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 flashcard question: net force

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

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

Two forces act on an object. The net force x-component on the object is:
(A) +3 N.
(B) +4 N.
(C) +5 N.
(D) +6 N.
(E) +7 N.
(F) +8 N.
(G) +11 N.
(H) (Unsure/guessing/lost/help!)

Sections 70854, 70855 (pre-)
(A) : 0 students
(B) : 5 students
(C) : 0 students
(D) : 0 students
(E) : 1 students
(F) : 16 students
(G) : 0 students
(H) : 13 students

This question was asked again after students were instructed to discuss with a neighbor how they chose their answer, and to convince each other why their answer is (in)correct.

Sections 70854, 70855 (post-)
(A) : 0 students
(B) : 4 students
(C) : 0 students
(D) : 0 students
(E) : 0 students
(F) : 20 students
(G) : 0 students
(H) : 14 students

Correct answer: (F)

Pre- to post- peer-interaction gains:
pre-interaction correct = 46%
post-interaction correct = 53%
Hake (normalized) gain <g> = 13%

Physics flashcard question: acceleration at highest point in trajectory

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

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

The velocity vector components of a projectile trajectory is shown. Air resistance is negligible.


At the highest point in the trajectory, which acceleration component is zero?
(A) a_x.
(B) a_y.
(C) (Both of the above choices.)
(D) (None of the above choices.)
(E) (Unsure/guessing/lost/help!)

Sections 70854, 70855 (pre-)
(A) : 15 students
(B) : 17 students
(C) : 3 students
(D) : 3 students
(E) : 4 students

This question was asked again after students were instructed to discuss with a neighbor how they chose their answer, and to convince each other why their answer is (in)correct.

Sections 70854, 70855 (post-)
(A) : 20 students
(B) : 7 students
(C) : 5 students
(D) : 1 student
(E) : 5 students

Correct answer: (A)

Pre- to post- peer-interaction gains:
pre-interaction correct = 36%
post-interaction correct = 53%
Hake (normalized) gain <g> = 27%

Physics clicker question: vector direction, from given components

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 3.21 (extended)

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

The x- and y-components of a vector are given: A_x = +1.4 m, A_y = –5.7 m. What is the direction of this vector, as measured counterclockwise from the +x axis?
(A) +76°.
(B) +104°.
(C) +256°.
(D) +284°.
(E) (I'm lost, and don't know how to answer this.)

Sections 70854, 70855
(A) : 3 students
(B) : 2 students
(C) : 5 students
(D) : 19 students
(E) : 11 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 70854, 70855
(A) : 0 students
(B) : 5 students
(C) : 2 students
(D) : 35 students
(E) : 1 students

Correct answer: (D)

The acute angle made by this vector with the x-axis is given by:

Arctan(A_y/A_x) = 76°,

which is not placed in the proper quadrant by most calculators; this must be done by inspection. From the fact that A_x is positive and A_y is negative, this vector is in the fourth quadrant, and thus the proper angle is 360° - 76° = +256, as measured counterclockwise with respect to the positive x-axis, as is done when using the unit circle in trigonometry.

Response (A) is the raw answer from Arctan(A_y/A_x), response (B) is 180° - 76°, while response (C) is 180° + 76°.

Note that from inspection, response (D) is the only vector angle that is in the fourth quadrant, and thus can be deduced as the correct angle without explicitly doing any calculations.

Pre- to post- peer-interaction gains:
pre-interaction correct =48%
post-interaction correct = 81%
Hake, or normalized gain = 65%

Physics clicker question: warming ice, freezing water

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.33 (extended)

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

500 g of ice at -10 degrees C is placed into 500 g of 0 degrees C water, in a container that is insulated from the environment. As this system reached thermal equilibrium:
(A) ice melted.
(B) water froze.
(C) (Both of the above choices.)
(D) (None of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 4 students
(B) : 15 students
(C) : 4 students
(D) : 7 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 2 students
(B) : 18 students
(C) : 5 students
(D) : 6 students
(E) : 0 students

Correct answer: (B)

Heat will flow from the water to the ice, which results in water freezing (as it is already at 0 degrees C), and ice warming up from its initial temperature. Ice does not melt during this process, as would have to be warmed up entirely to 0 degrees C, in which case is not consistent with heat leaving the water in this isolated system.

