Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.20
rva1945
youtu.be/1aDDlCHjewc
A submarine of mass 7.69×105 kg[*] and initially at rest ejects a torpedo of mass 1.5×103 kg[**]. The torpedo has a speed of 10 m/s as it leaves the submarine[***]. Neglect external friction/drag during this brief process. The recoil speed of the submarine is:
(A) 2.0×10–4 m/s.
(B) 2.0×10–3 m/s.
(C) 6.7×10–3 m/s.
(D) 2.0×10–2 m/s.
[*] wki.pe/German_Type_VII_submarine.
[**] wki.pe/G7a_torpedo.
[***] uboatarchive.net/KTBNotesArmament.htm.
Correct answer (highlight to unhide): (D)
Since there are no significant external impulses on the submarine-torpedo system during this brief process, total momentum is conserved, so:
0 = m1·(vf1 – v01) + m2·(vf2 – v02),
where for the submarine, m1 = 7.69×105 kg, v01 = 0, vf1 = ?; and for the torpedo, m2 = 1.5×103 kg, v02 = 0, vf2 = +10 m/s (in the forward direction). So solving for the final velocity vf1 of the submarine:
0 = m1·vf1 + m2·vf2,
–m1·vf1 = m2·vf2,
vf1 = –m2·vf2/m1 = –(1.5×103 kg)·(+10 m/s)/(7.69×105 kg) = –1.950585176×10–2 m/s,
or to two significant figures, the recoil speed of the submarine is the magnitude of its final velocity: 2.0×10–2 m/s. (Note even with neglecting drag/friction forces, the recoil speed of the submarine is miniscule compared to the torpedo's launch speed, due to the relatively larger mass of the submarine.)
(Response (A) is m2/(m1·vf2); response (B) is m2/m1; response (C) is vf2/m2.)
Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 9 students
(C) : 2 students
(D) : 34 students
Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73
Student responses
Section 72177
(A) : 0 students
(B) : 1 student
(C) : 0 students
(D) : 12 students
Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.20
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