41 students
(B) : 1 student
(C) : 6 students
(D) : 3 students

Success level: 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.64

Physics quiz question: laboratory cart collision

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.55

(Ignore friction, drag, and other external forces during this brief collision.) A laboratory cart moving to the right on a horizontal track collides elastically with a more massive cart moving to the left. Based solely on the information given, __________ must be conserved.
(A) momentum.
(B) kinetic energy.
(C) (Both of the above choices.)
(D) (None of the above choices.)

Correct answer (highlight to unhide): (C)

Negligible net external force and brief time duration for this collision makes the external impulse on this system zero, such that momentum is conserved. Since the carts are described as colliding elastically, then kinetic energy is also conserved.

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 2 students
(B) : 5 students
(C) : 44 students
(D) : 0 students

Success level: 86%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45

Physics quiz archive: energy conservation, momentum conservation

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1

Sections 70854, 70855 results
Exam code: quiz04iMpL

 0- 6 : **  [low = 6]
 7-12 : *****
13-18 : ******************
19-24 : ******************  [mean = 19.4 +/- 5.5]
25-30 : ********  [high = 30]

20111020

Physics presentation: collisions

Consider the collision of these two cars. Or rather, the "interaction" of these two cars, as they exchange momentum with each other, and convert kinetic energy to deformation. We'll be seeing a lot of interactions in this presentation. But don't worry, nobody gets hurt. (Video link: "2009 Chevrolet Malibu and 1959 Chevrolet Bel Air.")

We'll be emphasizing how to categorize collisions into one of three types, and determining which conservation laws apply to these collision types.

What will differentiate completely inelastic, inelastic, and elastic collisions is whether or not they stick to each other, and whether or not there is permanent damage.

In a completely inelastic collision between two objects, they will stick to each other. Also note the permanent deformation of the car afterwards. (Video link: "Insane Volvo brake test epic fail.")

For an inelastic collision (sometimes known as a partially inelastic collision, to distinguish it from a completely inelastic collision), the two objects rebound off of each other, but there is permanent deformation afterwards. (Video link: "Most Small Cars Aren't Economical for Crash Repairs Ford Foc.")

And for an elastic collision, the two objects rebound off of each other, and (ideally) there is no permanent deformation afterwards. (Note that even though there is no cosmetic damage in this collision, there was minor structural damage, but this is close to an ideal elastic collision.) (Video link: "Crash Test of Ford Explorer and Ford Taurus 2004.")

Let's organize the types of collisions in a flowchart that differentiates between them by asking whether or not the two objects stick to each other, and whether or not the two objects are permanently damaged. Note that the elastic collision can be considered the most restrictive case, being a collision that is selected against both criteria (not stuck together, no permanent deformation).

Consider which conservation laws apply for these types of collisions.

Let's revisit our collision type flowchart with kinetic energy conservation in mind.

If two cars stick to each other in a completely inelastic collision, then they convert as much kinetic energy as they can into permanent deformation, so kinetic energy is definitely not conserved.
If two cars deform and then rebound in a (partially) inelastic collision, then some kinetic energy goes into permanent deformation, while some of it is converted back to into kinetic energy (such that the cars will rebound off each other), so kinetic energy is also not conserved.
If two cars deform, and rebound completely without any permanent damage in an elastic collision, then no kinetic energy is lost, so kinetic energy is conserved for this type of collision.

As pointed out earlier, an elastic collision can be considered the most restrictive case, being a collision that is selected against in both criteria (not stuck together; no permanent deformation), and the only type of collision where kinetic energy is conserved, as none is lost to deformation.

Let's practice categorizing collisions, and determining whether kinetic energy is conserved or not.

Note there is no (cosmetic) damage to the bumper of this car in this slow-speed bumper test after it rebounds, which would make it a(n) ___________ collision, where kinetic energy __________ conserved. (Video link: "Bumper Crash Test: 2007 Toyota Camry.")

Answers (highlight to unhide): elastic; kinetic energy is conserved (ideally all of that energy went into temporarily "squishing" the bumper, which rebounded and returned that energy).

Note the damage to the bumper of this car in this slow-speed bumper test after it rebounds, which would make it a(n) ___________ collision, where kinetic energy __________ conserved. (Video link: "Bumper Crash Test: '07 Volkswagen Passat.")

Answers (highlight to unhide): (partially) inelastic (the cars are separated after the collision); kinetic energy is not conserved (as some of that energy went into damaging the bumper).

Note the extensive damage to the truck in this high-speed frontal offset collision test, which would make it a(n) ___________ collision, where kinetic energy __________ conserved. (Video link: "2001 Ford F-150 frontal offset test.")

Answers (highlight to unhide): completely inelastic (the truck is stuck to the collision barrier); kinetic energy is definitely not conserved (nearly all of that energy went into crumpling the front of the truck).

For this high-speed collision, ideally the vehicle crumple zones would absorb as much kinetic energy as possible, as opposed to the low-speed fender benders shown above, where the vehicle bumper would deform and rebound back, to minimize cosmetic damage.

What about momentum conservation? If collisions occurred on frictionless roads, and/or the drivers did not apply their brakes, then momentum would be conserved with no external forces/impulses. However, let's limit our discussion to looking at the initial state just before a collision, and the final state just after the collision--since vehicle collisions happen in fractions of a second, even if the external friction forces are large, their impulse would be negligible due to the relatively brief time that collisions take place in, typically fractions of a second. Thus we can set the left-hand side of this equation equal to zero, and the two objects during a collision merely exchange momentum: whatever one object loses in momentum must correspond to an increase in the other object's momentum. Consider this next time you are about to get into an accident...

