Physics midterm question: comparing relative amounts of energy changes

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

A 2.0 kg crate is attached to a block (of unknown mass) by use of an ideal rope and pulley. Starting from rest, the crate slides to the right while the block descends downwards, both with increasing speed. Ignore friction and drag. Determine which of these two energy forms undergoes a larger amount of change (increase or decrease) for this process, or if there is a tie:

translational kinetic energy of the crate;
gravitational potential energy of the block.

Explain your reasoning using the properties of energy forms, and conservation of energy.

[*] youtu.be/CgNlPOMOps0.

Solution and grading rubric:

• p:
  Correct. Applies energy forms and conservation concepts with:
  1. both the crate and the box are speeding up, such that their (same) final speed is faster than their initial speed (zero), making both their translational kinetic energy terms increase; and
  2. the box is going downwards, such that the final height is lower than the initial height, making its gravitational potential energy decrease; and
  3. sets up a transfer-balance energy conservation equation with the sum of the changes in translational kinetic energy of the crate, translational kinetic energy of the box, and gravitational potential energy of the box set to zero (as no energy is lost to non-conservative work); then
  4. since the decrease in gravitational potential energy of the box is "feeding" the increase in the translational kinetic energies of both the crate and the box; the gravitational potential energy of the box must undergo a larger change than the translational kinetic energy of the crate.
• 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. Correctly determines that no energy is lost to non-conservative work, but discusses how the decrease in gravitational potential energy of the block is transferred solely to the increase in translational kinetic energy of the crate (concluding that there is a tie in the amount of change of these two energy forms), while neglecting the increase in translational kinetic energy of the block.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at using properties of forces, work, energy forms and (non-)conservation of energy.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than using properties of forces, work, energy forms and (non-)conservation of energy.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 5 students
r: 1 student
t: 35 students
v: 9 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 1478):

Physics midterm question: comparing vertical forces supporting tilted beams

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

A force F₁ pulls up at the end of a uniform beam to hold it stationary at an angle of 80° above the horizontal, and a force F₂ pulls up at the end of an identical uniform beam to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beams.) Discuss why these forces 
 F₁ and F₂ have the same magnitude. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; and
  2. for each beam, the lever arm for the applied force F is twice the lever for the weight force w (2⋅ℓ_w = ℓ_F); and
  3. Newton's first law for rotations applies to both higher and lower beams, where the ccw force torque F⋅(ℓ_F) and cw weight torque w⋅(ℓ_w) must balance each other out, and so: F⋅(ℓ_F) = w⋅(ℓ_w), F = w⋅(ℓ_w/ℓ_F) = w⋅(ℓ_w/(2⋅ℓ_w)) = w/2; such that
  4. the applied forces on the higher and lower beam must be equal in magnitude, as they are both equal to one-half of the weight of the beam.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Does not explicitly note that the ℓ_F lever arm is always twice the ℓ_w lever arm for both situations. Instead, argues that since the ℓ_F and ℓ_w values for the higher beam are both bigger than the respective ℓ_F and ℓ_w values for the lower beam, then the higher beam F = w⋅(ℓ_w/ℓ_F) = w⋅(bigger/bigger) must be equal to the lower beam F = w⋅(ℓ_w/ℓ_F) = w⋅(smaller/smaller), but only implicitly demonstrates how the "bigger/bigger" ratio is exactly equal to the "smaller/smaller" ratio by use of a scaled drawing instead of using geometry/trigonometry, etc.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. As (r), but does not clearly/correctly show ℓ_F and ℓ_w lever arms for both situations. At least has two sets of Newton's first law for rotations, one for the higher beam and one for the the lower beam, setting the ccw torques equal to the cw torques.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 3 students
r: 18 students
t: 14 students
v: 14 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 5281):

Physics midterm question: comparing net force for afloat vs. submerged sinking block

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

A solid object is (a) partially submerged in water as it sinks with increasing speed, then while (b) completely underwater it still sinks with increasing speed. Discuss why the magnitude of the net force on the object is greater for case (a) than for case (b). Ignore friction and drag. Explain your reasoning using the properties of Newton's laws, Archimedes' principle (buoyant forces), and free-body diagrams.

