Physics quiz question: spring constant of slingshot elastic bands

Physics 205A Quiz 4, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.63

"Twice as long - twice as good?"
The Slingshot Channel
youtu.be/7TSrTF4Smak

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.

[*] youtu.be/7TSrTF4Smak

Correct answer: (D)

Starting with the energy balance equation:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where W_nc = 0 (no external gains/losses of mechanical energy), and ∆PE_grav = 0 (as there is no change in elevation of the ball as it travels horizontally), such that:

0 = ∆KE_tr + ∆PE_elas,

0 = (1/2)·m·∆(v²) + (1/2)·k·∆(x²),

0 = (1/2)·m·(v_f² – v₀²) + (1/2)·k·(x_f² – x₀²).

With initial parameters of v₀ = 0 (starting from rest) and x₀ = +0.75 m (stretched from equilibrium), and final parameters v_f = 69 m/s and x_f = 0 (elastic bands relaxed after releasing the ball), then:

0 = (1/2)·m·(v_f² – 0²) + (1/2)·k·(0² – x₀²),

0 = m·v_f² – k·x₀²,

k = m·v_f²/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·v_f²)/(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