20160705

Physics final exam question: reception from dipole antenna transmitter in different directions

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA

A radio transmitter broadcasts using two horizontal electric dipole antennae, with one mounted along the north-south direction, and the other mounted along the east-west direction. A Physics 205B student holding a horizontal receiver is located to the west of the transmitter, and walks to the south of the transmitter, but the receiver is aligned such that is always pointed directly towards the center of the transmitting antennae. Discuss whether the signal (if any) received by the Physics 205B student will increase, decrease, or not change as he walks from west to south. Explain your reasoning using the properties of light and polarization.

Solution and grading rubric:
  • p:
    Correct. Discusses/demonstrates that a constant zero signal will be received by the Physics 205B student's receiving antenna as he walks from west of to the south of the horizontal north-south and east-west transmitting antenna array because:
    1. the transmitting antenna array effectively transmits horizontally polarized radio waves in all horizontal directions;
    2. the receiving antenna axis always points longitudinally in towards the transmitting antenna array, thus is unable to receive any transversely polarized signal at all points along the horizon.
  • r:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. At least understands that no signal is received when west or south of the transmitting antenna array, but claims some signal is received when southwest of of the transmitting antenna array.
  • t:
    Nearly correct, but argument has conceptual errors, or is incomplete. At least understands that a receiving antenna must be held in the same transverse orientation as a transmitting antenna in order to receive a signal.
  • v:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying the properties of light (radio waves) and polarization.
  • x:
    Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying the properties of light (radio waves) and polarization.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 21 students
r: 6 students
t: 5 students
v: 6 students
x: 1 student
y: 0 students
z: 0 student

A sample "p" response (from student):

Another sample "p" response (from student):

Physics final exam question: increasing negative electric potential energy value

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA

Two point charges lie along the 
x-axis. A –2.0 nC charge is fixed at the origin, and a +1.0 nC charge is at x = +1.0 m. In order to make the electric potential energy of this system a smaller negative number, determine whether the +1.0 nC charge should be moved to the left, or to the right, or if this is not possible. Explain your reasoning by using the properties of charges, electric forces, electric potential, and electric potential energy.

Solution and grading rubric:
  • p:
    Correct. Understands that the electric potential energy of these charges of opposite sign is negative, and to make it a smaller negative number, the distance between them must increase, such that the +1.0 nC charge must be moved to the right. May also argue from doing work on the +1.0 nC to increase the electric potential energy of this system.
  • r:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
  • t:
    Nearly correct, but argument has conceptual errors, or is incomplete.
  • v:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying properties of charges, electric forces, electric potential, and electric potential energy.
  • x:
    Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying properties of charges, electric forces, electric potential, and electric potential energy.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 28 students
r: 3 students
t: 2 students
v: 1 student
x: 3 students
y: 2 students
z: 0 students

A sample "p" response (from student 3214):

Another sample "p" response (from student 3158):

Yet another sample "p" response (from student 5396):

Physics final exam question: invalid Feynman diagram

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA

Discuss why this Feynman diagram is invalid. Explain your reasoning using the properties of Feynman diagrams, particles and antiparticles, and interactions.

Solution and grading rubric:
  • p:
    Correct. Recognizes that the diagram is invalid by discussing:
    1. this was originally an electron capture process that has been rotated, but the proton path is reversed; or
    2. the "one line in, one line out" vertex rule is violated; or
    3. charge conservation is violated (net negative charge of electron and antineutrino in does not equal the net positive charge of neutron and proton out).
  • r:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
  • t:
    Nearly correct, but argument has conceptual errors, or is incomplete.
  • v:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying properties of Feynman diagrams.
  • x:
    Implementation/application of ideas, but credit given for effort rather than merit.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 26 students
r: 7 students
t: 2 students
v: 3 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 2691):

Another sample "p" response (from student 3158):

Physics final exam question: solidification age comparison

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA

Two samples currently have certain amounts of a radioactive isotope, embedded gaseous daughter element, and inert material (which is not involved in the decay process). Determine which sample has an older solidification age (as determined by radioactive dating), or if there is a tie. Explain your reasoning using properties of radioactive decay.

