Physics midterm problem: rolling uphill, with friction

Physics 8A Midterm 3, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q9.2

[20 points.] A 7.00 kg bowling ball of radius 0.150 m rolls up an inclined plane. When a Physics 8A student releases the bowling ball at the bottom of the inclined plane, it has an initial speed of 4.00 m/s. Assume that the bowling ball rolls up the inclined plane without slipping, and that work done by friction and air resistance against the bowling ball removes one-tenth of its initial energy by the time it reaches its maximum vertical height above the bottom of the inclined plane. Find the maximum vertical height increase of the bowling ball above the bottom of the inclined plane. Show your work and explain your reasoning.

(Cf. Young and Freeman, University Physics, 11/e, Problem 10.81.)

Solution and grading rubric:

• p = 20/20: Correct.
  The intial total mechanical energy is K_rot + K_trans =78.4 J. One-tenth of this energy is lost to friction, so the open-energy
  equation is
  
  -(1/10)*(78.4 J) = (1/2)*(I_sphere)*(0 - w_i^2) + (1/2)*m*(0 -v_i^2) + m*g*(y_f - 0),
  
  where w_i = v_i/r, thus y_f = 1.03 m. Or more simply, nine-tenths of 78.4 J is converted to U_grav, so m*g*y_f = (9/10)*(78.4 J).
  
  
• 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. Omits K_rot (but still has frictional losses), or includes both K_trans and K_rot, but does not account for frictional losses.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least attempts (open) energy conservation.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
p: 3 students
r: 3 students
t: 12 students
v: 12 students
x: 2 students
y: 0 students
z: 0 students

Astronomy clicker question: main sequence to giant/supergiant evolution

Astronomy 10, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal M3.1

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

[0.3 points.] According to the Stefan-Boltzmann law, how does the luminosity of a medium mass or massive main sequence star change as its outer layers expand and cool off, as it becomes a giant or supergiant?
(A) It becomes dimmer.
(B) It remains the same.
(C) It becomes brighter.
(D) (Any of the above (A)-(C) choices, depending on how old the star is.)

Correct answer: (B).

The evolutionary track of a main sequence star as it becomes a giant or supergiant can be approximated as a horizontal path from left-to-right across an H-R diagram, which means that its luminosity remains constant while its surface temperature decreases. From the Stefan-Boltzmann law, constant luminosity means that size must increase while temperature decreases.

Student responses
Section 4136
(A) : 13 students
(B) : 3 students
(C) : 9 students
(D) : 5 students

Section 5076
(A) : 8 students
(B) : 6 students
(C) : 4 students
(D) : 2 students

Astronomy quiz question: CNO cycle stars

Astronomy 10 Quiz 9, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.4

[3.0 points.] Which one of the following choices best explains what allows hydrogen to fuse together with much heavier elements in the cores of massive main sequence stars?
(A) Stronger gravitational forces.
(B) Stronger degenerate pressures.
(C) Continuous absorption and reemission of higher energy photons.
(D) Higher temperatures.
(E) Stronger convection currents.

Correct answer: (D)
Higher temperatures mean faster speeds. For hydrogen fusion, the temperature must be high enough for protons to approach close enough to fuse, in spite of the mutually repulsive force due to their positive charge. For hydrogen to fuse together with a heavier element (such as carbon, which has 14 protons), the temperature must be even higher for the hydrogen proton to approach the stronger repelling carbon nucleus.

Student responses
Section 4136
(A) : 18 students
(B) : 6 students
(C) : 3 students
(D) : 2 students
(E) : 10 students

Student responses
Section 5076
(A) : 5 student
(B) : 2 students
(C) : 3 students
(D) : 10 students
(E) : 1 students

Physics quiz question: ring and disk

Physics 8A Quiz 9, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q9.4

[3.0 points.] Consider a disk that rotates about a fixed vertical axis at a rate of 2.00 rad/s. The diameter of the disk is D, and its mass is M. A stationary thin, uniform ring with the same diameter and mass as the disk is dropped straight down onto the disk. The ring and disk reach a new final angular speed as they rotate together. Neglect angular impulses due to torques exerted by external friction and drag forces during this brief process.

Which one of the following choices best describes the quantities that were conserved during this process?
(A) Both angular momentum and rotational kinetic energy were conserved.
(B) Only angular momentum was conserved.
(C) Only rotational kinetic energy was conserved.
(D) Neither angular momentum nor rotational kinetic energy was conserved.

(Cf. Young and Freeman, University Physics, 11/e, Problem 10.43.)

Correct answer: (B)
No external angular impulses are exerted on this system, so the total angular momentum of the system is conserved. However, friction allows the disk to speed up the ring such that they rotate together with the same final angular speed, thus rotational kinetic energy is lost for this completely inelastic "angular collision." (The change in the total rotational kinetic energy during this process can be shown to be negative, once w_f is determined from applying angular momentum conservation.)

Student responses:
(A) : 9 students
(B) : 12 students
(C) : 6 students
(D) : 3 students

Bon mots: understanding trumps math

"I love only nature, and I hate mathematicians."
--Richard Feynman

"Nothing in life is to be feared. It is only to be understood."
--Marie Curie

Upon seeing the first quote in the class announcements, a student asked who Richard Feynmann was, although it was not clear whether this interest was due to a pro-nature or anti-math sentiment.

Physics quiz question: falling box, rotating disk

Physics 8A Quiz 9, spring semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Young and Freeman, University Physics, 11/e, Problem 9.86

A 3.0 kg solid disk of radius 0.80 m that is mounted on a horizontal frictionless axis through its center. A light string is wrapped around its rim, and a 1.0 kg box is attached to the other end of the string. The box is released from rest from an unknown height above the floor, and has a speed of 4.2 m/s just as it strikes the floor. Neglect friction and drag. The initial height of the box above the floor is:
(A) 0.45 m.
(B) 0.90 m.
(C) 1.4 m.
(D) 2.3 m.