So what conservation law(s) apply for collisions? Momentum can be considered to be conserved for any collision (whether completely inelastic, inelastic, or elastic) provided that it is sufficiently brief, where the initial and final states are just before and just after the collision. However, kinetic energy is only conserved for elastic collisions, as that is the only type of collision where none of it is permanently lost in deformation.

Let's take a look at more examples of brief collisions (such that momentum would be conserved), classify their collision type and consider whether kinetic energy is conserved or not.

What type of collision is this? Is kinetic energy conserved?

What type of collision is this? Is kinetic energy conserved? (Video link: "Top Gear - Car hit by train - Car safety message - BBC.")

Observe this bullet being fired into a baseball (and passing through it, exiting out the other side). What type of collision is this? Is kinetic energy conserved? (Video link: "Will a .22LR Bullet go THRU a baseball? In Real SLOMO.")

Now consider this elephant gun being fired (a slug exits the gun to the right, and is included in the gun and person system). What type of collision could this be considered the time-reverse of? Is kinetic energy conserved? (Video link: "...Shooting 700 nitro GunTest.MP4.")

20111019

Astronomy current events question: Vesta northern vs. southern hemispheres

Astronomy.com editors, "Dawn at Vesta: Massive Mountains, Rough Surface, and Old-Young Dichotomy in Hemispheres," October 4, 2011
http://www.astronomy.com/News-Observing/News/2011/10/Dawn%20at%20Vesta%20-%20massive%20mountains%20-%20rough%20surface%20-%20and%20old-young%20dichotomy%20in%20hemispheres.aspx

__________ observed by the NASA Dawn spacecraft is evidence that the southern hemisphere of the asteroid Vesta may be much younger than its northern hemisphere.
(A) Different color features.
(B) The number of craters per area.
(C) Warm thermal vents.
(D) Rotation rate fluctuations.
(E) Subterranean oceans.

Correct answer: (B)

Student responses
Sections 70178, 70186, 70200
(A) : 7 students
(B) : 33 students
(C) : 2 students
(D) : 1 student
(E) : 3 students

Astronomy current events question: 2011 Nobel Prize in Physics

Kelly Beatty, "Three Cosmologists Share Nobel Prize," October 4, 2011
http://www.skyandtelescope.com/news/131071118.html

Three astrophysicists shared the 2011 Nobel Prize in Physics for measuring type Ia supernovae brightnesses in distant galaxies to provide evidence that the:
(A) speed of light is broken by neutrinos.
(B) expansion of the universe is accelerating.
(C) universe will not expand forever.
(D) cosmic background radiation is older than the universe itself.
(E) number of multiverses may be infinite.

Correct answer: (B)

Student responses
Sections 70178, 70186, 70200
(A) : 0 students
(B) : 43 students
(C) : 0 students
(D) : 3 students
(E) : 0 students

Astronomy current events question: Earth's oceans from comets?

Charles Q. Choi, "Comets Created Earth's Oceans, Study Concludes," October 5, 2011
http://www.space.com/13185-comets-water-earth-oceans-source.html

The __________ comet Hartley 2 observed by the Herschel Space Observatory may be evidence that Earth's oceans came from comets.
(A) snowflake-like patterns of.
(B) water vapor jets from.
(C) hydrogen isotopes of.
(D) melting polar ice caps on.
(E) plankton and algae growth on.

Correct answer: (C)

Student responses
Sections 70178, 70186, 70200
(A) : 4 students
(B) : 5 students
(C) : 36 students
(D) : 1 student
(E) : 0 students

20111018

Physics flashcard question: three thrown balls

Physics 205A, fall semester 2011
Cuesta College, San Luis Obispo, CA

Students were asked near the end of their learning cycle the following think-pair-share question, to be answered using flashcards.

The ball thrown __________ would have more K_tr just before hitting the floor.
(A) upwards.
(B) downwards.
(C) to the right.
(D) (There is two-way tie.)
(E) (There is three-way tie.)
(F) (Unsure/guessing/lost/help!)

Sections 70854, 70855
(A) : 2 students (estimated)
(B) : 2 students (estimated)
(C) : 5 students (estimated)
(D) : 5 students (estimated)
(E) : 20 students (estimated)
(F) : 20 students (estimated)

Correct answer: (E)

From the energy balance equation:

W_nc = ∆K_tr + ∆U_grav + ∆U_elas,

where W_nc = 0 (no external gains/losses of mechanical energy), and ∆U_elas = 0 (no springs involved), such that for all three balls,

0 = ∆K_tr + ∆U_grav.

Each ball experiences the same initial-to-final loss of U_grav, so they must experience the same increase in K_tr, and since they all have the same initial K_tr, then they will have the same final K_tr.

Physics midterm question: cars braking

Physics 205A Midterm 1, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 2.11, 2.25

[10 points.] Two cars are initially traveling side-by-side with the same speed in the forward direction along a straight horizontal road. Both cars apply their brakes at the same time, such that they slow down with different accelerations until they each come to a complete stop. Describe/show how the v_x(t) graph at right should be changed such that it correctly describes this situation. Explain your reasoning by using the properties of position, displacement, velocity, and acceleration.

Solution and grading rubric:

p = 10/10:
Correct. Describes/shows two lines with different slopes, with same initial velocity at same initial time, and zero velocities at different times.
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. At least understands how initial and final velocity, acceleration and displacement information is represented on a v_x(t) graph.
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
Exam code: midterm01w4Sh
p: 48 students
r: 1 student
t: 3 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 4816):

Another sample "p" response (from student 5950), showing some of the (identical) behavior of the cars before brakes were applied:

Yet another sample "p" response (from student 6666), with perhaps a more realistic depiction of deceleration due to anti-lock brakes: