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Unformatted text preview: Chapter 1 – Student Solutions Manual
3. Using the given conversion factors, we find
(a) the distance d in rods to be
d = 4.0 furlongs = ( 4.0 furlongs )( 201.168 m furlong )
5.0292 m rod = 160 rods, (b) and that distance in chains to be
d = ( 4.0 furlongs )( 201.168 m furlong )
20.117 m chain = 40 chains. 5. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as
R = ( 6.37 × 106 m )(10 −3 km m ) = 6.37 × 103 km, its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km.
(b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2 (c) The volume of Earth is V = 4π 3 4π
R =
6.37 × 103 km
3
3 ( ) 3 = 1.08 × 1012 km3 . 17. None of the clocks advance by exactly 24 h in a 24h period but this is not the most
important criterion for judging their quality for measuring time intervals. What is
important is that the clock advance by the same amount in each 24h period. The clock
reading can then easily be adjusted to give the correct interval. If the clock reading jumps
around from one 24h period to another, it cannot be corrected since it would impossible
to tell what the correction should be. The following gives the corrections (in seconds) that
must be applied to the reading on each clock for each 24h period. The entries were
determined by subtracting the clock reading at the end of the interval from the clock
reading at the beginning.
CLOCK
A
B
C
D Sun.
Mon.
−16
−3
−58
+67 Mon.
Tues.
−16
+5
−58
+67 Tues.
Wed.
−15
−10
−58
+67 Wed.
Thurs.
−17
+5
−58
+67 Thurs.
Fri.
−15
+6
−58
+67 Fri.
Sat.
−15
−7
−58
+67 E +70 +55 +2 +20 +10 +10 Clocks C and D are both good timekeepers in the sense that each is consistent in its daily
drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and
predictable corrections. The correction for clock C is less than the correction for clock D,
so we judge clock C to be the best and clock D to be the next best. The correction that
must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range
from 5 s to +10 s, for clock E it is in the range from 70 s to 2 s. After C and D, A has
the smallest range of correction, B has the next smallest range, and E has the greatest
range. From best to worst, the ranking of the clocks is C, D, A, B, E.
21. We introduce the notion of density: ρ= m
V and convert to SI units: 1 g = 1 × 10−3 kg.
(a) For volume conversion, we find 1 cm3 = (1 × 10−2m)3 = 1 × 10−6m3. Thus, the density
in kg/m3 is
−3
3
⎛ 1 g ⎞ ⎛ 10 kg ⎞ ⎛ cm ⎞
3
3
1 g cm = ⎜ 3 ⎟ ⎜
⎟ ⎜ −6 3 ⎟ = 1 × 10 kg m .
⎝ cm ⎠ ⎝ g ⎠ ⎝ 10 m ⎠
3 Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide the mass of the water by the time taken to drain it. The mass is found from
M = ρV (the product of the volume of water and its density): ( M = 5700 m 3 ) (1 × 10 3 ) kg m 3 = 5.70 × 106 kg. The time is t = (10h)(3600 s/h) = 3.6 × 104 s, so the mass flow rate R is
M 5.70 × 106 kg
R=
=
= 158 kg s.
t
3.6 × 104 s 35. (a) Dividing 750 miles by the expected “40 miles per gallon” leads the tourist to
believe that the car should need 18.8 gallons (in the U.S.) for the trip.
(b) Dividing the two numbers given (to high precision) in the problem (and rounding off)
gives the conversion between U.K. and U.S. gallons. The U.K. gallon is larger than the
U.S gallon by a factor of 1.2. Applying this to the result of part (a), we find the answer
for part (b) is 22.5 gallons. 39. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 =
0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8 ×
4 × 4 = 128 ft3, which we convert (multiplying by 0.30483) to 3.6 m3. Therefore, one
cubic meter of wood corresponds to 1/3.6 ≈ 0.3 cord.
41. (a) The difference between the total amounts in “freight” and “displacement” tons,
(8 − 7)(73) = 73 barrels bulk, represents the extra M&M’s that are shipped. Using the
conversions in the problem, this is equivalent to (73)(0.1415)(28.378) = 293 U.S.
bushels.
(b) The difference between the total amounts in “register” and “displacement” tons,
(20 − 7)(73) = 949 barrels bulk, represents the extra M&M’s are shipped. Using the
conversions in the problem, this is equivalent to (949)(0.1415)(28.378) = 3.81 × 103 U.S.
bushels.
45. We convert meters to astronomical units, and seconds to minutes, using 1000 m = 1 km
1 AU = 1.50 × 108 km
60 s = 1 min .
Thus, 3.0 × 108 m/s becomes ⎛ 3.0 × 108 m ⎞ ⎛ 1 km ⎞ ⎛
⎞ ⎛ 60 s ⎞
AU
⎟⎜
⎟⎜
⎟⎜
⎜
⎟
⎟⎜
⎟⎜
⎜
⎟ ⎜1000 m ⎟ ⎜1.50 × 108 km ⎟ ⎜ min ⎟ = 0.12 AU min .
⎟⎝
⎟
⎜
⎟
⎟
⎟⎝
⎠
s
⎠⎝
⎠⎜
⎝
⎠
57. (a) When θ is measured in radians, it is equal to the arc length s divided by the radius
R. For a very large radius circle and small value of θ, such as we deal with in Fig. 1–9,
the arc may be approximated as the straight linesegment of length 1 AU. First, we
convert θ = 1 arcsecond to radians: ⎛ 1 arcminute ⎞ ⎛
⎞ ⎛ 2π radian ⎞
1°
⎟⎜
⎟⎜
⎟
⎝ 60 arcsecond ⎠ ⎝ 60 arcminute ⎠ ⎝ 360° ⎠ (1 arcsecond ) ⎜ which yields θ = 4.85 × 10−6 rad. Therefore, one parsec is
Ro = s θ = 1 AU
= 2.06 × 10 5 AU.
−6
4.85 × 10 Now we use this to convert R = 1 AU to parsecs: b R = 1 AU 1 pc
g FGH 2.06 × 10 5 IJ = 4.9 × 10
AU K −6 pc. (b) Also, since it is straightforward to figure the number of seconds in a year (about 3.16
× 107 s), and (for constant speeds) distance = speed × time, we have a fc h 1 ly = 186, 000 mi s 316 × 10 7 s 5.9 × 1012 mi
. which we convert to AU by dividing by 92.6 × 106 (given in the problem statement),
obtaining 6.3 × 104 AU. Inverting, the result is 1 AU = 1/6.3 × 104 = 1.6 × 10−5 ly. Chapter 2 – Student Solutions Manual
1. We use Eq. 22 and Eq. 23. During a time tc when the velocity remains a positive
constant, speed is equivalent to velocity, and distance is equivalent to displacement, with
Δx = v tc.
(a) During the first part of the motion, the displacement is Δx1 = 40 km and the time
interval is
t1 = (40 km)
= 133 h.
.
(30 km / h) During the second part the displacement is Δx2 = 40 km and the time interval is
t2 = (40 km)
= 0.67 h.
(60 km / h) Both displacements are in the same direction, so the total displacement is
Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km.
The total time for the trip is t = t1 + t2 = 2.00 h. Consequently, the average velocity is
vavg = (80 km)
= 40 km / h.
(2.0 h) (b) In this example, the numerical result for the average speed is the same as the average
velocity 40 km/h.
(c) As shown below, the graph consists of two contiguous line segments, the first having
a slope of 30 km/h and connecting the origin to (t1, x1) = (1.33 h, 40 km) and the second
having a slope of 60 km/h and connecting (t1, x1) to (t, x) = (2.00 h, 80 km). From the
graphical point of view, the slope of the dashed line drawn from the origin to (t, x)
represents the average velocity. 5. Using x = 3t – 4t2 + t3 with SI units understood is efficient (and is the approach we will
use), but if we wished to make the units explicit we would write
x = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3.
We will quote our answers to one or two significant figures, and not try to follow the
significant figure rules rigorously.
(a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0.
(b) With t = 2 s we get x = 3(2) – 4(2)2+(2)3 = –2 m.
(c) With t = 3 s we have x = 0 m.
(d) Plugging in t = 4 s gives x = 12 m.
For later reference, we also note that the position at t = 0 is x = 0.
(e) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement
Δx = 12 m.
(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the
displacement Δx = 14 m. Eq. 22, then, leads to
vavg = Δx 14 m
=
= 7 m/s.
2s
Δt (g) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood.
Not shown is a straight line drawn from the point at (t, x) = (2, –2) to the highest point
shown (at t = 4 s) which would represent the answer for part (f). 19. We represent its initial direction of motion as the +x direction, so that v0 = +18 m/s
and v = –30 m/s (when t = 2.4 s). Using Eq. 27 (or Eq. 211, suitably interpreted) we
find aavg = (−30 m/s) − (+1 m/s)
= − 20 m/s 2
2.4 s which indicates that the average acceleration has magnitude 20 m/s2 and is in the opposite
direction to the particle’s initial velocity.
25. The constant acceleration stated in the problem permits the use of the equations in
Table 21.
(a) We solve v = v0 + at for the time:
t= v − v0
=
a 1
10 (3.0 × 10 8 m / s)
= 31 × 10 6 s
.
9.8 m / s 2 which is equivalent to 1.2 months.
(b) We evaluate x = x0 + v0 t + 1 at 2 , with x0 = 0. The result is
2
x= 1
( 9.8 m/s2 ) (3.1×106 s)2 = 4.6 ×1013 m .
2 27. Assuming constant acceleration permits the use of the equations in Table 21. We
2
solve v 2 = v0 + 2a ( x − x0 ) with x0 = 0 and x = 0.010 m. Thus, a= 2
v 2 − v0 (5.7 ×105 m/s) 2 − (1.5 ×105 m/s) 2
=
= 1.62 × 1015 m/s 2 .
2x
2(0.010 m) 33. The problem statement (see part (a)) indicates that a = constant, which allows us to
use Table 21.
1
(a) We take x0 = 0, and solve x = v0t + 2 at2 (Eq. 215) for the acceleration: a = 2(x –
v0t)/t2. Substituting x = 24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find a= 2 ( 24.0m − ( 15.55m/s ) ( 2.00s ) 2 ( 2.00s ) ) = − 3.56m/s 2 , or  a = 3.56 m/s 2 . The negative sign indicates that the acceleration is opposite to the
direction of motion of the car. The car is slowing down.
(b) We evaluate v = v0 + at as follows: c v = 1555 m / s − 3.56 m / s2
. h b2.00 sg = 8.43 m / s which can also be converted to 30.3 km/h.
45. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as
the –y direction) for the duration of the fall. This is constant acceleration motion, which
justifies the use of Table 21 (with Δy replacing Δx).
2
(a) Starting the clock at the moment the wrench is dropped (v0 = 0), then v 2 = vo − 2 g Δy
leads to
(−24 m/s) 2
Δy = −
= −29.4 m
2(9.8 m/s 2 ) so that it fell through a height of 29.4 m.
(b) Solving v = v0 – gt for time, we find:
t= v0 − v 0 − ( −24 m/s)
=
= 2.45 s.
g
9.8 m/s 2 (c) SI units are used in the graphs, and the initial position is taken as the coordinate
origin. In the interest of saving space, we do not show the acceleration graph, which is a
horizontal line at –9.8 m/s2. 47. We neglect air resistance for the duration of the motion (between “launching” and
“landing”), so a = –g = –9.8 m/s2 (we take downward to be the –y direction). We use the
equations in Table 21 (with Δy replacing Δx) because this is a = constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking y0 = 0, we set v = 0 in
2
v 2 = v0 − 2 gy , and solve for the initial velocity: v0 = 2 gy . Since y = 50 m we find v0 =
31 m/s. (b) It will be in the air from the time it leaves the ground until the time it returns to the
ground (y = 0). Applying Eq. 215 to the entire motion (the rise and the fall, of total time t
> 0) we have
1
2v
y = v0t − gt 2 ⇒ t = 0
g
2
which (using our result from part (a)) produces t = 6.4 s. It is possible to obtain this
without using part (a)’s result; one can find the time just for the rise (from ground to
highest point) from Eq. 216 and then double it.
(c) SI units are understood in the x and v graphs shown. In the interest of saving space,
we do not show the graph of a, which is a horizontal line at –9.8 m/s2. 49. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as
the –y direction) for the duration of the motion. We are allowed to use Table 21 (with Δy
replacing Δx) because this is constant acceleration motion. We are placing the coordinate
origin on the ground. We note that the initial velocity of the package is the same as the
velocity of the balloon, v0 = +12 m/s and that its initial coordinate is y0 = +80 m.
1
(a) We solve y = y0 + v0t − 2 gt 2 for time, with y = 0, using the quadratic formula
(choosing the positive root to yield a positive value for t). t= b gb g 2
2
v0 + v0 + 2 gy0 12 + 12 + 2 9.8 80
=
= 5.4 s
g
9.8 (b) If we wish to avoid using the result from part (a), we could use Eq. 216, but if that is
not a concern, then a variety of formulas from Table 21 can be used. For instance, Eq. 211 leads to
v = v0 – gt = 12 – (9.8)(5.4) = – 41 m/s.
Its final speed is 41 m/s. 51. The speed of the boat is constant, given by vb = d/t. Here, d is the distance of the boat
from the bridge when the key is dropped (12 m) and t is the time the key takes in falling.
To calculate t, we put the origin of the coordinate system at the point where the key is
dropped and take the y axis to be positive in the downward direction. Taking the time to
be zero at the instant the key is dropped, we compute the time t when y = 45 m. Since the
initial velocity of the key is zero, the coordinate of the key is given by y = 1 gt 2 . Thus
2
2y
=
g t= 2(45 m)
= 3.03 s .
9.8 m / s2 Therefore, the speed of the boat is
vb = 12 m
= 4.0 m / s .
3.03 s 55. (a) We first find the velocity of the ball just before it hits the ground. During contact
with the ground its average acceleration is given by
aavg = Δv
Δt where Δv is the change in its velocity during contact with the ground and
Δt = 20.0 ×10 −3 s is the duration of contact. Now, to find the velocity just before contact,
we put the origin at the point where the ball is dropped (and take +y upward) and take t =
0 to be when it is dropped. The ball strikes the ground at y = –15.0 m. Its velocity there is
found from Eq. 216: v2 = –2gy. Therefore, v = − −2 gy = − −2(9.8)( −15.0) = −17.1 m / s
where the negative sign is chosen since the ball is traveling downward at the moment of
contact. Consequently, the average acceleration during contact with the ground is
aavg = 0 − ( −17.1)
= 857 m / s2 .
−3
20.0 × 10 89. Integrating (from t = 2 s to variable t = 4 s) the acceleration to get the velocity (and
using the velocity datum mentioned in the problem, leads to
1 v = 17 + 2 (5)(42 – 22) = 47 m/s.
91. We take +x in the direction of motion, so b v = 60 km / h g FGH 1000 ms//kmIJK = + 16.7 m / s
3600 h and a > 0. The location where it starts from rest (v0 = 0) is taken to be x0 = 0.
(a) Eq. 27 gives aavg = (v – v0)/t where t = 5.4 s and the velocities are given above. Thus,
aavg = 3.1 m/s2.
(b) The assumption that a = constant permits the use of Table 21. From that list, we
choose Eq. 217:
x= 1
1
( v0 + v ) t = (16.7 m/s )( 5.4 s ) = 45 m.
2
2 (c) We use Eq. 215, now with x = 250 m:
x= 2 ( 250 m )
1 2
2x
at ⇒ t =
=
a
2
3.1 m/s 2 which yields t = 13 s.
97. The (ideal) driving time before the change was t = Δx/v, and after the change it is t' =
Δx/v'. The time saved by the change is therefore
t − t ' = Δx FG 1 − 1 IJ = Δx FG 1 − 1 IJ = Δxb0.0028 h / mig
H v v'K H 55 65K which becomes, converting Δx = 700/1.61 = 435 mi (using a conversion found on the
inside front cover of the textbook), t – t' = (435)(0.0028) = 1.2 h. This is equivalent to 1 h
and 13 min.
99. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as
the –y direction) for the duration of the motion. We are allowed to use Table 21 (with Δy
replacing Δx) because this is constant acceleration motion. When something is thrown
straight up and is caught at the level it was thrown from (with a trajectory similar to that
shown in Fig. 231), the time of flight t is half of its time of ascent ta, which is given by
Eq. 218 with Δy = H and v = 0 (indicating the maximum point).
H = vt a + 1 2
gt a
2 ⇒ ta = 2H
g Writing these in terms of the total time in the air t = 2ta we have H= 1 2
gt
8 ⇒ 2H
.
g t=2 We consider two throws, one to height H1 for total time t1 and another to height H2 for
total time t2, and we set up a ratio: FG IJ
H K 2
1
H2 8 gt 2
t
= 1 2 = 2
H1 8 gt1
t1 2 from which we conclude that if t2 = 2t1 (as is required by the problem) then H2 = 22H1 =
4H1.
107. (a) The wording of the problem makes it clear that the equations of Table 21 apply,
the challenge being that v0, v, and a are not explicitly given. We can, however, apply x –
1
x0 = v0t + 2at2 to a variety of points on the graph and solve for the unknowns from the
simultaneous equations. For instance,
1
16 – 0 = v0(2.0) + 2 a(2.0)2
1
27 – 0 = v0(3.0) + 2 a(3.0)2
lead to the values v0 = 6.0 m/s and a = 2.0 m/s2.
(b) From Table 21,
1
1
x – x0 = vt – 2at2 ⇒ 27 – 0 = v(3.0) – 2 (2.0)(3.0)2
which leads to v = 12 m/s.
1 (c) Assuming the wind continues during 3.0 ≤ t ≤ 6.0, we apply x – x0 = v0t + 2at2 to this
interval (where v0 = 12.0 m/s from part (b)) to obtain
1
Δx = (12.0)(3.0) + 2 (2.0)(3.0)2 = 45 m . Chapter 3 – Student Solutions Manual
1. A vector a can be represented in the magnitudeangle notation (a, θ), where
2
2
a = ax + a y is the magnitude and ⎛ ay ⎞
⎟
⎝ ax ⎠ θ = tan − 1 ⎜
is the angle a makes with the positive x axis. (a) Given Ax = −25.0 m and Ay = 40.0 m, A = (− 25.0 m) 2 + (40.0 m) 2 = 47.2 m
(b) Recalling that tan θ = tan (θ + 180°), tan–1 [40/ (– 25)] = – 58° or 122°. Noting that
the vector is in the third quadrant (by the signs of its x and y components) we see that
122° is the correct answer. The graphical calculator “shortcuts” mentioned above are
designed to correctly choose the right possibility.
3. The x and the y components of a vector a lying on the xy plane are given by
ax = a cos θ , a y = a sin θ where a = a  is the magnitude and θ is the angle between a and the positive x axis.
(a) The x component of a is given by ax = 7.3 cos 250° = – 2.5 m.
(b) and the y component is given by ay = 7.3 sin 250° = – 6.9 m.
In considering the variety of ways to compute these, we note that the vector is 70° below
the – x axis, so the components could also have been found from ax = – 7.3 cos 70° and
ay = – 7.3 sin 70°. In a similar vein, we note that the vector is 20° to the left from the – y
axis, so one could use ax = – 7.3 sin 20° and ay = – 7.3 cos 20° to achieve the same
results.
7. The length unit meter is understood throughout the calculation.
(a) We compute the distance from one corner to the diametrically opposite corner: d = 3.002 + 3.702 + 4.302 = 6.42 . (b) The displacement vector is along the straight line from the beginning to the end point
of the trip. Since a straight line is the shortest distance between two points, the length of
the path cannot be less than the magnitude of the displacement.
(c) It can be greater, however. The fly might, for example, crawl along the edges of the
room. Its displacement would be the same but the path length would be
+ w + h = 11.0 m.
(d) The path length is the same as the magnitude of the displacement if the fly flies along
the displacement vector.
(e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to
be upward. Then the x component of the displacement is w = 3.70, the y component of
the displacement is 4.30, and the z component is 3.00. Thus d = 3.70 i + 4.30 j + 3.00 k .
An equally correct answer is gotten by interchanging the length, width, and height. (f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram.
Pretend there is a hinge where the front wall of the room joins the floor and lay the wall
down as shown on the lower diagram. The shortest walking distance between the lower
left back of the room and the upper right front corner is the dotted straight line shown on
the diagram. Its length is Lmin = (w + h) 2 + 2 = ( 3.70 + 3.00 ) 2 + 4.302 = 7.96 m . 9. We write r = a + b . When not explicitly displayed, the units here are assumed to be
meters. (a) The x and the y components of r are rx = ax + bx = (4.0 m) – (13 m) = –9.0 m and ry =
ay + by = (3.0 m) + (7.0 m) = 10 m, respectively. Thus r = ( −9.0 m) ˆ + (10 m) ˆ .
i
j
(b) The magnitude of r is
r = r = rx2 + ry2 = (−9.0 m) 2 + (10 m) 2 = 13 m .
(c) The angle between the resultant and the +x axis is given by θ = tan–1(ry/rx) = tan–1 [(10 m)/( –9.0 m)] = – 48° or 132°.
Since the x component of the resultant is negative and the y component is positive,
characteristic of the second quadrant, we find the angle is 132° (measured
counterclockwise from +x axis).
17. It should be mentioned that an efficient way to work this vector addition problem is
with the cosine law for general triangles (and since a , b and r form an isosceles triangle,
the angles are easy to figure). However, in the interest of reinforcing the usual systematic
approach to vector addition, we note that the angle b makes with the +x axis is 30°
+105° = 135° and apply Eq. 35 and Eq. 36 where appropriate.
(a) The x component of r is rx = (10.0 m) cos 30° + (10.0 m) cos 135° = 1.59 m.
(b) The y component of r is ry = (10.0 m) sin 30° + (10.0 m) sin 135° = 12.1 m.
(c) The magnitude of r is r =  r  = (1.59 m) 2 + (12.1 m) 2 = 12.2 m.
(d) The angle between r and the +x direction is tan–1[(12.1 m)/(1.59 m)] = 82.5°.
39. Since ab cos φ = axbx + ayby + azbz,
cos φ = a x bx + a y by + a zbz
ab . The magnitudes of the vectors given in the problem are a = a  = (3.00) 2 + (3.00) 2 + (3.00) 2 = 5.20
b =  b  = (2.00) 2 + (1.00) 2 + (3.00) 2 = 3.74.
The angle between them is found from cos φ = (3.00) (2.00) + (3.00) (1.00) + (3.00) (3.00)
= 0.926.
(5.20) (3.74) The angle is φ = 22°.
43. From the figure, we note that c ⊥ b , which implies that the angle between c and the
+x axis is 120°. Direct application of Eq. 35 yields the answers for this and the next few
parts.
(a) ax = a cos 0° = a = 3.00 m.
(b) ay = a sin 0° = 0.
(c) bx = b cos 30° = (4.00 m) cos 30° = 3.46 m.
(d) by = b sin 30° = (4.00 m) sin 30° = 2.00 m.
(e) cx = c cos 120° = (10.0 m) cos 120° = –5.00 m.
(f) cy = c sin 30° = (10.0 m) sin 120° = 8.66 m.
(g) In terms of components (first x and then y), we must have
−5.00 m = p (3.00 m) + q (3.46 m)
8.66 m = p (0) + q (2.00 m). Solving these equations, we find p = –6.67.
(h) And q = 4.33 (note that it’s easiest to solve for q first). The numbers p and q have no
units.
47. We apply Eq. 320 and Eq. 327.
(a) The scalar (dot) product of the two vectors is
a ⋅ b = ab cos φ = (10) (6.0) cos 60° = 30. (b) The magnitude of the vector (cross) product of the two vectors is
 a × b  = ab sin φ = (10) (6.0) sin 60° = 52.
→ → 51. Let A represent the first part of his actual voyage (50.0 km east) and C represent
→ → → the intended voyage (90.0 km north). We are looking for a vector B such that A + B
→ = C. (a) The Pythagorean theorem yields B = (50.0) 2 + (90.0) 2 = 103 km.
(b) The direction is tan − 1 (50.0 / 90.0) = 29.1 ° west of north (which is equivalent to
60.9° north of due west).
→ → → → 71. Given: A + B =6.0 ^ + 1.0 ^ and A – B = – 4.0 ^ + 7.0 ^ . Solving these
i
j
i
j
→ simultaneously leads to A =1.0 ^ + 4.0 ^. The Pythagorean theorem then leads to
i
j
A = (1.0) 2 + (4.0) 2 = 4.1 . Chapter 4 – Student Solutions Manual
11. We apply Eq. 410 and Eq. 416.
(a) Taking the derivative of the position vector with respect to time, we have, in SI units
(m/s),
v= d ˆ
ˆ
(i + 4t 2 ˆ + t k) = 8t ˆ + k .
j
j ˆ
dt (b) Taking another derivative with respect to time leads to, in SI units (m/s2),
a= d
(8t ˆ + k) = 8 ˆ .
j ˆ
j
dt 17. Constant acceleration in both directions (x and y) allows us to use Table 21 for the
motion along each direction. This can be handled individually (for Δx and Δy) or together
with the unitvector notation (for Δr). Where units are not shown, SI units are to be
understood.
(a) The velocity of the particle at any time t is given by v = v0 + at , where v0 is the
initial velocity and a is the (constant) acceleration. The x component is vx = v0x + axt =
3.00 – 1.00t, and the y component is vy = v0y + ayt = –0.500t since v0y = 0. When the
particle reaches its maximum x coordinate at t = tm, we must have vx = 0. Therefore, 3.00
– 1.00tm = 0 or tm = 3.00 s. The y component of the velocity at this time is
vy = 0 – 0.500(3.00) = –1.50 m/s;
this is the only nonzero component of v at tm.
(b) Since it started at the origin, the coordinates of the particle at any time t are given by
1
r = v 0 t + 2 at 2 . At t = tm this becomes ( ) ( ) 1
2
ˆ
r = 3.00i ( 3.00 ) + −1.00 ˆ − 0.50 ˆ ( 3.00 ) = (4.50 ˆ − 2.25 ˆ m.
i
j
i
j)
2
29. The initial velocity has no vertical component — only an x component equal to +2.00
m/s. Also, y0 = +10.0 m if the water surface is established as y = 0. (a) x – x0 = vxt readily yields x – x0 = 1.60 m.
1
(b) Using y − y0 = v0 y t − 2 gt 2 , we obtain y = 6.86 m when t = 0.800 s and v0y=0. 1
(c) Using the fact that y = 0 and y0 = 10.0, the equation y − y0 = v0 y t − 2 gt 2 leads to t = 2(10.0 m) / 9.80 m/s 2 = 1.43 s . During this time, the xdisplacement of the diver
is x – x0 = (2.00 m/s)(1.43 s) = 2.86 m.
31. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable. The coordinate origin is at ground level directly below
the release point. We write θ0 = –37.0° for the angle measured from +x, since the angle
given in the problem is measured from the –y direction. We note that the initial speed of
the projectile is the plane’s speed at the moment of release.
