## 20140930

### Physics quiz question: block stopped by compressing spring

Physics 205A Quiz 4, fall semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.58, Comprehensive Problem 6.82

An 0.080 kg block traveling horizontally at 5.5 m/s hits a horizontal spring (k = 320 N/m) that is initially uncompressed. Neglect friction and drag. As a result the block is brought to a rest when the spring is compressed by:
(A) 0.037 m.
(B) 0.061 m.
(C) 0.087 m.
(D) 1.2 m.

Correct answer (highlight to unhide): (C)

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 block as it travels horizontally to come to rest), 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 = +5.5 m/s (block is moving) and x0 = 0 (spring is relaxed/uncompressed), and final parameters vf = 0 (block at rest), then the distance xf that the spring has been compressed from equilibrium can now be solved for:

0 = (1/2)·m·(02v02) + (1/2)·k·(xf2 – 02),

0 = –(1/2)·m·v02 + (1/2)·k·xf2,

k·xf2 = m·v02,

xf = v0·(m/k)1/2,

xf = (5.5 m/s)·((0.080 kg)/(320 N/m))1/2 = 0.08696263565 m,

or to two significant figures, the spring is compressed by 0.087 m.

(Response (A) is (m·v0/k)1/2; response (B) is v0·(m/(2·k))1/2; response (D) is v0·(2·g)1/2.)

Sections 70854, 70855
Exam code: quiz04s7aT
(A) : 12 students
(B) : 16 students
(C) : 17 students
(D) : 3 students

Success level: 35%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.37