Solution and grading rubric:

• p:
  Correct. Recognizes that:
  1. each block (a) or (b) has two vertical forces acting on it:

     Weight force of Earth on block (downwards, magnitude w = m⋅g, same for both (a) and (b)),
     Buoyant force of water on block (upwards, magnitude F_B = ρ_water⋅g⋅V_sub, less for (a)); and

  2. block (a) has a downwards weight force, and an upwards buoyant force much less than the magnitude of the weight force; and
  3. block (b) has the same downwards weight force as (a), also with an upwards buoyant force less than the magnitude of the weight force, but with a magnitude greater than the magnitude of the buoyant force in (a) (as more volume is submerged); and
  4. from Newton's second law, the downwards net force for (a) has a greater magnitude than the downwards net force for (b), as demonstrated by either explicit comparison of vector lengths and/or comparing terms in ΣF = +F_B − w equations for each case.
  May either draw a free-body diagram, and/or discuss these forces and Newton's laws in words.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically does not explicitly demonstrate Newton's second law via vector addition (different up vectors drawn much less than, or a little less than the same down vector for each case; and/or comparing same/different quantities in ΣF = +F_B − w equations for each case).
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. At least recognizes that the object has a greater buoyant force once it is fully submerged.
• v:
  imited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some constructive attempt at relating the buoyant force to the density of the fluid and volume displaced (Archimedes' principle) and/or Newton's first law.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Appeals to some other properties of fluids and densities other than Archimedes' principle and Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 13 students
r: 12 students
t: 8 students
v: 15 students
x: 4 students
y: 0 students
z: 0 students

A sample "p" response (from student 3372):

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 midterm question: distance traveled vs. displacement

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

The v_x(t) graph of a Physics 205A student walking along a horizontal road is shown at right. The student started at x = 0 at t = 0. Discuss whether the magnitude of the distance traveled by the student is equal to or greater than the magnitude of the displacement. Explain your reasoning using the properties of velocity, position, time, distance traveled, and/or displacement.

Solution and grading rubric:

• p:
  Correct. Demonstrates that the distance traveled is equal to the magnitude of displacement, as the student always travels in the same (positive) direction (only positive horizontal velocity values), without reversing direction (no negative horizontal velocity values).
• 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. Conceptual understanding of displacement and distance traveled, but somehow misinterprets/misapplies/ignores the given information.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 38 students
r: 0 students
t: 5 students
v: 4 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 5808):

Physics midterm question: cargo-loaded truck vs. truck-loaded cargo

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student draws a (correct) free body diagram for a 600 kg cargo load resting on a stationary 11,000 kg truck with these two forces[*]:

Weight force of Earth on cargo load (5,800 N, downwards),

Normal force of truck on cargo load (5,800 N, upwards).

This student additionally claims that "this [free body diagram for the cargo load] would change if the truck was on top of the cargo load." Discuss why both the magnitude and direction of the normal force of truck on the cargo load would change if the truck were instead resting on top of the cargo load, and how you know this. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.


[*] waiferx.blogspot.com/2017/10/physics-midterm-question-proposed-test.html.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram(s), and discusses/demonstrates that when the truck is on top of the cargo load:
  1. the truck has two vertical forces acting on it:

     Weight force w of Earth on truck (m_truck·g = 107,800 N, downwards),
     Normal force N of cargo load on truck (107,800 N, upwards),

     and since there is no vertical motion, these two vertical forces must be equal in magnitude due to Newton's first law; and
  
  
  2. from Newton's third law, these two forces must have equal magnitudes and opposite directions:

     Normal force Nof cargo load on truck (107,800 N, upwards),
     Normal force Nof truck on cargo load (107,800 N, downwards),

     such that the normal force of truck on cargo load for the case where the truck is on top of the cargo load is both different in magnitude (107,800 N vs. 5,800 N) and direction (downwards vs. upwards) compared to the normal force of truck on cargo load in the case where the cargo load was on top of the truck.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically application of Newton's third law is problematic, or only implied, but still discusses how the normal force of truck on the cargo load is both different in magnitude and direction than in the previous case.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at analyzing forces using Newton's laws and free-body diagrams.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at using Newton's laws and free-body diagrams.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic attempt at using Newton's laws and free-body diagrams.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 17 students
r: 16 students
t: 18 students
v: 4 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1995):

Physics midterm question: pulled box pulling on table underneath

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student applies a force (magnitude of 32 N, directed to the left) to pull on a rope attached to a 12.0 kg box, which moves at constant speed to the left across a fixed, stationary table. Discuss why both the magnitude and direction of the kinetic friction force of the box on the table would also be 32 N, directed to the left. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram(s), and discusses/demonstrates:
  1. the box has two horizontal forces acting on it:

     Tension force T of student on box (32 N, to the left),
     Kinetic friction force  f_k of table on box (32 N, to the right),

     and since the box has a constant velocity ("constant speed to the left"), these two horizontal forces must be equal in magnitude due to Newton's first law; and
  
  
  2. from Newton's third law, these two forces must have equal magnitudes and opposite directions:

     Kinetic friction force  f_k of table on box (32 N, to the right),
     Kinetic friction force  f_k of box on table (32 N, to the left).

• 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. Some substantive attempt at analyzing forces using Newton's laws and free-body diagrams. Typically discusses Newton's first law for the forces acting on the box, but subsequent discussion of Newton's third law is omitted or merely implied.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at using Newton's laws and free-body diagrams.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic attempt at using Newton's laws and free-body diagrams.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 15 students
r: 5 students
t: 14 students
v: 15 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 6900):
Another sample "p" response (from student 2533; note that the static friction force of the ground on table pointing to the right is denoted as a tension force):
A sample "t" response (from student 1995), discussing Newton's first law for the box, demonstrating that the kinetic friction force of the table on the box points to the right with a magnitude of 32 N:

Physics midterm question: faster vertically swung ball

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student twirls a ball attached to a string in a vertical circle with constant speed. When the ball is swinging through the lowest part of the circle, the tension force on the ball is 1.8 N. Discuss why the magnitude of the tension force on this ball would have a larger magnitude if the ball had a faster constant speed while swinging through the lowest part of the circle (with the same radius). Explain your reasoning by using free-body diagram(s), the properties of forces and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. while the ball is swinging through the lowest part of the circle, there are two forces acting on the ball:

     Tension force T of hand on the ball (originally 1.8 N, upwards),
     Weight force w of Earth on the ball (constant magnitude of m⋅g, downwards);

  2. Newton's second law for uniform circular motion applies, such that while the ball is swinging through the lowest part of the circle, the net force ΣF (magnitude m⋅v²/r) must point in towards the center of the circular motion--which is vertically upwards; and
  3. increasing the speed v (while m and r are constant) would increase the required upwards net force for uniform circular motion, such that the upwards tension force will be greater than the original 1.8 N
• 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. Some substantive attempt at applying Newton's second law for uniform circular motion.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying Newton's first law or third law for uniform circular motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 21 students
r: 11 students
t: 9 students
v: 9 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 1478):

Physics midterm question: comparing same-energy, different potential capacitors

Physics 205B Midterm 2, spring semester 2019
Cuesta College, San Luis Obispo, CA

A capacitor is connected to a 1.5 V battery and a different capacitor is connected to a 6.0 V battery. Both capacitors store the same amount of electrical potential energy. Discuss why the capacitor connected to the 1.5 V battery has a larger capacitance than the capacitor connected to the 6.0 V battery. Explain your reasoning by using the properties of capacitors, charge, electric potential, and energy.