Correct answer: (D)

No external work is done on this system, so the decrease in gravitational potential energy of the box is equal to the sum of the increases of the translational kinetic energy of the box, and the rotational kinetic energy of the disk. The translational speed of the box v is related to the angular speed of the disk via v = r·ω. With initial conditions v_i = 0, and ω_i = 0; final conditions v_f = r·ω_f, and y_f = 0; and the moment of inertia of the disk is I_disk = (1/2)·m_disk·r², the energy conservation transfer-balance equation is:

W_nc = ∆K_tr + ∆K_rot + ∆U_grav + ∆U_elas,

0 = (1/2)·m_box·(v_f² - 0²) + (1/2)·I_disk·(ω_f² - 0²) + m_box·g·(0 - y_i) + 0,

0 = (1/2)·m_box·v_f² + (1/2)·((1/2)·m_disk·r²)·(v_f/r)² - m_box·g·y_i,

0 = (1/2)·m_box·v_f² + (1/4)·m_disk·v_f² - m_box·g·y_i,

Solving for y_i:

y_i = ((1/2)·m_box + (1/4)·m_disk)·v_f²/m_box·g = 2.25 m,

or to two significant figures, 2.3 m.

Choice (A) stems from a sign error in the kinetic energy terms. Neglecting the rotational kinetic energy of the disk results in (B); neglecting the translational kinetic energy of the box results in (C).

Student responses:
(A) : 6 students
(B) : 12 students
(C) : 10 students
(D) : 2 students

Astronomy in-class activity: OBAFGKM poetry slam, illustrated

Astronomy 10, spring semester 2007
Cuesta College, San Luis Obispo, CA

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.

Three favorites from this semester, by virtue of including illustrations (which were projected onto an overhead screen while the students read their poems):

Oh Because A Freaking Giraffe Kicked Me Right Near Something Critical.
--J. S.

Only Beautiful Astronomers Find Gorgeous Killer Moons.
--D. S.

Oh Bummer, Another Freaky Giant Killer Monkey Raided Nearby School Classes.
--S. P.

Previous post: OBAFGKM poetry slam (Spring Session 2007).

Plus a favorite from a past semester (Fall 2005):

Oh Beautiful Astronomical Friends Go Kiss Mr. Len.
--B. B.

Astronomy in-class activity: OBAFGKM poetry slam

Astronomy 10, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

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.

Two favorites from this semester:

Oops! Britney Attacked Federline's GMC. Kevin's Mad.
--C. K.

Only Batman Always Forgets Giant Kryptonite Meteorites Resting Near Superman's Car.
--K. P.

Astronomy clicker question: protostar to main sequence evolution

Astronomy 10, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q9.5

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

[0.3 points.] How is it possible for the luminosity of a protostar to remain (approximately) constant as it becomes a main sequence star?
(A) Its surface temperature gets hotter as its size gets smaller.
(B) Its surface temperature gets cooler as its size gets smaller.
(C) Its surface temperature gets hotter as its size gets larger.
(D) Its surface temperature gets cooler as its size gets larger.

Correct answer: (A).

The evolutionary track of a protostar as it becomes a main sequence star can be approximated as a horizontal path from right-to-left across an H-R diagram, which means that its luminosity remains constant while its surface temperature increases. From the Stefan-Boltzmann law, constant luminosity means that size must decrease while temperature increases.

Student responses
Section 4136
(A) : 19 students
(B) : 2 students
(C) : 7 students
(D) : 1 student

Student responses
Section 5076
(A) : 7 students
(B) : 0 students
(C) : 13 students
(D) : 1 student

Astronomy quiz question: applications of the Stefan-Boltzmann law

Astronomy 10 Quiz 8, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Astronomy 10 learning goal Q8.5

[Version 1]

[3.0 points.] Which one of the following statements best describes the relationship between a main sequence star and a supergiant that have the same luminosity?
(A) The main sequence star is cooler and smaller than the supergiant.
(B) The main sequence star is cooler and larger than the supergiant.
(C) The main sequence star is hotter and smaller than the supergiant.
(D) The main sequence star is hotter and larger than the supergiant.
(E) (None of the above choices (A)-(D), as it is not possible for a main sequence star to have the same luminosity as a supergiant.)

Correct answer: (C)
From an H-R diagram, a main sequence star must be hotter in order to have the same luminosity as a supergiant. From the Stefan-Boltzmann law, since luminosity is proportional to size and temperature^4, and with both stars having the same luminosity, the hotter star must be smaller in size.

Student responses
Section 4136
(A) : 6 students
(B) : 3 students
(C) : 22 students
(D) : 1 student
(E) : 0 students

[Version 2]
[3.0 points.] Which one of the following statements best describes the relationship between a main sequence star and a white dwarf that have the same luminosity?
(A) The main sequence star is cooler and smaller than the white dwarf.
(B) The main sequence star is cooler and larger than the white dwarf.
(C) The main sequence star is hotter and smaller than the white dwarf.
(D) The main sequence star is hotter and larger than the white dwarf.
(E) (None of the above choices (A)-(D), as it is not possible for a main sequence star to have the same luminosity as a white dwarf.)

Correct answer: (B)
From an H-R diagram, a main sequence star must be cooler in order to have the same luminosity as a supergiant. From the Stefan-Boltzmann law, since luminosity is proportional to size and temperature^4, and with both stars having the same luminosity, the cooler star must be larger in size.

Student responses
Section 5076
(A) : 3 students
(B) : 10 students
(C) : 1 student
(D) : 4 students
(E) : 2 students