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 p. 53: In the example solution. in two places the units of \Delta\vec{p} should be \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} \\ :
\langle 0, 0, 5.072 \rangle \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} = \vec{F}_{net} \Delta t\\
 \langle 0, 0, 5.072 \rangle \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} = \langle 0, 0, 1000 \rangle \mathrm{N} \Delta t\\

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 prob. P27b (p. 124 and A-1): Answer to part(b) is \langle 8.4 \times 10^{-4}, -1.4\times 10^{-3}, -1.12 \times 10^{-3}\ \mathrm{kg}\cdot \mathrm{m/s} \rangle .

Ch. 3 prob. P37e (p.125 and A-1): Answer to part (e) is \langle 1,91\times 10^{-16}, -1.27\times 10^{-16}, 0\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. 4 prob. P30 (p. 166): The answer is 30 N/m.

Ch. 4 prob. P33a (p. A-1):  The answer to part (a) is 2.28\times 10^{-10}\ \mathrm{m}.

Ch. 4 prob. P44 (p. 167): The answers are: (a) 44 N; (b) 29 N; (c) 74 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. P24b (p. 208): |\vec{p}|\hat{p}/dt should be |\vec{p}|d\hat{p}/dt.

Ch. 5 prob. P33e (p. 209, p. A-2 ): The answer to part (e) is \langle 492,0,0\rangle\ \mathrm{N}.

Ch. 5 prob. P37a (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. 6 prob. P73 (p. 281): The coordinates of B should be \langle -3,0,0\rangle, not \langle 0,-3,0\rangle.

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 Q7 (p. 377): The center of mass velocity is constant, thus the sketch should illustrate constant velocity like this one.

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

c09f063

Ch. 10 p. 411 (Q13 Background): The first fission reactor was constructed in 1942, not 1941.

Ch. 10 p. 414: In the program skeleton the electric charges for the gold nucleus (79e) and the alpha particle (2e) are reversed. The initial value of the alpha particle momentum was calculated from the kinetic energy of  the 10 MeV alpha particles used by Rutherford.

Ch. 11 p. 421, line 7: “The vector dot product” should be “The vector cross product”.

Ch. 13 p. 521, example entitled “Source Location”:  The absolute value of q must be used:

 |\vec{E}| =  \dfrac{1}{4\pi\epsilon_0}\dfrac{|q|}{|\vec{r}|^2}\hat{r}
|\vec{r}| = \sqrt{\dfrac{1}{4\pi\epsilon_0}\dfrac{|q|}{|\vec{E}|}}  

Ch.13 prob. P45c (p. 542): The answer to part (c) is |\vec{r}|  =  6.3 \times 10^{-8}\ \mathrm{m} ,

Ch. 13prob. P62 (p. 544): The electric field units should be N/C.

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

Ch 14 EXP6 part g (p. 575): The field is E = \dfrac{1}{2\epsilon_0}\left(\dfrac{Q}{A}\right)

Ch. 15 p. 605, Step 4: In both cases the denominator should be inverted. The fractions should be written \dfrac{\text{C/m}^2}{\text{C}^2/\text{(N}\cdot\text{m}^2)}, which yields \dfrac{\text{N}}{\text{C}}

Ch. 16, top of p. 641: The expression for \Delta V is missing an overall minus sign.

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. P23b (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. 21 prob. P30 (p. 901): In part (d) the location inside the cube should be <0.032, 0, 0> m.

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)