Errata

4th Edition Errata (earlier editions below)

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Ch. 1 Fig. 1.49 (p. 29): In the WileyPLUS/VitalSource version (but in no other version), the exponents in the first column should all be positive, not negative.

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 $\cdot$ m/s.

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

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

Ch. 1 prob. P63 (p. 42 and A-1): Answer is $3.6 \times 10^{-17}$ kg $\cdot$ m/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 prob. P27 (p. 84 and A-1): (a) 37.5 mi/h; (b) 45 mi/h; (c) A graph of v_x vs. t is not a straight line.

Ch. 2 prob.P33 (p. 85 and A-1): The pot fell from rest at 4.5 m above the top of the window.

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, section 4.14, subsection “A Constant-Density Model of the Atmosphere” (p. 158), first sentence: ( $1.3\times 10^{-3} \text{g/cm}^3$ )

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

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

Ch. 4 prob. P45 (p.167): The final position is $\langle 10.2, 2, -1 \rangle$ m.

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

Ch. 5 (p. 184): In the second equation in the Solution the derivative should be $\dfrac{dv_{\text{cm,}x}}{dt}$ .

Ch. 5 prob. P7 (p. 207): (a) You should show 3 arrows representing the forces acting on the box; one pointing down for the gravitational force, one pointing to the left for the force exerted by Rope 1, and one pointing up and to the right for the force exerted by Rope 2. (b) Zero.

Ch. 5 prob. P9 (p. 207): (a) < 195.34, 163.91, 0 > N;
(b) < -195.34, 326.09, 0 > N

Ch. 15 prob. P15 (p. 208): (a) 2.95 $\mathrm{m/s^2}$
(b) 50.2 $\mathrm{kg}\cdot\mathrm{m/s^2}$
(c) 14.8 N

Ch. 5 prob. P19a (p. 208): The answer to part (a) is $8.73 \times 10^{22}$ N

Ch. 5 prob. P27 (p. 209):
(a) <0, 0, 0> $\mathrm{kg}\cdot\mathrm{m/s^2}$
(b) 0.975 $\mathrm{kg}\cdot\mathrm{m/s^2}$ radially inward
(c) 0.975 N, radially inward
(d) The saddle of the wooden horse

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. P37b (p. 210 and A-2): Answer to part (b) is 871 kg $\cdot$ m/s/s.

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

Ch. 5 prob. P51 (p.212 and A-2): Speed is 1.64 m/s, period is 1.99 s.

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. P13 (p. 225 and A-2): (d) is 3.

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

Ch. 6 prob. P54 (p. 279 and A-2): (a) 9.8 J, 6.3 m/s; (b) 19.6 J, 6.3 m/s

Ch. 6 problems P52, P58, P59, P60, and P61 are in section 6.8 of the problems but belong in section 6.10 because there are references to escape speed. Also, P66 refers to making a graph and belongs in section 6.10.

Ch. 6 prob. P64 (p. 280): If you use the approximate value G = 6.7e-11 the orbit is indeed a bound state (total energy is negative), but if you use a more accurate value for G the total energy is positive, and the orbit is not bound (and not an ellipse).

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 p. 301, next to last paragraph: The last part of the sentence should say, “the energy of an open system need not be constant”. (The energy of a closed system will necessarily stay constant.)

Ch. 7 p. 313: To correctly detect when the oscillation of a spring-mass system subject to sliding friction should stop, it isn’t adequate to watch for the velocity to become very small, because that also occurs at the end points of the oscillation. The correct criterion for stopping is that the velocity is very small AND the spring force is less than the sliding friction force.

Ch. 7 prob. P25 (p. 318): (a) K = 1.1 eV, U = -1.3 eV, K+U = -0.2 eV.
(b) 0.2 eV. (c) $6a/r^{7}$ .

Ch. 7 prob.P29 (p. 318): F = 172 N; internal energy change = -516 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. 7 prob. P30 is in section 7.5 of the problems but belongs in section 7.4, since it does not involve Q.

Ch. 8 prob. P9 (p. 344): Answer is 0.9 eV.

Ch. 9 prob. P13 (p. 377): This problem belongs in section 9.2 because it involves moment of inertia. The answer is 24.6 J.

Ch. 9 prob. P43, Figure 9.63: The answer to part (b) is $\sqrt{2Fw/(MR^{2}+4mb^{2})}$

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. 411-412: Problems P18 and P20 involve 1D collisions and so belong with problems from earlier sections of Ch. 10.

