Physics midterm problem: basketball rolling up ramp

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

A basketball (mass 0.43 kg, radius 0.11 m) rolls without slipping with a constant initial velocity of 2.4 m/s across a horizontal floor. The basketball begins to roll up a ramp. Determine the highest vertical height that the basketball will reach before rolling back down the ramp. Ignore friction and drag. Show your work and explain your reasoning using the properties of rotational inertia, energy forms, and conservation of energy.

The basketball is a hollow sphere (I_{hollow sphere} = (2/3)·M·R².)


Solution and grading rubric:

• p:
  Correct. Sets up a transfer-balance energy conservation equation with the sum of the changes in translational kinetic energy of the basketball, rotational kinetic energy of the basketball, and gravitational potential energy of the basketball set to zero (as no energy is lost to non-conservative work), fills in all known/given values, and solves for the unknown. (May have ± sign errors for final terms subtracting initial terms, but in a somewhat consistent manner that still results in correct final height of basketball).
• r:
  Nearly correct, but includes minor math errors. Or multiple arithmetic errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically omits one of the energy terms, but attempts to apply energy conservation to the remaining two terms.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Calculations of some energy terms, but does not sufficiently tie them together in a transfer-balance energy conservation equation. Typically has only one energy term, or relates an energy term with a non-energy quantity (such as weight, momentum, moment of inertia, etc.).
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach involving methods other than energy conservation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 26 students
r: 16 students
t: 7 students
v: 3 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 8520):
Another sample "p" response (from student 1203):

Physics midterm problem: skateboard-launched rubber duck toy

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

"WRECKING BALL Vs. SEESAW from 45m! How High Will the Watermelon Go?"
How Ridiculous
youtu.be/1quHlRJLtgM

Brett Stanford, Derek Herron and Scott Gaunson for the "How Ridiculous" YouTube channel dropped a heavy ball on one end of a skateboard to launch a rubber duck toy from the other end. Video analysis shows that the toy was launched at an angle of 61° from the horizontal, and took 3.2 s from the moment it was launched from the ground to land back down on the ground. Determine the horizontal distance along the ground from where it was launched to where it landed. Neglect air resistance, and treat the toy as a point object that started on the ground. Show your work and explain your reasoning using properties of projectile motion.

[*] youtu.be/1quHlRJLtgM?t=335.

Solution and grading rubric:

• p:
  Correct. From the time of flight t = 3.2 s, solves for the initial vertical velocity component v_0y = +16 m/s. Next, using the launch angle of elevation θ = 61° finds the initial horizontal velocity component v_0x = +8.7 m/s, and subsequently uses that value and the time of flight t = 3.2 s to solve for the final horizontal position x = +28 m.
• r:
  Nearly correct, but includes minor math errors. At least successfully solves for the vertical v_0y and/or horizontal v_0x components of the initial velocity vector.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least some systematic attempt at using kinematic equations for projectile motion. May have made one or more erroneous assumptions about certain values, such as setting the final velocity components v_x = 0 and or v_y = 0, but still methodically solves for a (wrong) value of v_0y, and then (somehow) solves for a (wrong) value of v_0x using trigonometry to find a (wrong) value for the final horizontal position x.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at systematic use of kinematic equations for projectile motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at kinematic equations for projectile motion.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 15 students
r: 4 students
t: 9 students
v: 21 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 5832):
Another sample "p" response (from student 2342):

Physics midterm problem: brightness of light bulbs in circuit

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

An ideal 9.0 V emf source is connected to several light bulbs that all have the same resistance. Calculate the powers dissipated (in watts) for each of these light bulbs. Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of electrical power.

Solution and grading rubric:

• p:
  Correct. Solves for the powers dissipated by each light bulb by:
  1. finding equivalent resistance of the circuit by recognizing that the top light bulb is in series to the lower three parallel light bulbs);
  2. applying Ohm's law to determine the current of the equivalent circuit, which is the current flowing through the top light bulb;
  3. determines the power dissipated by the top light bulb;
  4. applies Kirchhoff's loop and/or junction rules to solve for the voltage difference used by and/or the current flowing through each of the lower three parallel light bulbs; and
  5. determines the power dissipated by each of the lower three parallel light bulbs.
• r:
  Nearly correct, but includes minor math errors. Typically incorrect calculation in (1) or in (5), but otherwise everything else is consistent with this error.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Multiple issues in (1)-(5), but still attempts to systematically analyze most of (1)-(5) even with wrong numerical values.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying Kirchhoff's rules, Ohm's law, and properties of electrical power.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying Kirchhoff's rules, Ohm's law, and properties of electrical power.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02u7aH
p: 10 students
r: 6 students
t: 7 students
v: 18 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 1982):
Another sample "p" response (from student 8812):