Solution and grading rubric:

• p:
  Correct. Discusses why the capacitor connected to the 1.5 V battery has a greater capacitance than the capacitor connected to the 6.0 V battery because:
  1. from EPE = (1/2)⋅Q⋅(ΔV), both capacitors store the same amount of electrical potential energy; such that the capacitor connected to the 1.5 V battery holds a larger charge than the capacitor connected to the 6.0 V battery; and
  2. from C = Q/ΔV, since the capacitor connected to the 1.5 V battery has a smaller potential difference and a larger charge than the capacitor connected to the 6.0 V battery; then the capacitor connected to the 1.5 V battery must have a larger capacitance.
• 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. Typically assumes that both capacitors have the same charge, and/or does not explicitly use the given fact that the capacitors store the same amount of electrical potential energy.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of capacitors, charge, electric potential, and energy.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying properties of capacitors, charge, electric potential, and energy.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 22 students
r: 1 student
t: 18 students
v: 0 students
x: 1 student
y: 1 student
z: 0 students

A sample "p" response (from student 2334):
Another sample "p" response (from student 1842), substituting in Q = C·ΔV into the electric potential energy equation:

Physics midterm question: comparing voltmeter readings

Physics 205B Midterm 2, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 9.0 V emf source is connected to a light bulb and two resistors, with a voltmeter connected to the light bulb, and another voltmeter connected to one of the resistors. Discuss why the two voltmeters have the same reading (in volts). Show your work and explain your reasoning using Kirchhoff's laws, Ohm's law, and properties of voltmeters.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates how the two voltmeters have the same reading because:
  1. due to Kirchhoff's junction rule, the current flowing through each of the 4.0 Ω resistors is one-half of the current flowing through the 2.0 Ω resistor; and
  2. since each voltmeter will read the voltage drop (I⋅R) of their respective resistors, the smaller current (factor of one-half) flowing through the 4.0 Ω resistor will be compensated by its larger resistance (factor of two).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of Kirchhoff's rules, Ohm's law, and properties of ammeters.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 28 students
r: 4 students
t: 2 students
v: 8 students
x: 0 students
y: 0 students
z: 1 student

A sample "p" response (from student 1950):

Physics midterm question: ammeter reading after switch is closed

Physics 205B Midterm, spring semester 2019
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to an ammeter, a resistor, a light bulb, and an open switch. When the switch is closed, determine whether the ammeter reading (in amps) will decrease, increase, or remain the same, and explain why. 
Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of ammeters.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates how ammeter reading will increase when the switch is closed because:
  1. when the switch is open, the equivalent resistance is 3.0 Ω, and the ammeter will read the current of this circuit I = ε_eq/R_eq = (6.0 V)/(3.0 Ω) = 2.0 A;
  2. when the switch is closed, no current will flow through the 0.5 Ω resistor (flowing only through the zero resistance path of the closed switch), such that the equivalent resistance decreases to 2.5 Ω, such that the ammeter will read a higher amount of current in this circuit I = ε_eq/R_eq = (6.0 V)/(2.5 Ω) = 2.4 A.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of Kirchhoff's rules, Ohm's law, and properties of ammeters.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 20 students
r: 4 students
t: 10 students
v: 6 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1001):
Another sample "p" response (from student 1810):

Physics midterm question: stationary loop near constant current wire

Physics 205B Midterm 2, spring semester 2019
Cuesta College, San Luis Obispo, CA

A square metal loop of resistance R is held stationary near a wire that carries a constant amount of current. Discuss whether or not there will be any induced current in the square metal loop, and explain why. Explain your reasoning using the properties of magnetic fields, forces, motional emf, Faraday's law and Lenz's law.

