Errata

4th Edition Errata (earlier editions below)

Ch. 1 answer to checkpoint 6 (p. 44): Part (a) is \langle 450, -300, -200 \rangle.

Ch. 1 answer to checkpoint 15 (p. 44): Part (a) is \langle -1.6 \times 10^{28}, 0, 1.6 \times 10^{28} \rangle kg \cdotm/s.

CH. 1 prob. P28 (p. 40): Answer to part (a) is  \langle -7\times10^{10}, 13\times10^{10}, -4\times10^{10}\rangle

CH. 1 prob. P36 (p. 40):  Answer to part (d) is \langle -0.2, -0.9, 0.4\rangle.

CH. 1 prob. P42 (p. 41):  Answer is \langle 11.96, 3.83, 4.2 \mathrm{\ m}\rangle.

Ch. 1 prob. P53 (p. 41 and A-1): Answer is \langle 0, 4.9, 0 \rangle kg \cdotm/s.

Ch. 1 prob. P61 (p. 42 and A-1): Answer is 9.4 \times 10^{-19} kg \cdotm/s.

Ch. 1 prob. P63 (p. 42 and A-1): Answer is 3.6 \times 10^{-17} kg \cdotm/s. Also, the problem statement states that you don’t need to calculate the speed, which is true for calculating gamma but not for calculating the momentum.

Ch. 1 prob. P65 (p. 42 and A-1): Answer is 35.4.

Ch. 2 pp. 58-59: In Time Step 1, \vec{r}_\text{future} = \langle 4, 4.43, 0 \rangle m. In Time Step 2 the calculation of the future position should read as follows:  \vec{r}_{\mathrm{future}}  =  \langle 4, 4.43, 0 \rangle \mathrm{ m}  +  \left(\dfrac{\langle 0.6, 0.130, 0 \rangle  \mathrm{  kg}\cdot\mathrm{m/s}} {0.06\mathrm{ kg}}\right) \times 0.4 \mathrm{ s} = \langle 8, 5.30, 0 \rangle \mathrm{ m} \\

Ch. 2 prob. P21 (p. 83 and A-1): Answers are  t=0.05\ \mathrm{s}: \langle 0.15, 2.18, 0 \rangle\ \mathrm{ m} , \langle3, 3.51, 0 \rangle \mathrm{ m/s}
  t=0.1\ \mathrm{s}: \langle 0.3, 2.33, 0 \rangle\ \mathrm{ m} , \langle 3, 3.02, 0 \rangle \mathrm{ m/s}
  t=0.15\ \mathrm{s}: \langle 0.45, 2.45, 0  \rangle\ \mathrm{ m} , \langle  3, 2.53, 0 \rangle \mathrm{ m/s}

Ch. 2 prob. P41 (p. 86 and A-1): Answer is 13 cm.

Ch. 3 prob. P13d (p. 124 and A-1): Answer is \langle -1.27 \times 10^{20}, 0, 1.27 \times 10^{20} \rangle N.

Ch. 3 answer to checkpoint 8 (p. 129): Units are m/s.

Ch. 4 answer to checkpoint 7 (p. 172): Part (a) is \langle 0, 0, 4 \rangle N.

Ch. 5 prob. P14 (p. 208): The coefficient of friction between the floor and the bottom block should be 0.5.

Ch. 5 prob. P37 (p. 210 and A-2): Answer to part (a) is 871 kg \cdotm/s/s.

Ch. 5 prob. P39 (p. 210 and A-2): Answer is 6.65 \times 10^{15} kg.

Ch. 5 checkpoint 4 answers (p. 214): For points A and D, the answers to part (b) are nonzero, and the answers to part (c) are zero.

Ch. 6 Figure 6.42 (p. 252): The labels “1” and “2” should be interchanged.

Ch. 6 checkpoint 4 answers (p. 283): (a) negative, decreases, (b) positive, increases, (c) negative, decreases, (d) positive, increases, (e) zero, stays the same

Ch. 6 prob. P29 (p. 277 and A-2): Answer is 66 J.

Ch. 7 summary (p. 316): The first sentence at the top right of the page should say, “Sliding friction force can be modeled in terms of the ball-and-spring model of solids.” (A more detailed treatment of friction is found in Ch. 9.)

Ch. 9 prob. P43, Figure 9.63: The figure may be hard to understand. Here is an improved figure:

c09f063

Ch. 13 answer to checkpoint 8 (p. 545): The second answer is missing, 3.2 \times 10^{-19} C, which is 2e.

Ch. 16, top of p. 653, 4th and 6th lines: pos=(…) should be pos=vector(…).

Ch. 16 Summary (p. 661): Field is negative gradient of potential; the correct field components should have minus signs:
  E_x = -\dfrac{\partial V}{\partial x}\mathrm{, }E_y = -\dfrac{\partial V}{\partial y}\mathrm{, }E_z = -\dfrac{\partial V}{\partial z}  

Ch. 17 p. 689: In the VPython code fragment, just after the statement while t\boldsymbol{<} tmax: insert the statement rate(30) .

Ch. 17 p. 690: In the VPython code fragment, delete the statement rate(1000), which is not needed, since no display is made in this loop.

Ch. 17 prob. P23 (p. 209 and A-4): Answer to part b for magnetic field is 0.166 J.

Ch. 18 by Figure 18.48 (p. 736): In the 2nd full paragraph, the electrons pile up at location 4. In the 3rd full paragraph, + charge accumulates at location 3.

Ch. 20 p. 819, Fig. 20.25: The caption should say “out of the page.”

Ch. 20 p. 820, Fig. 20.26: The caption should say “with an upward Coulomb electric field.”

Ch. 20 p. 822, Fig. 20.31: The magnetic field should point out of the page, as in the previous diagrams.

Ch. 20 prob. P72 (p. 863): The magnet is a cylinder with diameter 1 cm and length 3 cm.

Ch. 21 prob. P23 (p. 900 and A-5): Answer for direction is “into the page.”

Ch. 22 problems P40, P41, P42 (p. 938): These problems are associated with Section 22.7, not 22.6.

Errata for previous editions

Third edition (2011)

Second edition volume 1 (2007)

Second edition volume 2 (2007)

Version 1.5 volume 1 (2006)

Version 1.5 volume 2 (2006)

First edition volume 1 (2002)

First edition volume 2 (2002)