Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.63
The Slingshot Channel
A slingshot built by Jörg Sprave from "The Slingshot Channel" releases elastic potential energy from its bands to accelerate a stationary steel ball (mass 6.8×10–2 kg) to a final (horizontal) speed of 69 m/s (154 mph)[*]. Ignore friction/drag. Assuming that the "active draw length" of 0.75 m corresponds to the distance the elastic bands in the slingshot are stretched from equilibrium, the spring constant of these elastic bands (approximated as a single spring) is:
(A) 13 N/m.
(B) 1.1×102 N/m.
(C) 2.2×102 N/m.
(D) 5.8×102 N/m.
Correct answer: (D)
Starting with the energy balance equation:
Wnc = ∆KEtr + ∆PEgrav + ∆PEelas,
where Wnc = 0 (no external gains/losses of mechanical energy), and ∆PEgrav = 0 (as there is no change in elevation of the ball as it travels horizontally), such that:
0 = ∆KEtr + ∆PEelas,
0 = (1/2)·m·∆(v2) + (1/2)·k·∆(x2),
0 = (1/2)·m·(vf2 – v02) + (1/2)·k·(xf2 – x02).
With initial parameters of v0 = 0 (starting from rest) and x0 = +0.75 m (stretched from equilibrium), and final parameters vf = 69 m/s and xf = 0 (elastic bands relaxed after releasing the ball), then:
0 = (1/2)·m·(vf2 – 02) + (1/2)·k·(02 – x02),
0 = m·vf2 – k·x02,
k = m·vf2/x02,
k = (6.8×10–2 kg)·(69 m/s)2/(0.75 m)2 = 575.552 kg/s2,
or to two significant figures with more conventional units, the spring constant of the elastic bands is 5.8×102 N/m.
(Response (A) is g/x0; response (B) is g·x0/m; response (C) is (m·vf2)/(2·x0).)
Sections 70854, 70855, 73320
Exam code: quiz04w3Rc
(A) : 2 students
(B) : 16 students
(C) : 28 students
(D) : 19 students
Success level: 39%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53