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

Astronomy clicker question: why are white dwarfs small?

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

Astronomy 10 learning goal Q8.5

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

[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.

Correct answer: (A).

The fact that both stars have the same white-hot color tells you that they must have the same temperature (Wien's law). From the Stefan-Boltzmann law, luminosity is proportional to size and temperature^4, thus with both stars having the same temperature, the less luminous star is the smaller star.

Student responses
Section 4136
(A) : 9 students
(B) : 4 students
(C) : 7 students
(D) : 11 students

Section 5076
(A) : 2 students
(B) : 1 student
(C) : 10 students
(D) : 3 students

Physics quiz question: rope doing work on pulley and box

Physics 8A Quiz 9, Spring Semester 2004
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q9.2

[3.0 points.] Consider a pulley that turns without friction about a stationary horizontal axis through its center. A horizontal ideal massless, stretchless rope passes over the pulley, and has a 4.00 kg box suspended from its free end. There is no slipping between the rope and the rim of the pulley. The pulley is a uniform disk with radius 0.200 m and a moment of inertia of 0.700 kg*m^2. Initially this system is stationary, but a constant force pulls to the left on the rope, such that the box has a speed of 0.750 m/s after it has moved upwards by 0.100 m.

Which one of the following choices best describes the magnitude of the tension force in the rope?

(A) 39.2 N.
(B) 99.7 N.
(C) 127 N.
(D) 392 N.

(Cf. Young and Freeman, University Physics, 11/e, Problem 9.92.)

Correct answer: (B)
The external work done on this system is equal to the sum of the increase in gravitational potential energy of the box; increase in the translational kinetic energy of the box; and the increase in rotational kinetic energy of the pulley. Where d = +0.100 m, y_i = 0, w_f = v_f/r, and I_pulley = (1/2)*m_pulley*(r^2) then:

T_rope = ((m_box*g*y_f) + (1/2)*m_box*(v_f^2) + (1/2)*(I_pulley)*(w_f)^2)/d.

Somehow equating the tension in the rope to the weight of the box results in the incorrect response (A); the incorrect response (D) is the increase in gravitational energy of the box.

Student responses:
(A) : 17 students
(B) : 10 students
(C) : 1 student
(D) : 0 students

Physics quiz question: loaded spring, friction floor

Physics 8A Quiz 7, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q7.4

[3.0 points.] Consider a 0.600 kg box that is held against a horizontal spring of negligible mass, compressing the spring a distance of 0.120 m. When released, the box moves along a horizontal table for 1.00 m before coming to a rest. (This 1.00 m distance includes the distance covered by the box while the spring was still pushing it.) The coefficient of kinetic friction between the box and table is 0.221. Neglect air resistance.


Which one of the following choices best corresponds to the spring constant of the spring?

(A) 180 N/m.
(B) 408 N/m.
(C) 817 N/m.
(D) 3.70E+03 N/m.

(Cf. Young and Freeman, University Physics, 11/e, Problem 7.43.)

Correct answer: (A)
The decrease in spring potential energy of the unknown mass box is equal to the (non-conservative) work done by friction. Where d = +0.120 m, and x_f = 0 then:

k = (2*(µ_k)*m*g*d)/(x_i^2) = 180 N/m.

Somehow equating the decrease in spring potential energy to an increase in gravitational potential energy (?) results in the incorrect response (C):

k = 2*m*g*d/(x^2),

with another incorrect response (B) being similar, but without the factor of 2.

Student responses:
(A) : 10 students
(B) : 12 students
(C) : 8 students
(D) : 1 student

Physics quiz question: two-box system, frictionless table, ideal pulley

Physics 8A Quiz 7, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q7.2

[3.0 points.] Consider a 4.00 kg box on a frictionless table, connected by an ideal cable to a box of unknown mass. The pulley has negligible mass, and is frictionless. The system is released from rest, and after the 4.00 kg box has moved 0.155 m, it has a speed of 1.08 m/s. Neglect air resistance. Which one of the following choices best corresponds to the mass of the unknown box?

(A) 1.11 kg.
(B) 1.54 kg.
(C) 2.49 kg.
(D) 26.5 kg.