Solution and grading rubric:

• p:
  Correct. Explains how there would be no induced current in the square loop of wire because:
  1. from RHR2, the direction of current in the straight wire creates a magnetic field at the location of the square loop points into the page, creating a magnetic flux through the square loop that points into the page; and
  2. since the current in the straight wire is constant, the magnetic field it creates at the location of the square loop will have a constant magnitude (along with its constant direction), such that there is a constant, unchanging magnetic flux (magnitude and direction) through the square loop; so
  3. from Faraday's law and Lenz's law, since there is no change in magnetic flux through the square loop, there will be no induced emf and no induced current in the square loop.
  (May instead use RHR1 and discuss how the fictitious positive charges in each segment of the square loop are stationary with respect to the magnetic field of the wire, such that there is no force exerted on them to create an induced current.)
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes:
  4. did not clearly indicate the direction of the magnetic field/flux through the square loop; or
  5. argues that there is no induced current in the square loop because there is no magnetic flux through the square loop (when there is a magnetic flux through the square loop, but it is constant); or
  6. argues that there is an induced current in the square loop even though the magnetic flux through the square loop is constant.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying properties of magnetic fields, forces, motional emf, Faraday's law and Lenz's law.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at systematically applying properties of magnetic fields, forces, motional emf, Faraday's law and Lenz's law.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 14 students
r: 15 students
t: 4 students
v: 3 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 0691):
Another sample "p" response (from student 4692):

Physics midterm question: same transmitted fraction of unpolarized/polarized light

Physics 205B Midterm 1, spring semester 2019
Cuesta College, San Luis Obispo, CA

In case (a) unpolarized light is incident on a 45° diagonal polarizer, and a certain fraction of incident light is transmitted through it. For case (b) horizontal polarized light is incident on a polarizer with a transmission axis that can be rotated to any θ angle. Discuss what the θ angle should be in case (b) such that the same fraction of incident light is transmitted through it as in case (a). Explain your reasoning using the properties of light and polarization.

Solution and grading rubric:

• p:
  Discusses/demonstrates that:
  1. in case (a), the fraction of unpolarized light that passes through a polarizer (regardless of its transmission angle) is (1/2); and
  2. for case (b), the fraction of polarized light that passes through a polarizer is given by Malus' law: cos²(90° − θ), where the angle of interest is between the horizontally polarization of the incident light and the transmission angle of the polarizer; and
  3. for the fraction transmitted in case (b) to equal the fraction transmitted in case (a), sets cos²(90° − θ) = (1/2), and thus θ = 45°.
• 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. At least understands what happens in case (a), and has some systematic approach to match this fraction for case (b).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying the properties of light, polarizers, and polarization.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying the properties of light, polarizers, and polarization.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 16 students
r: 2 students
t: 14 students
v: 10 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7843):

Physics midterm question: comparing indices of refraction

Physics 205B Midterm 1, spring semester 2019
Cuesta College, San Luis Obispo, CA

Light in air with an incident angle of 19.0° is transmitted into acetone at an angle of 13.9°. Light in air with an incident angle of 44.0° is transmitted into turpentine at an angle of 28.2°. Show that the index of refraction of acetone is less than the index of turpentine. Explain your reasoning using the properties of light and refraction.

Solution and grading rubric:

• p:
  Correct. Uses Snell's law to solve for and compare the numerical values for the indices of refraction of acetone versus turpentine.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Snell's law, angles, and/or indices of refraction.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying Snell's law, angles, and/or indices of refraction.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 41 students
r: 0 students
t: 0 students
v: 1 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1348):

Physics midterm question: comparing diverging lens image sizes

Physics 205B Midterm 1, spring semester 2019
Cuesta College, San Luis Obispo, CA

An object 1.0 cm in height is placed 16.0 cm in front of a f = –15.0 cm diverging lens, producing an image. Show that moving the object slightly closer, such that it is 14.0 cm in front of this diverging lens will result in a slightly larger image than before. Explain your reasoning by using ray tracings and/or thin lens equations, properties of lenses, images, and magnification.