(a) We use Eq. 422 to find v0:
y − y0 = (v0 sin θ 0 ) t − 1 2
1
gt ⇒ 0 − 730 m = v0 sin(−37.0°)(5.00 s) − (9.80 m/s 2 )(5.00 s) 2
2
2 which yields v0 = 202 m/s.
(b) The horizontal distance traveled is x = v0tcos θ0 = (202 m/s)(5.00 s)cos(–37.0°) = 806 m.
(c) The x component of the velocity (just before impact) is
vx = v0cosθ0 = (202 m/s)cos(–37.0°) = 161 m/s.
(d) The y component of the velocity (just before impact) is
vy = v0 sin θ0 – gt = (202 m/s) sin (–37.0°) – (9.80 m/s2)(5.00 s) = –171 m/s.
39. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable. The coordinate origin is at the end of the rifle (the initial
point for the bullet as it begins projectile motion in the sense of § 45), and we let θ0 be
the firing angle. If the target is a distance d away, then its coordinates are x = d, y = 0.
The projectile motion equations lead to d = v0t cos θ0 and 0 = v0 t sin θ 0 − 1 gt 2 .
2
2
1
Eliminating t leads to 2v0 sin θ 0 cos θ 0 − gd = 0 . Using sin θ 0 cos θ 0 = 2 sin 2θ 0 , we
obtain b g 2
v0 sin (2θ 0 ) = gd ⇒ sin(2θ 0 ) = gd (9.80 m/s 2 )(45.7 m)
=
2
v0
(460 m/s) 2 which yields sin(2θ 0 ) = 2.11× 10−3 and consequently θ0 = 0.0606°. If the gun is aimed at a
point a distance above the target, then tan θ 0 = d so that
= d tan θ 0 = (45.7 m) tan(0.0606°) = 0.0484 m = 4.84 cm. 47. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable. The coordinate origin is at ground level directly below
impact point between bat and ball. The Hint given in the problem is important, since it
provides us with enough information to find v0 directly from Eq. 426.
(a) We want to know how high the ball is from the ground when it is at x = 97.5 m, which
requires knowing the initial velocity. Using the range information and θ0 = 45°, we use
Eq. 426 to solve for v0:
v0 = gR
=
sin 2θ 0 ( 9.8 m/s ) (107 m ) = 32.4 m/s.
2 1 Thus, Eq. 421 tells us the time it is over the fence: t= x
97.5 m
=
= 4.26 s.
v0 cos θ 0 ( 32.4 m/s ) cos 45° At this moment, the ball is at a height (above the ground) of b g y = y0 + v0 sin θ 0 t − 1 2
gt = 9.88 m
2 which implies it does indeed clear the 7.32 m high fence.
(b) At t = 4.26 s, the center of the ball is 9.88 m – 7.32 m = 2.56 m above the fence.
51. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable. The coordinate origin is at the point where the ball is
kicked. We use x and y to denote the coordinates of ball at the goalpost, and try to find
the kicking angle(s) θ0 so that y = 3.44 m when x = 50 m. Writing the kinematic
equations for projectile motion:
x = v0 cos θ 0 , y = v0t sin θ 0 − 1
2 gt 2 , we see the first equation gives t = x/v0 cos θ0, and when this is substituted into the second
the result is
y = x tan θ 0 − gx 2
.
2
2v0 cos2 θ 0 One may solve this by trial and error: systematically trying values of θ0 until you find the
two that satisfy the equation. A little manipulation, however, will give an algebraic
solution: Using the trigonometric identity 1 / cos2 θ0 = 1 + tan2 θ0, we obtain 1 gx 2
1 gx 2
tan 2 θ 0 − x tan θ 0 + y +
=0
2
2
2 v0
2 v0
which is a secondorder equation for tan θ0. To simplify writing the solution, we denote
2
2
2
c = 1 gx 2 / v0 = 1 ( 9.80 m/s 2 ) ( 50 m ) / ( 25 m/s ) = 19.6m. Then the secondorder
2
2
equation becomes c tan2 θ0 – x tan θ0 + y + c = 0. Using the quadratic formula, we obtain
its solution(s). tan θ 0 = x ± x2 − 4 ( y + c ) c
2c = 50 m ± (50 m) 2 − 4 ( 3.44 m + 19.6 m )(19.6 m )
2 (19.6 m ) . The two solutions are given by tan θ0 = 1.95 and tan θ0 = 0.605. The corresponding (firstquadrant) angles are θ0 = 63° and θ0 = 31°. Thus,
(a) The smallest elevation angle is θ0 = 31°, and
(b) The greatest elevation angle is θ0 = 63°.
If kicked at any angle between these two, the ball will travel above the cross bar on the
goalposts.
53. We denote h as the height of a step and w as the width. To hit step n, the ball must fall
a distance nh and travel horizontally a distance between (n – 1)w and nw. We take the
origin of a coordinate system to be at the point where the ball leaves the top of the
stairway, and we choose the y axis to be positive in the upward direction. The coordinates
1
of the ball at time t are given by x = v0xt and y = − 2 gt 2 (since v0y = 0). We equate y to
–nh and solve for the time to reach the level of step n:
t= 2nh
.
g The x coordinate then is
x = v0 x 2nh
2n(0.203 m)
= (1.52 m/s)
= (0.309 m) n .
g
9.8 m/s 2 The method is to try values of n until we find one for which x/w is less than n but greater
than n – 1. For n = 1, x = 0.309 m and x/w = 1.52, which is greater than n. For n = 2, x =
0.437 m and x/w = 2.15, which is also greater than n. For n = 3, x = 0.535 m and x/w =
2.64. Now, this is less than n and greater than n – 1, so the ball hits the third step. 67. To calculate the centripetal acceleration of the stone, we need to know its speed
during its circular motion (this is also its initial speed when it flies off). We use the
kinematic equations of projectile motion (discussed in §46) to find that speed. Taking
the +y direction to be upward and placing the origin at the point where the stone leaves its
circular orbit, then the coordinates of the stone during its motion as a projectile are given
1
by x = v0t and y = − 2 gt 2 (since v0y = 0). It hits the ground at x = 10 m and y = –2.0 m.
Formally solving the second equation for the time, we obtain t = −2 y / g , which we
substitute into the first equation:
v0 = x − b g g
= 10 m
2y − 9.8 m / s2
= 15.7 m / s.
2 −2.0 m b g Therefore, the magnitude of the centripetal acceleration is b g 15.7 m / s
v2
a=
=
15 m
.
r 2 = 160 m / s2 . 75. Relative to the car the velocity of the snowflakes has a vertical component of 8.0 m/s
and a horizontal component of 50 km/h = 13.9 m/s. The angle θ from the vertical is found
from
tan θ = vh 13.9 m/s
=
= 1.74
vv
8.0 m/s which yields θ = 60°.
77. Since the raindrops fall vertically relative to the train, the horizontal component of the
velocity of a raindrop is vh = 30 m/s, the same as the speed of the train. If vv is the vertical
component of the velocity and θ is the angle between the direction of motion and the
vertical, then tan θ = vh/vv. Thus vv = vh/tan θ = (30 m/s)/tan 70° = 10.9 m/s. The speed of
a raindrop is
2
v = vh + vv2 = ( 30 m / s) 2 + (10.9 m / s) 2 = 32 m / s . 91. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable.
(a) With the origin at the firing point, the y coordinate of the bullet is given by
1
y = − 2 gt 2 . If t is the time of flight and y = – 0.019 m indicates where the bullet hits the
target, then t= 2 ( 0.019 m )
9.8 m/s 2 = 6.2 × 10−2 s. (b) The muzzle velocity is the initial (horizontal) velocity of the bullet. Since x = 30 m is
the horizontal position of the target, we have x = v0t. Thus,
v0 = x
30 m
=
= 4.8 × 102 m/s.
t 6.3 × 10 −2 s 107. (a) Eq. 215 can be applied to the vertical (y axis) motion related to reaching the
maximum height (when t = 3.0 s and vy = 0):
1
ymax – y0 = vyt – 2gt2 .
1 With ground level chosen so y0 = 0, this equation gives the result ymax = 2 g(3.0 s)2 = 44
m.
(b) After the moment it reached maximum height, it is falling; at t = 2.5 s, it will have
fallen an amount given by Eq. 218
1
yfence – ymax = (0)(2.5 s) – 2 g(2.5 s)2
which leads to yfence = 13 m.
(c) Either the range formula, Eq. 426, can be used or one can note that after passing the
fence, it will strike the ground in 0.5 s (so that the total "falltime" equals the "risetime").
Since the horizontal component of velocity in a projectilemotion problem is constant
(neglecting air friction), we find the original xcomponent from 97.5 m = v0x(5.5 s) and
then apply it to that final 0.5 s. Thus, we find v0x = 17.7 m/s and that after the fence
Δx = (17.7 m/s)(0.5 s) = 8.9 m.
111. Since the x and y components of the acceleration are constants, we can use Table 21
for the motion along both axes. This can be handled individually (for Δx and Δy) or
together with the unitvector notation (for Δr). Where units are not shown, SI units are to
be understood.
(a) Since r0 = 0 , the position vector of the particle is (adapting Eq. 215)
r = v0t + ( ) ( ) ( ) ( ) 1 2
1
at = 8.0 ˆ t + 4.0 ˆ + 2.0 ˆ t 2 = 2.0t 2 ˆ + 8.0t + 1.0t 2 ˆ
j
i
j
i
j.
2
2 Therefore, we find when x = 29 m, by solving 2.0t2 = 29, which leads to t = 3.8 s. The y
coordinate at that time is y = (8.0 m/s)(3.8 s) + (1.0 m/s2)(3.8 s)2 = 45 m.
(b) Adapting Eq. 211, the velocity of the particle is given by
v = v0 + at . Thus, at t = 3.8 s, the velocity is ( ) v = (8.0 m/s) ˆ + (4.0 m/s 2 ) ˆ + (2.0 m/s 2 ) ˆ ( 3.8 s ) = (15.2 m/s) ˆ + ( 15.6 m/s) ˆ
j
i
j
i
j
which has a magnitude of
2
2
v = vx + v y = (15.2 m/s) 2 + (15.6 m/s) 2 = 22 m/s. 121. On the one hand, we could perform the vector addition of the displacements with a
vectorcapable calculator in polar mode ((75 ∠ 37°) + (65 ∠ − 90°) = (63 ∠ − 18°)),
but in keeping with Eq. 35 and Eq. 36 we will show the details in unitvector notation.
We use a ‘standard’ coordinate system with +x East and +y North. Lengths are in
kilometers and times are in hours.
(a) We perform the vector addition of individual displacements to find the net
displacement of the camel.
Δr1 = (75 km)cos(37°) ˆ + (75 km) sin(37°) ˆ
i
j
Δr =( − 65 km) ˆ
j
2 i
j
Δr = Δr1 + Δr2 = (60 km) ˆ − (20 km )ˆ .
If it is desired to express this in magnitudeangle notation, then this is equivalent to a
vector of length  Δr = (60 km) 2 +( − 20 km) 2 = 63 km .
(b) The direction of Δ r is θ = tan − 1[(− 20 km) /(60 km)] = − 18 ° , or 18 ° south of east.
(c) We use the result from part (a) in Eq. 48 along with the fact that Δt = 90 h. In unit
vector notation, we obtain
vavg = (60 ˆ − 20 ˆ km
i
j)
= (0.67 ˆ − 0.22 ˆ km/h.
i
j)
90 h This leads to  vavg  = 0.70 km/h. (d) The direction of vavg is θ = tan − 1[(− 0.22 km/h) /(0.67 km/h)] = − 18 ° , or 18 ° south
of east.
(e) The average speed is distinguished from the magnitude of average velocity in that it
depends on the total distance as opposed to the net displacement. Since the camel travels
140 km, we obtain (140 km)/(90 h) = 1.56 km/h ≈ 1.6 km/h .
(f) The net displacement is required to be the 90 km East from A to B. The displacement
from the resting place to B is denoted Δr3 . Thus, we must have Δr1 + Δr2 + Δr3 = (90 km) ˆ
i
ˆ
ˆ
which produces Δr3 = (30 km)i + (20 km)j in unitvector notation, or (36 ∠ 33° ) in
magnitudeangle notation. Therefore, using Eq. 48 we obtain  vavg  = 36 km
= 1.2 km/h.
(120 − 90) h (g) The direction of vavg is the same as r3 (that is, 33° north of east). Chapter 5 – Student Solutions Manual
5. We denote the two forces F1 and F2 . According to Newton’s second law,
F1 + F2 = ma , so F2 = ma − F1 . b g (a) In unit vector notation F1 = 20.0 N i and
a = − (12.0 sin 30.0° m/s 2 ) ˆ − (12.0 cos 30.0° m/s 2 ) ˆ = − ( 6.00 m/s 2 ) ˆ − (10.4m/s 2 ) ˆ
i
i
j.
j Therefore,
F2 = ( 2.00kg ) ( −6.00 m/s 2 ) ˆ + ( 2.00 kg ) ( −10.4 m/s 2 ) ˆ − ( 20.0 N ) ˆ
i
j
i
= ( −32.0 N ) ˆ − ( 20.8 N ) ˆ
i
j.
(b) The magnitude of F2 is
 F2 = F22x + F22y = (− 32.0) 2 + (− 20.8) 2 = 38.2 N.
(c) The angle that F2 makes with the positive x axis is found from
tan θ = (F2y/F2x) = [(–20.8)/(–32.0)] = 0.656.
Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y
components are negative, the correct result is 213°. An alternative answer is
213 ° − 360 ° = − 147 ° .
13. (a) – (c) In all three cases the scale is not accelerating, which means that the two
cords exert forces of equal magnitude on it. The scale reads the magnitude of either of
these forces. In each case the tension force of the cord attached to the salami must be the
same in magnitude as the weight of the salami because the salami is not accelerating.
Thus the scale reading is mg, where m is the mass of the salami. Its value is (11.0 kg) (9.8
m/s2) = 108 N.
19. (a) Since the acceleration of the block is zero, the components of the Newton’s
second law equation yield
T – mg sin θ = 0
FN – mg cos θ = 0.
Solving the first equation for the tension in the string, we find b gc h T = mg sin θ = 8.5 kg 9.8 m / s 2 sin 30° = 42 N . (b) We solve the second equation in part (a) for the normal force FN:
FN = mg cos θ = ( 8.5 kg ) ( 9.8 m/s 2 ) cos 30° = 72 N . (c) When the string is cut, it no longer exerts a force on the block and the block
accelerates. The x component of the second law becomes –mgsinθ =ma, so the
acceleration becomes
a = − g sin θ = −9.8 sin 30° = − 4.9 m/s 2 .
The negative sign indicates the acceleration is down the plane. The magnitude of the
acceleration is 4.9 m/s2.
25. In terms of magnitudes, Newton’s second law is F = ma, where F = Fnet , a =  a  ,
and m is the (always positive) mass. The magnitude of the acceleration can be found
using constant acceleration kinematics (Table 21). Solving v = v0 + at for the case where
it starts from rest, we have a = v/t (which we interpret in terms of magnitudes, making
specification of coordinate directions unnecessary). The velocity is v = (1600 km/h)
(1000 m/km)/(3600 s/h) = 444 m/s, so b F = 500 kg .
g 444. m s = 12 × 10
18 s 5 N. 29. The acceleration of the electron is vertical and for all practical purposes the only force
acting on it is the electric force. The force of gravity is negligible. We take the +x axis to
be in the direction of the initial velocity and the +y axis to be in the direction of the
electrical force, and place the origin at the initial position of the electron. Since the force
and acceleration are constant, we use the equations from Table 21: x = v0t and FG IJ
H K 1
1 F 2
y = at 2 =
t .
2
2 m The time taken by the electron to travel a distance x (= 30 mm) horizontally is t = x/v0 and
its deflection in the direction of the force is FG IJ
H K 1F x
y=
2 m v0 2 FG
H 1 4.5 × 10−16
=
2 9.11 × 10 −31 IJ FG 30 × 10 IJ
.
K H 12 × 10 K
−3
7 2 = 15 × 10−3 m .
. 35. The freebody diagram is shown next. FN is the normal force of the plane on the
block and mg is the force of gravity on the block. We take the +x direction to be down
the incline, in the direction of the acceleration, and the +y direction to be in the direction of the normal force exerted by the incline on the block. The x component of Newton’s
second law is then mg sin θ = ma; thus, the acceleration is a = g sin θ. (a) Placing the origin at the bottom of the plane, the kinematic equations (Table 21) for
2
motion along the x axis which we will use are v 2 = v0 + 2ax and v = v0 + at . The block
momentarily stops at its highest point, where v = 0; according to the second equation, this
occurs at time t = − v0 a . The position where it stops is
2
⎞
1 v0
1⎛
(−3.50 m/s) 2
⎟ = − 1.18 m ,
x= −
=− ⎜
2 a
2 ⎜ ( 9.8 m/s 2 ) sin 32.0° ⎟
⎝
⎠ or  x  = 1.18 m.
(b) The time is
t= v0
v0
−3.50m/s
=−
=−
= 0.674s.
a
g sin θ
( 9.8m/s2 ) sin 32.0° (c) That the returnspeed is identical to the initial speed is to be expected since there are
no dissipative forces in this problem. In order to prove this, one approach is to set x = 0
1
and solve x = v0 t + 2 at 2 for the total time (up and back down) t. The result is
t =− b g 2 −350 m / s
.
2 v0
2 v0
=−
=−
= 135 s .
.
a
g sin θ
9.8 m / s2 sin 32.0° c h The velocity when it returns is therefore b gb g .
.
.
v = v0 + at = v0 + gt sin θ = − 350 + 9.8 135 sin 32° = 350 m / s.
45. (a) The links are numbered from bottom to top. The forces on the bottom link are the
force of gravity mg , downward, and the force F2 on1 of link 2, upward. Take the positive
direction to be upward. Then Newton’s second law for this link is F2on1 – mg = ma. Thus, F2on1 = m(a + g) = (0.100 kg) (2.50 m/s2 + 9.80 m/s2) = 1.23 N.
(b) The forces on the second link are the force of gravity mg , downward, the force F1on 2
of link 1, downward, and the force F3on 2 of link 3, upward. According to Newton’s third
law F1on2 has the same magnitude as F2 on1 . Newton’s second law for the second link is
F3on2 – F1on2 – mg = ma, so
F3on2 = m(a + g) + F1on2 = (0.100 kg) (2.50 m/s2 + 9.80 m/s2) + 1.23 N = 2.46 N.
(c) Newton’s second for link 3 is F4on3 – F2on3 – mg = ma, so
F4on3 = m(a + g) + F2on3 = (0.100 N) (2.50 m/s2 + 9.80 m/s2) + 2.46 N = 3.69 N,
where Newton’s third law implies F2on3 = F3on2 (since these are magnitudes of the force
vectors).
(d) Newton’s second law for link 4 is F5on4 – F3on4 – mg = ma, so
F5on4 = m(a + g) + F3on4 = (0.100 kg) (2.50 m/s2 + 9.80 m/s2) + 3.69 N = 4.92 N,
where Newton’s third law implies F3on4 = F4on3.
(e) Newton’s second law for the top link is F – F4on5 – mg = ma, so
F = m(a + g) + F4on5 = (0.100 kg) (2.50 m/s2 + 9.80 m/s2) + 4.92 N = 6.15 N,
where F4on5 = F5on4 by Newton’s third law.
(f) Each link has the same mass and the same acceleration, so the same net force acts on
each of them:
Fnet = ma = (0.100 kg) (2.50 m/s2) = 0.250 N.
53. The freebody diagrams for part (a) are shown below. F is the applied force and f
is the force exerted by block 1 on block 2. We note that F is applied directly to block 1
and that block 2 exerts the force − f on block 1 (taking Newton’s third law into
account). (a) Newton’s second law for block 1 is F – f = m1a, where a is the acceleration. The
second law for block 2 is f = m2a. Since the blocks move together they have the same
acceleration and the same symbol is used in both equations. From the second equation we
obtain the expression a = f /m2, which we substitute into the first equation to get F – f =
m1f/m2. Therefore,
f = b gb g 3.2 N 12 kg
.
Fm2
=
= 11 N .
.
m1 + m2
2.3 kg + 12 kg
. (b) If F is applied to block 2 instead of block 1 (and in the opposite direction), the force
of contact between the blocks is
f = b gb g 3.2 N 2.3 kg
Fm1
=
= 2.1 N .
m1 + m2
2.3 kg + 12 kg
. (c) We note that the acceleration of the blocks is the same in the two cases. In part (a), the
force f is the only horizontal force on the block of mass m2 and in part (b) f is the only
horizontal force on the block with m1 > m2. Since f = m2a in part (a) and f = m1a in part
(b), then for the accelerations to be the same, f must be larger in part (b).
57. We take +y to be up for both the monkey and the package.
(a) The force the monkey pulls downward on the rope has magnitude F. According to
Newton’s third law, the rope pulls upward on the monkey with a force of the same
magnitude, so Newton’s second law for forces acting on the monkey leads to
F – mmg = mmam,
where mm is the mass of the monkey and am is its acceleration. Since the rope is massless
F = T is the tension in the rope. The rope pulls upward on the package with a force of
magnitude F, so Newton’s second law for the package is
F + FN – mpg = mpap, where mp is the mass of the package, ap is its acceleration, and FN is the normal force
exerted by the ground on it. Now, if F is the minimum force required to lift the package,
then FN = 0 and ap = 0. According to the second law equation for the package, this means
F = mpg. Substituting mpg for F in the equation for the monkey, we solve for am:
2
F − mm g ( m p − mm ) g (15 kg − 10 kg ) ( 9.8 m/s )
am =
=
=
= 4.9 m/s 2 .
mm
mm
10 kg (b) As discussed, Newton’s second law leads to F − m p g = m p a p for the package and
F − mm g = mm am for the monkey. If the acceleration of the package is downward, then
the acceleration of the monkey is upward, so am = –ap. Solving the first equation for F d i b F = m p g + a p = mp g − am g and substituting this result into the second equation, we solve for am:
am (m
= p − mm ) g m p + mm = (15 kg − 10 kg ) ( 9.8 m/s 2 )
15 kg + 10 kg = 2.0 m/s 2 . (c) The result is positive, indicating that the acceleration of the monkey is upward.
(d) Solving the second law equation for the package, we obtain
F = m p ( g − am ) = (15 kg ) ( 9.8 m/s 2 − 2.0 m/s 2 ) = 120N. 61. The forces on the balloon are the force of gravity mg (down) and the force of the air
Fa (up). We take the +y to be up, and use a to mean the magnitude of the acceleration
(which is not its usual use in this chapter). When the mass is M (before the ballast is
thrown out) the acceleration is downward and Newton’s second law is
Fa – Mg = –Ma.
After the ballast is thrown out, the mass is M – m (where m is the mass of the ballast) and
the acceleration is upward. Newton’s second law leads to
Fa – (M – m)g = (M – m)a.
The previous equation gives Fa = M(g – a), and this plugs into the new equation to give b g b g b g M g − a − M − m g = M− ma ⇒ m= 2 Ma
.
g+a 73. Although the full specification of Fnet = ma in this situation involves both x and y
axes, only the xapplication is needed to find what this particular problem asks for. We
note that ay = 0 so that there is no ambiguity denoting ax simply as a. We choose +x to the
right and +y up. We also note that the x component of the rope’s tension (acting on the
crate) is
Fx = Fcosθ = 450 cos 38° = 355 N,
and the resistive force (pointing in the –x direction) has magnitude f = 125 N.
(a) Newton’s second law leads to
Fx − f = ma ⇒ a = 355 N − 125 N
= 0.74m/s 2 .
310 kg (b) In this case, we use Eq. 512 to find the mass: m = W/g = 31.6 kg. Now, Newton’s
second law leads to
Tx − f = ma ⇒ a = 355 N − 125 N
= 7.3 m/s 2 .
31.6 kg 79. The “certain force” denoted F is assumed to be the net force on the object when it
gives m1 an acceleration a1 = 12 m/s2 and when it gives m2 an acceleration a2 = 3.3 m/s2.
Thus, we substitute m1 = F/a1 and m2 = F/a2 in appropriate places during the following
manipulations.
(a) Now we seek the acceleration a of an object of mass m2 – m1 when F is the net force
on it. Thus,
F
F
aa
a =
=
= 1 2
m2 − m1 ( F / a2 ) − (F / a1 ) a1 − a2
which yields a = 4.6 m/s2.
(b) Similarly for an object of mass m2 + m1:
a= F
F
aa
=
= 1 2
m2 + m1 ( F / a2 ) + (F / a1 ) a1 + a2 which yields a = 2.6 m/s2.
91. (a) The bottom cord is only supporting m2 = 4.5 kg against gravity, so its tension is
T2= m2g = (4.5 kg)(9.8 m/s2) = 44 N. (b) The top cord is supporting a total mass of m1 + m2 = (3.5 kg + 4.5 kg) = 8.0 kg against
gravity, so the tension there is
T1= (m1 + m2)g = (8.0 kg)(9.8 m/s2) = 78 N.
(c) In the second picture, the lowest cord supports a mass of m5 = 5.5 kg against gravity
and consequently has a tension of T5 = (5.5 kg)(9.8 m/s2) = 54 N.
(d) The top cord, we are told, has tension T3 =199 N which supports a total of (199
N)/(9.80 m/s2) = 20.3 kg, 10.3 kg of which is already accounted for in the figure. Thus,
the unknown mass in the middle must be m4 = 20.3 kg – 10.3 kg = 10.0 kg, and the
tension in the cord above it must be enough to support m4 + m5 = (10.0 kg + 5.50 kg) =
15.5 kg, so T4 = (15.5 kg)(9.80 m/s2) = 152 N. Another way to analyze this is to examine
the forces on m3; one of the downward forces on it is T4.
95. The freebody diagrams is shown on the right.
Note that Fm, ry and Fm, rx , respectively, and thought
of as the y and x components of the force Fm, r
exerted by the motorcycle on the rider.
(a) Since the net force equals ma, then the
magnitude of the net force on the rider is
(60.0 kg) (3.0 m/s2) = 1.8 × 102 N.
(b) We apply Newton’s second law to the x axis:
Fm,rx − mg sin θ = ma
where m = 60.0 kg, a = 3.0 m/s2, and θ = 10°. Thus, Fm, rx = 282 N Applying it to the y
axis (where there is no acceleration), we have
Fm,ry − mg cos θ = 0 which produces Fm, ry = 579 N . Using the Pythagorean theorem, we find
2
2
Fm ,rx + Fm , ry = 644 N. Now, the magnitude of the force exerted on the rider by the motorcycle is the same
magnitude of force exerted by the rider on the motorcycle, so the answer is 6.4 × 102 N.
99. The +x axis is “uphill” for m1 = 3.0 kg and “downhill” for m2 = 2.0 kg (so they both
accelerate with the same sign). The x components of the two masses along the x axis are
given by w1x = m1 g sin θ1 and w2 x = m2 g sin θ 2 , respectively. Applying Newton’s second law, we obtain
T − m1 g sin θ1 = m1a
m2 g sin θ 2 − T = m2 a
Adding the two equations allows us to solve for the acceleration: ⎛ m sin θ 2 − m1 sin θ1 ⎞
a=⎜ 2
⎟g
m2 + m1
⎝
⎠
With θ1 = 30 ° and θ 2 = 60 ° , we have a = 0.45 m/s2. This value is plugged back into
either of the two equations to yield the tension T = 16 N.