(Cf. Young and Freeman, University Physics, 11/e, Problem 6.82.)

Correct answer: (C)
The decrease in gravitational potential energy of the unknown mass box is equal to the sum of the increases in kinetic energies of the 4.00 kg box and the unknown mass box. Thus if y_f = 0, then:

m_unknown = ((4.00 kg)*v_f^2)/(2*g*y_i – v_f^2) = 2.49 kg.

Note that setting the decrease in gravitational potential energy of the unknown mass box equal to the increase in kinetic energies of the 4.00 kg box only, where again y_f = 0, results in the incorrect response (B):

m_unknown = ((4.00 kg)*v_f^2)/(2*g*y_i) = 1.54 kg.

Student responses:
(A) : 2 students
(B) : 20 students
(C) : 9 students
(D) : 0 students

Physics midterm problem: up-tugged yo-yo

Physics 8A Midterm 2, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q6.4

[20 points.] A thin ring of mass 2.40 kg and radius 0.100 m has a string wrapped around it that is pulled vertically upwards with an applied force 7.20 N in magnitude, and as a result it begins to roll (clockwise) to the right without slipping. Find the minimum coefficient of static friction that will allow this to happen.

(Cf. Young and Freeman, University Physics, 11/e, Problem 10.70.)

Solution and grading rubric:

• p = 20/20:
  Correct. Draws an extended-body diagram, correctly identifying the directions and locations of F_applied, w, n, and f_s (horizontally to the right, at the point of contact). Then applies N1 in the y-direction, and finds that n = 16.3 N.
  
  Applies N2 in the x-direction, and finds that
  
  a_x = f_s/m = u_s*n/m.
  
  Applies N2 rotational, and finds that
  
  -m*(r^2)*alpha = -F_applied*r + f_s*r.
  
  Constrains translational and rotational motion using a_x = r*alpha, and finds that
  
  u_s = f/(2*(m*g) - F_applied) = 0.221.
  
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least methodically applies Newton's laws for x, y, and rotations, and constrains translational and rotational motion, but typically demands n = mg, or has similar misapplications between using N1 or N2.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least some attempt at using Newton's laws, but with serious omissions or complications.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit. Typically sets up a diagram, and/or finds I = m*(r^2).
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
p: 1 student
r: 2 students
t: 12 students
v: 16 students
x: 5 students
y: 1 student
z: 1 student

Physics midterm problem: upward-sliding sponge

Physics 8A (currently Physics 208A) Midterm 2, spring semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Young and Freeman, University Physics, 11/e, Problem 3.63

A window washer pushes a sponge up a vertical window at constant speed by applying a force as shown at right. The sponge has a mass of 0.800 kg, and the coefficient of kinetic friction between the sponge and window is μ_k = 0.253. Determine the magnitude of the applied force, and the magnitude of the normal force exerted by the window on the sponge. Show your work and explain your reasoning using a free-body diagram, and the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Breaks up F_applied into x- and y-components. Applies Newton's first law in the x-direction, where:

  F_applied·cos(20.0°) = N.

  Then applies Newton's first law in the y-direction ("constant speed"), such that:

  F_applied·sin(20.0°) = m·g + μ_k·N.

  With two equations for two unknowns, solves for:

  F_applied = m·g/(sin(20.0°) - μ_k·cos(20°)) = 75.2 N,
  
  N = m·g·cos(20.0°)/(sin(20.0° - μ_k·cos(20.0°) = 70.6 N.

• r:
  Nearly correct, but includes minor math errors. And/or has correct numerical value for only one of the forces.
  
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Correctly identifies all forces acting on the sponge. At least methodically applies Newton's first law in the x- and y-directions, with F_applied broken up into x- and y-components.
  
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically omits f_k acting downwards, and/or demands N = m·g while applying Newton's laws.
  
• x:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y:
  Irrelevant discussion/effectively blank.
  
• z:
  Blank.
  

Grading distribution:
p: 6 students
r: 4 students
t: 7 students
v: 11 students
x: 10 students
y: 0 students
z: 0 students