Pre- to post- peer-interaction gains:
pre-interaction correct =50%
post-interaction correct = 58%
Hake, or normalized gain = 16%

Physics clicker question: temperature change during phase change

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Question 14.8 (extended)

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

The change in temperature of a material undergoing a phase change (such as melting/freezing, or boiling/condensing) is typically:
(A) 0.
(B) ∞.
(C) some finite number.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 12 students
(B) : 4 students
(C) : 14 students
(D) : 0 students
(E) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 14 students
(B) : 2 students
(C) : 14 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

Temperatures remain constant during phase changes; thus the change in temperature must be zero.

Pre- to post- peer-interaction gains:
pre-interaction correct =39%
post-interaction correct = 47%
Hake, or normalized gain = 13%

Physics clicker question: internal energy changes in an insulated system

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.33 (extended)

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

A 50 g iron sample at 75° C is placed into 1000 g of room temperature water (25° C), in a container that is insulated from the environment. Which component of this isolated system had the greatest change in internal energy once thermal equilibrium is reached?
(A) 50 g iron sample.
(B) 1000 g of water.
(C) (There is a tie.)
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 8 students
(B) : 11 students
(C) : 13 students
(D) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 13 students
(B) : 6 students
(C) : 14 students
(D) : 0 students

Correct answer: (C)

In this isolated system, the heat given up by the iron sample must be exactly equal to the heat taken in by the water. Thus the amount of internal energy changes of both components of this system must be equal.

Pre- to post- peer-interaction gains:
pre-interaction correct =39%
post-interaction correct = 42%
Hake, or normalized gain = 5%

Physics clicker question: internal energy of different mass samples

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Question 14.20 (extended)

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

A 50 g iron sample and a 10 g iron sample are both at room temperature (25 degrees C). Which sample has more internal energy?
(A) 50 g iron sample.
(B) 10 g iron sample.
(C) (There is a tie.)
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 14 students
(B) : 9 students
(C) : 4 students
(D) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 27 students
(B) : 0 students
(C) : 1 student
(D) : 0 students

Correct answer: (A)

Adding heat (Q) to an object increases its internal energy, resulting in increasing its temperature. If both samples could be at absolute zero, then they would possess the same (zero) internal energy. The larger sample would require more heat to be added in order to raise its temperature up to room temperature, as expressed as:

Q = delta(internal energy) = m*c*delta(T),

where c is the (mass) specific heat (same for both samples), and delta(T) is the same for both samples.

Pre- to post- peer-interaction gains:
pre-interaction correct =50%
post-interaction correct = 96%
Hake, or normalized gain = 93%

Physics clicker question: warmer air wavelength

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 12.1

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

Sound wave source frequency f = 1,500 Hz.
Sound wave speed v = 340 m/s (STP)

If the temperature of the air were to increase, the wavelength of the sound wave in air would:
(A) increase.
(B) remain constant.
(C) decrease.
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 25 students
(B) : 5 students
(C) : 4 students
(D) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 31 students
(B) : 1 student
(C) : 0 students
(D) : 0 students

Correct answer: (A)

The speed v of sound waves in air at a temperature T (in Kelvin) is:

v = v_0*sqrt(T/T_0),

where v_0 and T_0 are the speed of sound waves in air at an absolute temperature T_0.

The wavelength of a wave is given by:

lambda = v*T = v/f,

where T and f are the period and frequency of the wave, respectively; both are set by the source, which is presumed to be unchanged by the increase in air temperature. Thus increasing the temperature of the air would increase the speed of sound waves, which would result in longer wavelengths.

Pre- to post- peer-interaction gains:
pre-interaction correct = 74%
post-interaction correct = 97%
Hake, or normalized gain = 88%

Physics clicker question: standing wave on tightened string

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 11.46

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

A guitar's A-string of a length 0.80 m and is stretched to a tension of 40 N. It vibrates at a fundamental frequency of 220 Hz. If the tension in the string is increased by a factor of 2.0, the fundamental frequency of this string increases by a factor of:
(A) 1.4.
(B) 2.0.
(C) 4.0.
(D) (The fundamental frequency remains the same.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 19 students
(B) : 7 students
(C) : 4 students
(D) : 0 students
(E) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 21 students
(B) : 1 student
(C) : 3 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The speed v of transverse waves along a string is:

v = sqrt(F/mu),

where F is the tension in the string, while mu is the linear mass density (mass per unit length) of the string.

The fundamental frequency f_1 for a standing wave on a string is given by:

f_1 = v/(2*L),

where L is the length of the rope. Thus doubling the tension F would increase the wave speed v by a factor of sqrt(2) = 1.4, and thus the fundamental frequency f_1 would also increase by the same 1.4 factor.

