Physics quiz question: supersonic ping-pong ball vacuum cannon

Physics 205A Quiz 4, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.13(a)

A 0.0027 kg ping-pong ball at rest is ejected by a horizontally-mounted vacuum cannon[*] with a final speed of 400 m/s (900 mph). The vacuum cannon is 3.7 m (12 feet) in length. Neglect friction and drag. The magnitude of the average force exerted on the ping-pong ball by the vacuum cannon is:
(A) 0.098 N.
(B) 0.15 N.
(C) 58 N.
(D) 800 N.

[*] Brian Dodson, "Ping-Pong Gun Fires Balls at Supersonic Speeds," gizmag.com/how-to-build-a-supersonic-ping-pong-gun/26082/ (February 3, 2013).

Correct answer (highlight to unhide): (C)

The energy transfer-balance equation is given by:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where ∆PE_grav = 0 (the ping-pong ball travels horizontally), and ∆PE_elas = 0 (no springs involved). The non-conservative work done by the vacuum cannon (considered as an external agent outside of the ping-pong ball's energy systems) increases the ping-pong ball's translational kinetic energy:

W_nc = ∆KE_tr,

where the work done is the product of the average force exerted and the displacement, and the angle θ between the exerted force and the displacement is 0° (as the force is exerted in the same direction as the ping-pong ball traveling down the cannon):

W_nc = (F_av·cosθ)·s,

such that the average force is:

(F_av·cosθ)·s = ∆KE_tr,

F_av = ∆KE_tr/(s·cosθ),

F_av = ((1/2)·m·(v_f² – v₀²))/(s·cosθ),

F_av = (1/2)·(0.0027 kg)·((400 m/s)² – (0 m/s)²)/((3.7 m)·cos(0°)) = 58.378378378 N,

or to two significant figures, the magnitude of the force is 58 N.

(Response (A) is m·g·s; response (B) is (1/2)·m·v₀/s; response (D) is (1/2)·m·v₀²·s.)

Sections 70854, 70855, 73320
Exam code: quiz04iSs5
(A) : 7 students
(B) : 14 students
(C) : 38 students
(D) : 7 students

Success level: 58%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.61