Physics quiz question: impulse on rebounding tennis ball

Physics 205A Quiz 4, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Tennis"
PughPugh
flic.kr/p/acSs9R

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·(v_f – v₀).

The initial velocity vector is v₀ = –1.44 m/s (traveling downwards), and the final velocity vector is v_f = +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