101. We first analyze the forces on m1=1.0 kg. The +x direction is “downhill” (parallel to T ).
With the acceleration (5.5 m/s2) in the positive x direction for m1, then Newton’s second
law, applied to the x axis, becomes
T + m1 g sin β = m1 ( 5.5m/s 2 ) But for m2=2.0 kg, using the more familiar vertical y axis (with up as the positive
direction), we have the acceleration in the negative direction: F + T − m2 g = m2 ( −5.5m/s 2 ) where the tension comes in as an upward force (the cord can pull, not push).
(a) From the equation for m2, with F = 6.0 N, we find the tension T = 2.6 N.
(b) From the equation for m, using the result from part (a), we obtain the angle β = 17° . Chapter 6 – Student Solutions Manual
1. We do not consider the possibility that the bureau might tip, and treat this as a purely
horizontal motion problem (with the person’s push F in the +x direction). Applying
Newton’s second law to the x and y axes, we obtain
F − f s , max = m a
FN − mg = 0 respectively. The second equation yields the normal force FN = mg, whereupon the
maximum static friction is found to be (from Eq. 61) f s ,max = μ s mg . Thus, the first
equation becomes
F − μ s mg = ma = 0 where we have set a = 0 to be consistent with the fact that the static friction is still (just
barely) able to prevent the bureau from moving.
(a) With μ s = 0.45 and m = 45 kg, the equation above leads to F = 198 N. To bring the
bureau into a state of motion, the person should push with any force greater than this
value. Rounding to two significant figures, we can therefore say the minimum required
push is F = 2.0 × 102 N.
(b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly
F = 1.2 × 10 2 N .
3. We denote F as the horizontal force of the person exerted on the crate (in the +x
direction), f k is the force of kinetic friction (in the –x direction), FN is the vertical
normal force exerted by the floor (in the +y direction), and mg is the force of gravity.
The magnitude of the force of friction is given by fk = μkFN (Eq. 62). Applying Newton’s
second law to the x and y axes, we obtain F − f k = ma
FN − mg = 0
respectively.
(a) The second equation yields the normal force FN = mg, so that the friction is f k = μk mg = ( 0.35)( 55 kg ) (9.8 m/s 2 ) = 1.9 × 102 N .
(b) The first equation becomes
F − μ k mg = ma which (with F = 220 N) we solve to find
a= F
− μ k g = 0.56 m / s2 .
m 13. (a) The freebody diagram for the crate is shown on
the right. T is the tension force of the rope on the crate,
FN is the normal force of the floor on the crate, mg is the
force of gravity, and f is the force of friction. We take
the +x direction to be horizontal to the right and the +y
direction to be up. We assume the crate is motionless. The
equations for the x and the y components of the force
according to Newton’s second law are:
T cos θ – f = 0
T sin θ + FN − mg = 0
where θ = 15° is the angle between the rope and the horizontal. The first equation gives f
= T cos θ and the second gives FN = mg – T sin θ. If the crate is to remain at rest, f must
be less than μs FN, or T cos θ < μs (mg – T sinθ). When the tension force is sufficient to
just start the crate moving, we must have
T cos θ = μs (mg – T sin θ).
We solve for the tension: ( 0.50 ) ( 68 kg ) ( 9.8 m/s2 )
μ s mg
T=
=
= 304 N ≈ 3.0 ×102 N.
cos θ + μ s sin θ
cos 15° + 0.50 sin 15°
(b) The second law equations for the moving crate are
T cos θ – f = ma
FN + T sin θ – mg = 0.
Now f =μkFN, and the second equation gives FN = mg – Tsinθ, which yields
f = μ k (mg − T sin θ ) . This expression is substituted for f in the first equation to obtain
T cos θ – μk (mg – T sin θ) = ma,
so the acceleration is a= T ( cos θ + μ k sin θ )
− μk g .
m Numerically, it is given by
a= b304 Ngbcos15° + 0.35 sin 15°g − b0.35gc9.8 m / s h = 13 m / s .
.
2 2 68 kg 23. The freebody diagrams for block B and for the knot just above block A are shown
next. T1 is the tension force of the rope pulling on block B or pulling on the knot (as the
case may be), T2 is the tension force exerted by the second rope (at angle θ = 30°) on the
knot, f is the force of static friction exerted by the horizontal surface on block B, FN is
normal force exerted by the surface on block B, WA is the weight of block A (WA is the
magnitude of m A g ), and WB is the weight of block B (WB = 711 N is the magnitude of
mB g ).
B B For each object we take +x horizontally rightward and +y upward. Applying Newton’s
second law in the x and y directions for block B and then doing the same for the knot
results in four equations: T1 − f s ,max
FN − WB
T2 cos θ − T1
T2 sin θ − WA =
=
=
= 0
0
0
0 where we assume the static friction to be at its maximum value (permitting us to use Eq.
61). Solving these equations with μs = 0.25, we obtain WA = 103 N ≈ 1.0 × 10 2 N .
27. The freebody diagrams for the two blocks are shown next. T is the magnitude of the
tension force of the string, FNA is the normal force on block A (the leading block), FNB is
the normal force on block B, f A is kinetic friction force on block A, f B is kinetic friction
force on block B. Also, mA is the mass of block A (where mA = WA/g and WA = 3.6 N), and mB is the mass of block B (where mB = WB/g and WB = 7.2 N). The angle of the incline is
θ = 30°.
B B B B For each block we take +x downhill (which is toward the lowerleft in these diagrams)
and +y in the direction of the normal force. Applying Newton’s second law to the x and y
directions of both blocks A and B, we arrive at four equations:
WA sin θ − f A − T = mA a
FNA − WA cos θ = 0
WB sin θ − f B + T = mB a
FNB − WB cosθ = 0
which, when combined with Eq. 62 ( f A = μ kA FNA where μk A = 0.10 and f B = μ kB FNB fB
where μk B = 0.20), fully describe the dynamics of the system so long as the blocks have
the same acceleration and T > 0.
B (a) From these equations, we find the acceleration to be
⎛
⎞
⎛ μ W + μ k BWB ⎞
2
a = g ⎜ sin θ − ⎜ k A A
⎟ cos θ ⎟ = 3.5 m/s .
⎜
⎟
WA + WB
⎝
⎠
⎝
⎠ (b) We solve the above equations for the tension and obtain ⎛ WW ⎞
T = ⎜ A B ⎟ ( μ k B − μ k A ) cos θ = 0.21 N.
⎝ WA + WB ⎠
Simply returning the value for a found in part (a) into one of the above equations is
certainly fine, and probably easier than solving for T algebraically as we have done, but
the algebraic form does illustrate the μk B – μk A factor which aids in the understanding of
the next part.
35. We denote the magnitude of the frictional force αv , where α = 70 N ⋅ s m . We take
the direction of the boat’s motion to be positive. Newton’s second law gives −αv = m Thus, ∫ v
v0 dv
.
dt dv
α t
= − ∫ dt
v
m 0 where v0 is the velocity at time zero and v is the velocity at time t. The integrals are
evaluated with the result
⎛v⎞
αt
ln ⎜ ⎟ = −
m
⎝ v0 ⎠
We take v = v0/2 and solve for time:
t= m α ln 2 = 1000 kg
ln 2 = 9.9 s .
70 N ⋅ s/m 59. The freebody diagram for the ball is shown below. Tu is the tension exerted by the
upper string on the ball, T is the tension force of the lower string, and m is the mass of
the ball. Note that the tension in the upper string is greater than the tension in the lower
string. It must balance the downward pull of gravity and the force of the lower string. (a) We take the +x direction to be leftward (toward the center of the circular orbit) and +y
upward. Since the magnitude of the acceleration is a = v2/R, the x component of
Newton’s second law is
Tu cosθ + T cosθ = mv 2
,
R where v is the speed of the ball and R is the radius of its orbit. The y component is Tu sin θ − T sin θ − mg = 0.
The second equation gives the tension in the lower string: T = Tu − mg / sin θ . Since the
triangle is equilateral θ = 30.0°. Thus
T = 35.0 N − (1.34 kg)(9.80 m/s 2 )
= 8.74 N.
sin 30.0° (b) The net force has magnitude Fnet,str = (Tu + T ) cos θ = (35.0 N + 8.74 N) cos 30.0° = 37.9 N.
(c) The radius of the path is
R = ((1.70 m)/2)tan 30.0° = 1.47 m.
Using Fnet,str = mv2/R, we find that the speed of the ball is v= RFnet,str
m = (1.47 m)(37.9 N)
= 6.45 m/s.
1.34 kg (d) The direction of Fnet,str is leftward (“radially inward’’).
65. (a) Using F = μ s mg , the coefficient of static friction for the surface between the two
blocks is μ s = (12 N)/(39.2 N) = 0.31, where mt g = (4.0 kg)(9.8 m/s2)=39.2 N is the
weight of the top block. Let M = mt + mb = 9.0 kg be the total system mass, then the
maximum horizontal force has a magnitude Ma = Mμs g = 27 N.
(b) The acceleration (in the maximal case) is a = μsg =3.0 m/s2.
77. The magnitude of the acceleration of the cyclist as it moves along the horizontal
circular path is given by v2/R, where v is the speed of the cyclist and R is the radius of the
curve.
(a) The horizontal component of Newton’s second law is f = mv2/R, where f is the static
friction exerted horizontally by the ground on the tires. Thus,
f ( 85.0 kg )( 9.00 m/s )
=
25.0 m 2 = 275 N. (b) If FN is the vertical force of the ground on the bicycle and m is the mass of the bicycle
and rider, the vertical component of Newton’s second law leads to FN = mg = 833 N. The
magnitude of the force exerted by the ground on the bicycle is therefore
2
f 2 + FN = (275 N)2 + (833 N)2 = 877 N. 81. (a) If we choose “downhill” positive, then Newton’s law gives
mA g sinθ – fA – T = mA a
for block A (where θ = 30º). For block B we choose leftward as the positive direction and
write T – fB = mB a. Now
B fA = μk,incline FNA = μ′mA g cosθ
using Eq. 612 applies to block A, and
fB = μk FNB = μk mB g.
B In this particular problem, we are asked to set μ′ = 0, and the resulting equations can be
straightforwardly solved for the tension: T = 13 N.
(b) Similarly, finding the value of a is straightforward:
a = g(mA sinθ – μk mB )/(mA + mB) =1.6 m/s2.
B 85. The mass of the car is m = (10700/9.80) kg = 1.09 × 103 kg. We choose “inward”
(horizontally towards the center of the circular path) as the positive direction.
(a) With v = 13.4 m/s and R = 61 m, Newton’s second law (using Eq. 618) leads to
mv 2
fs =
= 3.21 × 103 N .
R
(b) Noting that FN = mg in this situation, the maximum possible static friction is found to
be f s ,max = μs mg = ( 0.35 )(10700 N ) = 3.75 × 103 N
using Eq. 61. We see that the static friction found in part (a) is less than this, so the car
rolls (no skidding) and successfully negotiates the curve.
91. We apply Newton’s second law (as Fpush – f = ma). If we find Fpush < fmax, we
conclude “no, the cabinet does not move” (which means a is actually 0 and f = Fpush), and if we obtain a > 0 then it is moves (so f = fk). For fmax and fk we use Eq. 61 and Eq. 62
(respectively), and in those formulas we set the magnitude of the normal force equal to
556 N. Thus, fmax = 378 N and fk = 311 N.
(a) Here we find Fpush < fmax which leads to f = Fpush = 222 N.
(b) Again we find Fpush < fmax which leads to f = Fpush = 334 N.
(c) Now we have Fpush > fmax which means it moves and f = fk = 311 N.
(d) Again we have Fpush > fmax which means it moves and f = fk = 311 N.
(e) The cabinet moves in (c) and (d).
99. Replace fs with fk in Fig. 65(b) to produce the appropriate force diagram for the first
part of this problem (when it is sliding downhill with zero acceleration). This amounts to
replacing the static coefficient with the kinetic coefficient in Eq. 613: μk = tanθ. Now
(for the second part of the problem, with the block projected uphill) the friction direction
is reversed from what is shown in Fig. 65(b). Newton’s second law for the uphill motion
(and Eq. 612) leads to
– m g sinθ – μk m g cosθ = m a.
Canceling the mass and substituting what we found earlier for the coefficient, we have
– g sinθ – tanθ g cosθ = a .
This simplifies to – 2 g sinθ = a. Eq. 216 then gives the distance to stop: Δx = –vo2/2a.
(a) Thus, the distance up the incline traveled by the block is Δx = vo2/(4gsinθ ).
(b) We usually expect μs > μk (see the discussion in section 61). Sample Problem 62
treats the “angle of repose” (the minimum angle necessary for a stationary block to start
sliding downhill): μs = tan(θrepose). Therefore, we expect θrepose > θ found in part (a).
Consequently, when the block comes to rest, the incline is not steep enough to cause it to
start slipping down the incline again.
105. Probably the most appropriate picture in the textbook to represent the situation in
this problem is in the previous chapter: Fig. 59. We adopt the familiar axes with +x
rightward and +y upward, and refer to the 85 N horizontal push of the worker as P (and
assume it to be rightward). Applying Newton’s second law to the x axis and y axis,
respectively, produces
P − f k = ma
FN − mg = 0. 2
Using v 2 = v0 + 2aΔx we find a = 0.36 m/s2. Consequently, we obtain fk = 71 N and FN =
392 N. Therefore, μk = fk/ FN = 0.18. Chapter 7 – Student Solutions Manual
2
3. (a) From Table 21, we have v 2 = v0 + 2aΔx . Thus, 2
v = v0 + 2aΔx = ( 2.4 ×10 7 ) ( ) m/s + 2 3.6 × 1015 m/s 2 ( 0.035 m ) = 2.9 × 107 m/s.
2 (b) The initial kinetic energy is
Ki = 2
1 2 1
mv0 = (1.67 × 10−27 kg )( 2.4 × 107 m/s ) = 4.8 × 10 −13 J.
2
2 The final kinetic energy is
Kf = 2
1 2 1
mv = (1.67 × 10 −27 kg )( 2.9 × 107 m/s ) = 6.9 × 10−13 J.
2
2 The change in kinetic energy is ΔK = 6.9 × 10–13 J – 4.8 × 10–13 J = 2.1 × 10–13 J.
17. (a) We use F to denote the upward force exerted by the cable on the astronaut. The
force of the cable is upward and the force of gravity is mg downward. Furthermore, the
acceleration of the astronaut is g/10 upward. According to Newton’s second law, F – mg
= mg/10, so F = 11 mg/10. Since the force F and the displacement d are in the same
direction, the work done by F is
WF = Fd = 11mgd 11 (72 kg)(9.8 m/s 2 )(15 m)
=
= 1.164 × 104 J
10
10 which (with respect to significant figures) should be quoted as 1.2 × 104 J.
(b) The force of gravity has magnitude mg and is opposite in direction to the
displacement. Thus, using Eq. 77, the work done by gravity is Wg = −mgd = − (72 kg)(9.8 m/s 2 )(15 m) = −1.058 × 104 J
which should be quoted as – 1.1 × 104 J.
(c) The total work done is W = 1164 × 104 J − 1.058 × 104 J = 1.06 × 103 J . Since the
.
astronaut started from rest, the workkinetic energy theorem tells us that this (which we
round to 1.1 × 103 J ) is her final kinetic energy.
1
(d) Since K = 2 mv 2 , her final speed is .
2K
2(106 × 103 J)
=
= 5.4 m / s.
72 kg
m v= 19. (a) We use F to denote the magnitude of the force of the cord on the block. This force
is upward, opposite to the force of gravity (which has magnitude Mg). The acceleration is
a = g / 4 downward. Taking the downward direction to be positive, then Newton’s
second law yields
g
Fnet = ma ⇒ Mg − F = M
4 FG IJ
H K so F = 3Mg/4. The displacement is downward, so the work done by the cord’s force is,
using Eq. 77,
WF = –Fd = –3Mgd/4.
(b) The force of gravity is in the same direction as the displacement, so it does work
Wg = Mgd .
(c) The total work done on the block is −3 M gd 4 + M gd = M gd 4 . Since the block
starts from rest, we use Eq. 715 to conclude that this M gd 4 is the block’s kinetic
energy K at the moment it has descended the distance d. b g (d) Since K = 1 Mv 2 , the speed is
2 v= 2K
2( Mgd / 4)
=
=
M
M gd
2 at the moment the block has descended the distance d.
29. (a) As the body moves along the x axis from xi = 3.0 m to xf = 4.0 m the work done by
the force is
xf xf xi xi W = ∫ Fx dx = ∫ −6 x dx = −3( x 2 − xi2 ) = −3 (4.02 − 3.02 ) = −21 J.
f According to the workkinetic energy theorem, this gives the change in the kinetic
energy:
W = ΔK = d 1
m v 2 − vi2
f
2 i where vi is the initial velocity (at xi) and vf is the final velocity (at xf). The theorem yields vf = 2W
2(−21 J)
+ vi2 =
+ (8.0 m/s)2 = 6.6 m/s.
m
2.0 kg (b) The velocity of the particle is vf = 5.0 m/s when it is at x = xf. The workkinetic energy
theorem is used to solve for xf. The net work done on the particle is W = −3 ( x 2 − xi2 ) , so
f
the theorem leads to d i −3 x 2 − xi2 =
f d i 1
m v 2 − vi2 .
f
2 Thus, xf = − m 2 2
( v f − vi ) + xi2 = − 2.0 kg ( (5.0 m/s)2 − (8.0 m/s)2 ) + (3.0 m)2 = 4.7 m.
6
6 N/m 35. (a) The graph shows F as a function of x assuming x0 is positive. The work is negative
as the object moves from x = 0 to x = x0 and positive as it moves from x = x0 to x = 2 x0 .
Since the area of a triangle is (base)(altitude)/2, the
work done from x = 0 to x = x0 is −( x0 )( F0 ) / 2 and
the work done from x = x0 to x = 2 x0 is
(2 x0 − x0 )( F0 ) / 2 = ( x0 )( F0 ) / 2
The total work is the sum, which is zero.
(b) The integral for the work is
W =∫ 2x0 0 ⎛ x
⎞
⎛ x2
⎞
− x⎟
F0 ⎜ − 1⎟ dx = F0 ⎜
⎝ x0 ⎠
⎝ 2 x0
⎠ 2 x0 = 0.
0 43. The power associated with force F is given by P = F ⋅ v , where v is the velocity
of the object on which the force acts. Thus,
P = F ⋅ v = Fv cos φ = (122 N)(5.0 m/s)cos37° = 4.9 × 102 W. 45. (a) The power is given by P = Fv and the work done by F from time t1 to time t 2 is
given by
W = z t2 t1 P dt = z t2 t1 Fv dt . Since F is the net force, the magnitude of the acceleration is a = F/m, and, since the
initial velocity is v0 = 0 , the velocity as a function of time is given by
v = v0 + at = ( F m) t . Thus
t2
1
2
W = ∫ ( F 2 / m)t dt = (F 2 / m)(t2 − t12 ).
t1
2 For t1 = 0 and t2 = 1.0s, 1 ⎛ (5.0 N)2 ⎞
2
W= ⎜
⎟ (1.0 s) = 0.83 J.
2 ⎝ 15 kg ⎠
(b) For t1 = 1.0s, and t2 = 2.0s, 1 ⎛ (5.0 N) 2 ⎞
2
2
W= ⎜
⎟ [(2.0 s) − (1.0 s) ] = 2.5 J.
2 ⎝ 15 kg ⎠
(c) For t1 = 2.0s and t2 = 3.0s, 1 ⎛ (5.0 N) 2 ⎞
2
2
W= ⎜
⎟ [(3.0 s) − (2.0 s) ] = 4.2 J.
2 ⎝ 15 kg ⎠
(d) Substituting v = (F/m)t into P = Fv we obtain P = F 2 t m for the power at any time t.
At the end of the third second P = FG (5.0 N) (3.0 s) IJ
H 15 kg K
2 = 5.0 W. 47. The total work is the sum of the work done by gravity on the elevator, the work done
by gravity on the counterweight, and the work done by the motor on the system:
WT = We + Wc + Ws.
Since the elevator moves at constant velocity, its kinetic energy does not change and
according to the workkinetic energy theorem the total work done is zero. This means We
+ Wc + Ws = 0. The elevator moves upward through 54 m, so the work done by gravity on
it is
We = − me gd = −(1200 kg)(9.80 m/s 2 )(54 m) = −6.35 × 105 J. The counterweight moves downward the same distance, so the work done by gravity on it
is Wc = mc gd = (950 kg)(9.80 m/s 2 )(54 m) = 5.03 ×105 J. Since WT = 0, the work done by the motor on the system is
Ws = −We − Wc = 6.35 ×105 J − 5.03 ×105 J = 1.32 ×105 J. This work is done in a time interval of Δt = 3.0 min = 180 s, so the power supplied by
the motor to lift the elevator is
Ws 1.32 × 105 J
P=
=
= 7.4 × 10 2 W.
Δt
180 s
63. (a) In 10 min the cart moves
mi ⎞ ⎛ 5280 ft/mi ⎞
⎛
d = ⎜ 6.0
⎟⎜
⎟ (10 min) = 5280 ft
h ⎠ ⎝ 60 min/h ⎠
⎝ so that Eq. 77 yields
W = Fdcos φ = (40 lb)(5280 ft) cos 30° = 1.8 ×105 ft ⋅ lb.
(b) The average power is given by Eq. 742, and the conversion to horsepower (hp) can
be found on the inside back cover. We note that 10 min is equivalent to 600 s.
Pavg = 1.8 ×105 ft ⋅ lb
= 305 ft ⋅ lb/s
600 s which (upon dividing by 550) converts to Pavg = 0.55 hp.
69. (a) Eq. 76 gives Wa = Fd = (209 N)(1.50 m) ≈ 314 J.
(b) Eq. 712 leads to Wg = (25.0 kg)(9.80 m/s2)(1.50 m)cos(115º) ≈ –155 J.
(c) The angle between the normal force and the direction of motion remains 90º at all
times, so the work it does is zero.
(d) The total work done on the crate is WT = 314 J – 155 J =158 J.
71. (a) Hooke’s law and the work done by a spring is discussed in the chapter. Taking
absolute values, and writing that law in terms of differences ΔF and Δx , we analyze the
first two pictures as follows:  Δ F  = k  Δ x
240 N − 110 N = k ( 60 mm − 40 mm) which yields k = 6.5 N/mm. Designating the relaxed position (as read by that scale) as xo
we look again at the first picture:
110 N = k ( 40 mm − xo ) which (upon using the above result for k) yields xo = 23 mm.
(b) Using the results from part (a) to analyze that last picture, we find
W = k (30 mm − xo ) = 45 N . 73. A convenient approach is provided by Eq. 748.
P = F v = (1800 kg + 4500 kg)(9.8 m/s2)(3.80 m/s) = 235 kW.
Note that we have set the applied force equal to the weight in order to maintain constant
velocity (zero acceleration).
1 77. (a) We can easily fit the curve to a concavedownward parabola: x = 10 t(10 – t), from
which (by taking two derivatives) we find the acceleration to be a = –0.20 m/s2. The
(constant) force is therefore F = ma = –0.40 N, with a corresponding work given by W =
2
Fx = 50 t(t – 10). It also follows from the x expression that vo = 1.0 m/s. This means that
Ki = 2 mv2 = 1.0 J. Therefore, when t = 1.0 s, Eq. 710 gives K = Ki + W = 0.64 J ≈ 0.6 J ,
1 where the second significant figure is not to be taken too seriously.
(b) At t = 5.0 s, the above method gives K = 0.
2 (c) Evaluating the W = 50 t(t – 10) expression at t = 5.0 s and t = 1.0 s, and subtracting,
yields –0.6 J. This can also be inferred from the answers for parts (a) and (b). Chapter 8 – Student Solutions Manual
1
5. The potential energy stored by the spring is given by U = 2 kx 2 , where k is the spring
constant and x is the displacement of the end of the spring from its position when the
spring is in equilibrium. Thus k= b g
b
g 2 25 J
2U
=
2
x
0.075 m 2 = 8.9 × 103 N m . 9. We neglect any work done by friction. We work with SI units, so the speed is
converted: v = 130(1000/3600) = 36.1 m/s.
(a) We use Eq. 817: Kf + Uf = Ki + Ui with Ui = 0, Uf = mgh and Kf = 0. Since
1
Ki = 2 mv 2 , where v is the initial speed of the truck, we obtain
1 2
mv = mgh
2 ⇒ v 2 (36.1 m/s) 2
h=
=
= 66.5 m .
2 g 2(9.8 m/s 2 ) If L is the length of the ramp, then L sin 15° = 66.5 m so that L = 66.5/sin 15° = 257 m.
Therefore, the ramp must be about 2.6 × 102 m long if friction is negligible.
(b) The answers do not depend on the mass of the truck. They remain the same if the
mass is reduced.
(c) If the speed is decreased, h and L both decrease (note that h is proportional to the
square of the speed and that L is proportional to h).
11. (a) If Ki is the kinetic energy of the flake at the edge of the bowl, Kf is its kinetic
energy at the bottom, Ui is the gravitational potential energy of the flakeEarth system
with the flake at the top, and Uf is the gravitational potential energy with it at the bottom,
then Kf + Uf = Ki + Ui.
Taking the potential energy to be zero at the bottom of the bowl, then the potential energy
at the top is Ui = mgr where r = 0.220 m is the radius of the bowl and m is the mass of the
flake. Ki = 0 since the flake starts from rest. Since the problem asks for the speed at the
1
bottom, we write mv 2 for Kf. Energy conservation leads to
2 Wg = Fg ⋅ d = mgh = mgL (1 − cos θ ) .
The speed is v = 2 gr = 2.08 m/s . (b) Since the expression for speed does not contain the mass of the flake, the speed would
be the same, 2.08 m/s, regardless of the mass of the flake.
(c) The final kinetic energy is given by Kf = Ki + Ui – Uf. Since Ki is greater than before,
Kf is greater. This means the final speed of the flake is greater.
31. We refer to its starting point as A, the point where it first comes into contact with the
spring as B, and the point where the spring is compressed x = 0.055 m as C. Point C is
our reference point for computing gravitational potential energy. Elastic potential energy
(of the spring) is zero when the spring is relaxed. Information given in the second
sentence allows us to compute the spring constant. From Hooke's law, we find
k= F 270 N
=
= 1.35 × 104 N m .
x 0.02 m (a) The distance between points A and B is Fg and we note that the total sliding distance
+ x is related to the initial height h of the block (measured relative to C) by h
= sin θ
+x
where the incline angle θ is 30°. Mechanical energy conservation leads to
K A + U A = KC + U C
1
0 + mgh = 0 + kx 2
2 which yields c hb 1.35 × 104 N m 0.055 m
kx 2
h=
=
2mg
2 12 kg 9.8 m s2 b gc h g 2 = 0.174 m . Therefore,
+x = h
0.174 m
=
= 0.35 m .
sin 30°
sin 30° (b) From this result, we find = 0.35 − 0.055 = 0.29 m , which means that
Δy = − sin θ = −0.15 m in sliding from point A to point B. Thus, Eq. 818 gives
ΔK + ΔU = 0
1 2
mv B + mgΔh = 0
2 b gb g which yields v B = −2 gΔh = − 9.8 −0.15 = 1.7 m s .
