20161125

Physics midterm question: speeds of different mass, translational kinetic energy bullets

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

The following information is given for legal requirements on bullets[*]:
In Denmark rifle ammunition used for hunting must have a translational kinetic energy of 2,700 J for a bullet with less mass; or alternatively a translational kinetic energy of 2,000 J for a bullet with more mass.
Discuss why the less massive bullet would have a faster speed than the more massive bullet. Ignore drag. Show your work and explain your reasoning using the properties of mass, speed, and translational kinetic energy.

[*] wki.pe/Muzzle_energy.

Solution and grading rubric:
  • p:
    Correct. Discusses/demonstrates understanding that:
    1. translational kinetic energy depends on both the mass and the (square of the) speed of a bullet (KEtrans = (1/2)⋅mv2); and
    2. since there is more translational kinetic energy for the less massive bullet, from v = √(KEtrans/(2⋅m)) the larger numerator and smaller denominator under the radical sign means that this bullet must have a faster speed.
    May have also put in relative or absolute assumed numbers for the two bullet masses to prove a faster speed for the less massive bullet.
  • r:
    Nearly correct, but includes minor math errors.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. May have assumed that both bullets have the same translational kinetic energy, or did not explicitly account for both the difference in masses and translational kinetic energy.
  • v:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some constructive attempt at applying translational kinetic energy parameters.
  • x:
    Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying translational kinetic energy parameters.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 41 students
r: 0 students
t: 10 students
v: 4 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 1420):

A sample "p" response (from student):

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