Physics midterm problem: total electric field direction to the side of two charges

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

A –3.0 nC point charge is held at the origin, and +2.0 nC point charge is held at x = +1 cm. Discuss why the electric field at x = +2 cm is directed to the right. Explain your reasoning using properties of electric forces, fields, and vector superposition.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates that the (total) electric field at x = +2 cm would point to the right by:
  1. calculating or comparing the individual electric field magnitudes for E₂ (larger magnitude, 1.8×10⁵ N/C) and E₁ (smaller magnitude, 6.8×10⁴ N/C) created by the source charges Q₁ and Q₂ at x = +2 cm; and
  2. determining the individual electric field directions for E₁ (to the right, out away from the positive Q₁ source charge) and E₂ (to the left, in towards the negative Q₂ source charge) at x = +2 cm; and
  3. explaining that the vector superposition of E₁ and E₂ at x = +2 cm will result in a total electric field that points to the right, as it (a) will have the direction of the larger of two opposite electric field vectors there, or (b) would be the result of the subtraction of a smaller E₁ magnitude (to the left) from the larger E₂ magnitude (to the right).
• 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 approach has conceptual errors, and/or major/compounded math errors. At least some attempt at finding electric field magnitudes and directions created by each source charge at x = +2 cm, and discusses result of the addition of these electric field vectors.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying electric forces, fields, and vector superposition. Typically discusses the force that these charges exert on each other.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying electric forces, fields, and vector superposition.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01Ft6G
p: 24 students
r: 3 students
t: 6 students
v: 7 students
x: 0 students
y: 0 students
z: 2 students

A sample "p" response (from student 1810):
Another sample "p" response (from student 7843), being more explicit about the different directions of the individual electric fields:
And another sample "p" response (from student 1982), ignoring the common factor of k, and nC and cm unit conversions for purposes of comparison:

Physics midterm question: rotational kinetic energies of basketball vs. tennis ball

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

A basketball (mass 0.43 kg, radius 0.11 m) and a tennis ball (mass 0.058 kg, radius 0.033 m) 
both roll without slipping across a horizontal floor with the same constant speed of 0.50 m/s. Discuss why the basketball will have more rotational kinetic energy than the tennis ball.



Both objects are hollow spheres (I = (2/3)·M·R²).

Solution and grading rubric:

• p:
  Correct. Numerically calculates for each ball the angular speed from the v = R⋅ω condition for rolling without slipping, and moment of inertia I = (2/3)·M·R², and then includes both these factors to compare the rotational kinetic energy KE_rot = (1/2)⋅M⋅ω^2 of both objects, such that the basketball has a larger numerical value for the rotational kinetic energy than the tennis ball.
• 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 shows that the basketball has a higher moment of inertia than the tennis ball, but claims that they have the same angular speed, or does not explicitly show that difference in angular speeds is much smaller than the difference in moments of inertia in determining that the basketball has a greater rotational kinetic energy.
• 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: midterm02r3iN
p: 17 students
r: 2 students
t: 37 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student):

Physics midterm problem: categorizing a cart collision

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

A 0.90 kg cart traveling in the +x direction at 0.45 m/s collides with a 0.15 kg cart that is initially at rest. The carts are not stuck together after the collision. After the collision, the 0.90 kg cart continues traveling in the +x direction at 0.35 m/s. Ignore friction, drag and other external forces during this brief collision. Find (a) the final velocity of the 0.15 kg cart, and (b) classify this collision as elastic, inelastic, or completely inelastic. Show your work and explain your reasoning using properties of collisions, energy (non-)conservation, and momentum conservation.


Solution and grading rubric:

• p:
  Correct. Finds final speed v_f2 = +0.60 m/s of the 0.15 kg cart, using conservation of momentum (as there is negligible drag/friction for this brief collision). It is not known whether the carts are permanently deformed and/or energy was lost to thermal/sound systems, so collision could be either inelastic or elastic (but cannot be completely inelastic because the carts are not stuck together after the collision). Explicitly tests for whether or not translational kinetic energy is conserved, and finds that since it is not conserved, this collision must be inelastic.
  
• r:
  Nearly correct, but includes minor math errors.
  