45. (a) The work done on the block by the force in the rope is, using Eq. 77,
W = Fd cos θ = (7.68 N)(4.06 m) cos15.0° = 30.1J. (b) Using f for the magnitude of the kinetic friction force, Eq. 829 reveals that the
increase in thermal energy is
ΔEth = fd = (7.42 N)(4.06 m) = 30.1J.
(c) We can use Newton's second law of motion to obtain the frictional and normal forces,
then use μk = f/FN to obtain the coefficient of friction. Place the x axis along the path of
the block and the y axis normal to the floor. The x and the y component of Newton's
second law are
x:
F cos θ – f = 0
y: FN + F sin θ – mg = 0,
where m is the mass of the block, F is the force exerted by the rope, and θ is the angle
between that force and the horizontal. The first equation gives
f = F cos θ = (7.68 N) cos15.0° = 7.42 N
and the second gives
FN = mg – F sin θ = (3.57 kg)(9.8 m/s2) – (7.68 N) sin 15.0° = 33.0 N.
Thus,
f
7.42 N
=
= 0.225 .
FN 33.0 N
47. (a) We take the initial gravitational potential energy to be Ui = 0. Then the final
gravitational potential energy is Uf = –mgL, where L is the length of the tree. The change
is
U f − U i = − mgL = − (25 kg) 9.8 m s2 (12 m) = −2.9 × 103 J . μk = d (b) The kinetic energy is K = i 1 2 1
mv = (25 kg)(5.6 m/s) 2 = 3.9 × 10 2 J .
2
2 (c) The changes in the mechanical and thermal energies must sum to zero. The change in
thermal energy is ΔEth = fL, where f is the magnitude of the average frictional force;
therefore, f =− ΔK + ΔU
3.9 × 102 J − 2.9 × 103 J
=−
= 2.1× 102 N
L
12 m 69. There is the same potential energy change in both circumstances, so we can equate
the kinetic energy changes as well:
1
1
1
1
ΔK2 = ΔK1 ⇒ 2 m vB2 – 2 m(4.00)2 = 2 m(2.60)2 – 2 m(2.00)2
which leads to vB = 4.33 m/s.
B 75. We note that if the larger mass (block B, mB = 2 kg) falls d = 0.25 m, then the smaller
mass (blocks A, mA = 1 kg) must increase its height by h = d sin 30° . Thus, by
mechanical energy conservation, the kinetic energy of the system is
K total = mB gd − mA gh = 3.7 J .
83. The initial height shown in the figure is the y = 0 level in our computations of Ug, and
in parts (a) and (b) the heights are ya = 0.80 sin 40° = 0.51 m and yb = 1.00 sin 40° = 0.64
m, respectively.
(a) The conservation of energy, Eq. 817, leads to
1
Ki + Ui = Ka + Ua ⇒ 16 + 0 = Ka + mgya + 2k(0.20)2
from which we obtain Ka = 16 – 5.0 – 4.0 = 7.0 J.
(b) Again we use the conservation of energy
1
Ki + Ui = Kb + Ub ⇒ Ki + 0 = 0 + mgyb + 2 k(0.40)2
from which we obtain Ki = 6.0 + 16 = 22 J.
87. Since the speed is constant ΔK = 0 and Eq. 833 (an application of the energy
conservation concept) implies
Wapplied = ΔE th = ΔE th b cube g + ΔE th b floor g . Thus, if Wapplied = (15)(3.0) = 45 J, and we are told that ΔEth (cube) = 20 J, then we conclude
that ΔEth (floor) = 25 J.
109. The connection between angle θ (measured from vertical) and height h (measured
from the lowest point, which is our choice of reference position in computing the gravitational potential energy mgh) is given by h = L(1 – cos θ ) where L is the length of
the pendulum.
(a) Using this formula (or simply using intuition) we see the initial height is h1 = 2L, and
of course h2 = 0. We use energy conservation in the form of Eq. 817.
K1 + U 1 = K2 + U 2
0 + mg (2 L) = 1 2
mv + 0
2 This leads to v = 2 gL . With L = 0.62 m, we have
v = 2 (9.8 m/s 2 )(0.62 m) = 4.9 m/s . (b) The ball is in circular motion with the center of the circle above it, so a = v 2 / r
upward, where r = L. Newton's second law leads to FG
H IJ
K 4 gL
v2
T − mg = m ⇒ T = m g +
= 5 mg.
r
L
With m = 0.092 kg, the tension is given by T = 4.5 N.
(c) The pendulum is now started (with zero speed) at θ i = 90° (that is, hi = L), and we
look for an angle θ such that T = mg. When the ball is moving through a point at angle θ,
then Newton's second law applied to the axis along the rod yields
v2
T − mg cosθ = m
r
which (since r = L) implies v2 = gL(1 – cos θ ) at the position we are looking for. Energy
conservation leads to
Ki + U i = K + U
1
0 + mgL = mv 2 + mgL (1 − cosθ )
2
1
gL = ( gL(1 − cosθ )) + gL (1 − cosθ )
2
where we have divided by mass in the last step. Simplifying, we obtain
⎛1⎞
⎝ ⎠ θ = cos − 1 ⎜ ⎟ = 71 ° .
3 (d) Since the angle found in (c) is independent of the mass, the result remains the same if
the mass of the ball is changed.
111. (a) At the top of its flight, the vertical component of the velocity vanishes, and the
horizontal component (neglecting air friction) is the same as it was when it was thrown.
Thus,
2
1 2 1
Ktop = mv x = 0.050 kg 8.0 m s cos 30° = 1.2 J.
2
2 b g cb g h (b) We choose the point 3.0 m below the window as the reference level for computing the
potential energy. Thus, equating the mechanical energy when it was thrown to when it is
at this reference level, we have (with SI units understood)
mgy0 + K0 = K b gb g b g 1
1
2
m 9.8 3.0 + m 8.0 = mv 2
2
2 which yields (after canceling m and simplifying) v = 11 m/s.
(c) As mentioned, m cancels — and is therefore not relevant to that computation.
(d) The v in the kinetic energy formula is the magnitude of the velocity vector; it does not
depend on the direction.
119. (a) During the final d = 12 m of motion, we use
K1 + U1 = K 2 + U 2 + f k d
1 2
mv + 0 = 0 + 0 + f k d
2
where v = 4.2 m/s. This gives fk = 0.31 N. Therefore, the thermal energy change is
f k d = 3.7 J.
(b) Using fk = 0.31 N we obtain fkdtotal = 4.3 J for the thermal energy generated by
friction; here, dtotal = 14 m.
(c) During the initial d' = 2 m of motion, we have
K0 + U 0 + Wapp = K1 + U 1 + f k d ′ ⇒ 0 + 0 + Wapp = 1 2
mv + 0 + f k d ′
2 which essentially combines Eq. 831 and Eq. 833. This leads to the result Wapp = 4.3 J,
and — reasonably enough — is the same as our answer in part (b). 121. We use Eq. 820.
(a) The force at x = 2.0 m is
F =− dU
−(17.5) − ( −2.8)
≈−
= 4.9 N.
dx
4.0 − 1.0 (b) The force points in the +x direction (but there is some uncertainty in reading the graph
which makes the last digit not very significant).
(c) The total mechanical energy at x = 2.0 m is
E= 1 2
1
mv + U ≈ (2.0)( −15) 2 − 7.7 = −5.5
.
2
2 in SI units (Joules). Again, there is some uncertainty in reading the graph which makes
the last digit not very significant. At that level (–5.5 J) on the graph, we find two points
where the potential energy curve has that value — at x ≈ 1.5 m and x ≈ 13.5 m.
Therefore, the particle remains in the region 1.5 < x < 13.5 m. The left boundary is at x =
1.5 m.
(d) From the above results, the right boundary is at x = 13.5 m.
(e) At x = 7.0 m, we read U ≈ –17.5 J. Thus, if its total energy (calculated in the previous
part) is E ≈ –5.5 J, then we find 1 2
2
mv = E − U ≈ 12 J ⇒ v =
( E − U ) ≈ 35 m s
.
2
m
where there is certainly room for disagreement on that last digit for the reasons cited
above.
123. Converting to SI units, v0 = 8.3 m s and v = 111 m s . The incline angle is θ = 5.0° .
.
The height difference between the car's highest and lowest points is (50 m) sin θ = 4.4 m.
We take the lowest point (the car's final reported location) to correspond to the y = 0
reference level.
(a) Using Eq. 831 and Eq. 833, we find
f k d = − ΔK − Δ U ⇒ f k d = c h 1
2
m v0 − v 2 + mgy0 .
2 Therefore, the mechanical energy reduction (due to friction) is fk d = 2.4 × 104 J.
(b) With d = 50 m, we solve for fk and obtain 4.7 × 102 N. 127. (a) When there is no change in potential energy, Eq. 824 leads to
Wapp = ΔK = c h 1
2
m v 2 − v0 .
2 Therefore, ΔE = 6.0 × 103 J .
(b) From the above manipulation, we see Wapp = 6.0 × 103 J. Also, from Chapter 2, we
know that Δt = Δv a = 10 s . Thus, using Eq. 742,
Pavg = W 6.0 × 103
=
= 600 W .
Δt
10 (c) and (d) The constant applied force is ma = 30 N and clearly in the direction of motion,
so Eq. 748 provides the results for instantaneous power
P = F ⋅v = R 300 W
S 900 W
T for v = 10 m s
for v = 30 m s We note that the average of these two values agrees with the result in part (b).
131. The power generation (assumed constant, so average power is the same as
instantaneous power) is
P= mgh (3 / 4)(1200 m3 )(103 kg / m3 )(9.8 m / s 2 )(100 m)
=
= 8.80 × 108 W.
t
1.0s 133. (a) Sample Problem 83 illustrates simple energy conservation in a similar situation,
and derives the frequently encountered relationship: v = 2gh . In our present problem,
the height change is equal to the rod length L. Thus, using the suggested notation for the
speed, we have vo = 2gL .
(b) At B the speed is (from Eq. 817)
2
v = v0 + 2 gL = 4 gL . The direction of the centripetal acceleration (v2/r = 4gL/L = 4g) is upward (at that
moment), as is the tension force. Thus, Newton’s second law gives
T – mg = m(4g) ⇒ T = 5mg. (c) The difference in height between C and D is L, so the “loss” of mechanical energy
(which goes into thermal energy) is –mgL.
(d) The difference in height between B and D is 2L, so the total “loss” of mechanical
energy (which all goes into thermal energy) is –2mgL. Chapter 9 – Student Solutions Manual
15. We need to find the coordinates of the point where the shell explodes and the velocity
of the fragment that does not fall straight down. The coordinate origin is at the firing
point, the +x axis is rightward, and the +y direction is upward. The y component of the
velocity is given by v = v0 y – gt and this is zero at time t = v0 y/g = (v0/g) sin θ0, where v0
is the initial speed and θ0 is the firing angle. The coordinates of the highest point on the
trajectory are ( 20 m/s ) sin 60° cos 60° = 17.7 m
v2
x = v0 x t = v0t cos θ 0 = 0 sin θ0 cos θ0 =
g
9.8 m/s 2
2 and b g 2 2
1 2 1 v0
1 20 m / s
y = v0 y t − gt =
sin 2 θ 0 =
sin 2 60° = 15.3 m.
2
2
2 g
2 9.8 m / s Since no horizontal forces act, the horizontal component of the momentum is conserved.
Since one fragment has a velocity of zero after the explosion, the momentum of the other
equals the momentum of the shell before the explosion. At the highest point the velocity
of the shell is v0 cosθ0, in the positive x direction. Let M be the mass of the shell and let
V0 be the velocity of the fragment. Then Mv0cosθ0 = MV0/2, since the mass of the
fragment is M/2. This means b g V0 = 2v0 cosθ 0 = 2 20 m / s cos 60° = 20 m / s. This information is used in the form of initial conditions for a projectile motion problem
to determine where the fragment lands. Resetting our clock, we now analyze a projectile
launched horizontally at time t = 0 with a speed of 20 m/s from a location having
1
coordinates x0 = 17.7 m, y0 = 15.3 m. Its y coordinate is given by y = y0 − 2 gt 2 , and
when it lands this is zero. The time of landing is t = 2 y0 / g and the x coordinate of the
landing point is b g b g 2 15.3 m
2 y0
= 17.7 m + 20 m / s
= 53 m.
9.8 m / s2
g
23. The initial direction of motion is in the +x direction. The magnitude of the average
force Favg is given by
x = x0 + V0t = x0 + V0 Favg = J
32.4 N ⋅ s
=
= 1.20 × 103 N
Δ t 2.70 × 10− 2 s The force is in the negative direction. Using the linear momentumimpulse theorem
stated in Eq. 931, we have
–FavgΔt = mvf – mvi. where m is the mass, vi the initial velocity, and vf the final velocity of the ball. Thus,
vf = mvi − Favg Δt
m ( 0.40 kg )(14 m s ) − (1200 N ) ( 27 ×10−3 s ) = 0.40 kg = −67 m s. (a) The final speed of the ball is  v f  = 67 m/s.
(b) The negative sign indicates that the velocity is in the –x direction, which is opposite to
the initial direction of travel.
(c) From the above, the average magnitude of the force is Favg = 1.20 ×103 N .
(d) The direction of the impulse on the ball is –x, same as the applied force.
35. (a) We take the force to be in the positive direction, at least for earlier times. Then the
impulse is J =∫ 3.0 × 10− 3 0 Fdt = ∫ 3.0 × 10− 3 0 ⎡(6.0 ×106 ) t − (2.0 ×109 )t 2 ⎤ dt
⎣
⎦ 1
⎡1
⎤
= ⎢ (6.0 × 106 )t 2 − (2.0 × 109 )t 3 ⎥
3
⎣2
⎦ 3.0 × 10− 3 0 = 9.0 N ⋅ s.
(b) Since J = Favg Δt, we find
Favg J
9.0 N ⋅ s
= 3.0 × 103 N.
=
Δt
3.0 × 10 −3 s (c) To find the time at which the maximum force occurs, we set the derivative of F with
respect to time equal to zero – and solve for t. The result is t = 1.5 × 10–3 s. At that time
the force is c hc h c hc Fmax = 6.0 × 106 15 × 10−3 − 2.0 × 109 15 × 10−3
.
. h 2 = 4.5 × 103 N. (d) Since it starts from rest, the ball acquires momentum equal to the impulse from the
kick. Let m be the mass of the ball and v its speed as it leaves the foot. Then,
v = p J 9.0 N ⋅ s
= =
= 20 m/s.
m m 0.45 kg 39. No external forces with horizontal components act on the manstone system and the
vertical forces sum to zero, so the total momentum of the system is conserved. Since the
man and the stone are initially at rest, the total momentum is zero both before and after
the stone is kicked. Let ms be the mass of the stone and vs be its velocity after it is kicked;
let mm be the mass of the man and vm be his velocity after he kicks the stone. Then
msvs + mmvm = 0 → vm = –msvs/mm.
We take the axis to be positive in the direction of motion of the stone. Then vm = − ( 0.068 kg )( 4.0 m/s ) = −3.0 ×10−3 m/s ,
91 kg or  vm = 3.0 × 10−3 m/s . The negative sign indicates that the man moves in the direction
opposite to the direction of motion of the stone.
47. Our notation is as follows: the mass of the original body is M = 20.0 kg; its initial
velocity is v0 = 200 i in SI units (m/s); the mass of one fragment is m1 = 10.0 kg; its
velocity is v1 = −100 j in SI units; the mass of the second fragment is m2 = 4.0 kg; its
velocity is v2 = −500 i in SI units; and, the mass of the third fragment is m3 = 6.00 kg.
(a) Conservation of linear momentum requires Mv0 = m1v1 + m2 v2 + m3v3 , which (using the
above information) leads to v3 = (1.00 ×103 ˆ − 0.167 ×103 ˆ m/s
i
j)
in SI units. The magnitude of v3 is v3 = 10002 + ( −167) 2 = 1.01 × 103 m / s . It points at
tan–1 (–167/1000) = –9.48° (that is, at 9.5° measured clockwise from the +x axis).
(b) We are asked to calculate ΔK or FG 1 m v
H2 2
1 1 IJ
K 1
1
1
2
2
2
+ m2 v2 + m3v3 − Mv0 = 3.23 × 106 J.
2
2
2 61. (a) Let m1 be the mass of the cart that is originally moving, v1i be its velocity before
the collision, and v1f be its velocity after the collision. Let m2 be the mass of the cart that
is originally at rest and v2f be its velocity after the collision. Then, according to Eq. 967,
v1 f = m1 − m2
v1i .
m1 + m2 Using SI units (so m1 = 0.34 kg), we obtain m2 = v1i − v1 f ⎛ 1.2 m/s − 0.66 m/s ⎞
m1 = ⎜
⎟ (0.34 kg) = 0.099 kg.
v1i + v1 f
1.2 m/s + 0.66 m/s ⎠
⎝ (b) The velocity of the second cart is given by Eq. 968: v2 f = ⎛
⎞
2m1
2(0.34 kg)
v1i = ⎜
⎟ (1.2 m/s) = 1.9 m/s.
m1 + m2
⎝ 0.34 kg + 0.099 kg ⎠ (c) The speed of the center of mass is
vcom = .
m1v1i + m2 v2i (0.34) (12) + 0
=
= 0.93 m s.
0.34 + 0.099
m1 + m2 Values for the initial velocities were used but the same result is obtained if values for the
final velocities are used.
63. (a) Let m1 be the mass of the body that is originally moving, v1i be its velocity before
the collision, and v1f be its velocity after the collision. Let m2 be the mass of the body that
is originally at rest and v2f be its velocity after the collision. Then, according to Eq. 967,
v1 f = m1 − m2
v1i .
m1 + m2 We solve for m2 to obtain
m2 = v1i − v1 f
v1 f + v1i m1 . b g We combine this with v1 f = v1i / 4 to obtain m2 = 3m1 5 = 3 2.0 5 = 12 kg .
.
(b) The speed of the center of mass is
vcom = b gb g 2.0 4.0
m1v1i + m2 v2i
=
= 2.5 m s .
m1 + m2
2.0 + 12
. 77. (a) The thrust of the rocket is given by T = Rvrel where R is the rate of fuel
consumption and vrel is the speed of the exhaust gas relative to the rocket. For this
problem R = 480 kg/s and vrel = 3.27 × 103 m/s, so b gc h T = 480 kg s 3.27 × 103 m s = 157 × 106 N .
. (b) The mass of fuel ejected is given by Mfuel = RΔt , where Δt is the time interval of the
burn. Thus, Mfuel = (480 kg/s)(250 s) = 1.20 × 105 kg. The mass of the rocket after the
burn is
Mf = Mi – Mfuel = (2.55 × 105 kg ) – (1.20 × 105 kg) = 1.35 ×105 kg.
(c) Since the initial speed is zero, the final speed is given by v f = vrel ln h FGH IJ
K Mi
2.55 × 105
= 3.27 × 103 ln
= 2.08 × 103 m s .
5
Mf
135 × 10
. c 79. (a) We consider what must happen to the coal that lands on the faster barge during
one minute (Δt = 60s). In that time, a total of m = 1000 kg of coal must experience a
change of velocity Δv = 20 km h − 10 km h = 10 km h = 2.8 m s , where rightwards is
considered the positive direction. The rate of change in momentum for the coal is
therefore
Δp mΔv (1000 kg )( 2.8 m/s )
=
=
= 46 N
Δt
Δt
60 s
which, by Eq. 923, must equal the force exerted by the (faster) barge on the coal. The
processes (the shoveling, the barge motions) are constant, so there is no ambiguity in
Δp
dp
with
equating
.
Δt
dt
(b) The problem states that the frictional forces acting on the barges does not depend on
mass, so the loss of mass from the slower barge does not affect its motion (so no extra
force is required as a result of the shoveling).
91. (a) If m is the mass of a pellet and v is its velocity as it hits the wall, then its
momentum is p = mv = (2.0 × 10–3 kg)(500 m/s) = 1.0 kg · m/s, toward the wall.
(b) The kinetic energy of a pellet is
K= c g hb 1 2 1
2
mv = 2.0 × 10 −3 kg 500 m s = 2.5 × 102 J .
2
2 (c) The force on the wall is given by the rate at which momentum is transferred from the
pellets to the wall. Since the pellets do not rebound, each pellet that hits transfers p =
1.0 kg · m/s. If ΔN pellets hit in time Δt, then the average rate at which momentum is
transferred is
Favg = pΔ N
= 10 kg ⋅ m s 10 s−1 = 10 N.
.
Δt b gc h The force on the wall is in the direction of the initial velocity of the pellets.
(d) If Δt is the time interval for a pellet to be brought to rest by the wall, then the average
force exerted on the wall by a pellet is
Favg = p 1.0 kg ⋅ m s
= 1.7 × 103 N.
=
Δt 0.6 × 10 −3 s The force is in the direction of the initial velocity of the pellet.
(e) In part (d) the force is averaged over the time a pellet is in contact with the wall, while
in part (c) it is averaged over the time for many pellets to hit the wall. During the
majority of this time, no pellet is in contact with the wall, so the average force in part (c)
is much less than the average force in part (d).
93. (a) The initial momentum of the car is b gb g b g pi = mvi = 1400 kg 5.3 m s j = 7400 kg ⋅ m s j b g
its momentum: J = p − p = b7400 N ⋅ sg e i − jj.
(b) The initial momentum of the car is p = b7400 kg ⋅ m sg i and the final momentum is and the final momentum is p f = 7400 kg ⋅ m s i. The impulse on it equals the change in
f i i ˆ
p f = 0. The impulse acting on it is J = p f − pi = (− 7.4 ×103 N ⋅ s)i.
(c) The average force on the car is Favg = b ge j b 7400 kg ⋅ m s i − j
Δp J
=
=
= 1600 N i − j
Δt Δt
4.6s ge j and its magnitude is Favg = (1600 N ) 2 = 2.3 ×103 N.
(d) The average force is Favg = b g c −7400 kg ⋅ m s i
J
=
= −2.1 × 104 N i
−3
Δt
350 × 10 s and its magnitude is Favg = 2.1 × 104 N. h (e) The average force is given above in unit vector notation. Its x and y components have
equal magnitudes. The x component is positive and the y component is negative, so the
force is 45° below the positive x axis.
97. Let mF be the mass of the freight car and vF be its initial velocity. Let mC be the mass
of the caboose and v be the common final velocity of the two when they are coupled.
Conservation of the total momentum of the twocar system leads to mFvF = (mF + mC)v,
so v = v F mF mF + mC . The initial kinetic energy of the system is b g Ki = 1
2
mF v F
2 and the final kinetic energy is Kf = b g b gb 2 2
mF v F
1
1
mF + mC v 2 = mF + mC
2
2
mF + mC g 2 = 2 2
1 mF v F
.
2 mF + mC b g Since 27% of the original kinetic energy is lost, we have Kf = 0.73Ki. Thus, b g FGH
g IJ
K 2 2
1 mF v F
1
2
= 0.73
mF v F .
2 mF + mC
2 b b g Simplifying, we obtain mF mF + mC = 0.73, which we use in solving for the mass of the
caboose:
mC = b gc h 0.27
mF = 0.37mF = 0.37 318 × 104 kg = 118 × 104 kg .
.
.
0.73 101. The mass of each ball is m, and the initial speed of one of the balls is v1i = 2.2 m s.
We apply the conservation of linear momentum to the x and y axes respectively. mv1i = mv1 f cosθ1 + mv2 f cosθ 2
0 = mv1 f sin θ1 − mv2 f sin θ 2
The mass m cancels out of these equations, and we are left with two unknowns and two
equations, which is sufficient to solve.
(a) The ymomentum equation can be rewritten as, using θ 2 = 60 ° and v2 f = 1.1 m/s ,
v1 f sin θ1 = (1.1 m/s) sin 60° = 0.95 m/s. and the xmomentum equation yields v1 f cos θ1 = (2.2 m/s) − (1.1 m/s) cos 60° = 1.65 m/s. Dividing these two equations, we find tanθ1= 0.576 which yields θ1 = 30°. We plug the
value into either equation and find v1 f ≈ 1.9 m/s.
(b) From the above, we have θ1 = 30°.
(c) One can check to see if this an elastic collision by computing
2K f
2 Ki
2
= v12i and
= v12f + v2 f
m
m
and seeing if they are equal (they are), but one must be careful not to use roundedoff
values. Thus, it is useful to note that the answer in part (a) can be expressed “exactly” as
v1 f = 1 v1i 3 (and of course v2 f = 1 v1i “exactly” — which makes it clear that these two
2
2
kinetic energy expressions are indeed equal).
107. (a) Noting that the initial velocity of the system is zero, we use Eq. 919 and Eq. 215 (adapted to two dimensions) to obtain
→ → ^ ^ →
1 ⎛ F + F2 ⎞
1 ⎛–2i + j ⎞
d = 2 ⎜ m1 + m ⎟ t2 = 2 ⎜ 0.006 ⎟ (0.002)2
⎝ 1
⎝
⎠
2⎠ which has a magnitude of 0.745 mm.
→ (b) The angle of d is 153° counterclockwise from +xaxis.
(c) A similar calculation using Eq. 211 (adapted to two dimensions) leads to a center of
→
mass velocity of v = 0.7453 m/s at 153°. Thus, the center of mass kinetic energy is
Kcom = 1
2 (m1 + m2)v2 = 0.00167 J. 113. By conservation of momentum, the final speed v of the sled satisfies b2900 kggb250 m / sg = b2900 kg + 920 kggv
which gives v = 190 m/s.
115. (a) We locate the coordinate origin at the center of Earth. Then the distance rcom of
the center of mass of the EarthMoon system is given by rcom = mM rM
mM + mE where mM is the mass of the Moon, mE is the mass of Earth, and rM is their separation.
These values are given in Appendix C. The numerical result is
rcom ( 7.36 ×10
=
7.36 ×10 22 22 kg )( 3.82 ×108 m )
kg + 5.98 ×10 24 kg = 4.64 × 106 m ≈ 4.6 ×103 km. (b) The radius of Earth is RE = 6.37 × 106 m, so rcom / RE = 0.73 = 73 % .
117. (a) The thrust is Rvrel where vrel = 1200 m/s. For this to equal the weight Mg where
M = 6100 kg, we must have R = (6100) (9.8)/1200 ≈ 50 kg/s.
(b) Using Eq. 942 with the additional effect due to gravity, we have
Rv rel − Mg = Ma so that requiring a = 21 m/s2 leads to R = (6100)(9.8 + 21)/1200 = 1.6 × 102 kg/s.