Pre- to post- peer-interaction gains:
pre-interaction correct = 61%
post-interaction correct = 84%
Hake, or normalized gain = 59%

Physics clicker question: changing tension of a rope wave

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Questions 11.5, 11.6, 11.8

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

Cord tension F = 75 N
Wave speed v = 140 m/s
Frequency f = 20 Hz (from a vibrating source)

Which wave parameter(s) increase if the cord were made more taut?
(A) Wave speed v.
(B) Wavelength lambda.
(C) Period T.
(D) (More than one of above choices.)
(E) (None of above choices.)
(F) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 7 students
(B) : 5 students
(C) : 2 students
(D) : 19 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 4 students
(B) : 0 students
(C) : 0 students
(D) : 27 students
(E) : 0 students

Correct answer: (D)

The frequency f of a wave is determined solely by the properties of the source, while the wave speed v is determined solely by the properties of the medium (thus frequency and wave speed are said to be the independent wave parameters). For rope waves:

v = sqrt(F/mu),

where F is the tension in the rope, while mu is the linear mass density (mass per unit length) of the rope.

The wavelength is the dependent wave parameter that is determined by both the source and medium, given by:

lambda = v/f or v*T,

where T is the period (which is the inverse of frequency). So if tension F increases, the wave speed v will increase (while the frequency remains constant), and also the wavelength will increase as result of the increase in v.

Pre- to post- peer-interaction gains:
pre-interaction correct = 58%
post-interaction correct = 87%
Hake, or normalized gain = 70%

Physics clicker question: changing frequency of a wave source

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Questions 11.5, 11.6, 11.8

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

A wave has a frequency and wavelength of 200 Hz and 0.75 m, respectively. If the frequency of the wave source is changed, which wave parameters(s) will change as a result?
(A The wavelength.
(B) The speed of the wave.
(C) (Both of the above choices.)
(D) (None of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 20 students
(B) : 4 students
(C) : 9 students
(D) : 1 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 25 students
(B) : 2 students
(C) : 6 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The frequency f of a wave is determined solely by the properties of the source, while the wave speed v is determined solely by the properties of the medium (thus frequency and wave speed are said to be the independent wave parameters). The wavelength is the dependent wave parameter that is determined by both the source and medium, given by:

lambda = v/f or v*T,

where T is the period (which is the inverse of frequency). So if frequency changes, the wave speed will still remain constant, but the wavelength will as a result change.

Pre- to post- peer-interaction gains:
pre-interaction correct = 59%
post-interaction correct = 76%
Hake, or normalized gain = 41%

Physics clicker question: maximum SHM speed

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.30

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

A 0.40 kg object on a 99 N/m spring oscillates left to right on a frictionless surface, with an amplitude of 0.10 m. What is the speed of the object at the equilibrium point?
(A) 0 m/s.
(B) 0.40 m/s.
(C) 1.57 m/s.
(D) 24.8 m/s.
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 2 students
(B) : 14 students
(C) : 9 students
(D) : 3 students
(E) : 4 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 1 student
(B) : 19 students
(C) : 0 students
(D) : 0 students
(E) : 5 students

Correct answer: (C)

From energy conservation, the potential energy of the mass when the spring is fully compressed or stretched at its maximum displacement from equilibrium x = +/- A can be set equal to the kinetic energy of the mass as it passes through equilibrium at x = 0:

U_max = K_max,
(1/2)*k*A^2 = (1/2)*m*v_max^2,

such that v_max = sqrt(k/m). Response (B) is the period T = 2*pi*sqrt(m/k) of this mass-spring system, and (D) is the maximum acceleration a_max = (k/m*)*A.

Pre- to post- peer-interaction gains:
pre-interaction correct = 28%
post-interaction correct = 0%
Hake, or normalized gain = -39% (!)

Physics clicker question: partially- and fully-submerged objects

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

(Adapted from Fig. 2.1, p. 11, Mazur, Peer Instruction: A User's Manual, Prentice-Hall, Inc., 1997)

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

Two objects (same volumes, same densities) are held either half- or fully-submerged in a tank of water. It is not known whether these two objects are both lighter or denser than water.

Which object has the greatest bouyant force exerted on it?
(A) Object X.
(B) Object Y.
(C) (There is a tie.)
(D) (Depends on whether both these two objects are lighter or denser than water.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 5 students
(B) : 25 students
(C) : 2 students
(D) : 1 student
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 1 student
(B) : 30 students
(C) : 1 student
(D) : 1 student
(E) : 0 students

Correct answer: (A)

The bouyant force on an object is given by:

F_B = rho*g*V,

where rho is the density of the fluid, and V is the volume displaced by the object--the volume of the object that is submerged under the surface of the liquid. Thus regardless of the density of the objects, and whether the hands are lifting or pushing down on these objects, the submerged object experiences a greater bouyant force than the partially submerged object.