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Has correct v_f2 from momentum conservation, and at least some attempt at testing for translational kinetic energy conservation.
  
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Application of momentum conservation, but v_f2 is incorrect, with little or no test of kinetic energy conservation.
  
• 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: midterm02r3iN
p: 28 students
r: 7 students
t: 13 students
v: 4 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 1408):

Physics midterm problem: plausible cliff height for ATV jump

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

In 2005, Chip Gaines reportedly drove a four-wheeled all-terrain vehicle over an embankment at the edge of a cliff, and became airborne:

I gunned it and launched that four-wheeler straight off the other side of the hill—over a sheer cliff that dropped a good twenty feet to the ground... In a matter of two seconds, the four-wheeler and I...face-planted into the dirt from nearly twenty feet up... And that's how I wound up with this awesome scar.[*]

While Chip Gaines claims that the cliff was 20 ft high (6.0 m), his wife Joanna Gaines recalls that the cliff was only 10 ft high (3.0 m).

Determine which cliff height (6.0 m or 3.0 m) was more plausible for Chip Gaines to be airborne for two seconds after launching himself on his four-wheeler (presumably a 2003 Kawasaki KVF 360 4⨉4[**]) with a speed of 38 mph (17 m/s) at an angle of 30° above the horizontal. Neglect air resistance, and treat Chip Gaines as a point object. Show your work and explain your reasoning using properties of projectile motion.

[*] Chip Gaines, Capital Gaines: Smart Things I Learned Doing Stupid Stuff, W Publishing (2017), pp. 44-47.
[**] kawasakimotorcycle.org/forum/kawasaki-atv-mule/30810-top-speed-360-a.html.

Solution and grading rubric:

• p:
  Correct. From the initial speed of v₀ = 17 m/s and direction of 30° above the horizontal, finds the y-component of initial velocity v_0y = v₀·sinθ = +8.5 m/s; applies projectile motion equations to determine that at t = 2 s, Chris Gaines would be at a final height of y = –2.6 m (thus 2.6 m below his starting point of y₀ = 0), thus making Joanna Gaines' estimate for the cliff height (3.0 m) more plausible that Chris Gaines' estimate of 6.0 m. May instead started with y = –3.0 m and y = –6.0 m and used the quadratic equation to solve for the expected times to reach the bottom of these cliffs, and found that the time to reach a final position of y = –3.0 m is closer to the given t = 2 s.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least some systematic attempt at using kinematic equations for projectile motion.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at systematic use of kinematic equations for projectile motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at kinematic equations for projectile motion. Primarily applies trigonometry to find distances rather than velocity components.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 31 students
r: 8 students
t: 8 students
v: 3 students
x: 7 students
y: 0 students
z: 1 student

A sample "p" response (from student 0517):

Physics midterm problem: diverging lens and converging lens

Physics 205B Midterm 1, spring semester 2018
Cuesta College, San Luis Obispo, CA

An object 1.0 cm in height is placed 5.0 cm in front of a f = –20.0 cm diverging lens, producing an upright image. This same 1.0 cm high object is then placed an unknown distance in front of a f = +20.0 cm converging lens, producing an inverted image that is the same size as the upright image originally produced by the diverging lens.

Determine (a) the size of the image produced by the diverging lens, and (b) the distance of this object in front of the f = +20.0 cm converging lens.

Show your work and explain your reasoning by using ray tracings and/or thin lens equations, properties of lenses, images, and magnification.

Solution and grading rubric:

• p:
  Correct. Identifies relevant parameters to methodically use the thin lens equation and linear magnification equation, in order to determine:
  1. that for the diverging lens, h_o = +1.0 cm, d_o = +5.0 cm, f = −20.0 cm, and uses thin lens equation to either find d_i = −4 cm (a virtual image) to find image height h_i = +0.8 cm (upright image) from the linear magnification equation, or may eliminate d_i in both equations to solve for h_i directly; and
  2. for the converging lens, h_o = +1.0 cm (the same object), d_o and d_i are both unknown, f = +20.0 cm, and h_i = −0.8 cm (inverted image that is the same size as the upright image produced by the diverging lens), and eliminates d_i in both equations to find d_o = +45 cm. May include very minor math errors with handling fractions and inverses.
• r:
  Nearly correct, but includes minor math conceptual errors, typically overlooking the fact that the image produced by the converging lens is inverted.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least solves for (1) successfully, and still attempts to methodically use this information in (2) to solve for the object distance for the converging lens. Typically makes multiple conceptual errors, such as overlooking the fact that the image produced by the converging lens is inverted; claiming that the image distance for the converging lens is the same as the image distance for the diverging lens, etc.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at ray tracings and/or thin lens equations, the properties of lenses, images, and magnifications.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying ray tracings and/or thin lens equations, the properties of lenses, images, and magnifications.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01cVdP
p: 4 students
r: 7 students
t: 24 students
v: 0 students
x: 0 students
y: 0 students
z: 0 student