129. Using Eq. 968 with m1 = 3.0 kg, v1i = 8.0 m/s and v2f = 6.0 m/s, then
v2 f = leads to m2 = M = 5.0 kg. ⎛ 2v
⎞
2m1
v1i ⇒ m2 = m1 ⎜ 1i − 1⎟
⎜v
⎟
m1 + m2
⎝ 2f
⎠ Chapter 10 – Student Solutions Manual
13. We take t = 0 at the start of the interval and take the sense of rotation as positive.
1
Then at the end of the t = 4.0 s interval, the angular displacement is θ = ω 0t + 2 αt 2 . We
solve for the angular velocity at the start of the interval: ω0 = θ − 1 α t2
2
t = ( 120 rad − 1 3.0 rad/s 2
2 ) ( 4.0 s ) 4.0 s 2 = 24 rad/s. We now use ω = ω0 + α t (Eq. 1012) to find the time when the wheel is at rest:
t=− ω0
24 rad / s
=−
= −8.0 s.
α
3.0 rad / s2 That is, the wheel started from rest 8.0 s before the start of the described 4.0 s interval.
21. (a) We obtain ω= (200 rev / min)(2 π rad / rev)
= 20.9 rad / s.
60 s / min (b) With r = 1.20/2 = 0.60 m, Eq. 1018 leads to v = rω = (0.60)(20.9) = 12.5 m/s.
(c) With t = 1 min, ω = 1000 rev/min and ω0 = 200 rev/min, Eq. 1012 gives α= ω −ωo
t = 800 rev / min 2 . (d) With the same values used in part (c), Eq. 1015 becomes θ= b g 1
1
ω o + ω t = (200 + 100)(1) = 600 rev.
2
2 29. (a) Earth makes one rotation per day and 1 d is (24 h) (3600 s/h) = 8.64 × 104 s, so the
angular speed of Earth is
2π rad
ω=
= 7.3 × 10− 5 rad/s.
4
8.64 × 10 s
(b) We use v = ω r, where r is the radius of its orbit. A point on Earth at a latitude of 40°
moves along a circular path of radius r = R cos 40°, where R is the radius of Earth (6.4 ×
106 m). Therefore, its speed is
v = ω ( R cos 40°) = (7.3 × 10−5 rad/s)(6.4 × 106 m)cos40° = 3.5 × 102 m/s. (c) At the equator (and all other points on Earth) the value of ω is the same (7.3 × 10–5
rad/s).
(d) The latitude is 0° and the speed is
v = ω R = (7.3 ×10−5 rad/s)(6.4 × 106 m) = 4.6 × 102 m/s.
1
33. The kinetic energy (in J) is given by K = 2 Iω 2 , where I is the rotational inertia (in kg ⋅ m2 ) and ω is the angular velocity (in rad/s). We have ω= (602 rev / min)(2 π rad / rev)
= 63.0 rad / s.
60 s / min Consequently, the rotational inertia is
I= 2K ω 2 = 2(24400 J)
= 12.3 kg ⋅ m2 .
2
(63.0 rad / s) 35. We use the parallel axis theorem: I = Icom + Mh2, where Icom is the rotational inertia
about the center of mass (see Table 102(d)), M is the mass, and h is the distance between
the center of mass and the chosen rotation axis. The center of mass is at the center of the
meter stick, which implies h = 0.50 m – 0.20 m = 0.30 m. We find
I com = b gb g 1
1
2
ML2 =
.
0.56 kg 10 m = 4.67 × 10−2 kg ⋅ m2 .
12
12 Consequently, the parallel axis theorem yields b gb g 2 I = 4.67 × 10−2 kg ⋅ m2 + 0.56 kg 0.30 m = 9.7 × 10−2 kg ⋅ m2 .
1
37. Since the rotational inertia of a cylinder is I = 2 MR 2 (Table 102(c)), its rotational
kinetic energy is K= 1 2 1
Iω = MR 2ω 2 .
2
4 1
(a) For the smaller cylinder, we have K = 4 (1.25)( 0.25) 2 ( 235) 2 = 11 × 103 J.
.
1
(b) For the larger cylinder, we obtain K = 4 (1.25)( 0.75) 2 ( 235) 2 = 9.7 × 103 J. 41. We use the parallelaxis theorem. According to Table 102(i), the rotational inertia of
a uniform slab about an axis through the center and perpendicular to the large faces is
given by
M 2
I com =
a + b2 .
12 c h A parallel axis through the corner is a distance h = ba / 2g + bb / 2g
2 2 from the center. Therefore,
I = I com + Mh 2 =
= M
12 (a 2 + b2 ) + M
4 (a 2 + b2 ) = M
3 (a 2 + b2 ) 0.172 kg
[(0.035 m) 2 +(0.084 m) 2 ] = 4.7 × 10 −4 kg ⋅ m 2 .
3 45. Two forces act on the ball, the force of the rod and the force of gravity. No torque
about the pivot point is associated with the force of the rod since that force is along the
line from the pivot point to the ball. As can be seen from the diagram, the component of
the force of gravity that is perpendicular to the rod is mg sin θ. If is the length of the
rod, then the torque associated with this force has magnitude τ = mg sin θ = (0.75)(9.8)(1.25) sin 30° = 4.6 N ⋅ m .
For the position shown, the torque is counterclockwise. .
47. We take a torque that tends to cause a counterclockwise rotation from rest to be
positive and a torque tending to cause a clockwise rotation to be negative. Thus, a
positive torque of magnitude r1 F1 sin θ1 is associated with F1 and a negative torque of
magnitude r2F2 sin θ2 is associated with F2 . The net torque is consequently τ = r1 F1 sin θ 1 − r2 F2 sin θ 2 .
Substituting the given values, we obtain τ = (1.30 m)(4.20 N) sin 75° − (2.15 m)(4.90 N) sin 60° = −3.85 N ⋅ m.
49. (a) We use the kinematic equation ω = ω 0 + αt , where ω0 is the initial angular
velocity, ω is the final angular velocity, α is the angular acceleration, and t is the time.
This gives α= ω −ω0 = t 6.20 rad / s
= 28.2 rad / s2 .
−3
220 × 10 s (b) If I is the rotational inertia of the diver, then the magnitude of the torque acting on her
is c hc h τ = Iα = 12.0 kg ⋅ m2 28.2 rad / s2 = 3.38 × 102 N ⋅ m.
63. We use to denote the length of the stick. Since its center of mass is / 2 from
1
either end, its initial potential energy is 2 mg , where m is its mass. Its initial kinetic
1
energy is zero. Its final potential energy is zero, and its final kinetic energy is 2 Iω 2 ,
where I is its rotational inertia about an axis passing through one end of the stick and ω is
the angular velocity just before it hits the floor. Conservation of energy yields 1
1
mg
mg = Iω 2 ⇒ ω =
.
2
2
I
The free end of the stick is a distance
floor is (from Eq. 1018) from the rotation axis, so its speed as it hits the v =ω = mg 3
.
I Using Table 102 and the parallelaxis theorem, the rotational inertial is I = 1 m 2 , so
3 c hb g v = 3g = 3 9.8 m / s2 1.00 m = 5.42 m / s.
69. We choose positive coordinate directions (different choices for each item) so that
each is accelerating positively, which will allow us to set a2 = a1 = Rα (for simplicity, we
denote this as a). Thus, we choose rightward positive for m2 = M (the block on the table),
downward positive for m1 = M (the block at the end of the string) and (somewhat
unconventionally) clockwise for positive sense of disk rotation. This means that we
interpret θ given in the problem as a positivevalued quantity. Applying Newton’s second
law to m1, m2 and (in the form of Eq. 1045) to M, respectively, we arrive at the following
three equations (where we allow for the possibility of friction f2 acting on m2).
m1 g − T1 = m1a1
T2 − f 2 = m2 a2
T1 R − T2 R = Iα (a) From Eq. 1013 (with ω0 = 0) we find 1
2 θ = ω0t + α t 2 ⇒ α = 2θ 2(1.30 rad)
=
= 314 rad/s 2 .
2
2
t
(0.0910 s) (b) From the fact that a = Rα (noted above), we obtain
a= 2 Rθ 2(0.024 m)(1.30 rad)
=
= 7.54 m/s 2 .
t2
(0.0910 s) 2 (c) From the first of the above equations, we find
2 Rθ
⎛
T1 = m1 ( g − a1 ) = M ⎜ g − 2
t
⎝ ⎛
2(0.024 m)(1.30 rad) ⎞
⎞
2
⎟ = 14.0 N.
⎟ = (6.20 kg) ⎜ 9.80 m/s −
(0.0910 s) 2
⎠
⎝
⎠ (d) From the last of the above equations, we obtain the second tension:
T2 = T1 − Iα
2 Rθ
⎛
= M ⎜g− 2
R
t
⎝ (7.40 × 10−4 kg ⋅ m 2 )(314 rad/s 2 )
⎞ 2 Iθ
− 2 = 14.0 N −
= 4.36 N.
⎟
0.024 m
⎠ Rt 87. With rightward positive for the block and clockwise negative for the wheel (as is
conventional), then we note that the tangential acceleration of the wheel is of opposite
sign from the block’s acceleration (which we simply denote as a); that is, at = – a.
Applying Newton’s second law to the block leads to P − T = ma , where m = 2.0 kg.
Applying Newton’s second law (for rotation) to the wheel leads to −TR = Iα , where
I = 0.050 kg ⋅ m 2 .
Noting that Rα = at = – a, we multiply this equation by R and obtain
−TR 2 = − Ia ⇒ T = a I
.
R2 Adding this to the above equation (for the block) leads to P = (m + I / R 2 )a.
Thus, a = 0.92 m/s2 and therefore α = – 4.6 rad/s2 (or α = 4.6 rad/s2 ), where the negative
sign in α should not be mistaken for a deceleration (it simply indicates the clockwise
sense to the motion).
89. We assume the sense of initial rotation is positive. Then, with ω0 > 0 and ω = 0 (since
it stops at time t), our angular acceleration is negativevalued.
(a) The angular acceleration is constant, so we can apply Eq. 1012 (ω = ω0 + αt). To
obtain the requested units, we have t = 30/60 = 0.50 min. Thus, α =− 33.33 rev/min
= − 66.7 rev/min 2 ≈ − 67 rev/min 2 .
0.50 min 1
1
(b) We use Eq. 1013: θ = ω 0t + αt 2 = ( 33.33) ( 0.50) + ( −66.7) ( 0.50) 2 = 8.3 rev.
2
2 91. (a) According to Table 102, the rotational inertia formulas for the cylinder (radius R)
and the hoop (radius r) are given by
IC = 1
MR 2 and I H = Mr 2 .
2 Since the two bodies have the same mass, then they will have the same rotational inertia
if
2
R 2 / 2 = RH → R H = R / 2 .
(b) We require the rotational inertia to be written as I = Mk 2 , where M is the mass of the
given body and k is the radius of the “equivalent hoop.” It follows directly that
k= I/M .
115. We employ energy methods in this solution; thus, considerations of positive versus
negative sense (regarding the rotation of the wheel) are not relevant.
(a) The speed of the box is related to the angular speed of the wheel by v = Rω, so that Kbox = 1
2 Kbox
mbox v 2 ⇒ v =
= 141 m / s
.
2
mbox implies that the angular speed is ω = 1.41/0.20 = 0.71 rad/s. Thus, the kinetic energy of
1
rotation is 2 Iω 2 = 10.0 J.
(b) Since it was released from rest at what we will consider to be the reference position
for gravitational potential, then (with SI units understood) energy conservation requires K0 + U 0 = K + U
Therefore, h = 16.0/58.8 = 0.27 m. ⇒ 0 + 0 = ( 6.0 + 10.0 ) + mbox g ( −h ) . Chapter 11 – Student Solutions Manual
5. By Eq. 1052, the work required to stop the hoop is the negative of the initial kinetic
1
1
energy of the hoop. The initial kinetic energy is K = 2 Iω 2 + 2 mv 2 (Eq. 115), where I =
mR2 is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the
speed of its center of mass. Eq. 112 relates the angular speed to the speed of the center of
mass: ω = v/R. Thus, K= ⎛ v2 ⎞ 1
1
2
mR 2 ⎜ 2 ⎟ + mv 2 = mv 2 = (140 kg )( 0.150 m/s )
2
⎝R ⎠ 2 which implies that the work required is – 3.15 J.
23. If we write r = x i + yj + zk, then (using Eq. 330) we find r × F is equal to d yF − zF i i + bzF − xF g j + d xF − yF i k.
z y x z y x ˆ
ˆ
(a) Plugging in, we find τ = ⎡( 3.0 m )( 6.0 N ) − ( 4.0 m )( −8.0 N ) ⎤ k = (50 N ⋅ m) k.
⎣
⎦ (b) We use Eq. 327,  r × F  = rF sin φ , where φ is the angle between r and F . Now
r = x 2 + y 2 = 5.0 m and F = Fx2 + Fy2 = 10 N. Thus, b gb g rF = 5.0 m 10 N = 50 N ⋅ m, the same as the magnitude of the vector product calculated in part (a). This implies sin φ
= 1 and φ = 90°.
29. (a) We use = mr × v , where r is the position vector of the object, v is its velocity
vector, and m is its mass. Only the x and z components of the position and velocity
vectors are nonzero, so Eq. 330 leads to r × v = − xvz + zvz j. Therefore, b g = m ( − xvz + zvx ) ˆ = ( 0.25 kg ) ( − ( 2.0 m )( 5.0 m s ) + ( −2.0 m )( −5.0 m s ) ) ˆ = 0.
j
j (b) If we write r = x i + yj + zk, then (using Eq. 330) we find r × F is equal to d yF − zF i i + bzF − xF g j + d xF − yF i k.
z y x z y x With x = 2.0, z = –2.0, Fy = 4.0 and all other components zero (and SI units understood)
the expression above yields e j τ = r × F = 8.0 i + 8.0 k N ⋅ m.
33. If we write (for the general case) r = x i + yj + zk, then (using Eq. 330) we find r × v
is equal to d yv − zv i i + bzv
z y x g d i − xvz j + xv y − yv x k. (a) The angular momentum is given by the vector product = mr × v , where r is the
position vector of the particle, v is its velocity, and m = 3.0 kg is its mass. Substituting
(with SI units understood) x = 3, y = 8, z = 0, vx = 5, vy = –6 and vz = 0 into the above
expression, we obtain
ˆ
ˆ
= ( 3.0 ) [(3.0)(−6.0) − (8.0)(5.0)]k = (−1.7 ×102 kg ⋅ m 2 s)k.
(b) The torque is given by Eq. 1114, τ = r × F. We write r = x i + yj and F = Fx i and
obtain
τ = x i + yj × Fx i = − yFx k e j e j since i × i = 0 and j × i = − k. Thus, we find
ˆ
ˆ
τ = − ( 8.0m )( −7.0N ) k = (56 N ⋅ m)k.
(c) According to Newton’s second law τ = d dt , so the rate of change of the angular
momentum is 56 kg ⋅ m2/s2, in the positive z direction.
37. (a) Since τ = dL/dt, the average torque acting during any interval Δ t is given by
τ avg = L f − Li Δt , where Li is the initial angular momentum and Lf is the final angular d i momentum. Thus, τ avg = 0.800 kg ⋅ m 2 s − 3.00 kg ⋅ m 2 s
= −1.47 N ⋅ m ,
1.50s or τ avg  = 1.47 N ⋅ m . In this case the negative sign indicates that the direction of the
torque is opposite the direction of the initial angular momentum, implicitly taken to be
positive.
(b) The angle turned is θ = ω0t + α t 2 / 2. If the angular acceleration α is uniform, then so
is the torque and α = τ/I. Furthermore, ω0 = Li/I, and we obtain 2
Li t + τ t 2 / 2 ( 3.00 kg ⋅ m s ) (1.50s ) + ( −1.467 N ⋅ m )(1.50s ) / 2
=
= 20.4 rad.
θ=
0.140 kg ⋅ m 2
I
2 (c) The work done on the wheel is W = τθ = ( −1.47 N ⋅ m )( 20.4 rad ) = −29.9 J
where more precise values are used in the calculation than what is shown here. An
equally good method for finding W is Eq. 1052, which, if desired, can be rewritten as
W = ( L2f − L2 ) 2 I .
i
(d) The average power is the work done by the flywheel (the negative of the work done
on the flywheel) divided by the time interval:
43. (a) No external torques act on the system consisting of the man, bricks, and platform,
so the total angular momentum of the system is conserved. Let Ii be the initial rotational
inertia of the system and let If be the final rotational inertia. Then Iiωi = Ifωf and
⎛ Ii
⎜
⎝ If ωf = ⎜ ⎞
⎛ 6.0 kg ⋅ m 2 ⎞
1.2 rev s ) = 3.6 rev s.
⎟ ωi = ⎜
2 ⎟(
⎟
⎝ 2.0 kg ⋅ m ⎠
⎠ (b) The initial kinetic energy is Ki =
and their ratio is Kf
Ki = 1
1
I iω i2 , the final kinetic energy is K f = I f ω 2f ,
2
2 ( 2.0 kg ⋅ m ) ( 3.6 rev s )
=
/ 2 ( 6.0 kg ⋅ m ) (1.2 rev s )
2 I iω 2
i 2 2 I f ω2 / 2
f 2 /2 = 3.0. /2 (c) The man did work in decreasing the rotational inertia by pulling the bricks closer to
his body. This energy came from the man’s store of internal energy.
45. (a) No external torques act on the system consisting of the two wheels, so its total
angular momentum is conserved. Let I1 be the rotational inertia of the wheel that is
originally spinning at ω i and I2 be the rotational inertia of the wheel that is initially at b rest. Then I1 ω i g
= b I + I gω
1 2 f and ωf = I1
ωi
I1 + I 2 where ω f is the common final angular velocity of the wheels. Substituting I2 = 2I1 and ω i = 800 rev min, we obtain ω f = 267 rev min. 1
(b) The initial kinetic energy is Ki = 2 I1ω i2 and the final kinetic energy is Kf = 1
2 b I + I gω
1 2 2
f . We rewrite this as gFGH I I+ω2 I IJK b 1
K f = I1 + 2 I1
2 d Therefore, the fraction lost, Ki − K f
1− Kf
Ki 1 1 iK =1− i i 2 = 1 1 2
Iω i .
6 is I ωi2 / 6 2
= = 0.667.
I ωi2 / 2 3 49. No external torques act on the system consisting of the train and wheel, so the total
angular momentum of the system (which is initially zero) remains zero. Let I = MR2 be
the rotational inertia of the wheel. Its final angular momentum is
→ Lf = Iωk = − M R 2 ω k,
where k is up in Fig. 1147 and that last step (with the minus sign) is done in recognition
that the wheel’s clockwise rotation implies a negative value for ω. The linear speed of a
point on the track is ωR and the speed of the train (going counterclockwise in Fig. 1147
with speed v ′ relative to an outside observer) is therefore v ′ = v − ω R where v is its
speed relative to the tracks. Consequently, the angular momentum of the train is
m v − ω R R k . Conservation of angular momentum yields c h c h 0 = − MR 2 ω k + m v − ω R Rk.
When this equation is solved for the angular speed, the result is  ω = mvR
v
(0.15 m/s)
=
=
= 0.17 rad/s.
2
( M + m ) R ( M / m + 1) R (1.1+1)(0.43 m) 67. (a) If we consider a short time interval from just before the wad hits to just after it hits
and sticks, we may use the principle of conservation of angular momentum. The initial
angular momentum is the angular momentum of the falling putty wad. The wad initially
moves along a line that is d/2 distant from the axis of rotation, where d = 0.500 m is the
length of the rod. The angular momentum of the wad is mvd/2 where m = 0.0500 kg and
v = 3.00 m/s are the mass and initial speed of the wad. After the wad sticks, the rod has
angular velocity ω and angular momentum Iω, where I is the rotational inertia of the
system consisting of the rod with the two balls and the wad at its end. Conservation of
angular momentum yields mvd/2 = Iω where I = (2M + m)(d/2)2
and M = 2.00 kg is the mass of each of the balls. We solve b gb g mvd 2 = 2 M + m d 2 ω
2 for the angular speed: ω= 2 ( 0.0500 kg )( 3.00 m/s )
2mv
=
= 0.148 rad s.
( 2M + m ) d ( 2 ( 2.00 kg ) + 0.0500 kg ) ( 0.500 m ) 1
1
(b) The initial kinetic energy is Ki = 2 mv 2 , the final kinetic energy is K f = 2 Iω 2 , and b b g g their ratio is K f Ki = Iω 2 mv 2 . When I = 2 M + m d 2 4 and ω = 2mv 2 M + m d
are substituted, this becomes
Kf
Ki = 0.0500 kg
m
=
= 0.0123.
2 M + m 2 ( 2.00 kg ) + 0.0500 kg (c) As the rod rotates, the sum of its kinetic and potential energies is conserved. If one of
the balls is lowered a distance h, the other is raised the same distance and the sum of the
potential energies of the balls does not change. We need consider only the potential
energy of the putty wad. It moves through a 90° arc to reach the lowest point on its path,
gaining kinetic energy and losing gravitational potential energy as it goes. It then swings
up through an angle θ, losing kinetic energy and gaining potential energy, until it
momentarily comes to rest. Take the lowest point on the path to be the zero of potential
energy. It starts a distance d/2 above this point, so its initial potential energy is Ui =
mgd/2. If it swings up to the angular position θ, as measured from its lowest point, then
its final height is (d/2)(1 – cos θ) above the lowest point and its final potential energy is b gb g U f = mg d 2 1 − cosθ .
The initial kinetic energy is the sum of that of the balls and wad:
Ki = 1 2 1
2
I ω = ( 2 M + m )( d 2 ) ω 2 .
2
2 At its final position, we have Kf = 0. Conservation of energy provides the relation: b gFGH IJK ω d 1
d
mg + 2 M + m
2 2
2 2 2 = mg When this equation is solved for cos θ, the result is b g d
1 − cosθ .
2 1 ⎛ 2M + m ⎞ ⎛ d ⎞ 2
1 ⎛ 2 ( 2.00 kg ) + 0.0500 kg ⎞ ⎛ 0.500 m ⎞
2
⎟⎜
cos θ = − ⎜
⎟ ⎜ ⎟ω = − ⎜
⎟ ( 0.148 rad s )
2
2 ⎝ mg ⎠ ⎝ 2 ⎠
2 ⎜ ( 0.0500 kg ) 9.8 m s ⎟ ⎝
2
⎠
⎝
⎠
= −0.0226. ( ) Consequently, the result for θ is 91.3°. The total angle through which it has swung is 90°
+ 91.3° = 181°.
73. (a) The diagram below shows the particles and their lines of motion. The origin is
marked O and may be anywhere. The angular momentum of particle 1 has magnitude
1 = mvr1 sin θ1 = mv ( d + h ) and it is into the page. The angular momentum of
particle 2 has magnitude
2 = mvr2 sin θ 2 = mvh and it is out of the page. The net angular
momentum has magnitude
L = mv(d + h) − mvh = mvd
= (2.90 ×10−4 kg)(5.46 m/s)(0.042 m)
= 6.65 ×10−5 kg ⋅ m 2 /s. and is into the page. This result is independent of the location of the origin.
(b) As indicated above, the expression does not change.
(c) Suppose particle 2 is traveling to the right. Then
L = mv(d + h) + mvh = mv(d + 2h).
This result depends on h, the distance from the origin to one of the lines of motion. If the
origin is midway between the lines of motion, then h = − d 2 and L = 0.
(d) As we have seen in part (c), the result depends on the choice of origin.
77. As the wheelaxel system rolls down the inclined plane by a distance d, the decrease
in potential energy is ΔU = mgd sin θ . This must be equal to the total kinetic energy
gained:
1
1
mgd sin θ = mv 2 + Iω 2 .
2
2 Since the axel rolls without slipping, the angular speed is given by ω = v / r , where r is
the radius of the axel. The above equation then becomes ⎛ mr 2 ⎞
⎛ mr 2 ⎞
1
mgd sin θ = Iω 2 ⎜
+ 1⎟ = K rot ⎜
+ 1⎟
2
⎝ I
⎠
⎝ I
⎠
(a) With m=10.0 kg, d = 2.00 m, r = 0.200 m, and I = 0.600 kg m2, mr2/I =2/3, the
rotational kinetic energy may be obtained as 98 J = K rot (5 / 3) , or K rot = 58.8 J .
(b) The translational kinetic energy is K trans = (98 − 58.8)J = 39.2 J.
85. (a) In terms of the radius of gyration k, the rotational inertia of the merrygoround is
I = Mk2. We obtain
I = (180 kg) (0.910 m)2 = 149 kg ⋅ m2.
(b) An object moving along a straight line has angular momentum about any point that is
not on the line. The magnitude of the angular momentum of the child about the center of
the merrygoround is given by Eq. 1121, mvR, where R is the radius of the merrygoround. Therefore, b gb gb g Lchild = 44.0 kg 3.00 m s 1.20 m = 158 kg ⋅ m2 / s. (c) No external torques act on the system consisting of the child and the merrygoround,
so the total angular momentum of the system is conserved. The initial angular momentum
is given by mvR; the final angular momentum is given by (I + mR2) ω, where ω is the
final common angular velocity of the merrygoround and child. Thus mvR = I + mR 2 ω c h and ω= mvR
158 kg ⋅ m2 s
=
I + mR 2 149 kg ⋅ m2 + 44.0 kg 120 m
. b gb g 2 = 0.744 rad s . 87. This problem involves the vector cross product of vectors lying in the xy plane. For
such vectors, if we write r ′ = x ′ i + y ′j , then (using Eq. 330) we find d i r ′ × v = x ′v y − y ′v x k. (a) Here, r ′ points in either the + i or the − i direction (since the particle moves along
the x axis). It has no y′ or z ′ components, and neither does v , so it is clear from the
above expression (or, more simply, from the fact that i × i = 0 ) that
this case. b g = m r ′ × v = 0 in (b) The net force is in the − i direction (as one finds from differentiating the velocity
expression, yielding the acceleration), so, similar to what we found in part (a), we obtain
τ = r′× F = 0.
(c) Now, r ′ = r − ro where ro = 2.0 i + 5.0 j (with SI units understood) and points from
(2.0, 5.0, 0) to the instantaneous position of the car (indicated by r which points in either
the +x or –x directions, or nowhere (if the car is passing through the origin)). Since
r × v = 0 we have (plugging into our general expression above) b g b g b g eb2.0gb0g − b5.0gc−2.0t hj k = m r ′ × v = − m ro × v = − 3.0
which yields c 3 h = −30t 3 k in SI units kg ⋅ m2 s . (d) The acceleration vector is given by a = dv = −6.0t 2 i in SI units, and the net force on
dt
the car is ma . In a similar argument to that given in the previous part, we have b b g g b g eb2.0gb0g − b5.0gc−6.0t hj k τ = m r ′ × a = −m ro × a = − 3.0 2 which yields τ = −90t 2 k in SI units (N ⋅ m) .