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. 11 prob. P21 (p. 412): The answer to part (b) is < 2.82, 0, 17.6 > m/s.

Ch. 11 prob. P35 (p. 462): (a) +0.308 $\mathrm{kg}\cdot \mathrm{m}/ \mathrm{s}^2$ ; (b) +0.308 $\mathrm{kg}\cdot \mathrm{m}/ \mathrm{s}^2$ ; (c) +0.0717 rad/s

Ch. 11 prob. P59 (p. 466): -2.33 rad/s

Ch. 11 prob. P60 (p. 466): Delete the symbols (a) and (b).

Ch. 12 p. 489, 2nd paragraph: The specific heat of iron is 0.45 J/K/g.

Ch. 12 prob. P25 (p. 508): (c) $1.23\times10^{-22}$ J/K

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 P33 (p. 541): The charge is negative, so the field is <225,0,0> N/C.

Ch. 13 P37 (p. 542): The start should read “At location <0, 0, 0> in a room…”

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

Ch 13 P61 (p. 544): (a) left end.

Ch. 13 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. 14 P47 (p. 583): The correct answer is c.

Ch. 14 P57 (p. 584): The answer to part (b) is $1.82\times 10^{-2}$ N/C.

Ch. 15 p. 604: It is not clearly stated that here the analysis is confined to the field between the plates; the analysis of the field outside the plates is carried out on p. 605. Also, in the calculation on p. 605 of $E_1$ , the sign of $z$ should be reversed.

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. 15 P39 (p. 620): The answer to part (a) in the back of the book is backwards; the left side of the metal foil is charged positive, and the right side is charged negative.

Ch 15 P51 (p. 621): At r = 3 cm the field is zero.

Ch. 15 P62 (p. 624): in part b the units of polarizability are C·m/(N/C).

Ch. 15 P65 (p. 624), second line: Missing charge units (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 P49 (p. 666): (a) Change $q^{2}$ to q; (c) $E_x = \dfrac{1}{4\pi\epsilon_0}\dfrac{2qs}{r^3}$

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. Also, due to changes in VPython, the calculations of dl and loc should be written like this:

dl = wire.point(i+1).pos – wire.point(i).pos
loc = wire.point(i).pos + dl/2

Ch. 17 prob. P23b (p. 709): Answer to part b for magnetic field is 0.166 T.

Ch. 17 prob. P25 (p. 709): (a) $7.5\times 10^{19}$ electrons/s. (b) To the left. (c) $3.15\times 10^{-5}$ m/s

Ch. 17 prob. P28 (p. 709): “long bulb” should be “light bulb”.

Ch. 17 prob. P37 (p. 710): (b) $\vec{r} = <-w,y,0> - <x,0,0>$
$\vec{B} = \dfrac{\mu_0}{4\pi}I\int\limits_{0}^{d} \dfrac{<1,o,o>\times\hat{r}}{r^2}dx$
(c) +z.

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. 18 prob. P33(p. 758): The answer to the second question is $6\times 10^{18}$ electrons/s.

Ch. 19 (p. 766): In the second paragraph of the subsection “Discharging a Capacitor” the second sentence should read “We’ll think about it step by step, iteratively, choosing a time step that is short compared to the time required to discharge the capacitor (several seconds), but long compared to the time it takes surface charge to arrange itself on the wire and battery (nanoseconds).”

Ch. 19 prob. P63 (p. 800): (b) 10.4 degrees; (d) $R_2$ and $R_3$ have the same length; (g) $2.4\times10^{-19}$ J.

Ch. 19 prob. P77d (p. 802): The answer is 3.78 V.

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. P33 (p. 856): The answer to part (a) is that the direction is -z (into the page) and the deflection angle is 4.49 degrees; in part (b) the force is < 0, -8e-18, 0 > N.

Ch. 20 p. 860, Fig. 20.118: The upper voltmeter reading should be -0.25 V.

Ch. 20 prob. P71 (p. 863): The units are N $\cdot$ m.

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

Ch. 21 first paragraph on p. 885: The first sentence should read like this: If $\Delta \vec{l}_1$ and $\Delta \vec{l}_2$ are short enough that the angle between the two radial lines is small, then the length $\Delta l_\parallel = \Delta l \cos{\theta}$ gets larger proportional to $r$ , but $B$ decreases proportional to $1/r$ , so the magnitude of the product $B\Delta l_\parallel$ is the same at both locations.