A sample "p" response (from student 7164):

Physics midterm problem: American Airlines Flight 1498 bird strike

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

"Bird slams into Miami-bound flight on approach to airport"
WPLG Local 10 News
local10.com/travel/bird-slams-into-miami-bound-flight-on-approach-to-airport-1-1

In November 2017, an Airbus 319 collided with a bird that embedded itself into the nose cone of the airplane.[*] Assume that the Airbus 319 had a mass of 62,600 kg and a horizontal speed of 70 m/s while landing[**][***], as it hit the bird (presumably a turkey buzzard[****][*****][******]) of 2.0 kg, flying away from the plane at its maximum speed of 27 m/s. No one was injured on the plane, but the bird was mostly likely instantaneously killed. Ignore friction, drag, and external forces.



Determine (a) the final speed of the plane and bird together after the impact, and (b) demonstrate that the bird experienced a crash severity (Δv = v_f – v₀) with a greater magnitude than the crash severity of the plane.



Show your work and explain your reasoning using properties of collisions, velocities, and momentum conservation.

[*] miamiherald.com/news/local/community/miami-dade/article184658633.html.
[**] wiki.pe/Airbus_A319.
[***] answers.com/Q/What_is_the_landing_speed_of_an_Airbus_A319?#slide=2.
[****] felid.org/activities/page_24.htm.
[*****] wiki.pe/Turkey_vulture#Description.
[******] kern.audubon.org/tvfacts.htm.

Solution and grading rubric:

• p:
  Correct. Determines/discusses:
  1. the final velocity of the airplane (and bird) in this completely inelastic collision from applying momentum conservation; and
  2. the relatively large crash severity magnitude of the bird (essentially 43 m/s), compared to the minuscule crash severity magnitude of the airplane (essentially zero).
• r:
  Nearly correct, but includes minor math errors. Correct final velocity of airplane/bird after collision, but does not find/compare relative crash severity magnitues; or sets up but does not successfully complete momentum conservation for this completely inelastic collision due to math errors, but attempts to use final velocity values to compare relative crash severity magnitudes.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically has at least one final velocity set to zero, and/or does not realize that the final velocities of the airplane and the bird are identical in setting up momentum conservation.
• 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 momentum conservation.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying momentum conservation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02bu2Z
p: 37 students
r: 7 students
t: 5 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 4632):
Another sample "p" response (from student 5203), with an added "whoops" for emphasis:

Physics midterm problem: world-record washing machine throw

Physics 205A Midterm 1, fall semester 2017
Cuesta College, San Luis Obispo, CA

"Washing Machine Throwing Showdown"
Guinness World Records
youtu.be/YC0oj7BcWiI

In 2017, Zydrunas Savickas set a world record throwing a 46 kg (101 lb) washing machine that landed a horizontal distance of 4.13 m from its starting position atop his head. Savickas' height is 1.91 m, and the washing machine was airborne for 1.84 s starting from just off the top of his head to just before hitting the ground[*].

Find both the horizontal and vertical components (v_0x, v_0y) of the initial velocity vector for the washing machine, as it was thrown and released from just above the top of Savickas' head. Neglect air resistance, and treat the washing machine as a point object. Show your work and explain your reasoning using properties of projectile motion.

[*] Rachel Swatman, "Watch Game of Thrones Star Take on World’s Strongest Man Winner in Washing Machine Throwing Showdown" (January 13, 2017), guinnessworldrecords.com/news/2017/1/watch-game-of-thrones-star-take-on-world%E2%80%99s-strongest-man-winner-in-washing-machi-458290.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates:
  1. uses given values of t = 1.84 s and x = +4.13 m to solve for the initial (and constant) horizontal velocity v_0x (where t₀ = 0, x₀ = 0); and
  2. uses given values of t = 1.84 s and y = −1.91 m to solve for the initial vertical velocity v_0y (where t₀ = 0, y₀ = 0).
• r:
  Nearly correct, but includes minor math errors. May have intentionally or unintentionally used y = +1.91 m or y = 0 instead of y = −1.91 m.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has one initial velocity component correct, but other component has errors in addition to those listed in (r), such as setting v_y = 0 in y = (1/2)⋅(v_y0 + v_y)⋅t to solve for v_y0, or setting v_x = 0 in x = (1/2)⋅(v_0x + v_x)⋅t to solve for v_x0, etc.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized 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: midterm01mOoL
p: 20 students
r: 16 students
t: 12 students
v: 3 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1956):

