Physics midterm question: comparing compression of rod in different orientations

Physics 205A Midterm 2, fall semester 2019
Cuesta College, San Luis Obispo, CA

A 0.40 m long copper rod with a square profile (0.10 m × 0.10 m) can be oriented standing up, or laid down on its side on a floor. If the same amount of downwards force is applied to the top surface in each case, discuss whether the standing-up rod or the laid-down rod will compress a greater ∆L amount (or if there will be a tie). Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:

• p:
  Correct. Applies Hooke's law in a systematic manner by:
  1. recognizing that the applied force F and Young's modulus Y are the same for both rods; and
  2. as a result ΔL depends only on the original length L divided by cross-sectional area A; and
  3. since the standing-up rod has a longer original length (L = 0.40 m) and a smaller cross-sectional area (A = 0.010 m²), it will compress more than the laid-down rod with a shorter original length (L = 0.10 m) and a greater cross-sectional area (A = 0.040 m²).
• 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. Considers only difference in cross-sectional areas (neglecting the difference in original lengths), or vice versa; or recognizes both differences but somehow argues that the rods will still compress by the same amount.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least some systematic attempt at using Hooke's law quantities.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of relating strain (force per unit area), Young's modulus, and strain using Hooke's law.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 18 students
r: 1 student
t: 28 students
v: 3 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 2586):
A sample "t" response (from student 6672), recognizing that the original length L changes with different orientation, but claims the cross-sectional area A is the same:
A sample "t" response (from student 2875), recognizing that the cross-sectional area A changes with different orientation, but claims the original length L is the same:

Physics quiz question: Young's modulus of steel

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA

A sample of "#4" steel reinforcing bar has a length of 4.9 m and a cross-sectional area of 1.3×10^–4 m². A force of 1.1×10⁵ N is applied to stretch this steel bar by 0.020 m. The Young's modulus of steel is:
(A) 3.5×10⁶ N/m².
(B) 8.5×10⁸ N/m².
(C) 4.2×10¹⁰ N/m².
(D) 2.1×10¹¹ N/m².

[*] webcivil.com/usrcrebar.aspx.

Correct answer (highlight to unhide): (D)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the Young's modulus would be:

(F/A)·(L/∆L) = Y = ((1.1×10⁵ N)/(1.3×10^–4 m²))·((4.9 m)/(0.020 m)),

Y = 207,307,692,307.692 N/m²,

or to two significant figures, the Young's modulus for steel is: 2.1×10¹¹ N/m².

(Response (A) is (F/A)·(∆L/L); response (B) is F/A; response (C) is F/(A·∆L).)

Sections 70854, 70855
Exam code: quiz06co6O
(A) : 5 students
(B) : 2 students
(C) : 2 students
(D) : 43 students

Success level: 83%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06co6O



Sections 70854, 70855 results

0- 6 :	* [low = 3]
7-12 :	****
13-18 :	*************
19-24 :	**************** [mean = 22.1 +/- 6.1]
25-30 :	****************** [high = 30]

Physics midterm question: comparing Young's moduli of fishing lines

Physics 205A Midterm 2, fall semester 2018
Cuesta College, San Luis Obispo, CA

A long 1.00 m fishing line and a short 0.50 m fishing line (same cross-sectional area) are each strung horizontally over a pulley, and are attached to a 100 g mass and a 50 g mass, respectively. As a result both fishing lines stretch the same amount from their original lengths. It is not known if these fishing lines are made of the same material. Discuss which material has the greater Young's modulus value (or if there is a tie), and why. Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:

• p:
  Correct. Applies Hooke's law in a systematic manner by:
  1. recognizing that they stretch the same amount ΔL and have the same cross-sectional area A; and
  2. the longer L fishing line has a greater tension force F applied to it than the shorter L fishing line with a lesser tension force F; and
  3. since Young's modulus Y = (F⋅L)/(A⋅ΔL), the longer L fishing line with the greater tension force F will have a larger Young's modulus (specifically four times larger) than the shorter fishing line with the lesser tension force.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically has clerical errors (mislabeling "long" versus "short" labels), and so concludes that Young's modulus must be the same for both fishing lines.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Typically only recognizes length L or tension force F has having an affect on Young's modulus Y; or has recognizes both quantities as having an affect on Y, but someone argues that these cancel each other out, such that the fishing lines have the same Y value.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least some systematic attempt at using Hooke's law quantities.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of relating strain (force per unit area), Young's modulus, and strain using Hooke's law.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02r3iN
p: 37 students
r: 6 students
t: 11 students
v: 2 students
x: 0 students
y: 0 students
z: 1 student

