Cuesta College, San Luis Obispo, CA
In an experiment performed at the Naval Postgraduate School laboratory, a 2.9 g projectile hit a wall with a speed of 910 m/s, passed through and exited with a speed of 870 m/s[*]. As it passed through the aluminum wall, the magnitude of the impulse exerted on the projectile was: (A) 0 N·s.
(B) 0.040 N·s.
(C) 0.12 N·s.
(D) 5.0 N·s.
[*] Kenneth Scott Bates, Jr., "Aircraft Fuel Tank Entry Wall-Projectile Interaction Studies," Naval Postgraduate School thesis (June 1973), Fig. 14, p. 39, dtic.mil/dtic/tr/fulltext/u2/765667.pdf.
Correct answer (highlight to unhide): (C)
The impulse J can be calculated as the initial-to-final change in momentum:
J = ∆p = m·∆v,
where ∆v = vf – v0.
The initial velocity vector is v0 = +910 m/s (traveling to the right, in the +x direction), and the final velocity vector is vf = +870 m/s (still traveling to the right, with a slower speed). Then with a projectile mass m = 2.9 g or 2.9×10–3 kg:
J = (2.9×10–3 kg)·((+870 m/s) – (+910 m/s)) = –0.116 N·s,
or to two significant figures, the magnitude of the impulse is 0.12 N·s (and the "–" sign indicates that it is directed to the left, opposite the motion of the projectile).
(Response (A) would be true if there was no change in the projectile's motion; response (B) is the ∆v change in velocity divided by a factor of 1,000; response (D) is m·(vf + v0).)
Sections 70854, 70855
Exam code: quiz04w33N
(A) : 5 students
(B) : 4 students
(C) : 38 students
(D) : 1 student
Success level: 79%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.44
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