Physics final exam problem: microscope object distance

Physics 205B Final Exam, spring semester 2017
Cuesta College, San Luis Obispo, CA

A NASA education guide for high school students[*] details plans for a simple microscope using an f = +28 mm objective lens and an f = +46 mm eyepiece lens, separated by a lens-to-lens distance of 160 mm. Determine the distance (in mm) that an object should be placed at in front of the objective lens for viewing through this microscope. Show your work and explain your reasoning by using ray tracings and/or thin lens equations, properties of lenses, images, and magnification.

[*] nasa.gov/pdf/350502main_Optics_Building_a_Microscope.pdf.

Solution and grading rubric:

• p:
  Correct. Determines the object distance d_o1 for the objective lens by discussing:
  1. that the lens-to-lens distance is equal to the objective image distance d_i1 plus the eyepiece focal length f₂ (as the image produced by the objective is placed at the focal point of the eyepiece), such that d_i1 = 160 mm − 46 mm = 114 mm; and
  2. knowing the focal point f₁ = +28 mm and the image distance d_i1 = 114 mm for the objective lens, uses the thin lens equation to find the object distance d_o1 for the objective lens.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least recognizes the placement of the intermediate image produced by the objective lens between the objective lens and eyepiece lens.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Garbled attempt at applying ray tracings and/or thin lens equations, properties of lenses, images, and magnification.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying ray tracings and/or thin lens equations, properties of lenses, images, and magnification.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: finalmR3x
p: 1 student
r: 1 student
t: 5 students
v: 14 students
x: 3 students
y: 1 student
z: 1 student

A sample "p" response (from student 7117):

Physics final exam problem: destructive interference angles in one quadrant

Physics 205B Final Exam, spring semester 2017
Cuesta College, San Luis Obispo, CA

Two vertical radio transmitters broadcast in phase at the same wavelength of 1.2 m, and are spaced a certain apart along the east-west direction. A Physics 205B student holding a receiver starts from due south of the transmitters, and detects three different locations with destructive interference signals before finally reaching due east of the transmitters. Determine a plausible separation distance (in m) between the transmitters. Explain your reasoning using the properties of source phases, path lengths, and interference.

Solution and grading rubric:

• p:
  Correct. Discusses/demonstrates that three minima locations will be found in the range θ = 0° (due south) to 90° (due west) by using one of two approaches:
  1. using the destructive interference condition d⋅sinθ = (m + 1/2)⋅λ, where m = 0, 1, 2, ..., finds a plausible separation distance d such that the third minima (m = 2) will be within θ = 90°, but the fourth minima (m = 3) is outside of θ = 90° (i.e., 3.0 m ≤ d ≤ 4.8 m); or
  2. using the constructive interference condition d⋅sinθ = m⋅λ, where m = 0, 1, 2, ..., finds the separation distance d such that the third maxima (m = 3) will be at θ = 90°; which allows for the m = 0, 1, and 2 minima to exist within that range (i.e., d = 3.6 m).
• r:
  Nearly correct, but includes minor math errors. May have claimed equally spaced minima angles at θ = 30°, 60° and 90° to find a plausible separation distance d using θ = 30° for the first minima angle.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Garbled attempt at applying properties of source phases, path lengths, and interference.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying properties of source phases, path lengths, and interference.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: finalmR3x
p: 3 students
r: 4 students
t: 6 students
v: 7 students
x: 4 students
y: 2 students
z: 0 students

A sample "p" response (from student 0428), finding the maximum possible separation distance:

Physics final exam problem: ammeter and voltmeter readings

Physics 205B Final Exam, spring semester 2017
Cuesta College, San Luis Obispo, CA

A lithium battery with an emf of 3.6 V and an internal resistance of r = 0.45 Ω is connected to two light bulbs (each with different resistances), an ammeter, and a voltmeter. Determine (a) the ammeter reading (in amps) and (b) the voltmeter reading (in volts). Show your work and explain your reasoning using the properties of voltmeters, Kirchhoff's rules and Ohm's law.