A sample "p" response (from student 5683):
A sample "p" response (from student 8812):
A sample "p" response (from student 1113):

Physics quiz question: marshmallow compression

Physics 205A Quiz 6, fall semester 2018
Cuesta College, San Luis Obispo, CA

A marshmallow has a height of 3.8×10^–2 m, a circular cross-sectional area of 5.1×10^–4 m², and a Young's modulus of 2.9×10⁴ N/m².[*][**] A downwards force of 10 N is applied evenly onto the top of the marshmallow. As a result, the marshmallow is compressed by:
(A) 6.7×10^–9 m.
(B) 1.3×10^–5 m.
(C) 2.6×10^–2 m.
(D) 2.0×10⁴ m.

[*] amazon.com/ask/questions/Tx21SLNIPLG0WI4.
[**] physics.info/elasticity/.

Correct answer (highlight to unhide): (C)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the amount that the marshmallow would be compressed is:

∆L = (F·L)/(A·Y),

∆L = ((10 N)·(3.8×10^–2 m))/((5.1×10^–4 m²)·(2.9×10⁴ N/m²)),

∆L = 0.02569303584... m,

or to two significant figures, the marshmallow would compress by 2.6×10^–2 m.

(Response (A) is (F·L·A)/Y; response (B) is F·L/Y; response (D) is the stress F/A.)

Sections 70854, 70855
Exam code: quiz06POr7
(A) : 4 students
(B) : 3 students
(C) : 43 students
(D) : 2 students

Success level: 83%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2018
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06POr7



Sections 70854, 70855 results

0- 6 :
7-12 :	**** [low = 9]
13-18 :	********
19-24 :	********************* [mean = 23.3 +/- 5.6]
25-30 :	******************** [high = 30]

Physics midterm question: comparing the stretching of fishing lines

Physics 205A Midterm 2, fall semester 2017
Cuesta College, San Luis Obispo, CA

A long, thin fishing line and a short, thick fishing line (both made of the same material) are strung horizontally over a pulley, and attached to a 100 g mass and a 50 g mass, respectively. Discuss why the thin fishing line will stretch a greater amount than the thick fishing line. Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:

• p:
  Correct. Applies Hooke's law in a systematic manner by:
  1. recognizing that the Young's modulus (Y) is the same for both fishing lines ("made of the same material"); and
  2. applying Hooke's law to show that the long, thin fishing line will stretch a greater amount ΔL = (F⋅L)/(A⋅Y) due to the combined contribution of three separate factors:
     1. a greater force F applied to it (which increases ΔL); and
     2. a smaller cross-sectional area A (which increases ΔL); and
     3. a longer, original unstretched length L (which increases ΔL).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically missing one of the three factors (a)-(c) above, but correctly discusses the effect of the two remaining factors.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Typically missing or misinterprets one of the three factors (a)-(c) above, but does not explicitly discuss how these factors affect ΔL in Hooke's law.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least some systematic attempt at using Hooke's law quantities. Typically missing two of three factors (a)-(c) above, and does not explicitly discuss how these factors affect ΔL in Hooke's law.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of relating strain (force per unit area), Young's modulus, and strain using Hooke's law.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02bu2Z
p: 15 students
r: 12 students
t: 2 students
v: 3 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 4532):

Physics quiz question: concrete sample compression

Physics 205A Quiz 6, fall semester 2017
Cuesta College, San Luis Obispo, CA

A concrete sample was (non-destructively) tested by compressing it with a stress of 9.9×10⁶ N/m²[*]. The Young's modulus of this concrete is 2.8×10¹⁰ N/m²[**]. If the concrete sample started with a height of 0.305 m, during testing it was compressed by:
(A) 1.1×10^–11 m.
(B) 1.1×10^–4 m.
(C) 3.5×10^–4 m.
(D) 5.5×10^–3 m.

[*] youtu.be/iCWsDHhbi9g.
[**] engineeringtoolbox.com/concrete-properties-d_1223.html.