(e) In this situation, r ′ = r − ro where ro = 2.0 i − 5.0 j (with SI units understood) and
points from (2.0, –5.0, 0) to the instantaneous position of the car (indicated by r which
points in either the +x or –x directions, or nowhere (if the car is passing through the
origin)). Since r × v = 0 we have (plugging into our general expression above) b g b g b g eb2.0gb0g − b−5.0gc−2.0t hj k = m r ′ × v = − m ro × v = − 3.0
which yields c 3 h = 30t 3 k in SI units kg ⋅ m2 s . (f) Again, the acceleration vector is given by a = −6.0t 2 i in SI units, and the net force on
the car is ma . In a similar argument to that given in the previous part, we have b g b g eb2.0gb0g − b−5.0gc−6.0t hj k
which yields τ = 90t k in SI units b N ⋅ mg.
g b τ = m r ′ × a = −m ro × a = − 3.0 2 2 95. We make the unconventional choice of clockwise sense as positive, so that the
angular acceleration is positive (as is the linear acceleration of the center of mass, since
we take rightwards as positive). (a) We approach this in the manner of Eq. 113 (pure rotation about point P) but use
torques instead of energy. The torque (relative to point P) is τ = I Pα , where
IP = 1
3
MR 2 + MR 2 = MR 2
2
2 with the use of the parallelaxis theorem and Table 102(c). The torque is due to the
Fapp = 12 N force and can be written as τ = Fapp (2 R ) . In this way, we find
⎛3 ⎞ τ = I Pα = ⎜ MR 2 ⎟ α = 2 RFapp
⎝2
⎠
which leads to α= 2 RFapp
2 3MR /2 = 4 Fapp
3MR = 4 (12 N )
= 16 rad/s 2 .
3(10 kg)(0.10 m) Hence, acom = Rα = 1.6 m/s2.
(b) As shown above, α = 16 rad/s2. b g (c) Applying Newton’s second law in its linear form yields 12 N − f = Macom .
Therefore, f = –4.0 N. Contradicting what we assumed in setting up our force equation,
ˆ
the friction force is found to point rightward with magnitude 4.0 N, i.e., f = (4.0 N)i . Chapter 12 – Student Solutions Manual
5. Three forces act on the sphere: the tension force T of the rope
(acting along the rope), the force of the wall FN (acting horizontally
away from the wall), and the force of gravity mg (acting
downward). Since the sphere is in equilibrium they sum to zero. Let
θ be the angle between the rope and the vertical. Then Newton’s
second law gives
vertical component :
horizontal component: T cos θ – mg = 0
FN – T sin θ = 0. (a) We solve the first equation for the tension: T = mg/ cos θ. We
substitute cosθ = L / L2 + r 2 to obtain
T= mg L2 + r 2 (0.85 kg)(9.8 m/s 2 ) (0.080 m) 2 + (0.042 m) 2
=
= 9.4 N .
L
0.080 m (b) We solve the second equation for the normal force: FN = T sin θ .
Using sinθ = r / L2 + r 2 , we obtain FN = Tr
L2 + r 2 = mg L2 + r 2
L r
L2 + r 2 = mgr (0.85 kg)(9.8 m/s 2 )(0.042 m)
=
= 4.4 N.
L
(0.080 m) 7. We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the
diving board. We take the force of the right pedestal to be F2 and denote its position as x
= d. W is the weight of the diver, located at x = L. The following two equations result
from setting the sum of forces equal to zero (with upwards positive), and the sum of
torques (about x2) equal to zero:
F1 + F2 − W = 0
F1d + W ( L − d ) = 0
(a) The second equation gives F1 = − ⎛ 3.0 m ⎞
L−d
W = −⎜
⎟ (580 N)= − 1160 N
d
⎝ 1.5 m ⎠ which should be rounded off to F1 = −1.2 × 103 N . Thus,  F1 = 1.2 × 103 N.
(b) Since F1 is negative, indicating that this force is downward. (c) The first equation gives F2 = W − F1 = 580 N+1160 N=1740 N
which should be rounded off to F2 = 1.7 × 103 N . Thus,  F2 = 1.7 × 103 N.
(d) The result is positive, indicating that this force is upward.
(e) The force of the diving board on the left pedestal is upward (opposite to the force of
the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is
being compressed.
11. The x axis is along the meter stick, with the origin at the
zero position on the scale. The forces acting on it are shown
on the diagram below. The nickels are at x = x1 = 0.120 m,
and m is their total mass. The knife edge is at x = x2 = 0.455 m
and exerts force F . The mass of the meter stick is M, and the
force of gravity acts at the center of the stick, x = x3 = 0.500
m. Since the meter stick is in equilibrium, the sum of the
torques about x2 must vanish:
Mg(x3 – x2) – mg(x2 – x1) = 0.
Thus, M= ⎛ 0.455 m − 0.120 m ⎞
x2 − x1
m=⎜
⎟ (10.0g)=74.4 g.
x3 − x2
⎝ 0.500 m − 0.455 m ⎠ 21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force
on the wheel, and the only forces acting are the force F applied horizontally at the axle,
the force of gravity mg acting vertically at the center of the wheel, and the force of the
step corner, shown as the two components fh and fv. If the minimum force is applied the
wheel does not accelerate, so both the total force and the total torque acting on it are zero. We calculate the torque around the step corner. The second diagram indicates that the
distance from the line of F to the corner is r – h, where r is the radius of the wheel and h
is the height of the step.
The distance from the line of mg to the corner is b g r2 + r − h 2 = 2rh − h 2 . Thus, b g F r − h − mg 2rh − h 2 = 0 . The solution for F is
2(6.00 ×10−2 m)(3.00 × 10−2 m) − (3.00 × 10−2 m) 2
2rh − h 2
F=
mg =
(0.800 kg)(9.80 m/s 2 )
−2
−2
r−h
(6.00 × 10 m) − (3.00 × 10 m)
= 13.6 N.
33. We examine the box when it is about to tip. Since it will rotate about the lower right
edge, that is where the normal force of the floor is exerted. This force is labeled FN on
the diagram below. The force of friction is denoted by f, the applied force by F, and the
force of gravity by W. Note that the force of gravity is applied at the center of the box.
When the minimum force is applied the box does not accelerate, so the sum of the
horizontal force components vanishes: F – f = 0, the sum of the vertical force components
vanishes: FN − W = 0 , and the sum of the torques vanishes:
FL – WL/2 = 0.
Here L is the length of a side of the box and the origin was chosen to be at the lower right
edge. (a) From the torque equation, we find
F= W 890 N
=
= 445 N.
2
2 (b) The coefficient of static friction must be large enough that the box does not slip. The
box is on the verge of slipping if μs = f/FN. According to the equations of equilibrium FN
= W = 890 N and f = F = 445 N, so μs = 445 N
= 0.50.
890 N (c) The box can be rolled with a smaller applied force if the force points upward as well
as to the right. Let θ be the angle the force makes with the horizontal. The torque
equation then becomes
FL cos θ + FL sin θ – WL/2 = 0,
with the solution
F= W
.
2(cos θ + sin θ ) We want cosθ + sinθ to have the largest possible value. This occurs if θ = 45º, a result we
can prove by setting the derivative of cosθ + sinθ equal to zero and solving for θ. The
minimum force needed is
F= W
890 N
=
= 315 N.
4 cos 45° 4 cos 45° 43. (a) The shear stress is given by F/A, where F is the magnitude of the force applied
parallel to one face of the aluminum rod and A is the cross–sectional area of the rod. In
this case F is the weight of the object hung on the end: F = mg, where m is the mass of
the object. If r is the radius of the rod then A = πr2. Thus, the shear stress is
F mg (1200 kg) (9.8 m/s2 )
= 2 =
= 6.5 × 106 N/m2 .
2
A πr
π (0.024 m)
(b) The shear modulus G is given by
G= F/A
Δx / L where L is the protrusion of the rod and Δx is its vertical deflection at its end. Thus, Δx = ( F / A) L (6.5 × 106 N/m2 ) (0.053m)
=
= 1.1× 10−5 m.
10
2
G
3.0 × 10 N/m 55. (a) The forces acting on bucket are the force of gravity, down, and the tension force
of cable A, up. Since the bucket is in equilibrium and its weight is
WB = mB g = ( 817kg ) ( 9.80m/s 2 ) = 8.01×103 N , the tension force of cable A is TA = 8.01 × 103 N .
(b) We use the coordinates axes defined in the diagram. Cable A makes an angle of θ2 =
66.0º with the negative y axis, cable B makes an angle of 27.0º with the positive y axis,
and cable C is along the x axis. The y components of the forces must sum to zero since
the knot is in equilibrium. This means TB cos 27.0º – TA cos 66.0º = 0 and
B TB = cos 66.0°
⎛ cos 66.0° ⎞
3
3
TA = ⎜
⎟ (8.01×10 N) = 3.65 ×10 N.
cos 27.0°
⎝ cos 27.0° ⎠ (c) The x components must also sum to zero. This means TC + TB sin 27.0º – TA sin 66.0º
= 0 and
B TC = TA sin 66.0° − TB sin 27.0° = (8.01×103 N)sin 66.0° − (3.65 × 103 N)sin 27.0°
= 5.66 ×103 N.
61. We denote the mass of the slab as m, its density as ρ , and volume as V = LTW . The
angle of inclination is θ = 26° .
(a) The component of the weight of the slab along the incline is
F1 = mg sin θ = ρVg sin θ
= (3.2×103 kg/m3 )(43m)(2.5 m)(12 m)(9.8 m/s 2 ) sin 26° ≈ 1.8 ×107 N.
(b) The static force of friction is
f s = μ s FN = μ s mg cos θ = μ s ρVg cos θ
= (0.39)(3.2×103 kg/m3 )(43m)(2.5 m)(12 m)(9.8 m/s 2 ) cos 26° ≈ 1.4 ×107 N.
(c) The minimum force needed from the bolts to stabilize the slab is
F2 = F1 − f s = 1.77 × 107 N − 1.42 × 107 N = 3.5 × 106 N. If the minimum number of bolts needed is n, then F2 / nA ≤ 3.6 × 108 N/m 2 , or n≥ 3.5 × 106 N
= 15.2
(3.6 × 108 N/m 2 )(6.4 × 10−4 m 2 ) Thus 16 bolts are needed.
63. Analyzing forces at the knot (particularly helpful is a graphical view of the vector
right–triangle with horizontal “side” equal to the static friction force fs and vertical “side”
equal to the weight mBg of block B), we find fs = mBg tan θ where θ = 30°. For fs to be at
its maximum value, then it must equal μsmAg where the weight of block A is mAg= (10
kg)(9.8 m/s2). Therefore,
B B μ s mA g = mB g tan θ ⇒ μ s = 5.0
tan 30° = 0.29.
10 65. With the pivot at the hinge, Eq. 129 leads to
– mg sinθ1 2 + T L sin(180° – θ1 – θ2) = 0 .
L where θ1 = 60° and T = mg/2. This yields θ2 = 60°.
81. When it is about to move, we are still able to apply the equilibrium conditions, but (to
obtain the critical condition) we set static friction equal to its maximum value and picture
the normal force FN as a concentrated force (upward) at the bottom corner of the cube,
directly below the point O where P is being applied. Thus, the line of action of FN passes
through point O and exerts no torque about O (of course, a similar observation applied to
the pull P). Since FN = mg in this problem, we have fsmax = μmg applied a distance h
away from O. And the line of action of force of gravity (of magnitude mg), which is best
pictured as a concentrated force at the center of the cube, is a distance L/2 away from O.
Therefore, equilibrium of torques about O produces
⎛L⎞ L (8.0 cm) μ mgh = mg ⎜ ⎟ ⇒ μ =
=
= 0.57
2h 2(7.0 cm)
⎝2⎠
for the critical condition we have been considering. We now interpret this in terms of a
range of values for μ.
(a) For it to slide but not tip, a value of μ less than that derived above is needed, since
then — static friction will be exceeded for a smaller value of P, before the pull is strong
enough to cause it to tip. Thus, μ < L/2h = 0.57 is required.
(b) And for it to tip but not slide, we need μ greater than that derived above is needed,
since now — static friction will not be exceeded even for the value of P which makes the cube rotate about its front lower corner. That is, we need to have μ > L/2h = 0.57 in this
case.
85. We choose an axis through the top (where the ladder comes into contact with the
wall), perpendicular to the plane of the figure and take torques that would cause
counterclockwise rotation as positive. Note that the line of action of the applied force
F intersects the wall at a height of ( 8.0 m) / 5 = 1.6 m ; in other words, the moment arm
for the applied force (in terms of where we have chosen the axis) is
r⊥ = (4 / 5)(8.0 m) = 6.4 m . The moment arm for the weight is half the horizontal distance
from the wall to the base of the ladder; this works out to be
(10 m) 2 − (8 m) 2 / 2 = 3.0 m . Similarly, the moment arms for the x and y components of d i the force at the ground Fg are 8.0 m and 6.0 m, respectively. Thus, with lengths in
meters, we have
∑ τ z = F (6.4 m) + W (3.0 m) + Fgx (8.0 m) − Fgy (6.0 m) = 0. In addition, from balancing the vertical forces we find that W = Fgy (keeping in mind that
the wall has no friction). Therefore, the above equation can be written as
∑ τ z = F (6.4 m) + W (3.0 m) + Fgx (8.0 m) − W (6.0 m) = 0. (a) With F = 50 N and W = 200 N, the above equation yields Fgx = 35 N. Thus, in unit
vector notation we obtain
ˆ
ˆ
Fg = (35 N)i+(200 N)j.
(b) With F = 150 N and W = 200 N, the above equation yields Fgx = –45 N. Therefore, in
unit vector notation we obtain
ˆ
ˆ
Fg = (−45 N)i+(200 N)j.
(c) Note that the phrase “start to move towards the wall” implies that the friction force is
pointed away from the wall (in the − i direction). Now, if f = –Fgx and FN = Fgy = 200 N
are related by the (maximum) static friction relation (f = fs,max = μs FN) with μs = 0.38,
then we find Fgx = –76 N. Returning this to the above equation, we obtain
F= (200 N) (3.0 m) + (76 N) (8.0 m)
= 1.9 × 102 N.
6.4 m Chapter 13 – Student Solutions Manual
1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2,
where m1 and m2 are the masses, r is their separation, and G is the universal gravitational
constant. We solve for r:
Gm1m2
r=
=
F ( 6.67 ×10 −11 N ⋅ m 2 / kg 2 ) ( 5.2kg )( 2.4kg )
2.3 × 10−12 N = 19 m 7. At the point where the forces balance GM e m / r12 = GM s m / r22 , where Me is the mass of
Earth, Ms is the mass of the Sun, m is the mass of the space probe, r1 is the distance from
the center of Earth to the probe, and r2 is the distance from the center of the Sun to the
probe. We substitute r2 = d − r1, where d is the distance from the center of Earth to the
center of the Sun, to find
Me
Ms
=
.
2
2
r1
( d − r1 ) Taking the positive square root of both sides, we solve for r1. A little algebra yields r1 = (150 ×10 m )
9 d Me
Ms + Me = 5.98 ×1024 kg 1.99 ×10 kg + 5.98 ×10 kg
30 24 = 2.60 ×108 m. Values for Me, Ms, and d can be found in Appendix C.
17. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of Earth
and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius
of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h and obtain
h = GM / ag − R . According to Appendix C, R = 6.37 × 106 m and M = 5.98 × 1024 kg,
so
h= ( 6.67 ×10 −11 m 3 / s 2 ⋅ kg )( 5.98 × 1024 kg ) ( 4.9m / s )
2 − 6.37 × 106 m = 2.6 × 106 m. 29. (a) The density of a uniform sphere is given by ρ = 3M/4πR3, where M is its mass and
R is its radius. The ratio of the density of Mars to the density of Earth is
3 3
⎛ 0.65 × 104 km ⎞
ρ M M M RE
=
= 0.11⎜
⎟ = 0.74.
3
3
ρ E M E RM
⎝ 3.45 × 10 km ⎠ (b) The value of ag at the surface of a planet is given by ag = GM/R2, so the value for
Mars is
2 2
⎛ 0.65 × 104 km ⎞
M RE
2
2
ag M = M 2 a g E = 0.11⎜
⎟ ( 9.8 m/s ) = 3.8 m/s .
3
M E RM
⎝ 3.45 × 10 km ⎠ (c) If v is the escape speed, then, for a particle of mass m 1 2
mM
mv = G
2
R ⇒ v= 2GM
.
R For Mars, the escape speed is v= 2(6.67 × 10 −11 m3 /s 2 ⋅ kg) ( 0.11) ( 5.98 × 1024 kg )
3.45 × 10 m
6 = 5.0 × 103 m/s. 37. (a) We use the principle of conservation of energy. Initially the particle is at the
surface of the asteroid and has potential energy Ui = −GMm/R, where M is the mass of
the asteroid, R is its radius, and m is the mass of the particle being fired upward. The
initial kinetic energy is 1 2 mv 2 . The particle just escapes if its kinetic energy is zero when
it is infinitely far from the asteroid. The final potential and kinetic energies are both zero.
Conservation of energy yields −GMm/R + ½mv2 = 0. We replace GM/R with agR, where
ag is the acceleration due to gravity at the surface. Then, the energy equation becomes
−agR + ½v2 = 0. We solve for v:
v= 2a g R = 2(3.0 m/s 2 ) (500 × 103 m) = 1.7 × 103 m/s. (b) Initially the particle is at the surface; the potential energy is Ui = −GMm/R and the
kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface when it
momentarily comes to rest. The final potential energy is Uf = −GMm/(R + h) and the final
kinetic energy is Kf = 0. Conservation of energy yields
− GMm 1 2
GMm
.
+ mv = −
2
R
R+h We replace GM with agR2 and cancel m in the energy equation to obtain
− ag R + The solution for h is a R2
1 2
.
v =− g
2
( R + h) h= 2a g R 2
2a g R − v 2 −R = 2(3.0 m/s 2 ) (500 × 103 m) 2
− (500 × 103 m)
2(3.0 m/s 2 ) (500 × 103 m) − (1000 m/s) 2 = 2.5 × 105 m. (c) Initially the particle is a distance h above the surface and is at rest. Its potential energy
is Ui = −GMm/(R + h) and its initial kinetic energy is Ki = 0. Just before it hits the
asteroid its potential energy is Uf = −GMm/R. Write 1 2 mv 2 for the final kinetic energy.
f
Conservation of energy yields
− GMm
GMm 1 2
=−
+ mv .
R+h
R
2 We substitute agR2 for GM and cancel m, obtaining
− ag R 2
R+h = − ag R + 1 2
v .
2 The solution for v is
v = 2ag R − 2a g R 2
R+h = 2(3.0 m/s 2 ) (500 × 103 m) − 2(3.0 m/s 2 )(500 × 103 m) 2
(500 × 103 m) + (1000 × 103 m) = 1.4 × 103 m/s.
39. (a) The momentum of the twostar system is conserved, and since the stars have the
same mass, their speeds and kinetic energies are the same. We use the principle of
conservation of energy. The initial potential energy is Ui = −GM2/ri, where M is the mass
of either star and ri is their initial centertocenter separation. The initial kinetic energy is
zero since the stars are at rest. The final potential energy is Uf = −2GM2/ri since the final
separation is ri/2. We write Mv2 for the final kinetic energy of the system. This is the sum
of two terms, each of which is ½Mv2. Conservation of energy yields
GM 2
2GM 2
−
=−
+ Mv 2 .
ri
ri
The solution for v is
v= GM
=
ri (6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg)
= 8.2 × 104 m/s.
1010 m (b) Now the final separation of the centers is rf = 2R = 2 × 105 m, where R is the radius of
either of the stars. The final potential energy is given by Uf = −GM2/rf and the energy
equation becomes −GM2/ri = −GM2/rf + Mv2. The solution for v is ⎛1
⎛
1⎞
1
1 ⎞
v = GM ⎜ − ⎟ = (6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg) ⎜
− 10 ⎟
5
⎜r
ri ⎟
⎝ 2 × 10 m 10 m ⎠
⎝ f
⎠
= 1.8 × 107 m/s. 45. Let N be the number of stars in the galaxy, M be the mass of the Sun, and r be the
radius of the galaxy. The total mass in the galaxy is N M and the magnitude of the
gravitational force acting on the Sun is F = GNM2/r2. The force points toward the galactic
center. The magnitude of the Sun’s acceleration is a = v2/R, where v is its speed. If T is
the period of the Sun’s motion around the galactic center then v = 2πR/T and a = 4π2R/T2.
Newton’s second law yields GNM2/R2 = 4π2MR/T2. The solution for N is
N = 4π 2 R 3
.
GT 2 M The period is 2.5 × 108 y, which is 7.88 × 1015 s, so
N = 4π 2 (2.2 × 1020 m)3
= 5.1 × 1010.
(6.67 × 10−11 m3 / s 2 ⋅ kg) (7.88 × 1015 s) 2 (2.0 × 1030 kg) 47. (a) The greatest distance between the satellite and Earth’s center (the apogee distance)
is Ra = (6.37 × 106 m + 360 × 103 m) = 6.73 × 106 m. The least distance (perigee distance)
is Rp = (6.37 × 106 m + 180 × 103 m) = 6.55 × 106 m. Here 6.37 × 106 m is the radius of
Earth. From Fig. 1313, we see that the semimajor axis is
a= Ra + R p
2 = 6.73 × 106 m + 6.55 × 106 m
= 6.64 × 106 m.
2 (b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and
Rp = a(1 − e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain Ra − Rp
= 2ae. Thus,
e= Ra − R p
2a = Ra − R p
Ra + R p = 6.73 × 106 m − 6.55 × 106 m
= 0.0136.
6.73 × 106 m + 6.55 × 106 m 61. (a) We use the law of periods: T2 = (4π2/GM)r3, where M is the mass of the Sun (1.99
× 1030 kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of
Earth’s orbit: r = 2re = 2(150 × 109 m) = 300 × 109 m. Thus,
T = 4π 2 r 3
4π 2 (300 × 109 m)3
=
= 8.96 × 107 s.
−11
3
2
30
GM
(6.67 × 10 m / s ⋅ kg) (1.99 × 10 kg) Dividing by (365 d/y) (24 h/d) (60 min/h) (60 s/min), we obtain T = 2.8 y.
(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given by
K = GMm/2r, where m is the mass of the asteroid or planet. We note that it is
proportional to m and inversely proportional to r. The ratio of the kinetic energy of the
asteroid to the kinetic energy of Earth is K/Ke = (m/me) (re/r). We substitute m = 2.0 ×
10−4me and r = 2re to obtain K/Ke = 1.0 × 10−4.
75. (a) Using Kepler’s law of periods, we obtain ⎛ 4π 2 ⎞ 3
4
T = ⎜
⎟ r = 2.15 × 10 s .
GM ⎠
⎝
(b) The speed is constant (before she fires the thrusters), so vo = 2πr/T = 1.23 × 104 m/s.
(c) A two percent reduction in the previous value gives v = 0.98vo = 1.20 × 104 m/s.
(d) The kinetic energy is K = ½mv2 = 2.17 × 1011 J.
(e) The potential energy is U = −GmM/r = −4.53 × 1011 J.
(f) Adding these two results gives E = K + U = −2.35 × 1011 J.
(g) Using Eq. 1342, we find the semimajor axis to be
a= −GMm
= 4.04 × 107 m .
2E (h) Using Kepler’s law of periods for elliptical orbits (using a instead of r) we find the
new period is ⎛ 4π 2 ⎞ 3
4
T′ = ⎜
⎟ a = 2.03 × 10 s .
⎝ GM ⎠
This is smaller than our result for part (a) by T − T´ = 1.22 × 103 s.
(i) Elliptical orbit has a smaller period.
79. We use F = Gmsmm/r2, where ms is the mass of the satellite, mm is the mass of the
meteor, and r is the distance between their centers. The distance between centers is r = R
+ d = 15 m + 3 m = 18 m. Here R is the radius of the satellite and d is the distance from
its surface to the center of the meteor. Thus, ( 6.67 ×10
F= −11 N ⋅ m 2 / kg 2 ) ( 20kg )( 7.0kg ) (18m ) 2 = 2.9 × 10−11 N. 83. (a) We write the centripetal acceleration (which is the same for each, since they have
identical mass) as rω2 where ω is the unknown angular speed. Thus,
G (M ) (M ) ( 2r ) 2 = GM 2
= Mrω 2
2
4r which gives ω = 1 2 MG / r 3 = 2.2 × 10−7 rad/s.
(b) To barely escape means to have total energy equal to zero (see discussion prior to Eq.
1328). If m is the mass of the meteoroid, then 1 2 GmM GmM
mv −
−
=0 ⇒ v=
2
r
r 4GM
= 8.9 × 104 m/s .
r 87. We apply the workenergy theorem to the object in question. It starts from a point at
the surface of the Earth with zero initial speed and arrives at the center of the Earth with
final speed vf. The corresponding increase in its kinetic energy, ½mvf2, is equal to the
work done on it by Earth’s gravity: ∫ F dr = ∫ (− Kr )dr (using the notation of that Sample
Problem referred to in the problem statement). Thus,
1 2
mv f =
2 ∫ 0 R F dr = ∫ 0 R (− Kr ) dr = 1
KR 2
2 where R is the radius of Earth. Solving for the final speed, we obtain vf = R K / m . We
note that the acceleration of gravity ag = g = 9.8 m/s2 on the surface of Earth is given by
ag = GM/R2 = G(4πR3/3)ρ/R2, where ρ is Earth’s average density. This permits us to write
K/m = 4πGρ/3 = g/R. Consequently, vf = R K
g
=R
= gR = (9.8 m/s 2 ) (6.37 × 106 m) = 7.9 × 103 m/s .
m
R 93. The magnitude of the net gravitational force on one of the smaller stars (of mass m) is GMm Gmm Gm
+
= 2
2
r2
r
( 2r ) m⎞
⎛
⎜ M + ⎟.
4⎠
⎝ This supplies the centripetal force needed for the motion of the star: Gm
r2 m⎞
v2
⎛
M + ⎟=m
⎜
r
4⎠
⎝ where v = 2pr
.
T Plugging in for speed v, we arrive at an equation for period T: T = 2π r 3 2
.
G ( M + m / 4) Chapter 14 – Student Solutions Manual
1. The pressure increase is the applied force divided by the area: Δp = F/A = F/πr2, where
r is the radius of the piston. Thus
Δp = (42 N)/π(0.011 m)2 = 1.1 × 105 Pa.
This is equivalent to 1.1 atm.
3. The air inside pushes outward with a force given by piA, where pi is the pressure inside
the room and A is the area of the window. Similarly, the air on the outside pushes inward
with a force given by poA, where po is the pressure outside. The magnitude of the net
force is F = (pi – po)A. Since 1 atm = 1.013 × 105 Pa,
F = (1.0 atm − 0.96 atm)(1.013 × 105 Pa/atm)(3.4 m)(2.1 m) = 2.9 × 104 N.
19. When the levels are the same the height of the liquid is h = (h1 + h2)/2, where h1 and
h2 are the original heights. Suppose h1 is greater than h2. The final situation can then be
achieved by taking liquid with volume A(h1 – h) and mass ρA(h1 – h), in the first vessel,
and lowering it a distance h – h2. The work done by the force of gravity is
W = ρA(h1 – h)g(h – h2).