Solution and grading rubric:

• p:
  Correct. Determines the ammeter and voltmeter readings by:
  1. finding the equivalent resistance of the circuit, and then uses Ohm's law to determine the current passing through the 0.45 Ω resistor; and
  2. knowing the current and the resistance, uses Ohm's law to determine the drop in voltage across the 0.45 Ω resistor; and
  3. knowing the voltage rise of the emf and the voltage drop across the 0.45 Ω resistor, uses Kirchhoff's loop rule and Ohm's law to determine the current passing through the 1.2 Ω resistor, which is the ammeter reading; then
  4. knowing the voltage rise of the emf and the voltage drop across the 0.45 Ω resistor, uses Kirchoff's loop rule to determine the voltage drop across the 2.2 Ω resistor, which is the voltmeter reading.
• r:
  Nearly correct, but includes minor math errors. Has determined at least one of (1)-(2), but only one of (3)-(4) is complete/correct.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least only one of (1)-(2) is complete/correct.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Garbled attempt at applying Kirchhoff's rules, Ohm's law, and properties of ammeters.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at 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: finalmR3x
p: 2 students
r: 2 students
t: 4 students
v: 2 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student):

Physics midterm problem: pencil lead variable resistor

Physics 205B Midterm 2, spring semester 2017
Cuesta College, San Luis Obispo, CA

A real battery with an emf of 6.0 V and an internal resistance of r = 1.2 Ω is attached to an ideal voltmeter, and is connected to an ideal ammeter and a pencil lead that acts as a variable resistor. If the amount of pencil lead between the contacts is shortened such that its resistance is reduced from 8.0 Ω to 1.0 Ω, discuss why the voltmeter reading will decrease while the ammeter reading will increase. Show your work and explain your reasoning using Kirchhoff's rules, Ohm's law, and properties of ammeters and voltmeters.

Solution and grading rubric:

• p:
  Correct. Explains why the voltmeter reading will decrease while the ammeter reading will increase as the amount of pencil lead between the contacts is shortened by discussing:
  1. the decrease in the resistance of the pencil lead resistor will reduce the equivalent resistance of the circuit (pencil lead and internal resistance are in series), such that from applying Ohm's law the amount of current passing everywhere through the circuit will increase, resulting in a higher ammeter reading; and
  2. the voltmeter measures the potential difference of the 6.0 V rise from the emf and the voltage drop I⋅r from the internal resistance, such that an increase in current will result in a lower voltage reading ΔV = +ε − I⋅r.
(May instead discuss how the voltmeter is equivalently measuring the voltage drop ΔV = −I⋅R across the pencil lead resistor, but must clearly show that the eight-fold decrease in the resistance (from 8.0 Ω to 1.0 Ω) will be larger than the corresponding approximate four-fold increase in current (0.65 A to 2.7 A) to result in a lower voltage reading.)
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least numerically or qualitatively demonstrates how current would increase, but does not definitely show why voltmeter reading would decrease.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying Kirchhoff's rules, Ohm's law, and equivalent resistance.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Kirchhoff's rules, Ohm's law, and properties of ammeters and voltmeters.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 30882, 30883
Exam code: midterm02GruT
p: 12 students
r: 0 students
t: 8 students
v: 8 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1412):
Another sample "p" response (from student 2643):
A sample "x" response (from student 9319):

Physics midterm problem: extending telescope length

Physics 205B Midterm 1, spring semester 2017
Cuesta College, San Luis Obispo, CA

Two converging lenses, with focal lengths of +40.0 cm (for the objective lens) and +2.5 cm (for the eyepiece) are used to make a telescope. The length of the telescope (measured from lens-to-lens) is adjusted by sliding cardboard tubes in or out. Starting with the telescope used to look at an object very far away (essentially at infinity), determine how much the length must be extended in order to look at a closer object 10 m away. Show your work and explain your reasoning by using ray tracings and/or thin lens equations, properties of lenses, images, and magnification.


[*] Alan M. MacRobert, "Astronomy with a $5 Telescope," Sky & Telescope, vol. 79 no. 4 (April 1990), p. 384.

