20181012

Physics midterm question: net forces on car and bus driving over hills

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

A 1,800 kg car and a 10,000 kg bus both drive over hills with the same circular radius. Both the car and the bus are still in contact with their hills. Discuss whether the car should drive at a faster or slower speed than the bus in order for them to have the same magnitude net force at the top of their hills. 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 diagrams are optional, but primarily discusses/demonstrates:
    1. Newton's second law for uniform circular motion applies, such that for both the car and bus, the net force ΣF (the resultant of the upwards normal force and the downwards weight force) must have a magnitude mv2/r) and point in towards the center of the circular motion--which is vertically downwards; and
    2. since both the car and the bus drive over hills with the same circular radius, then in order for them to have the same magnitude net force, the (less massive) car must drive at a faster speed, while the (more massive) bus must drive at a slower speed.
  • 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: midterm01g4iN
p: 26 students
r: 7 students
t: 11 students
v: 9 students
x: 3 students
y: 0 students
z: 2 students

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

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