## 20141101

### 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

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·(vf2v02) + (1/2)·k·(xf2x02).

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·(02x02),

0 = m·vf2k·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