20071029

Physics clicker questions: payload and toy car sequence

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 7.40

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) near the end of their learning cycle:

Object 1 = toy car
Object 2 = payload
i = toy car moving to the right, before payload is dropped down onto it
f = toy car after payload has been dropped on it


[0.6 participation points.] What is the final speed of the toy car? (Floor is frictionless.)
(A) Slower than its initial speed.
(B) Same as its initial speed.
(C) Faster than its initial speed.
(D) (Not enough information is given to determine this.)
(E) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 27 students
(B) : 5 students
(C) : 4 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

Since there are no external horizontal impulses, momentum is conserved, such that v_12,f = (m_1*v_1,i)/(m_1 + m2).



Object 1 = toy car
Object 2 = payload
i = toy car moving to the right, carrying payload
f = toy car after payload has been dropped (down into hole in floor)


[0.6 participation points.] What is the final speed of the toy car? (Floor is frictionless.)
(A) Slower than its initial speed.
(B) Same as its initial speed.
(C) Faster than its initial speed.
(D) (Not enough information is given to determine this.)
(E) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 4 students
(B) : 19 students
(C) : 13 students
(D) : 0 students
(E) : 0 students

Correct answer: (B)

Although not classifiable as a collision, after the payload separates from the toy car (but before it hits the ground) it will have the same velocity as the car, as there are no external horizontal impulses.

Many students will misapply the definition of momentum to a conservation of momentum situation.

p = m*v.
  • For the payload dropped down onto the toy car: p = m*v, where m increases, thus v decreases. (Seems correct?)
  • For the payload dropped down from the toy car: p = m*v, where m decreases, thus v increases. (No!)
More correctly, conservation of momentum should be applied properly as:

0 = delta(p_1) + delta(p_2).
  • For the payload dropped down onto the toy car, the payload gains horizontal momentum, thus the car loses horizontal momentum.
  • For the payload dropped down from the toy car, the payload retains the same horizontal momentum, thus the car's horizontal momentum remains the same!

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