Pre- to post- peer-interaction gains:
pre-interaction correct = 76%
post-interaction correct = 91%
Hake, or normalized gain = 63%

Physics clicker question: bag of chips, from sea level to the mountains

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Question 9.9

Gas Bag
Posted by witch532
http://www.youtube.com/watch?v=wRFZXy-999I

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

An unopened bag of potato chips is purchased at sea level, and then is brought up on a car trip into the mountains. As a result, the bag "inflates" and becomes rigid. This is because of a(n):
(A) increase in pressure inside the bag.
(B) decrease in pressure outside the bag.
(C) (Both of the above choices.)
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 5 students
(B) : 22 students
(C) : 5 students
(D) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 0 students
(B) : 31 students
(C) : 3 students
(D) : 0 students

Correct answer: (B)

The principal cause of the bag "inflating" is the decrease in pressure in the air surrounding the bag as it is brought to a higher elevation. The pressure inside the bag then is greater than the ambient pressure outside. The absolute pressure inside the bag could increase due to an increase in temperature (using the ideal gas law), which is what some students discussed in choosing (C).

Pre- to post- peer-interaction gains:
pre-interaction correct = 69%
post-interaction correct = 91%
Hake, or normalized gain <g> = 72%

Physics clicker question: Newton's third law (interaction) pairs

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 4.53

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

A book rests on the surface of a (horizontal) table.

__________ tells you that these two forces are equal in magnitude and opposite in direction:

The force of the Earth pulling on the book.
The force of the table pushing on the book.

(A) Newton's first law.
(B) Newton's second law.
(C) Newton's third law.
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 16 students
(B) : 2 students
(C) : 17 students
(D) : 4 students

This question was asked again after displaying the tallied results with the lack of consensus, in order to see if students were going to be able to discuss and determine the correct answer among themselves, with the following results. Note that in contrast to the usual practice of no comment, the instructor answered many clarification questions from the students ("Is the force of the Earth pulling on the book the weight force?" "Is the book also pushing on the table?").

Sections 30880, 30881
(A) : 28 students
(B) : 1 student
(C) : 5 students
(D) : 1 student

Correct answer: (A)

The upwards normal force of the table on the book must be equal in magnitude to the downwards weight force of the Earth on the book, due to Newton's first law, as these forces exerted on the stationary book must sum to zero. These two forces cannot be a Newton's third law pair (an "interaction pair") because they fail the last two parts of the three-part "POF-OST-ITO" checklist ("Pair of Opposite Forces; Of Same Type; Involving Two Objects," Benjamin Crowell, Fullteron College, CA) fails.

Pre- to post- peer-interaction gains:
pre-interaction correct = 41%
post-interaction correct = 80%
Hake, or normalized gain <g> = 66%

[Follow-up question.]

__________ tells you that these two forces are equal in magnitude and opposite in direction:

The force of the book pushing on the table.
The force of the table pushing on the book.

(A) Newton's first law.
(B) Newton's second law.
(C) Newton's third law.
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 8 students
(B) : 2 students
(C) : 29 students
(D) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 7 students
(B) : 0 students
(C) : 27 students
(D) : 0 students

Correct answer: (C)

The upwards normal force of the table on the book must be equal in magnitude to the downwards normal force of the book on the table, due to Newton's third law, as these forces satisfy all three parts of the "POF-OST-ITO" checklist ("Pair of Opposite Forces; Of Same Type; Involving Two Objects").

Pre- to post- peer-interaction gains:
pre-interaction correct = 74%
post-interaction correct = 79%
Hake, or normalized gain <g> = 20%

Physics clicker question: accelerating elevator force magnitudes

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 4.22

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

A 2,010 kg elevator moves with an upward acceleration of 1.50 m/s^2.

Which force has the least magnitude?
(A) Net force on the elevator.
(B) Weight of the elevator.
(C) Tension force of cable on the elevator.
(D) (There is a tie.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 29 students
(B) : 11 students
(C) : 0 students
(D) : 0 students
(E) : 1 student
(F) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 37 students
(B) : 6 students
(C) : 0 students
(D) : 0 students
(E) : 0 students
(F) : 0 students

Correct answer: (A)

The free-body diagram for the elevator is discussed in a previous post:

Physics clicker question: accelerating elevator free-body diagrams

The weight force is given by m*g = 1.97e4 N, upwards. The net force, from Newton's second law, is m*a = 3.02e3 N, upwards. Thus the magnitude of the upwards tension force must be greater than the weight force by 3.02e3 N, which must then be 2.27e4 N, making it the force with the greatest magnitude, and of these three choices, net force has the least magnitude.