Correct answer (highlight to unhide): (B)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

where the compressive stress (F/A) is given as 9.9×10⁶ N/m². The amount that sample would be compressed by will be:

∆L = (F/A)·(L/Y),

∆L = (9.9×10⁶ N/m²)·((0.305 m)/(2.8×10¹⁰ N/m²)),

∆L = 0.0001078392857 m,

or to two significant figures, the concrete sample would compress by 1.1×10^–4 m.

(Response (A) is L/Y; response (C) is the strain (∆L/L) = (F/A)/Y; response (D) is the volume of the concrete cylinder, which cannot be determined from the values given above.)

Sections 70854, 70855
Exam code: quiz06Ho0k
(A) : 4 students
(B) : 40 students
(C) : 1 student
(D) : 0 students

Success level: 89%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.27

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2017
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06Ho0k


Sections 70854, 70855 results

0- 6 :	* [low = 6]
7-12 :	*
13-18 :	*********
19-24 :	************* [mean = 23.3 +/- 5.8]
25-30 :	********************* [high = 30]

Physics midterm question: strain in lengthening loaded cable

Physics 205A Midterm 2, fall semester 2016
Cuesta College, San Luis Obispo, CA

A boom crane suspends a load from a 2.0 m long vertical cable. The load is lowered and held at a lower height by lengthening the cable to 4.0 m long. The weight of the cable is negligible compared to the weight of the load. Discuss why the strain in the cable does not change for this process. Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:

• p:
  Correct. Applies Hooke's law in a systematic manner by:
  1. recognizing that the load (F) applied to the cable does not change, as well as the Young's modulus (Y) and cross-sectional area (A) of the cable; and
  2. since strain is the (unitless) ratio (ΔL/L), then since (F/A) = Y⋅(ΔL/L), then from (ΔL/L) = F/(Y⋅A) it can be seen that both cables must experience the same strain.
• 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. At least some systematic attempt at using Hooke's law quantities.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of relating strain (force per unit area), Young's modulus, and strain using Hooke's law.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 25 students
r: 4 students
t: 8 students
v: 12 students
x: 6 students
y: 1 student
z: 0 students

A sample "p" response (from student 2217):

Physics quiz question: steamrolling a golf ball

Physics 205A Quiz 6, fall semester 2016
Cuesta College, San Luis Obispo, CA

"Steam Roller vs Golf Balls"
Crush
youtu.be/JGHcOd1kozc

Assume that a golf ball can be considered to be a cube with a cross-sectional area of 1.4×10^–3 m². A steam roller then applies 3.9×10⁴ N of downwards force on the golf ball, compressing it down to half of its original vertical height.[*][**] The overall Young's modulus for this golf ball is:
(A) 55 N/m².
(B) 2.0×10⁴ N/m².
(C) 2.8×10⁷ N/m².
(D) 5.6×10⁷ N/m².

[*] youtu.be/JGHcOd1kozc.
[**] wki.pe/Golf_ball.

Correct answer (highlight to unhide): (D)

Hooke's law for elastic materials is given by:

(F/A) = Y·(∆L/L),

where the cross-sectional area A of the wire is 1.4×10^–3 m², the force F applied is 3.9×10⁴ N, and the ratio ∆L/L that the golf ball is compressed to is 1/2. The (overall) Young's modulus of the golf ball can then be solved for:

Y = (F/A)/(∆L/L),

Y = ((3.9×10⁴ N)/(1.4×10^–3 m²))/(0.5),

Y = 5.5714285714×10⁷ N/m²,

or to two significant figures, Y = 5.6×10⁷ N/m².

(Response (A) is F·A; response (B) is F/2; and response (C) is the applied stress F/A.)

Sections 70854, 70855, 73320
Exam code: quiz06rn3T
(A) : 3 student
(B) : 6 students
(C) : 9 students
(D) : 37 students

Success level: 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73

Physics quiz archive: simple harmonic motion

Physics 205A Quiz 6, fall semester 2016
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06rn3T



Sections 70854, 70855, 73320 results

0- 6 :
7-12 :	** [low = 9]
13-18 :	*********
19-24 :	*************************** [mean = 22.9 +/- 4.6]
25-30 :	**************** [high = 30]

Physics quiz question: femur compression

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

An average femur bone (Young's modulus 9.4×10⁹ N/m²) has a length of 0.48 m and an approximate cross-sectional area of 1.72×10^–3 m², and reportedly can support a maximum 2.4×10⁴ N of force.[*][**] When this force is applied, the femur would compress by:
(A) 2.1×10^–9 m.
(B) 9.1×10^–9 m.
(C) 7.2×10^–8 m.
(D) 7.1×10^–4 m.

