Physics quiz question: cart on track

Physics 205A Quiz 4, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 6.10, 6.26

A 2.20 kg cart moving to the right has a speed of 12.0 m/s on the lower horizontal part of this track, and travels to the upper horizontal part of this track. Neglect friction and drag.

[3.0 points.] What is the kinetic energy of the cart on the lower horizontal part of this track?
(A) 16.2 J.
(B) 26.4 J.
(C) 142 J.
(D) 158 J.

Correct answer: (D)

This is a simple calculation of K_i = (1/2)*m*(v_i)^2. Response (A) is m*g*y_f; response (B) is m*v_i; response (C) is the final kinetic energy of the cart at the top of the ramp.

Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 10 students
(C) : 3 students
(D) : 26 students

"Difficulty level": 61%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47

[3.0 points.] What is the speed of the cart on the upper horizontal part of this track?
(A) 3.83 m/s.
(B) 7.35 m/s.
(C) 11.4 m/s.
(D) 12.6 m/s.

Correct answer: (C)

From before, K_i = 158 J. The increase in gravitational potential energy as it climbs up to the upper horizontal part of the track is delta(U) = m*g*delta(y) = 16.2 J. Thus the final kinetic energy will be 158 J - 16.2 J = 142 J; and the final speed will be v_f = sqrt(2*K_f/m) = 11.4 m/s.

Response (A) is sqrt(2*g*delta(y)), which is the speed that the cart would have at the lower part of the track, if it had started at rest from the upper part of the track. Response (B) is the maximum height that the cart can reach uphill (with the wrong units). Response (D) is sqrt((v_i)^2 + 2*g*delta(y)), which has a sign error inside of the square root, and results in a final speed greater than the initial speed.

Student responses
Sections 70854, 70855
(A) : 12 students
(B) : 8 students
(C) : 17 students
(D) : 4 students

"Difficulty level": 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.57