Pre- to post- peer-interaction gains:
pre-interaction correct = 71%
post-interaction correct = 86%
Hake, or normalized gain = 52%

Physics clicker question: accelerating elevator free-body diagrams

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 4.22

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

A 2,010 kg elevator moves with an upward acceleration of 1.50 m/s^2.

Which free-body diagram shows the tension and weight forces acting on the elevator, and the resulting net force?


(N.b.: (F) is the "I'm lost, and don't know how to answer this" response.)

Sections 30880, 30881
(A) : 25 students
(B) : 7 students
(C) : 6 students
(D) : 2 students
(E) : 3 students
(F) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 39 students
(B) : 0 students
(C) : 0 students
(D) : 1 student
(E) : 0 students
(F) : 0 students

Correct answer: (A)

The elevator is accelerating upwards, thus net force must point upwards as well, from Newton's second law. Thus FDBs (B) and (E) are eliminated for having downwards net forces, and FBD (C) is eliminated for having a zero net force. FBD (D) has the downwards weight force more than the upwards tension force, which is not consistent with its upwards net force, and thus be eliminated for being internally inconsistent. (Similarly, FBD (E) is also "impossible.")

FBD (A) correctly shows an upwards net force, and the corresponding upwards net force as having a greater magnitude than the downwards weight force.

Pre- to post- peer-interaction gains:
pre-interaction correct = 58%
post-interaction correct = 98%
Hake, or normalized gain = 94%

Physics clicker question: average acceleration from initial and final velocity vectors

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 3.41

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

Suppose a car travels one-quarter around a circle in a time of 1.0 s at a constant speed of 31.4 m/s, as shown at right.

Which red vector is delta(v) = v_f - v_i for this 1.0 s time interval?



Sections 30880, 30881
(A) : 18 students
(B) : 20 students
(C) : 1 student
(D) : 2 students
(E) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 10 students
(B) : 32 students
(C) : 0 students
(D) : 2 students
(E) : 0 students

Correct answer: (B)

The vector operation v_f - v_i can be interpreted as tail-to-head vector addition if the direction of v_i is reversed, such that delta(v) = v_f + (-v_i).

Pre- to post- peer-interaction gains:
pre-interaction correct = 48%
post-interaction correct = 73%
Hake, or normalized gain = 48%

Subsequently, students were asked to submit numerical answers to the following clicker question:

Note: a_av = delta(v)/delta(t) = (v_f - v_i)/delta(t).

The average acceleration magnitude of the car for this one-quarter around trip is _____ m/s^2.

Sections 30880, 30881
"-5" : 1 student
"-1.75" : 1 student
"-1" : 2 students
"0" : 19 students
"2" : 3 students
"4" : 1 student
"7.85" : 1 student
"8" : 2 students
"31.4" : 9 students
"44.4" : 2 students

Correct response: 44.4 m/s^2.

Due to too few students being able to answer this correctly, this question was not asked again for peer discussion. Instead, the instructor facilitated a whole-class discussion to determine why students chose certain answers.

"0" m/s^2 was chosen because the speed of the car was constant. However, the direction is changing, and this by itself requires a non-zero acceleration.

"2," "4," "7.85" and "8" m/s^2 were probably chosen due to reading the graph scales figuratively (even though no such scale was given).

"-1" m/s^2 was chosen because it was the "slope" of the a_av vector, thus confusing vector operations with kinematic graphs.

The length of the resultant delta(v) vector is 44.4 m/s (from applying the Pythagorean theorem), and thus a_av = delta(v)/delta(t) = 44.4 m/s^2.

Physics clicker question: free fall acceleration

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Question 2.1

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) in the middle of their learning cycle (after being asked about the acceleration while the ball is traveling downwards, and while traveling upwards):

A ball is thrown and released such that it initially travels upwards. While it is momentarily at its highest point, its acceleration is __________. (Up = +y direction; neglect air resistance ("free fall").)
(A) +9.80 m/s^2.
(B) 0 m/s^2.
(C) –9.80 m/s^2.
(D) (More than one of the above choices.)
(E) (None of the above choices.)
(F) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 0 students
(B) : 30 students
(C) : 13 students
(D) : 0 students
(E) : 0 students
(F) : 0 students

Correct answer: (C)

A whole-class discussion on velocity versus time graphs for objects in free fall immediately followed the tallying of results. All v_y(t) graphs for free fall may have different initial v_i,y values, but would all have the same negative slopes, and thus acceleration values of -9.80 m/s^2.