[*] "30 times the weight of an adult," orthopaedicsone.com/display/Review/Femur.
[**] "Weight of average adult: 178 lbs," wolfr.am/8dc2Ohi7.

Correct answer (highlight to unhide): (D)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the amount that the femur would be compressed is:

∆L = (F·L)/(A·Y),

∆L = ((2.4×10⁴ N)·(0.48 m))/((1.72×10^–3 m²)·(9.4×10⁹ N/m²)),

∆L = 0.0007125185552 m,

or to two significant figures, the femur would compress by 7.1×10^–4 m.

(Response (A) is (F·L·A)/Y; response (B) is (F·A)/(L·Y); response (C) is A/F.)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 67 students

Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.17

Physics quiz archive: simple harmonic motion

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06m45S



Sections 70854, 70855, 73320 results

0- 6 :
7-12 :	** [low = 12]
13-18 :	**********
19-24 :	***************************** [mean = 24.2 +/- 4.7]
25-30 :	******************************** [high = 30]

Physics quiz question: stretching fishing lines

Physics 205A Quiz 6, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 11.4, Problems 11.1, 11.3

"Untitled"
Eileen Delhi
flic.kr/p/fGR8Fi

Trilene® XL Super Strong fishing line (Young's modulus 2.0×10⁹ N/m²) and Eagle Claw® Sportfisher fishing line (Young's modulus 3.1×10⁹ N/m²) have the same 10.0 m length [*]. The Trilene® fishing line has a cross-sectional area 1.8 times that of the Eagle Claw®. Both fishing lines are stretched with a tension force of 98 N. The __________ fishing line will stretch more.
(A) Trilene®.
(B) Eagle Claw®.
(C) (There is a tie.)
(D) (Not enough information is given.)

[*] S. Ottolini, G. Halpin, P. LaBruzzo, "Tensile Strength of Fishing Line," santarosa.edu/~yataiiya/E45/PROJECTS/Tensile%20Strength%20of%20Fishing%20Line%20Power%20Point.ppt.

Correct answer (highlight to unhide): (B)

Hooke's law for the Trilene® and Eagle Claw® fishing lines are given by:

(F/A_Tri) = Y_Tri·(∆L_Tri/L),
(F/A_Eagle) = Y_Eagle·(∆L_Eagle/L),

where tension F and the original, unstretched length L are the same for both fishing lines. The Trilene® fishing line has a cross-sectional area 1.8× that of the Eagle Claw® fishing line:

A_Tri = 1.8·A_Eagle.

The amount that the Trilene® fishing line will be stretched is given by:

∆L_Tri = (F·L)/(A_Tri·Y_Tri),

∆L_Tri = ((98 N)·(10.0 m))/((1.8·A_Eagle)·(2.0×10⁹ N/m²)),

∆L_Tri = (2.7×10^–7 m³)/A_Eagle.

Similarly, the amount that the Eagle Claw® fishing line will be stretched is given by:

∆L_Eagle = (F·L)/(A_Eagle·Y_Eagle),

∆L_Eagle = ((98 N)·(10.0 m))/((A_Eagle)·(3.1×10⁹ N/m²)),

∆L_Eagle = (3.1×10^–7 m³)/A_Eagle.

Thus this sample of Eagle Claw® fishing line will stretch more than the Trilene® fishing line sample.

Sections 70854, 70855, 73320
Exam code: quiz06eAg7
(A) : 13 students
(B) : 49 students
(C) : 2 students
(D) : 0 students

Success level: 77%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2014
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06eAg7



Sections 70854, 70855, 73320 results

0- 6 :
7-12 :	***** [low = 12]
13-18 :	**********
19-24 :	********************** [mean = 23.3 +/- 5.3]
25-30 :	*************************** [high = 30]

Physics quiz question: Miley Cyrus' "Wrecking Ball"

Physics 205A Quiz 6, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 10.4

Miley Cyrus, "Wrecking Ball"
Terry Richardson (director)
Vevo, September 9, 2013

In a recent music video[*], pop singer Miley Cyrus sits on a (fake) demolition wrecking ball, together approximated as a 120 kg point mass hanging from a steel cable. While stationary, the steel cable stretches by 1.2 mm when 120 kg is hanging from it. If 240 kg were suspended from a steel cable of the same length with twice the radius, then it would stretch by __________ 1.2 mm.
(A) less than.
(B) exactly.
(C) more than.
(D) (Not enough information is given.)

