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:
Wnc = ∆
KEtr + ∆
PEgrav + ∆
PEelas,
where ∆
PEgrav = 0 (the ping-pong ball travels horizontally), and ∆
PEelas = 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:
Wnc = ∆
KEtr,
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):
Wnc = (
Fav·cosθ)·
s,
such that the average force is:
(
Fav·cosθ)·
s = ∆
KEtr,
Fav = ∆
KEtr/(
s·cosθ),
Fav = ((1/2)·
m·(
vf2 –
v02))/(
s·cosθ),
Fav = (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