Solution and grading rubric:

• p:
  The eyepiece must be moved back by approximately 2 cm because:
  1. the object at d_o1 = +∞ for the objective creates a real image at d_i1 = f₁ = +40.0 cm, which becomes the object at a distance d_o2 = f₂ = +2.5 cm for the eyepiece, thus the telescope length (lens-to-lens distance) is 40.0 cm + 2.5 cm = 42.5 cm;
  2. the object at d_o1 = +10 m for the objective creates a real image at a slightly farther distance of d_i1 = +41.7 cm, which becomes the object at the same distance d_o2 = f2 = +2.5 cm for the eyepiece, thus the telescope length (lens-to-lens distance) is now slightly longer: 41.7 cm + 2.5 cm = 44.2 cm;
  3. thus the slight increase (approximately 2 cm) in the objective image distance d_i1 requires the eyepiece to be moved back by the same amount in order for this image to be placed at its front focal point.
• 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 approach has conceptual errors, and/or major/compounded math errors. At least understands that the telescope length is f₁ + f₂ when focused at ∞, and some attempt at finding the telescope length d_i1 + f₂ when focused at a finite d_o1.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at ray tracings and/or thin lens equations, the properties of lenses, images, and magnifications. May have used microscope magnification equation to find length between lenses.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. No clear attempt at applying ray tracings and/or thin lens equations, the properties of lenses, images, and magnifications.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 30882, 30883
Exam code: midterm01AhC4
p: 5 students
r: 0 students
t: 7 students
v: 17 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1412):

Physics final exam problem: rubber vs. aluminum bullets shot at wooden blocks

Physics 205A Final Exam, fall semester 2016
Cuesta College, San Luis Obispo, CA

On an online discussion board[*] a claim was made that a stationary wood block will have a faster final speed if a rubber bullet fired at it bounces off of it, compared to a slower final speed for the wooden block if an aluminum bullet fired at it fully embeds itself inside it. Assume the rubber bullet[**] and the aluminum bullet have the same mass of 0.075 kg and the same initial horizontal speed of 150 m/s, fired at stationary wood blocks that each have the same mass of 8.0 kg. Verify this expected result by determining (a) the final speed of the wood block, after the rubber bullet rebounds with a speed of 143 m/s; and (b) the final speed of the wood block, after the aluminum bullet is fully embedded in the block. Ignore friction, drag, and external forces. Show your work and explain your reasoning using properties of collisions, impulse, momentum, and momentum conservation.




[*] answers.yahoo.com/question/index?qid=20121011172202AAobQUe.
[*] litfld.net/starlight-less-lethal-ammo-product-specifications/.

Solution and grading rubric:

• p:
  Correct. Determines/discusses:
  1. the final velocity of the wood block in the (partially) inelastic collision with the rubber bullet from applying momentum conservation;
  2. the final velocity of the wood block in the completely inelastic collision with the aluminum bullet from applying momentum conservation; and makes some argument that the results are consistent with the original claim.
• r:
  Nearly correct, but includes minor math errors. (Typically has correct final velocity of the block for the completely inelastic collision, but neglects to give the bullet a negative final velocity for the partially inelastic collision.)
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at applying momentum conservation to find the final velocity of the blocks.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach involving methods other than momentum conservation (typically kinetic energy conservation, when neither of these collisions are elastic)
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finali0w4
p: 10 students
r: 19 students
t: 1 student
v: 5 students
x: 10 student
y: 4 students
z: 0 students

A sample "p" response (from student):

Physics final exam problem: USS Iowa passing through Panama Canal

Physics 205A Final Exam, fall semester 2016
Cuesta College, San Luis Obispo, CA

"USS Iowa Pedro Miguel Locks"
National Archive #NN33300514 2005-06-30 PH1 JEFF HILTON
commons.wikimedia.org/wiki/File:USS_Iowa_Pedro_Miguel_Locks.jpg

One of the widest ships to pass through the Panama Canal was the USS Iowa, which has an outside width of 32.97 m. The narrowest portion of the Panama Canal is through the Pedro Miguel locks, which has an inside width of 33.53 m. Assume that the USS Iowa is entirely made of steel (coefficient of thermal expansion 12×10^–6 K^–1), the Pedro Miguel locks are entirely made of concrete (coefficient of thermal expansion 9.8×10^–6 K^–1), and that these widths were measured at 20° C. Determine whether or not it is plausible for this ship to still pass through the locks, if the ship and the canal are both at the highest reported temperature for this area of 39° C. [*][**][***][****]

[*] wki.pe/USS_Iowa_(BB-61).
[**] wki.pe/Panamax.
[***] engineeringtoolbox.com/linear-expansion-coefficients-d_95.html.
[****] wki.pe/Panama_City.