[*] Don't bother watching it. Also the chain in the video is simplified here as a uniform cable.

Correct answer (highlight to unhide): (A)

Hooke's law for the thin and thick cables are given by:

(F_thin/A_thin) = Y·(∆L_thin/L),
(F_thick/A_thick) = Y·(∆L_thick/L),

where the Young's modulus Y and the original, unstretched length L are the same for the thin and thick cables (being both made of steel). The thick cable has twice the load of the thin cable:

F_thick = 2·F_thin.

The radii and thus the cross-sectional areas of the thin and the thick cables are also different:

A_thin = π·r_thin²,
A_thick = π·r_thick².

The thick cable has a radius twice that of the thin cable (r_thick = 2·r_thin), which will give it a cross-sectional area of four times that of the thin cable:

A_thick = π·r_thick² = π·(2·r_thin)² = 4·π·r_thin² = 4·A_thin.

Then setting the ratio of Y/L for the thin and thick cables equal to each other:

Y/L = Y/L,

F_thin/(A_thin·∆L_thin) = F_thick/(A_thick·∆L_thick),

and substituting in F_thick = 2·F_thin and A_thick = 4·A_thin:

F_thin/(A_thin·∆L_thin) = 2·F_thin/(4·A_thin·∆L_thick),

1/∆L_thin = 1/(2·∆L_thick),

thus:

∆L_thick = (1/2)·∆L_thin,

such that the thick cable will stretch less than the thin cable.

Sections 70854, 70855, 73320
Exam code: quiz06wR3k
(A) : 12 students
(B) : 32 students
(C) : 20 students
(D) : 0 students

Success level: 19%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.08

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2013
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06wR3k



Sections 70854, 70855, 73320 results

0- 6 :	* [low = 3]
7-12 :	**************
13-18 :	********************************** [mean = 15.5 +/- 5.2]
19-24 :	**********
25-30 :	** [high = 27]

Physics midterm question: longer single cable vs. doubled-up shorter cables

Physics 205A Midterm 2, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Questions 10.4, 10.13

A 10 kg mass is suspended from a 1.0 m cable attached to the ceiling. Another 10 kg mass is suspended from two 0.50 m cables. All of these cables are made of the same material, have the same diameter, and all lengths are measured before the cables are stretched. Discuss why the single cable will stretch more than either of the double cables. Explain your reasoning using the properties of stress, strain, and Hooke's law.

Solution and grading rubric:

• p:
  Correct. Applies Hooke's law in a quantitative manner to argue that the single long cable will stretch more than the two shorter cables due to (a) the longer length, and (b) smaller cross-sectional area (may also say that there is twice as much force applied to the single long cable compared to one of the two shorter cables).
• 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. Describes only one of the two contributions (a)-(b) that makes the single long cable stretch more than the two shorter cables.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02gL0u
p: 11 students
r: 8 students
t: 29 students
v: 6 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7582), using a "plug-and-chug" approach:

Physics quiz question: stretching a wire

Physics 205A Quiz 6, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.3

A "000" AWG-gauge copper wire[*] with Young's modulus of 1.2×10¹¹ Pa is 1.0 m long and has a cross-sectional area of 85.0 mm². If a weight of 300 N is hung from the wire, it will stretch:
(A) 3.6×10^–9 m.
(B) 2.1×10^–7 m.
(C) 2.8×10^–7 m.
(D) 2.9×10^–5 m.

[*] wki.pe/American_wire_gauge.

Correct answer (highlight to unhide): (D)

Hooke's law for elastic materials is given by:

(F/A) = Y·(∆L/L),

where the cross-sectional area of the wire is 85.0 mm² = 8.50×10^–5 m², such that the length that the amount the wire stretches is:

∆L = (F·L)/(A·Y)

∆L = ((300 N)·(1.0 m))/((8.50×10^–5 m²)·(1.2×10¹¹ Pa) = 2.941176471×10^–5 m,

or to two significant figures, ∆L = 2.9×10^–5 m.

(Response (A) is F·L·(1.2×10^–11 Pa); response (B) is F·L·(85.0 m²)/Y; and response (C) is A/F.)

Sections 70854, 70855
Exam code: quiz06How3
(A) : 2 students
(B) : 5 students
(C) : 9 students
(D) : 30 students

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.52