Solution and grading rubric:

• p:
  Correct. Proves that the object should be as close to the diverging lens as possible in order to obtain the largest image (largest linear magnification factor) using either of these two methods:
  1. calculating the image distances produced by the different object distances, and finds the resulting respective image sizes and/or linear magnification factors; or
  2. drawing two carefully, properly scaled ray tracing diagrams.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May have misplaced values, but consistently interprets resulting numbers.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Problematic algebra (combines fractions by combining denominators, forgets to invert (1/f – 1/d_o) to solve for d_i, etc.).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying properties of lenses, images, and linear magnification. Typically has problematic algebra as in (t), but does not use (erroneous) d_i values to find image heights or linear magnification factors for comparison.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying properties of lenses, images, and angular magnification.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 25 students
r: 6 students
t: 4 students
v: 4 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 1810), using two sets of thin lens equations:
Another sample "p" response (from student 0691), using two ray tracings:

Physics midterm question: destructively interfering out-of-phase radio transmitters

Physics 205B Midterm 1, spring semester 2019
Cuesta College, San Luis Obispo, CA

Two vertical radio transmitters broadcast at the same wavelength, and are spaced 6.0 m apart along the east-west direction. A receiver held by a Physics 205B student located to the east of both transmitters detects a destructive interference signal, and a receiver held by another Physics 205B student located to the north also detects a destructive signal. Show (a) why the transmitters must be out-of-phase sources, and (b) find a plausible numerical value for the wavelength. Explain your reasoning using the properties of source phases, path lengths, and interference.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates:
  1. that the radio transmitters are out of phase, as there is no path length difference for the student located to the north, and destructive interference (as is the case here) can only occur if the sources are out of phase; and
  2. for the student located to the east, for these out of phase sources to interfere destructively the path length difference must be a whole number of wavelengths, and since the path length difference is 6.0 m, plausible (non-zero) wavelength values would be 6.0 m, 3.0 m, 1.5 m, etc.
• 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. At least discusses why (1) transmitters are out of phase, but (2) does not use the correct destructive interference condition (whole number of wavelengths) for out of phase sources for the student located to the east.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at discussing source phases, path lengths, and interference.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach other than that of discussing source phases, path lengths, and interference.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 13 students
r: 4 students
t: 13 students
v: 12 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 5250), identifying the correct path length difference condition for two sources that are out of phase:
Another sample "p" response (from student 1810), with a graphical representation of the interfering waves:

Physics midterm question: comparing horizontal forces supporting tilted beams

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

A horizontal force is applied to hold a uniform beam stationary at an angle of 80° above the horizontal, and another horizontal force is applied to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beam.) Discuss why less force required to hold the beam when it is at the higher 80° angle. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; but
  2. the lever arms ℓ_F are not the same, where ℓ_F is longer for the higher beam, and shorter for the lower beam; and
  3. the lever arms ℓ_w are not the same, where ℓ_w is shorter for the higher beam, and longer for the lower beam; and
  4. Newton's first law for rotations applies to both higher and lower beams, where the ccw force τ = F⋅ℓ_F and cw weight τ = w⋅ℓ_w must balance each other out, and so:

     F⋅ℓ_F = w⋅ℓ_w,
     
     F = w⋅(ℓ_w/ℓ_F);

     such that
  5. for the higher beam, the shorter ℓ_w in the numerator and longer ℓ_F in the denominator means that the applied force F is smaller than for the lower beam (where it has a longer ℓ_w in the numerator and a shorter ℓ_F in the denominator).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. At least demonstrates that for the higher beam ℓ_w is shorter and ℓ_F is longer, but typically discusses only how one of these contributes to making the applied force smaller.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Typically argues that F is smaller for the higher beam because ℓ_F is larger (while claiming ℓ_w is the same for both beams); or F is smaller for the higher beam because ℓ_w is smaller (while claiming ℓ_F is the same for both beams).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02r3iN
p: 17 students
r: 4 students
t: 21 students
v: 10 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 5250):