Solution and grading rubric:
  • p:
    Correct. The samples have the same solidification age, as they have the same ratio of radioactive isotopes today compared to when they first started, as the inert material does not factor into the decay process.
  • r:
    As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
  • t:
    Nearly correct, but argument has conceptual errors, or is incomplete.
  • v:
    Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying properties of radioactive decay.
  • x:
    Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of applying properties of radioactive decay.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 36 students
r: 0 students
t: 0 students
v: 2 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 9950):

Physics final exam problem: smartphone projector

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA

An inexpensive magnifying glass mounted in a shoebox is used with a smartphone as a light source to project images on a wall in a darkened room.[*] Shown at right is a view of the enlarged, upright image of the smartphone inside the projector when looking through the magnifying glass. Discuss what adjustments must be made (if any) to the location and/or orientation of the smartphone in order to properly display an enlarged, upright image on a wall. Show your work and explain your reasoning by using the properties of lenses, thin lens equations and/or ray tracings.

[*] content.photojojo.com/diy/turn-your-phone-into-a-photo-projector-for-1.

Solution and grading rubric:
  • p:
    Correct. Since the smartphone's upright, enlarged virtual image cannot be projected onto a wall, and in order to produce an "upright" image projected on a wall, the smartphone must instead be:
    1. turned upside-down; and
    2. pushed further away from the magnifying glass (such that it is placed just outside the focal point).
  • r:
    Nearly correct, but includes minor math errors. Neglects to mention that the phone must be turned upside-down.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
  • v:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. May claim that the smartphone already projects an image onto a wall, or that only the magnifying glass to wall distance needs to be changed.
  • x:
    Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying properties of lenses, thin lens equations, and ray tracings.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 6 students
r: 2 students
t: 4 students
v: 16 students
x: 9 students
y: 2 students
z: 0 student

A sample "p" response (from student 0001):

Another sample "p" response (from student 1614):

A sample "x" response (from student 1959):

Physics final exam problem: shunted voltmeter readings

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA


A 4.5 V emf source is connected to two light bulbs (each with different resistances), two voltmeters, and a switch. All of these components are ideal. Discuss why the voltmeters have equal readings while the switch is closed, and have unequal readings after the switch is opened. Show your work and explain your reasoning using the properties of voltmeters, Kirchhoff's rules and Ohm's law.

Solution and grading rubric:
  • p:
    Correct. Explicitly explains/demonstrates that the voltmeters in the closed-switch circuit have equal readings, and the voltmeters in the open-switch circuit have unequal readings by:
    1. applying Kirchhoff's loop rule to both circuits; and/or
    2. explicitly calculating the voltage drops for each of the voltmeters in both circuits.
  • r:
    Nearly correct, but includes minor math errors. For the closed-switch circuit, may instead compare the different amount of currents flowing through each light bulb, instead of through each ammeter.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
  • 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. No clear attempt at applying Kirchhoff's rules, Ohm's law, and properties of voltmeters.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 11 students
r: 8 students
t: 7 students
v: 7 students
x: 4 students
y: 0 students
z: 2 students

A sample "p" response (from student 1157):

Physics final exam problem: shunted ammeter readings

Physics 205B Final Exam, spring semester 2016
Cuesta College, San Luis Obispo, CA


A 4.5 V emf source is connected to two light bulbs (each with different resistances), two ammeters, and a switch. All of these components are ideal. Discuss why the ammeters have unequal readings while the switch is closed, and have equal readings after the switch is opened. Show your work and explain your reasoning using the properties of ammeters, Kirchhoff's rules and Ohm's law.

Solution and grading rubric:
  • p:
    Correct. Explicitly explains/demonstrates that the ammeters in the closed-switch circuit have unequal readings, and the ammeters in the open-switch circuit have equal readings by:
    1. applying Kirchhoff's junction rule to both circuits; and/or
    2. explicitly calculating the currents that flow through each of the ammeters in both circuits.
  • r:
    Nearly correct, but includes minor math errors. For the closed-switch circuit, may instead compare the different amount of currents flowing through each light bulb, instead of through each ammeter.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
  • v:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  • x:
    Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 30882, 30883
Exam code: finalm4u1
p: 12 students
r: 14 students
t: 5 students
v: 3 students
x: 2 students
y: 2 students
z: 1 student

A sample "p" response (from student 1157):

20160624

Physics quiz question: thermal contraction of James Webb Space Telescope mirror segment

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Reflections on JWST - May 4, 2016"
James Webb Space Telescope (JWST)
youtu.be/R-ldmlCkAoU

The James Webb Space Telescope will use beryllium mirror segments (linear expansion coefficient 7.0×10–6 K–1), each 1.32 m wide.[*][**] During testing it is cooled from room temperature (293 K) down to its operational temperature of 53 K. The width of a mirror segment will contract by:
(A) 2.2×10–3 m.
(B) 2.7×10–3 m.
(C) 3.2×10–3 m.
(D) 4.2×10–3 m.

