Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 7.4, 7.13
A golf club[*] exerts a horizontal force of 8,900 N for 5.0×10–4 s on an initially stationary golf ball[**] (mass 4.593×10–2 kg). As a result of the impact with the golf club, the golf ball has a final speed of:
(A) 4.4 m/s.
(B) 97 m/s.
(C) 410 m/s.
(D) 910 m/s.
[*] real-world-physics-problems.com/physics-of-golf.html .
Correct answer (highlight to unhide): (B)
The impulse exerted on the projectile is given by:
J = ∆p,
where the left side of the impulse-momentum theorem yields the impulse (directed to the right, in the positive +x direction):
J = (ΣF)·∆t = (+8,900 N)·(5.0×10–4 s) = +4.45 N·s,
of which only two figures are significant. The right side of the impulse-momentum theorem is the resulting change in momentum:
∆p = m·(vf – v0) = (0.04593 kg)·((vf) – (0)),
∆p = (0.04593 kg)·vf.
By equating the left and right-hand sides of the impulse-momentum theorem with the known quantities entered, the final velocity of the golf ball can be solved for:
J = ∆p,
+4.45 N·s = (0.04593 kg)·vf,
+96.886566514 m/s = vf,
which is also directed to the right, in the positive +x direction, and the magnitude of the final velocity (or speed) of the golf ball, to two significant figures is 97 m/s.
(Response (A) is √(2g); response (C) is the magnitude of (ΣF·m; response (D) is the magnitude of (ΣF)/g.)
Sections 30880, 30881
Exam code: quiz04w3Rc
(A) : 3 students
(B) : 54 students
(C) : 7 students
(D) : 1 student
Success level: 83%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.57