Physics midterm question: floating ebony-balsa wood cubes

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

A wooden cube is made by gluing ebony (denser) and balsa (less dense) pieces together. Both pieces have the same volume. The total density of the cube is less than that of water. The cube is carefully placed into water such that it floats "top-heavy" (ebony on top of balsa). The cube is then turned over such that it floats "bottom-heavy" (balsa on top of ebony). Discuss which orientation will float higher (or if there is tie), and why. (Ignore any water that may soak into the wood pieces, and the thin layer of glue between the two wood pieces.) Explain your reasoning using the properties of densities, volumes, forces, Newton's laws, Archimedes' principle (buoyant forces), and free-body diagrams.

Solution and grading rubric:

• p:
  Correct. Recognizes that:
  1. each block ("bottom-heavy" or "top-heavy") has two vertical forces acting on it:

     Weight force of Earth on block (downwards, magnitude w = m⋅g),
     Buoyant force of water on block (upwards, magnitude F_B = ρ_water⋅g⋅V_sub);

     and
  2. because each block ("bottom-heavy" or "top-heavy") is stationary in the vertical direction, then its downwards weight force must have the same magnitude as its upwards buoyant force, due to Newton's first law; and
  3. since the mass of each block ("bottom-heavy" or "top-heavy") does not matter which type of wood is stacked above the other, the magnitude of the weight is the same, making the magnitudes of the buoyant forces the same; such that
  4. the amount submerged volume underwater for both blocks must be the same.
  
  Thus the buoyant forces on each block ("bottom-heavy" or "top-heavy") are equal, and thus the amount of volume submerged for either block must be the same.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May somehow claim that the cube will float differently when "bottom-heavy" or "top-heavy," or does not explicitly conclude that the cube will float at the same water level whether "bottom-heavy" or "top-heavy."
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. At least recognizes that the weight force on the block is unchanged whether "bottom-heavy" or "top-heavy," but somehow has different buoyant forces acting (thus Newton's first law would not apply to at least one of the blocks); or has different weights and different buoyant forces acting on the blocks, but for each block these forces are balanced via Newton's first law.
• v:
  imited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some constructive attempt at relating the buoyant force to the density of the fluid and volume displaced (Archimedes' principle) and/or Newton's first law.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Appeals to some other properties of fluids and densities other than Archimedes' principle and Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
p: 21 students
r: 17 students
t: 13 students
v: 6 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 0921):

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 midterm question: position of vertically-launched ball

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A Physics 205A student slings a ball straight upwards. The v_y(t) graph of this ball is shown at right, starting from when the ball was released at y = 0 at t = 0. Neglect air resistance. Choose up to be the +y direction. At t = 6 s, discuss why the ball is at a position higher than its release point. Explain your reasoning using the properties of velocity, position, distance traveled, and/or displacement.

Solution and grading rubric:

• p:
  Correct. Supports claim that the ball is still above its release point at t = 6 s by discussing at least one of the following explanations:
  1. displacement is the bounded area between the velocity function and the time axis, and since there is a greater bounded area above the time axis (corresponding to a positive displacement for t = 0 to 4 s) than the bounded area below the time axis (corresponding to a negative displacement for t = 4 s to 6 s), the ball has traveled farther up from its starting point to its highest height, than traveling downwards from its highest height to its final position at t = 6 s; or
  2. the ball slows down from an upwards velocity of +40 m/s at t = 0 to zero velocity at its highest point at t = 4 s, and from symmetry, the ball should fall back down to its starting point with a downwards speed of –40 m/s at t = 8 s, such that the ball is still somewhere above its starting point at t = 6 s; or
  3. uses kinematic equations for constant motion to show that the final position at t = 6 s is still positive.
• 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.
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 24 students
r: 5 students
t: 21 students
v: 8 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1414):
Another sample "p" response (from student 3691):