Solution and grading rubric:

• p:
  Correct. Determines/discusses:
  1. the expansion and/or new width of ship at the warmer temperature; and
  2. the expansion and/or new width of the locks at the warmer temperature; then
  3. compares either the relative difference in expansion, or the new expanded widths, and concludes that the expanded size of the ship will still fit within the expanded size of the locks.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least enough steps are shown that would theoretically result in a complete answer, multiple errors notwithstanding.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at comparing linear thermal expansion of the ship and the locks.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach involving methods other than linear thermal expansion.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finali0w4
p: 28 students
r: 9 students
t: 1 student
v: 9 students
x: 1 student
y: 0 students
z: 1 student

A sample "p" response (from student):

Physics final exam problem: more dangerous collision type

Physics 205A Final Exam, fall semester 2015
Cuesta College, San Luis Obispo, CA

A 2016 Fiat 500X car[*] (mass 1.2×10³ kg) driving at 2.0 m/s collides with a stationary Ford F-150 pick-up truck[**] (mass 1.8×10³ kg). It is claimed that an elastic collision is more dangerous for the passengers than a completely inelastic collision[***].

1. For an elastic collision, the car would rebound with a speed of 0.40 m/s in the reverse direction (while the truck would move forward after the collision).
2. For a completely inelastic collision between the car and truck, they would stick and both move together in the forward direction.

Solve for (a) the final speed of the truck after the elastic collision, (b) the final speed of the truck after the completely inelastic collision, and (c) discuss whether this claim is plausible or implausible. Ignore friction, drag, and external forces. Show your work and explain your reasoning using properties of collisions, energy (non-)conservation, and momentum conservation.

[*] caranddriver.com/fiat/500.
[**] buyersguide.caranddriver.com/ford/f-150/specs#features.
[***] James Cunningham, Norman Herr, Hands-On Physics Activities with Real-Life Applications: Easy-to-Use Labs and Demonstrations for Grades 8-12, Wiley (1994), p. 323.

Solution and grading rubric:

• p:
  Correct. Determines/discusses:
  1. the final velocity of the truck in the elastic collision from applying momentum conservation;
  2. the final velocity of the truck in the completely inelastic collision from applying momentum conservation; and
  3. makes some reasonable interpretation that the claim of the elastic collision being more dangerous than a completely inelastic collision is plausible, given that the initial-to-final velocity changes for the truck (and the car) are greater for the elastic collision than for the inelastic collision.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. At least has one correct final velocity for the truck, the other result is problematic but at least momentum conservation was applied.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying momentum conservation to find the final velocity of the truck for one collision, but the other collision is incomplete or conceptually problematic (such as applying kinetic energy conservation for the completely inelastic collision, claiming zero final velocities, etc.).
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach involving methods other than momentum conservation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: final7rUk
p: 24 students
r: 3 students
t: 14 students
v: 16 students
x: 6 students
y: 5 students
z: 2 students

A sample "p" response (from student 1793):

Physics midterm problem: Resident Evil: Apocalypse rope descent

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

Resident Evil: Apocalypse
(Constantin Film, 2004)

In the movie Resident Evil: Apocalypse (Constantin Film, 2004) Milla Jovovich (mass 59 kg[*]) vertically descends 99.5 m down the east tower of the Toronto City Hall[**] starting from rest, reaching a final speed of 5.0 m/s using a device that continuously exerts a constant friction force on the rope controlling her descent. Consider her and the rope to be a single object as she descends. Ignore drag, stretching in the rope, the mass of the rope, and any contact between her feet and the side of the building. For this process, determine (a) how much energy was lost to friction, and (b) the magnitude of the friction force on the rope. Show your work and explain your reasoning using the properties of forces, work, energy forms and (non-)conservation of energy.

[*] healthyceleb.com/milla-jovovich-height-weight-body-statistics/1252.
[**] wki.pe/Toronto_City_Hall.

Solution and grading rubric:

• p:
  Correct. Determines:
  1. the total work done by friction against the motion of Milla Jovovich, by setting up an energy/transfer equation, and solving for the net change in her (increasing) translational kinetic energy and (decreasing) gravitational potential energy;
  2. solves for the magnitude of the friction force by dividing the work done by friction by the displacement (and the cosine of the angle between these directions when drawn tail-to-tail).
• r:
  Nearly correct, but includes minor math errors. Typically finds the total work done by friction, some attempt at finding the magnitude of the friction force with conceptual errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically finds the total work done by friction, no substantive approach to finding the magnitude of the friction force.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least some systematic attempt at using energy changes and work.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying energy changes and work.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 19 students
r: 11 students
t: 20 students
v: 5 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1317):
Another sample "p" response (from student 6969):