[*] P. Hidnert, W. T. Sweeney, "Thermal Expansion of Beryllium and Aluminum Beryllium Alloys," Scientific Papers of the Bureau of Standards, vol. 22 (1927), p. 533, dx.doi.org/10.6028/nbsscipaper.251.
[**] jwst.nasa.gov/mirrors.html.

Correct answer (highlight to unhide): (A)

The relation between the change ∆L of a linear dimension L (here, width) due to a temperature change ∆T is given by:

α·∆T = ∆L/L,

such that:

L = L·α·T = (1.32 m)·(7.0×10–6 K–1)·(53 K – 293 K) = -2.2176×10–3 m,

or to two significant figures, the width of the mirror segment decreases by 2.2×10–3 m.

(Response (B) is L·α·(293 K); response (C) is L·α·(53 K + 293 K); response (D) is 1/∆T.)

Sections 70854, 70855, 73320
Exam code: quiz07zSOL
(A) : 62 students
(B) : 5 students
(C) : 1 student
(D) : 1 student

Success level: 90%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.22

Physics quiz question: rate of thermal loss through sleeping pad

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Therm-a-Rest® Closed-Cell Mattresses"
Therm-a-Rest
youtu.be/SwWevyXpibg

A Thermarest® Z Lite SOL™ polyethylene insulating sleeping pad[*] 
(area of 0.933 m2 and thickness of 0.020 m) has a thermal resistance of 0.43 K/watt. If there is a 20° C temperature difference between the top (in contact with a sleeping person) and bottom (in contact with the cold ground) of the sleeping pad, the rate of heat per time conducted through the pad will be:
(A) 1.1×10–3 watts.
(B) 2.2×10–2 watts.
(C) 8.6 watts.
(D) 47 watts.

[*] cascadedesigns.com/therm-a-rest/mattresses/fast-and-light/z-lite/product.

Correct answer (highlight to unhide): (D)

The rate of heat per time conducted is given by:

Power = ∆T/R,

where the temperature difference ∆T and the thermal resistance R of the sleeping pad are given, such that:

Power = (20° C)/(0.43 K/watt) = 46.511627907 watts,

or to two significant figures, the rate of heat conducted through the sleeping pad is 47 watts.

(Response (A) is d/(A·∆T); response (B) is R/∆T; response (C) is R·∆T.)

Sections 70854, 70855, 73320
Exam code: quiz07zSOL
(A) : 3 students
(B) : 4 students
(C) : 8 students
(D) : 54 students

Success level: 78%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

20160622

Physics quiz question: comparing surface temperatures of Bellatrix and Pollux

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

Stars can be modeled as spherical blackbodies that radiate heat out to an environment assumed to have a temperature of 0 K. Pollux is a star that radiates less heat per time, but is larger in size than Bellatrix. Pollux has a surface temperature __________ Bellatrix.
(A) cooler than.
(B) same as.
(C) hotter than.
(D) (Not enough information given.)

Correct answer: (A)

The net power radiated by a star with an environment assumed to be at absolute zero is given by:

Power = –e·σ·A·((T)4 – (Tenv)4) = –e·σ·A·((T)4 – (0 K)4).

where the emissitivity e = 1 (for an ideal blackbody), the Stefan-Boltzmann constant is σ = 5.670×10–8 watts/(m2·K4), A is the surface area, and the surface temperature T is unknown, and can be solved for:

(T)4 = –Power/(e·σ·A),

T = (–Power/(e·σ·A))(1/4).

This allows a ratio to be set up to compare the surface temperatures of Bellatrix and Pollux:

TBellatrix = (–PowerBellatrix/(e·σ·ABellatrix))(1/4),

TPollux = (–PowerPollux/(e·σ·APollux))(1/4),

such that:

(TBellatrix/TPollux) = (PowerBellatrix/PowerPollux))·(APollux/ABellatrix)(1/4).

Since Pollux radiates less heat per time than Bellatrix, then the ratio (PowerBellatrix/PowerPollux)) is greater than 1; and since Pollux is larger in size than Bellatrix, the ratio (APollux/ABellatrix)(1/4) is also greater than 1. Thus the ratio (TBellatrix/TPollux) must then be greater than 1, so Pollux has a cooler surface temperature than Bellatrix.

Sections 70854, 70855
Exam code: quiz07zSOL
(A) : 50 students
(B) : 12 students
(C) : 7 students
(D) : 0 students

Success level: 72%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70