20111022

Physics quiz question: rebounding truck crash

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.7

"Crash Test Ford Ranger 2012..."
MZBTZOO M
youtu.be/44UXH4AvOnY

An unmanned 2003 Ford Ranger XL Short Bed 2WD truck[*] (mass 1,980 kg) crashes into a stationary concrete wall at a speed of 11.0 m/s, and rebounds off the wall with a speed of 2.2 m/s in the opposite direction. The magnitude of the impulse exerted on the truck is:
(A) 4.4×103 N·s.
(B) 1.7×104 N·s.
(C) 2.18×104 N·s.
(D) 2.61×104 N·s.

[*] "Standard GVWR: 4360 lb," autos.msn.com/research/vip/spec_engines.aspx?year=2003&make=Ford&model=Ranger&trimid=95693.

Correct answer (highlight to unhide): (D)

The impulse J can be calculated as the initial-to-final change in momentum:

J = ∆p = m·∆v,

where ∆v = vfv0.

The initial velocity vector is v0 = +11.0 m/s (traveling in the forwards direction), and the final velocity vector is vf = –2.2 m/s (rebounding in the reverse direction). Then:

J = (1,980 kg)·((–2.2 m/s) – (+11.0 m/s)) = (1,980 kg)·(–13.2 m/s) = –26,136 N·s,

or to two significant figures, the magnitude of the impulse is 2.61×104 N·s (and the "–" sign indicates that it is in the backwards direction).

(Response (A) is the magnitude of the initial momentum, response (B) is the magnitude of the final momentum, response (C) is (1,980 kg)((–2.2 m/s) + (+11.0 m/s)).)

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 1 student
(B) : 18 students
(C) : 11 students
(D) : 20 students

Success level: 32%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.64

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