Physics midterm problem: hydrostatic weighing

Physics 205A Midterm 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 9.37, 9.38

[20 points.] The average density of a person can be found by first weighing that person in air and then finding the scale reading for the person completely immersed in water (while suspended from the scale). If a person has a weight 550 N in air and has an average density of 1.05×10³ kg/m³, what would be the scale reading for the person completely immersed in water? The density of water is ρwater = 1.00×10³ kg/m³. Show your work and explain your reasoning.

Solution and grading rubric:

• p = 20/20:
  Correct. Finds volume of person, given density and mass (from weight in air), and calculates bouyant force on person while submerged in water. The scale reading is then the difference between the weight and the bouyant force. Proper use of Newton's first law, definition of bouyant force, and relation between mass and weight.
  
• r = 16/20:
  Nearly correct, but includes minor math errors. May have conflated mass and weight, or density values, but still clearly shows methodical process.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At some attempt at using Newton's laws, definition of bouyant force, and relation between mass and weight.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit. Discussion not based on methodical application Newton's laws, definition of bouyant force, and relation between mass and weight.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02fR3q
p: 24 students
r: 4 students
t: 5 students
v: 8 students
x: 7 students
y: 0 students
z: 0 students

A sample "p" response (from student 3737), rounding to two significant figures during each step:
Another sample "p" response (from student 3737), rounding to two significant figures only for the very last calculation:

Physics midterm problem: copper cylinder suspended in air, and in oil

Physics 205A Midterm 2, Fall Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.35

[20 points.] A copper cylinder (density 8.92e+3 kg/m^3) weighs 12.5 N when suspended from a scale in air. When this same cylinder is completely submerged in oil while suspended from a scale, the scale reading is 9.5 N. Find (a) the volume of the cylinder, and (b) the density of the oil. Show your work and explain your reasoning.

Solution and grading rubric:

• p = 20/20:
  Correct. Applies Newton's first law for the cylinder in air equating tension and weight to find mass, then finds volume V = m/rho_Cu = 1.4e-4 m^3. Applies Newton's first law for the cylinder in oil equating the upwards forces T = 9.5 N and F_B = rho_oil*g*V with the downwards weight (which is still 12.5 N), to solve for rho_oil = (w - T)/(g*V) = 2.1e+3 kg/m^3.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Solves for volume using F_B = rho*g*V where F_B is 3.0 N, but rho is the density of water, or copper(!).
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Applies specific gravity definition rho_object/rho_fluid = V_submerged/V_total to solve for the density of the oil and/or the cylinder volume, or uses P_2 = P_1 + rho*g*d.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 72177
p: 4 students
r: 1 students
t: 3 students
v: 5 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7575):

Open, flat, and closed universes

081104-1060413
http://www.flickr.com/photos/waiferx/3003896971/
Originally uploaded by Waifer X

Models representing a negative curvature "open" universe, a zero curvature "flat" universe, and a positive curvature "closed" universe. The closed universe is constructed by removing two triangles from a flat sheet, and taping the resulting cutout edges together. These triangles are then taped into slits cut into another flat sheet, producing the open universe. The black-and-white inverted version of the same photo is shown below.

081104-1060413-inverted
http://www.flickr.com/photos/waiferx/3004952710/
Originally uploaded by Waifer X