Cuesta College, San Luis Obispo, CA
PughPugh
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When dropped from a set height above a horizontal surface, a Type 2 tennis ball (mass of 5.8×10–2 kg[*]) hits the surface with an initial speed of 1.44 m/s, and rebounds upwards off the surface with a speed of 0.77 m/s. The Type 2 tennis ball experiences an impulse of magnitude __________ while rebounding off the surface.
(A) 3.9×10–2 N·s.
(B) 4.5×10–2 N·s.
(C) 8.4×10–2 N·s.
(D) 0.13 N·s.
[*] "Tennis ball dropped from 254 cm: Type 2 ball rebound is 135-147 cm," itftennis.com/technical/balls/approval-tests.aspx.
Correct answer (highlight to unhide): (D)
From the impulse-momentum theorem, the impulse J on an object causes its initial-to-final change in momentum ∆p:
J = ∆p,
where ∆p = m·(vf – v0).
The initial velocity vector is v0 = –1.44 m/s (traveling downwards), and the final velocity vector is vf = +0.77 (traveling upwards). Then:
J = (5.8×10–2 kg)·((+0.77 m/s) – (–1.44 m/s)) = (5.8×10–2 kg)·(+2.21 m/s) = +0.12818 N·s,
or to two significant figures, the magnitude of the impulse is 0.13 N·s (and the "+" sign indicates that it is in the upwards direction).
(Response (A) is (5.8×10–2 kg)·((+0.77 m/s) – (+1.44 m/s)), response (B) is the magnitude of the final momentum, response (C) is the magnitude of the initial momentum.)
Sections 70854, 70855
Exam code: quiz04w04K
(A) : 54 students
(B) : 5 students
(C) : 0 students
(D) : 13 students
Success level: 18%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.38
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