We substitute h = (h1 + h2)/2 to obtain
1
1
2
ρ gA ( h1 − h2 ) = (1.30 × 103 kg/m3 )(9.80 m/s 2 )(4.00 × 10−4 m 2 )(1.56 m − 0.854 m) 2
.
4
4
= 0.635 J W= 27. (a) We use the expression for the variation of pressure with height in an
incompressible fluid: p2 = p1 – ρg(y2 – y1). We take y1 to be at the surface of Earth, where
the pressure is p1 = 1.01 × 105 Pa, and y2 to be at the top of the atmosphere, where the
pressure is p2 = 0. For this calculation, we take the density to be uniformly 1.3 kg/m3.
Then,
p
1.01 × 105 Pa
y2 − y1 = 1 =
= 7.9 × 103 m = 7.9 km .
3
2
ρ g (1.3 kg/m ) (9.8 m/s )
(b) Let h be the height of the atmosphere. Now, since the density varies with altitude, we
integrate p2 = p1 − ∫ h 0 ρ g dy . Assuming ρ = ρ0 (1  y/h), where ρ0 is the density at Earth’s surface and g = 9.8 m/s2 for
0 ≤ y ≤ h, the integral becomes p2 = p1 − ∫ h 0 ⎛
⎝ y⎞
1
⎟ dy = p1 − ρ 0 gh.
h⎠
2 ρ 0 g ⎜1 − Since p2 = 0, this implies
h= 2 p1
2(1.01 × 105 Pa)
=
= 16 × 103 m = 16 km.
3
2
ρ 0 g (1.3 kg/m ) (9.8 m/s ) 31. (a) The anchor is completely submerged in water of density ρw. Its effective weight is
Weff = W – ρw gV, where W is its actual weight (mg). Thus, V = W − Weff
200 N
=
= 2.04 × 10−2 m3 .
3
2
ρw g
(1000 kg/m ) ( 9.8 m/s ) (b) The mass of the anchor is m = ρV, where ρ is the density of iron (found in Table
141). Its weight in air is W = mg = ρVg = ( 7870 kg/m3 ) (2.04 × 10−2 m3 ) ( 9.80 m/s 2 ) = 1.57 × 103 N .
35. (a) Let V be the volume of the block. Then, the submerged volume is Vs = 2V/3. Since
the block is floating, the weight of the displaced water is equal to the weight of the block,
so ρw Vs = ρb V, where ρw is the density of water, and ρb is the density of the block. We
substitute Vs = 2V/3 to obtain ρb = 2ρw/3 = 2(1000 kg/m3)/3 ≈ 6.7 ×102 kg/m3.
(b) If ρo is the density of the oil, then Archimedes’ principle yields ρo Vs = ρbV. We
substitute Vs = 0.90V to obtain ρo = ρb/0.90 = 7.4 ×102 kg/m3.
37. (a) The downward force of gravity mg is balanced by the upward buoyant force of the
liquid: mg = ρg Vs. Here m is the mass of the sphere, ρ is the density of the liquid, and Vs
is the submerged volume. Thus m = ρVs. The submerged volume is half the total volume
of the sphere, so Vs = 1 ( 4π 3) ro3 , where ro is the outer radius. Therefore,
2
m= 2π
⎛ 2π ⎞
ρ ro3 = ⎜ ⎟ ( 800 kg/m3 ) (0.090 m)3 = 1.22 kg.
3
⎝ 3 ⎠ (b) The density ρm of the material, assumed to be uniform, is given by ρm = m/V, where m
is the mass of the sphere and V is its volume. If ri is the inner radius, the volume is
V = 4π 3
4π
(ro − ri3 ) =
3
3 ( ( 0.090 m ) 3 − ( 0.080 m ) 3 ) = 9.09 × 10 −4 m3 . The density is ρm = 1.22 kg
= 1.3 × 103 kg/m3 .
9.09 × 10−4 m 3 49. We use the equation of continuity. Let v1 be the speed of the water in the hose and v2
be its speed as it leaves one of the holes. A1 = πR2 is the crosssectional area of the hose.
If there are N holes and A2 is the area of a single hole, then the equation of continuity
becomes
A1
R2
v1 A1 = v2 ( NA2 ) ⇒ v2 =
v1 =
v1
NA2
Nr 2
where R is the radius of the hose and r is the radius of a hole. Noting that R/r = D/d (the
ratio of diameters) we find (1.9 cm ) 0.91m s = 8.1m s.
D2
v2 =
v =
)
2 (
2 1
Nd
24 ( 0.13cm )
2 53. Suppose that a mass Δm of water is pumped in time Δt. The pump increases the
potential energy of the water by Δmgh, where h is the vertical distance through which it is
lifted, and increases its kinetic energy by 1 Δmv 2 , where v is its final speed. The work it
2
does is ΔW = Δmgh + 1 Δmv 2 and its power is
2
P= ΔW Δm ⎛
1 2⎞
=
⎜ gh + v ⎟ .
Δt
Δt ⎝
2 ⎠ Now the rate of mass flow is Δm/ Δt = ρwAv, where ρw is the density of water and A is the
area of the hose. The area of the hose is A = πr2 = π(0.010 m)2 = 3.14 × 10–4 m2 and ρwAv = (1000 kg/m3) (3.14 × 10–4 m2) (5.00 m/s) = 1.57 kg/s.
Thus,
2
⎛
( 5.0 m s ) ⎞ = 66 W.
1 2⎞
⎛
2
⎟
P = ρ Av ⎜ gh + v ⎟ = (1.57 kg s ) ⎜ 9.8 m s ( 3.0 m ) +
⎜
⎟
2 ⎠
2
⎝
⎝
⎠ ( ) 55. (a) We use the equation of continuity: A1v1 = A2v2. Here A1 is the area of the pipe at
the top and v1 is the speed of the water there; A2 is the area of the pipe at the bottom and
v2 is the speed of the water there. Thus
v2 = (A1/A2)v1 = [(4.0 cm2)/(8.0 cm2)] (5.0 m/s) = 2.5m/s.
(b) We use the Bernoulli equation: 2
p1 + 1 ρ v12 + ρ gh1 = p2 + 1 ρ v2 + ρ gh2 ,
2
2 where ρ is the density of water, h1 is its initial altitude, and h2 is its final altitude. Thus ( ) 1
2
ρ v12 − v2 + ρ g ( h1 − h2 )
2
1
⎡
= 1.5 × 105 Pa + (1000 kg m3 ) ⎣ (5.0 m s) 2 − (2.5 m s) 2 ⎤ + (1000 kg m 3 )(9.8 m/s 2 )(10 m)
⎦
2
= 2.6 × 105 Pa. p2 = p1 + 2
59. (a) We use the Bernoulli equation: p1 + 1 ρ v12 + ρ gh1 = p2 + 1 ρ v2 + ρ gh2 , where h1 is
2
2
the height of the water in the tank, p1 is the pressure there, and v1 is the speed of the water
there; h2 is the altitude of the hole, p2 is the pressure there, and v2 is the speed of the water
there. ρ is the density of water. The pressure at the top of the tank and at the hole is
atmospheric, so p1 = p2. Since the tank is large we may neglect the water speed at the top;
it is much smaller than the speed at the hole. The Bernoulli equation then becomes
2
ρ gh1 = 1 ρ v2 + ρ gh2 and
2 ( v2 = 2 g ( h1 − h2 ) = 2 9.8 m s 2 ) ( 0.30 m ) = 2.42 m s. The flow rate is A2v2 = (6.5 × 10–4 m2)(2.42 m/s) = 1.6 × 10–3 m3/s.
(b) We use the equation of continuity: A2v2 = A3v3, where A3 = 1 A2 and v3 is the water
2
speed where the area of the stream is half its area at the hole. Thus
v3 = (A2/A3)v2 = 2v2 = 4.84 m/s.
The water is in free fall and we wish to know how far it has fallen when its speed is
doubled to 4.84 m/s. Since the pressure is the same throughout the fall,
2
2
1
1
2 ρ v2 + ρ gh2 = 2 ρ v3 + ρ gh3 . Thus
v 2 − v 2 ( 4.84 m s ) − ( 2.42 m s )
h2 − h3 = 3 2 =
= 0.90 m.
2g
2 9.8 m s 2
2 ( 2 ) 67. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields
Δp + 1 ρ v 2 = 1 ρV 2 , where Δp = p1 – p2. The first equation gives V = (A/a)v. We use this
2
2
to substitute for V in the second equation, and obtain Δp + 1 ρ v 2 = 1 ρ ( A a ) v 2 . We
2
2
solve for v. The result is
2 v= 2 Δp
=
ρ ( A / a)2 − 1 ( ) 2 a 2 Δp
.
ρ A2 − a 2 ( ) (b) We substitute values to obtain
v= 2(32 × 10 −4 m 2 ) 2 (55 × 103 Pa − 41 × 103 Pa)
= 3.06 m/s.
(1000 kg / m 3 ) (64 × 10−4 m 2 ) 2 − (32 × 10−4 m 2 ) 2 ( ) Consequently, the flow rate is
Av = (64 × 10−4 m 2 ) (3.06 m/s) = 2.0 × 10−2 m3 / s.
75. If we examine both sides of the Utube at the level where the lowdensity liquid (with
ρ = 0.800 g/cm3 = 800 kg/m3) meets the water (with ρw = 0.998 g/cm3 = 998 kg/m3), then
the pressures there on either side of the tube must agree:
ρgh = ρwghw
where h = 8.00 cm = 0.0800 m, and Eq. 149 has been used. Thus, the height of the
water column (as measured from that level) is hw = (800/998)(8.00 cm) = 6.41 cm. The
volume of water in that column is therefore πr2hw = π(1.50 cm)2(6.41 cm) = 45.3 cm3. Chapter 15 – Student Solutions Manual
3. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm.
(b) The maximum speed vm is related to the amplitude xm by vm = ωxm, where ω is the
angular frequency. Since ω = 2πf, where f is the frequency,
vm = 2π fxm = 2π (120 Hz ) (1.0 ×10 −3 m ) = 0.75 m/s. (c) The maximum acceleration is
am = ω 2 xm = ( 2π f ) xm = ( 2π (120 Hz ) ) (1.0 ×10−3 m ) = 5.7 ×102 m/s 2 .
2 2 7. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s.
(b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) = 2.00 Hz.
(c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s.
(d) The angular frequency is related to the spring constant k and the mass m by
ω = k m . We solve for k and obtain
k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.
(e) Let xm be the amplitude. The maximum speed is
vm = ωxm = (12.6 rad/s)(0.350 m) = 4.40 m/s.
(f) The maximum force is exerted when the displacement is a maximum and its
magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N.
9. The magnitude of the maximum acceleration is given by am = ω2xm, where ω is the
angular frequency and xm is the amplitude.
(a) The angular frequency for which the maximum acceleration is g is given by
ω = g / x m , and the corresponding frequency is given by ω
1
f=
=
2π 2π g
1
=
xm 2π 9.8 m/s 2
= 498 Hz.
1.0 ×10−6 m (b) For frequencies greater than 498 Hz, the acceleration exceeds g for some part of the
motion. 17. The maximum force that can be exerted by the surface must be less than μsFN or else
the block will not follow the surface in its motion. Here, µs is the coefficient of static
friction and FN is the normal force exerted by the surface on the block. Since the block
does not accelerate vertically, we know that FN = mg, where m is the mass of the block. If
the block follows the table and moves in simple harmonic motion, the magnitude of the
maximum force exerted on it is given by
F = mam = mω2xm = m(2πf)2xm,
where am is the magnitude of the maximum acceleration, ω is the angular frequency, and
f is the frequency. The relationship ω = 2πf was used to obtain the last form. We
substitute F = m(2πf)2xm and FN = mg into F < µsFN to obtain m(2πf)2xm < µsmg. The
largest amplitude for which the block does not slip is xm b0.50gc9.8 m / s h = 0.031 m.
=
=
b2πf g b2π × 2.0 Hzg
μsg 2 2 2 A larger amplitude requires a larger force at the end points of the motion. The surface
cannot supply the larger force and the block slips.
19. (a) Let
x1 = FG IJ
H K A
2 πt
cos
2
T be the coordinate as a function of time for particle 1 and
x2 = FG
H 2 πt π
A
+
cos
6
2
T IJ
K be the coordinate as a function of time for particle 2. Here T is the period. Note that since
the range of the motion is A, the amplitudes are both A/2. The arguments of the cosine
functions are in radians. Particle 1 is at one end of its path (x1 = A/2) when t = 0. Particle
2 is at A/2 when 2πt/T + π/6 = 0 or t = –T/12. That is, particle 1 lags particle 2 by onetwelfth a period. We want the coordinates of the particles 0.50 s later; that is, at t = 0.50 s,
x1 = and
x2 = A
⎛ 2π × 0.50 s ⎞
cos ⎜
⎟ = −0.25 A
2
⎝ 1.5 s ⎠ A
⎛ 2π × 0.50 s π ⎞
cos ⎜
+ ⎟ = −0.43 A.
2
6⎠
⎝ 1.5 s Their separation at that time is x1 – x2 = –0.25A + 0.43A = 0.18A. (b) The velocities of the particles are given by
v1 =
and
v2 = F I
H K dx1 πA
2 πt
=
sin
dt
T
T F
H I
K dx2 πA
2 πt π
=
sin
+ .
dt
T
T
6 We evaluate these expressions for t = 0.50 s and find they are both negativevalued,
indicating that the particles are moving in the same direction.
27. When the block is at the end of its path and is momentarily stopped, its displacement
is equal to the amplitude and all the energy is potential in nature. If the spring potential
energy is taken to be zero when the block is at its equilibrium position, then
E= ha c 1 2 1
kx m = 1.3 × 10 2 N / m 0.024 m
2
2 f 2 = 3.7 × 10 −2 J. 2
29. The total energy is given by E = 1 kxm , where k is the spring constant and xm is the
2
amplitude. We use the answer from part (b) to do part (a), so it is best to look at the
solution for part (b) first. (a) The fraction of the energy that is kinetic is
K E −U
U
1 3
=
=1 − =1 − = = 0.75
4 4
E
E
E where the result from part (b) has been used.
2
(b) When x = 1 x m the potential energy is U = 1 kx 2 = 1 kx m . The ratio is
2
8
2 2
U kxm / 8 1
= 2
= = 0.25.
E kxm / 2 4 2
2
2
(c) Since E = 1 kxm and U = 1 kx 2 , U/E = x 2 xm . We solve x 2 xm = 1/2 for x. We should
2
2
get x = xm / 2 . 39. (a) We take the angular displacement of the wheel to be θ = θm cos(2πt/T), where θm
is the amplitude and T is the period. We differentiate with respect to time to find the
angular velocity: Ω = –(2π/T)θmsin(2πt/T). The symbol Ω is used for the angular
velocity of the wheel so it is not confused with the angular frequency. The maximum
angular velocity is Ωm = a fa f 2 πθ m
2 π π rad
=
= 39.5 rad / s.
0.500 s
T (b) When θ = π/2, then θ/θm = 1/2, cos(2πt/T) = 1/2, and sin ( 2π t T ) = 1 − cos 2 ( 2π t T ) = 1 − (1 2 ) = 3 2
2 where the trigonometric identity cos2θ + sin2θ = 1 is used. Thus,
Ω=− F I F
H K H Ia
K 2π
2 πt
2π
θ msin
=−
π rad
T
T
0.500 s fFGH 23 IJK = −34.2 rad / s. During another portion of the cycle its angular speed is +34.2 rad/s when its angular
displacement is π/2 rad.
(c) The angular acceleration is
d 2θ
⎛ 2π ⎞
⎛ 2π ⎞
α = 2 = − ⎜ ⎟ θ m cos ( 2π t / T ) = − ⎜ ⎟ θ .
dt
⎝ T ⎠
⎝ T ⎠
2 2 When θ = π/4, ⎛ 2π ⎞ ⎛ π ⎞
2
α = −⎜
⎟ ⎜ ⎟ = −124 rad/s ,
⎝ 0.500 s ⎠ ⎝ 4 ⎠
2 or  α  = 124 rad/s 2 .
43. (a) A uniform disk pivoted at its center has a rotational inertia of 1 Mr 2 , where M is
2
its mass and r is its radius. The disk of this problem rotates about a point that is displaced
from its center by r+ L, where L is the length of the rod, so, according to the parallelaxis
theorem, its rotational inertia is 1 Mr 2 + 1 M ( L + r ) 2 . The rod is pivoted at one end and
2
2
has a rotational inertia of mL2/3, where m is its mass. The total rotational inertia of the
disk and rod is
1
1
I = Mr 2 + M ( L + r ) 2 + mL2
2
3
1
1
= (0.500kg)(0.100m) 2 + (0.500kg)(0.500m + 0.100m) 2 + (0.270kg)(0.500m) 2
2
3
2
= 0.205kg ⋅ m .
(b) We put the origin at the pivot. The center of mass of the disk is d = L + r = 0.500 m + 0.100 m = 0.600 m away and the center of mass of the rod is r = L / 2 = (0.500 m) / 2 = 0.250 m away, on
the same line. The distance from the pivot point to the center of mass of the diskrod
system is d= M +m
M+m
d r = a0.500 kgfa0.600 mf + a0.270 kgfa0.250 mf = 0.477 m.
0.500 kg + 0.270 kg (c) The period of oscillation is
I
0.205 kg ⋅ m 2
T = 2π
= 2π
= 1.50 s .
(0.500 kg + 0.270 kg)(9.80 m/s 2 )(0.447 m)
( M + m ) gd 51. If the torque exerted by the spring on the rod is proportional to the angle of rotation of
the rod and if the torque tends to pull the rod toward its equilibrium orientation, then the
rod will oscillate in simple harmonic motion. If τ = –Cθ, where τ is the torque, θ is the
angle of rotation, and C is a constant of proportionality, then the angular frequency of
oscillation is ω = C / I and the period is
T = 2π / ω = 2π I / C , where I is the rotational inertia of the rod. The plan is to find the torque as a function of θ
and identify the constant C in terms of given quantities. This immediately gives the
period in terms of given quantities. Let 0 be the distance from the pivot point to the wall.
This is also the equilibrium length of the spring. Suppose the rod turns through the angle
θ, with the left end moving away from the wall. This end is now (L/2) sin θ further from
the wall and has moved a distance (L/2)(1 – cos θ) to the right. The length of the spring is
now = ( L / 2)2 (1 − cos θ )2 + [ 0 + ( L / 2)sin θ ]2 . If the angle θ is small we may approximate cos θ with 1 and sin θ with θ in radians. Then
the length of the spring is given by ≈ 0 + Lθ / 2 and its elongation is Δx = Lθ/2. The
force it exerts on the rod has magnitude F = kΔx = kLθ/2. Since θ is small we may
approximate the torque exerted by the spring on the rod by τ = –FL/2, where the pivot
point was taken as the origin. Thus τ = –(kL2/4)θ. The constant of proportionality C that
relates the torque and angle of rotation is C = kL2/4. The rotational inertia for a rod
pivoted at its center is I = mL2/12, where m is its mass. See Table 102. Thus the period
of oscillation is T = 2π I
mL2 / 12
m
= 2π
= 2π
.
2
C
kL / 4
3k With m = 0.600 kg and k = 1850 N/m, we obtain T = 0.0653 s.
57. (a) We want to solve e–bt/2m = 1/3 for t. We take the natural logarithm of both sides to
obtain –bt/2m = ln(1/3). Therefore, t = –(2m/b) ln(1/3) = (2m/b) ln 3. Thus, t= a f 2 1.50 kg
ln3 = 14.3 s.
0.230 kg / s (b) The angular frequency is ω′ = a f 8.00 N / m 0.230 kg / s
k
b2
−
=
−
2
2
1.50 kg
m 4m
4 1.50 kg a f 2 = 2.31 rad / s. The period is T = 2π/ω´ = (2π)/(2.31 rad/s) = 2.72 s and the number of oscillations is
t/T = (14.3 s)/(2.72 s) = 5.27. a 75. (a) The frequency for small amplitude oscillations is f = 1 / 2 π
the length of the pendulum. This gives a f = 1 / 2π f f g / L , where L is (9.80 m / s 2 ) / (2.0 m ) = 0.35 Hz. (b) The forces acting on the pendulum are the tension force T of the rod and the force of
gravity mg . Newton’s second law yields T + mg = ma , where m is the mass and a is the
acceleration of the pendulum. Let a = ae + a ′, where ae is the acceleration of the elevator
and a ′ is the acceleration of the pendulum relative to the elevator. Newton’s second law
can then be written m( g − ae ) + T = ma′ . Relative to the elevator the motion is exactly
the same as it would be in an inertial frame where the acceleration due to gravity is
g − ae . Since g and ae are along the same line and in opposite directions we can find the
frequency for small amplitude oscillations by replacing g with g + ae in the expression
f = (1 / 2π) g / L . Thus
f = 1
2π g + ae
1 9.8 m / s 2 + 2.0 m / s 2
=
= 0.39 Hz.
L
2π
2.0 m (c) Now the acceleration due to gravity and the acceleration of the elevator are in the
same direction and have the same magnitude. That is, g − ae = 0. To find the frequency
for small amplitude oscillations, replace g with zero in f = (1 / 2π) g / L . The result is
zero. The pendulum does not oscillate. 83. We use vm = ωxm = 2πfxm, where the frequency is 180/(60 s) = 3.0 Hz and the
amplitude is half the stroke, or xm = 0.38 m. Thus,
vm = 2π(3.0 Hz)(0.38 m) = 7.2 m/s.
89. (a) The spring stretches until the magnitude of its upward force on the block equals
the magnitude of the downward force of gravity: ky = mg, where y = 0.096 m is the
elongation of the spring at equilibrium, k is the spring constant, and m = 1.3 kg is the
mass of the block. Thus
k = mg/y = (1.3 kg)(9.8 m/s2)/(0.096 m) = 1.33 ×102 N/m.
(b) The period is given by T= m
1 2π
1.3 kg
=
= 2π
= 2π
= 0.62 s.
f
k
ω
133 N / m (c) The frequency is f = 1/T = 1/0.62 s = 1.6 Hz.
(d) The block oscillates in simple harmonic motion about the equilibrium point
determined by the forces of the spring and gravity. It is started from rest 5.0 cm below the
equilibrium point so the amplitude is 5.0 cm.
(e) The block has maximum speed as it passes the equilibrium point. At the initial
position, the block is not moving but it has potential energy
1
1
2
U i = − mgyi + kyi2 = − (1.3 kg ) ( 9.8 m/s 2 ) ( 0.146 m ) + (133 N / m )( 0.146 m ) = −0.44 J.
2
2 When the block is at the equilibrium point, the elongation of the spring is y = 9.6 cm and
the potential energy is
1
1
2
U f = − mgy + ky 2 = − (1.3 kg ) ( 9.8 m/s 2 ) ( 0.096 m ) + (133 N / m )( 0.096 m ) = −0.61 J.
2
2 We write the equation for conservation of energy as Ui = U f + 1 mv 2 and solve for v:
2
v= 2 (U i − U f )
m = 2 ( −0.44 J + 0.61J )
= 0.51 m/s.
1.3kg 91. We note that for a horizontal spring, the relaxed position is the equilibrium position
(in a regular simple harmonic motion setting); thus, we infer that the given v = 5.2 m/s at
x = 0 is the maximum value vm (which equals ωxm where ω = k / m = 20 rad / s ). (a) Since ω = 2π f, we find f = 3.2 Hz.
(b) We have vm = 5.2 m/s = (20 rad/s)xm, which leads to xm = 0.26 m.
(c) With meters, seconds and radians understood,
x = (0.26 m) cos(20t + φ )
v = −(5.2 m/s) sin(20t + φ ). The requirement that x = 0 at t = 0 implies (from the first equation above) that either φ =
+π/2 or φ = –π/2. Only one of these choices meets the further requirement that v > 0 when
t = 0; that choice is φ = –π/2. Therefore, FG
H x = 0.26 cos 20t − IJ
K π
= 0.26 sin 20t .
2 b g Chapter 16 – Student Solutions Manual
15. The wave speed v is given by v = τ μ , where τ is the tension in the rope and μ is
the linear mass density of the rope. The linear mass density is the mass per unit length of
rope: μ = m/L = (0.0600 kg)/(2.00 m) = 0.0300 kg/m.
Thus, v= 500 N
= 129 m s.
0.0300 kg m 17. (a) The amplitude of the wave is ym=0.120 mm.
(b) The wave speed is given by v = τ μ , where τ is the tension in the string and μ is the
linear mass density of the string, so the wavelength is λ = v/f = τ μ /f and the angular
wave number is k= 2π
μ
0.50 kg m
= 2 πf
= 2π (100 Hz )
= 141m −1.
λ
10 N
τ (c) The frequency is f = 100 Hz, so the angular frequency is ω = 2πf = 2π(100 Hz) = 628 rad/s.
(d) We may write the string displacement in the form y = ym sin(kx + ωt). The plus sign is
used since the wave is traveling in the negative x direction. In summary, the wave can be
expressed as ( ) ( ) y = ( 0.120 mm ) sin ⎡ 141m −1 x + 628s −1 t ⎤ .
⎣
⎦
21. (a) We read the amplitude from the graph. It is about 5.0 cm.
(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm
and again with the same slope at about x = 55 cm, so
λ = (55 cm – 15 cm) = 40 cm = 0.40 m.
(c) The wave speed is v = τ / μ , where τ is the tension in the string and μ is the linear
mass density of the string. Thus, v= 3.6 N
= 12 m/s.
25 × 10−3 kg/m (d) The frequency is f = v/λ = (12 m/s)/(0.40 m) = 30 Hz and the period is
T = 1/f = 1/(30 Hz) = 0.033 s.
(e) The maximum string speed is
um = ωym = 2πfym = 2π(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s.
(f) The angular wave number is k = 2π/λ = 2π/(0.40 m) = 16 m–1.
(g) The angular frequency is ω = 2πf = 2π(30 Hz) = 1.9×102 rad/s
(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10–2 m. The
formula for the displacement gives y(0, 0) = ym sin φ. We wish to select φ so that 5.0 ×
10–2 sin φ = 4.0 × 10–2. The solution is either 0.93 rad or 2.21 rad. In the first case the
function has a positive slope at x = 0 and matches the graph. In the second case it has
negative slope and does not match the graph. We select φ = 0.93 rad.
(i) The string displacement has the form y (x, t) = ym sin(kx + ωt + φ). A plus sign appears
in the argument of the trigonometric function because the wave is moving in the negative
x direction. Using the results obtained above, the expression for the displacement is
y ( x, t ) = ( 5.0 × 10−2 m ) sin ⎡ (16 m −1 ) x + (190s −1 )t + 0.93⎤ .
⎣
⎦ 31. The displacement of the string is given by
y = ym sin(kx − ω t ) + ym sin(kx − ω t + φ ) = 2 ym cos ( 1 φ ) sin ( kx − ω t + 1 φ ) ,
2
2
where φ = π/2. The amplitude is A = 2 ym cos ( 1 φ ) = 2 ym cos(π / 4) = 1.41ym .
2
35. The phasor diagram is shown below: y1m and y2m represent the original waves and ym
represents the resultant wave. The phasors corresponding to the two constituent waves
make an angle of 90° with each other, so the triangle is a right triangle. The Pythagorean
theorem gives
2
2
ym = y12m + y2 m = (3.0 cm) 2 + (4.0 cm) 2 = (25cm) 2 . Thus ym = 5.0 cm. 41. Possible wavelengths are given by λ = 2L/n, where L is the length of the wire and n is
an integer. The corresponding frequencies are given by f = v/λ = nv/2L, where v is the
wave speed. The wave speed is given by v = τ μ = τ L / M , where τ is the tension in
the wire, μ is the linear mass density of the wire, and M is the mass of the wire. μ = M/L
was used to obtain the last form. Thus
fn = n τL n
=
2L M 2 τ
LM = n
250 N
= n (7.91 Hz).
2 (10.0 m) (0.100 kg) (a) The lowest frequency is f1 = 7.91 Hz.
(b) The second lowest frequency is f 2 = 2(7.91 Hz) = 15.8 Hz.
(c) The third lowest frequency is f3 = 3(7.91 Hz) = 23.7 Hz.
43. (a) The wave speed is given by v = τ μ , where τ is the tension in the string and μ is
the linear mass density of the string. Since the mass density is the mass per unit length, μ
= M/L, where M is the mass of the string and L is its length. Thus
v= τL
M = (96.0 N) (8.40 m)
= 82.0 m/s.
0.120 kg (b) The longest possible wavelength λ for a standing wave is related to the length of the
string by L = λ/2, so λ = 2L = 2(8.40 m) = 16.8 m.
(c) The frequency is f = v/λ = (82.0 m/s)/(16.8 m) = 4.88 Hz.
47. (a) The resonant wavelengths are given by λ = 2L/n, where L is the length of the
string and n is an integer, and the resonant frequencies are given by f = v/λ = nv/2L,
where v is the wave speed. Suppose the lower frequency is associated with the integer n.
Then, since there are no resonant frequencies between, the higher frequency is associated with n + 1. That is, f1 = nv/2L is the lower frequency and f2 = (n + 1)v/2L is the higher.
The ratio of the frequencies is
f2 n + 1
.
=
f1
n The solution for n is
n= f1
315 Hz
=
= 3.
f 2 − f1 420 Hz − 315 Hz The lowest possible resonant frequency is f = v/2L = f1/n = (315 Hz)/3 = 105 Hz.
(b) The longest possible wavelength is λ = 2L. If f is the lowest possible frequency then
v = λf = 2Lf = 2(0.75 m)(105 Hz) = 158 m/s. 53. (a) The waves have the same amplitude, the same angular frequency, and the same
angular wave number, but they travel in opposite directions. We take them to be
y1 = ym sin(kx – ωt), y2 = ym sin(kx + ωt).
The amplitude ym is half the maximum displacement of the standing wave, or 5.0 × 10–3
m.
(b) Since the standing wave has three loops, the string is three halfwavelengths long: L =
3λ/2, or λ = 2L/3. With L = 3.0m, λ = 2.0 m. The angular wave number is k = 2π/λ =
2π/(2.0 m) = 3.1 m–1.
(c) If v is the wave speed, then the frequency is
f = v 3v 3 (100 m s )
=
=
= 50 Hz.
λ 2L
2 ( 3.0 m ) The angular frequency is the same as that of the standing wave, or ω = 2π f = 2π(50 Hz)
= 314 rad/s.
(d) The two waves are
y1 = ( 5.0 × 10−3 m ) sin ⎡ ( 3.14 m −1 ) x − ( 314s −1 ) t ⎤
⎣
⎦
and
y2 = ( 5.0 × 10−3 m ) sin ⎡( 3.14 m −1 ) x + ( 314s−1 ) t ⎤ .
⎣
⎦ Thus, if one of the waves has the form y ( x, t ) = ym sin(kx + ω t ) , then the other wave must
have the form y '( x, t ) = ym sin(kx − ω t ) . The sign in front of ω for y '( x, t ) is minus.
61. (a) The phasor diagram is shown here: y1, y2, and y3 represent the original waves and
ym represents the resultant wave. The horizontal component of the resultant is ymh = y1 – y3 = y1 – y1/3 = 2y1/3. The vertical
component is ymv = y2 = y1/2. The amplitude of the resultant is
2 ym = y 2
mh +y 2
mv 2 5
⎛ 2y ⎞ ⎛ y ⎞
= ⎜ 1 ⎟ + ⎜ 1 ⎟ = y1 = 0.83 y1.
6
⎝ 3 ⎠ ⎝2⎠ (b) The phase constant for the resultant is ⎛ ymv ⎞
−1 ⎛ y1 2 ⎞
−1 ⎛ 3 ⎞
⎟ = tan ⎜
⎟ = tan ⎜ ⎟ = 0.644 rad = 37°.
⎝4⎠
⎝ 2 y1 3 ⎠
⎝ ymh ⎠ φ = tan −1 ⎜ (c) The resultant wave is
y= 5
y1 sin ( kx − ω t + 0.644 rad).
6 The graph below shows the wave at time t = 0. As time goes on it moves to the right with
speed v = ω/k. 69. (a) We take the form of the displacement to be y (x, t) = ym sin(kx – ωt). The speed of
a point on the cord is u (x, t) = ∂y/∂t = –ωym cos(kx – ωt) and its maximum value is um =
ωym. The wave speed, on the other hand, is given by v = λ/T = ω/k. The ratio is
um ω ym
2 πy m
=
= kym =
.
λ
v ω /k (b) The ratio of the speeds depends only on the ratio of the amplitude to the wavelength.
Different waves on different cords have the same ratio of speeds if they have the same
amplitude and wavelength, regardless of the wave speeds, linear densities of the cords,
and the tensions in the cords.
77. (a) The wave speed is v= τ
120 N
=
= 144 m/s.
8.70 × 10−3 kg /1.50 m
μ (b) For the oneloop standing wave we have λ1 = 2L = 2(1.50 m) = 3.00 m.
(c) For the twoloop standing wave λ2 = L = 1.50 m.
(d) The frequency for the oneloop wave is f1 = v/λ1 = (144 m/s)/(3.00 m) = 48.0 Hz.
(e) The frequency for the twoloop wave is f2 = v/λ2 = (144 m/s)/(1.50 m) = 96.0 Hz.
2
78. We use P = 1 μνω 2 ym ∝ vf 2 ∝ τ f 2 .
2 (a) If the tension is quadrupled, then P2 = P
1 τ2
4τ1
=P
= 2P .
1
1
τ1
τ1
2 2 1
⎛ f ⎞
⎛ f /2⎞
(b) If the frequency is halved, then P2 = P ⎜ 2 ⎟ = P ⎜ 1 ⎟ = P .
1
1
1
4
⎝ f1 ⎠
⎝ f1 ⎠ 87. (a) From the frequency information, we find ω = 2πf = 10π rad/s. A point on the
rope undergoing simple harmonic motion (discussed in Chapter 15) has maximum speed
as it passes through its "middle" point, which is equal to ymω. Thus,
5.0 m/s = ymω ⇒ ym = 0.16 m .
(b) Because of the oscillation being in the fundamental mode (as illustrated in Fig. 1623(a) in the textbook), we have λ = 2L = 4.0 m. Therefore, the speed of waves along the
rope is v = fλ = 20 m/s. Then, with μ = m/L = 0.60 kg/m, Eq. 1626 leads to
τ
⇒ τ = μ v2 = 240 N ≈ 2.4 ×102 N .
μ v = (c) We note that for the fundamental, k = 2π/λ = π/L, and we observe that the antinode
having zero displacement at t = 0 suggests the use of sine instead of cosine for the simple
harmonic motion factor. Now, if the fundamental mode is the only one present (so the
amplitude calculated in part (a) is indeed the amplitude of the fundamental wave pattern)
then we have
⎛πx⎞
y = (0.16 m) sin ⎜ 2 ⎟ sin (10πt) = (0.16 m) sin[(1.57 m −1 ) x]sin[(31.4 rad/s)t ]
⎝ ⎠
89. (a) The wave speed is
v= F μ = kΔ
=
m /( + Δ ) kΔ ( + Δ )
.
m (b) The time required is t=
Thus if / Δ
t 2 π( + Δ )
2 π( + Δ )
m
=
= 2π
1+ .
v
k
Δ
kΔ ( + Δ ) / m 1 , then t ∝ 2π m / k = const. / Δ ∝ 1/ Δ ; and if / Δ 1 , then Chapter 17 – Student Solutions Manual
5. Let tf be the time for the stone to fall to the water and ts be the time for the sound of the
splash to travel from the water to the top of the well. Then, the total time elapsed from
dropping the stone to hearing the splash is t = tf + ts. If d is the depth of the well, then the
kinematics of free fall gives d = 1 gt 2 , or t f = 2d / g . The sound travels at a constant
f
2
speed vs, so d = vsts, or ts = d/vs. Thus the total time is t = 2d / g + d / vs . This equation is
to be solved for d. Rewrite it as 2d / g = t − d / vs and square both sides to obtain 2d/g = t2 – 2(t/vs)d + (1 + vs2 )d2.
Now multiply by g vs2 and rearrange to get
gd2 – 2vs(gt + vs)d + g vs2 t2 = 0.
This is a quadratic equation for d. Its solutions are
2vs ( gt + vs ) ± 4vs2 ( gt + vs ) − 4 g 2 vs2t 2
2 d= 2g . The physical solution must yield d = 0 for t = 0, so we take the solution with the negative
sign in front of the square root. Once values are substituted the result d = 40.7 m is
obtained.
7. If d is the distance from the location of the earthquake to the seismograph and vs is the
speed of the S waves then the time for these waves to reach the seismograph is ts. = d/vs.
Similarly, the time for P waves to reach the seismograph is tp = d/vp. The time delay is
Δt = (d/vs) – (d/vp) = d(vp – vs)/vsvp,
so
d= vs v p Δt
(v p − vs ) = (4.5 km/s)(8.0 km/s)(3.0 min)(60s /min)
= 1.9 × 103 km.
8.0 km/s − 4.5 km/s We note that values for the speeds were substituted as given, in km/s, but that the value
for the time delay was converted from minutes to seconds.
9. (a) Using λ = v/f, where v is the speed of sound in air and f is the frequency, we find
λ= 343m/s
= 7.62 × 10−5 m.
4.50 × 106 Hz (b) Now, λ = v/f, where v is the speed of sound in tissue. The frequency is the same for
air and tissue. Thus
λ = (1500 m/s)/(4.50 × 106 Hz) = 3.33 × 10–4 m.
19. Let L1 be the distance from the closer speaker to the listener. The distance from the
2
other speaker to the listener is L2 = L1 + d 2 , where d is the distance between the
speakers. The phase difference at the listener is φ = 2π(L2 – L1)/λ, where λ is the
wavelength. For a minimum in intensity at the listener, φ = (2n + 1)π, where n is an integer. Thus λ =
2(L2 – L1)/(2n + 1). The frequency is
f = v
=
λ 2 ( (2n + 1)v
2
L1 + d 2 − L1 = ) (
2 (2n + 1)(343m/s)
(3.75m) 2 + (2.00 m) 2 − 3.75m ) = (2n + 1)(343Hz). Now 20,000/343 = 58.3, so 2n + 1 must range from 0 to 57 for the frequency to be in the
audible range. This means n ranges from 0 to 28.
(a) The lowest frequency that gives minimum signal is (n = 0) f min,1 = 343 Hz.
(b) The second lowest frequency is (n = 1) f min,2 = [2(1) + 1]343 Hz = 1029 Hz = 3 f min,1.
Thus, the factor is 3.
(c) The third lowest frequency is (n=2) f min,3 = [2(2) + 1]343 Hz = 1715 Hz = 5 f min,1. Thus,
the factor is 5.
For a maximum in intensity at the listener, φ = 2nπ, where n is any positive integer. Thus
λ = (1/ n ) ( ) 2
L1 + d 2 − L1 and f = v
=
λ nv
L + d − L1
2
1 2 = n(343m/s)
(3.75 m) + (2.00 m) 2 − 3.75 m
2 = n (686 Hz). Since 20,000/686 = 29.2, n must be in the range from 1 to 29 for the frequency to be
audible.
(d) The lowest frequency that gives maximum signal is (n =1) f max,1 = 686 Hz.
(e) The second lowest frequency is (n = 2) f max,2 = 2(686 Hz) = 1372 Hz = 2 f max,1. Thus,
the factor is 2. (f) The third lowest frequency is (n = 3) f max,3 = 3(686 Hz) = 2058 Hz = 3 f max,1. Thus, the
factor is 3.
25. The intensity is the rate of energy flow per unit area perpendicular to the flow. The
rate at which energy flow across every sphere centered at the source is the same,
regardless of the sphere radius, and is the same as the power output of the source. If P is
the power output and I is the intensity a distance r from the source, then P = IA = 4πr2I,
where A (= 4πr2) is the surface area of a sphere of radius r. Thus
P = 4π(2.50 m)2 (1.91 × 10–4 W/m2) = 1.50 × 10–2 W.
29. (a) Let I1 be the original intensity and I2 be the final intensity. The original sound
level is β1 = (10 dB) log(I1/I0) and the final sound level is β2 = (10 dB) log(I2/I0), where I0
is the reference intensity. Since β2 = β1 + 30 dB which yields
(10 dB) log(I2/I0) = (10 dB) log(I1/I0) + 30 dB,
or
(10 dB) log(I2/I0) – (10 dB) log(I1/I0) = 30 dB.
Divide by 10 dB and use log(I2/I0) – log(I1/I0) = log(I2/I1) to obtain log(I2/I1) = 3. Now
use each side as an exponent of 10 and recognize that 10log( I 2 I1 ) = I 2 / I1 . The result is I2/I1
= 103. The intensity is increased by a factor of 1.0×103.
(b) The pressure amplitude is proportional to the square root of the intensity so it is
increased by a factor of 1000 = 32.
43. (a) When the string (fixed at both ends) is vibrating at its lowest resonant frequency,
exactly onehalf of a wavelength fits between the ends. Thus, λ = 2L. We obtain
v = fλ = 2Lf = 2(0.220 m)(920 Hz) = 405 m/s.
(b) The wave speed is given by v = τ / μ , where τ is the tension in the string and μ is
the linear mass density of the string. If M is the mass of the (uniform) string, then μ =
M/L. Thus τ = μv2 = (M/L)v2 = [(800 × 10–6 kg)/(0.220 m)] (405 m/s)2 = 596 N.
(c) The wavelength is λ = 2L = 2(0.220 m) = 0.440 m.
(d) The frequency of the sound wave in air is the same as the frequency of oscillation of
the string. The wavelength is different because the wave speed is different. If va is the
speed of sound in air the wavelength in air is λa = va/f = (343 m/s)/(920 Hz) = 0.373 m.
45. (a) Since the pipe is open at both ends there are displacement antinodes at both ends
and an integer number of halfwavelengths fit into the length of the pipe. If L is the pipe
length and λ is the wavelength then λ = 2L/n, where n is an integer. If v is the speed of
sound then the resonant frequencies are given by f = v/λ = nv/2L. Now L = 0.457 m, so
f = n(344 m/s)/2(0.457 m) = 376.4n Hz.
To find the resonant frequencies that lie between 1000 Hz and 2000 Hz, first set f = 1000
Hz and solve for n, then set f = 2000 Hz and again solve for n. The results are 2.66 and
5.32, which imply that n = 3, 4, and 5 are the appropriate values of n. Thus, there are 3
frequencies.
(b) The lowest frequency at which resonance occurs is (n = 3) f = 3(376.4 Hz) = 1129 Hz.
(c) The second lowest frequency at which resonance occurs is (n = 4)
f = 4(376.4 Hz) = 1506 Hz.
47. The string is fixed at both ends so the resonant wavelengths are given by λ = 2L/n,
where L is the length of the string and n is an integer. The resonant frequencies are given
by f = v/λ = nv/2L, where v is the wave speed on the string. Now v = τ / μ , where τ is
the tension in the string and μ is the linear mass density of the string. Thus
f = (n / 2 L) τ / μ . Suppose the lower frequency is associated with n = n1 and the higher
frequency is associated with n = n1 + 1. There are no resonant frequencies between so
you know that the integers associated with the given frequencies differ by 1. Thus
f1 = (n1 / 2 L) τ / μ and
f2 = n1 + 1 τ
n τ
1 τ
1 τ
= 1
+
= f1 +
.
2L μ 2L μ 2L μ
2L μ This means f 2 − f1 = (1/ 2 L) τ / μ and τ = 4 L2 μ ( f 2 − f1 ) 2 = 4(0.300 m) 2 (0.650 × 10−3 kg/m)(1320 Hz − 880 Hz) 2 = 45.3 N.
53. Each wire is vibrating in its fundamental mode so the wavelength is twice the length
of the wire (λ = 2L) and the frequency is f = v / λ = (1/ 2 L) τ / μ , where v = τ / μ is
the wave speed for the wire, τ is the tension in the wire, and μ is the linear mass density
of the wire. Suppose the tension in one wire is τ and the oscillation frequency of that wire
is f1. The tension in the other wire is τ + Δτ and its frequency is f2. You want to calculate Δτ/τ for f1 = 600 Hz and f2 = 606 Hz. Now, f1 = (1/ 2 L) τ / μ and f 2 = (1/ 2 L) (τ + Δτ / μ , so
f 2 / f1 = (τ + Δτ ) / τ = 1 + ( Δτ / τ ).
This leads to Δτ / τ = ( f 2 / f1 ) 2 − 1 = [(606 Hz) /(600 Hz)]2 − 1 = 0.020.
65. (a) The expression for the Doppler shifted frequency is
f′= f v ± vD
,
v ∓ vS where f is the unshifted frequency, v is the speed of sound, vD is the speed of the detector
(the uncle), and vS is the speed of the source (the locomotive). All speeds are relative to
the air. The uncle is at rest with respect to the air, so vD = 0. The speed of the source is vS
= 10 m/s. Since the locomotive is moving away from the uncle the frequency decreases
and we use the plus sign in the denominator. Thus f′= f v
= (500.0 Hz)
v + vS ⎛
⎞
343m/s
⎜
⎟ = 485.8 Hz.
⎝ 343m/s + 10.00 m/s ⎠ (b) The girl is now the detector. Relative to the air she is moving with speed vD = 10.00
m/s toward the source. This tends to increase the frequency and we use the plus sign in
the numerator. The source is moving at vS = 10.00 m/s away from the girl. This tends to
decrease the frequency and we use the plus sign in the denominator. Thus (v + vD) =
(v + vS) and f′ = f = 500.0 Hz.
(c) Relative to the air the locomotive is moving at vS = 20.00 m/s away from the uncle.
Use the plus sign in the denominator. Relative to the air the uncle is moving at vD =
10.00 m/s toward the locomotive. Use the plus sign in the numerator. Thus f′= f ⎛ 343m/s + 10.00 m/s ⎞
v + vD
= (500.0 Hz) ⎜
⎟ = 486.2 Hz.
v + vS
⎝ 343m/s + 20.00 m/s ⎠ (d) Relative to the air the locomotive is moving at vS = 20.00 m/s away from the girl and
the girl is moving at vD = 20.00 m/s toward the locomotive. Use the plus signs in both the
numerator and the denominator. Thus (v + vD) = (v + vS) and f′ = f = 500.0 Hz.
77. The siren is between you and the cliff, moving away from you and towards the cliff.
Both “detectors” (you and the cliff) are stationary, so vD = 0 in Eq. 17–47 (and see the
discussion in the textbook immediately after that equation regarding the selection of ±
signs). The source is the siren with vS = 10 m/s. The problem asks us to use v = 330 m/s
for the speed of sound. (a) With f = 1000 Hz, the frequency fy you hear becomes ⎛ v+0
fy = f ⎜
⎝ v + vS ⎞
2
⎟ = 970.6 Hz ≈ 9.7 ×10 Hz.
⎠ (b) The frequency heard by an observer at the cliff (and thus the frequency of the sound
reflected by the cliff, ultimately reaching your ears at some distance from the cliff) is ⎛ v+0 ⎞
3
fc = f ⎜
⎟ = 1031.3Hz ≈ 1.0 ×10 Hz.
⎝ v − vS ⎠
(c) The beat frequency is fc – fy = 60 beats/s (which, due to specific features of the human
ear, is too large to be perceptible).
81. (a) With r = 10 m in Eq. 17–28, we have
I= P
4 πr 2 ⇒ P = 10 W. (b) Using that value of P in Eq. 17–28 with a new value for r, we obtain
I= P
4 π ( 5.0 ) 2 = 0.032 W
.
m2 Alternatively, a ratio I′ /I = (r/r′ )2 could have been used.
(c) Using Eq. 17–29 with I = 0.0080 W/m2, we have β = 10 log I
= 99 dB
I0 where I0 = 1.0 × 10–12 W/m2.
2
85. (a) The intensity is given by I = 1 ρvω 2 sm , where ρ is the density of the medium, v is
2
the speed of sound, ω is the angular frequency, and sm is the displacement amplitude. The
displacement and pressure amplitudes are related by Δpm = ρvωsm, so sm = Δpm/ρvω and I
= (Δpm)2/2ρv. For waves of the same frequency the ratio of the intensity for propagation
in water to the intensity for propagation in air is 2 I w ⎛ Δpmw ⎞ ρ a va
=⎜
,
⎟
I a ⎝ Δpma ⎠ ρ w vw where the subscript a denotes air and the subscript w denotes water. Since Ia = Iw,
Δpmw
=
Δpma ρ w vw
(0.998 × 103 kg/m3 )(1482 m/s)
=
= 59.7.
ρ a va
(1.21kg/m 3 )(343m/s) The speeds of sound are given in Table 171 and the densities are given in Table 151.
(b) Now, Δpmw = Δpma, so
I w ρ a va
(1.21kg/m 3 )(343m/s)
=
=
= 2.81 × 10−4.
I a ρ w vw (0.998 × 103 kg/m3 )(1482 m/s)
87. (a) When the right side of the instrument is pulled out a distance d the path length for
sound waves increases by 2d. Since the interference pattern changes from a minimum to
the next maximum, this distance must be half a wavelength of the sound. So 2d = λ/2,
where λ is the wavelength. Thus λ = 4d and, if v is the speed of sound, the frequency is
f = v/λ = v/4d = (343 m/s)/4(0.0165 m) = 5.2 × 103 Hz.
(b) The displacement amplitude is proportional to the square root of the intensity (see Eq.
17–27). Write I = Csm , where I is the intensity, sm is the displacement amplitude, and C
is a constant of proportionality. At the minimum, interference is destructive and the
displacement amplitude is the difference in the amplitudes of the individual waves: sm =
sSAD – sSBD, where the subscripts indicate the paths of the waves. At the maximum, the
waves interfere constructively and the displacement amplitude is the sum of the
amplitudes of the individual waves: sm = sSAD + sSBD. Solve 100 = C ( sSAD − sSBD ) and 900 = C ( sSAD − sSBD )
for sSAD and sSBD. Add the equations to obtain
sSAD = ( 100 + 900 / 2C = 20 / C ,
then subtract them to obtain
sSBD = ( 900 − 100) / 2C = 10 / C .
The ratio of the amplitudes is sSAD/sSBD = 2.
(c) Any energy losses, such as might be caused by frictional forces of the walls on the air
in the tubes, result in a decrease in the displacement amplitude. Those losses are greater
on path B since it is longer than path A. 101. (a) The blood is moving towards the right (towards the detector), because the
Doppler shift in frequency is an increase: Δf > 0.
(b) The reception of the ultrasound by the blood and the subsequent remitting of the
signal by the blood back toward the detector is a two step process which may be
compactly written as ⎛ v + vx ⎞
f + Δf = f ⎜
⎟
⎝ v − vx ⎠ where vx = vblood cos θ . If we write the ratio of frequencies as R = (f + Δf)/f, then the solution of the above
equation for the speed of the blood is
vblood = ( R − 1) v
( R + 1) cosθ = 0.90 m/s where v = 1540 m/s, θ = 20°, and R = 1 + 5495/5 × 106.
(c) We interpret the question as asking how Δf (still taken to be positive, since the
detector is in the “forward” direction) changes as the detection angle θ changes. Since
larger θ means smaller horizontal component of velocity vx then we expect Δf to decrease
towards zero as θ is increased towards 90°. Chapter 18 – Student Solutions Manual
8. The change in length for the aluminum pole is
Δ = α A1ΔT = (33m)(23 × 10 −6 / C°)(15 °C) = 0.011m. 0 15. If Vc is the original volume of the cup, αa is the coefficient of linear expansion of
aluminum, and ΔT is the temperature increase, then the change in the volume of the cup
is ΔVc = 3αa Vc ΔT. See Eq. 1811. If β is the coefficient of volume expansion for
glycerin then the change in the volume of glycerin is ΔVg = βVc ΔT. Note that the original
volume of glycerin is the same as the original volume of the cup. The volume of glycerin
that spills is ΔVg − ΔVc = ( β − 3α a ) Vc ΔT = ⎡( 5.1× 10−4 / C° ) − 3 ( 23 × 10−6 / C° ) ⎤ (100 cm3 ) ( 6.0 °C )
⎣
⎦
3
= 0.26 cm .
21. Consider half the bar. Its original length is 0 = L0 / 2 and its length after the
temperature increase is = 0 + α 0 ΔT . The old position of the halfbar, its new position,
and the distance x that one end is displaced form a right triangle, with a hypotenuse of
length , one side of length 0 , and the other side of length x. The Pythagorean theorem
yields x 2 = 2 − 2 = 2 (1 + αΔT ) 2 − 2 . Since the change in length is small we may
0
0
0
2
approximate (1 + α ΔT ) by 1 + 2α ΔT, where the small term (α ΔT )2 was neglected.
Then,
x 2 = 2 + 2 2α ΔT − 2 = 2 2α ΔT
0
0
0
0
and
x= 0 2α ΔT = 3.77 m
2 ( 25 × 10−6 /C° ) ( 32° C ) = 7.5 × 10−2 m.
2 25. The melting point of silver is 1235 K, so the temperature of the silver must first be
raised from 15.0° C (= 288 K) to 1235 K. This requires heat Q = cm(T f − Ti ) = (236 J/kg ⋅ K)(0.130 kg)(1235°C − 288°C) = 2.91 × 104 J.
Now the silver at its melting point must be melted. If LF is the heat of fusion for silver
this requires
Q = mLF = ( 0.130 kg ) (105 × 103 J/kg ) = 1.36 × 104 J. The total heat required is ( 2.91 × 104 J + 1.36 × 104 J ) = 4.27 × 104 J. 27. The mass m = 0.100 kg of water, with specific heat c = 4190 J/kg·K, is raised from an
initial temperature Ti = 23°C to its boiling point Tf = 100°C. The heat input is given by Q
= cm(Tf – Ti). This must be the power output of the heater P multiplied by the time t; Q =
Pt. Thus,
t= Q cm (T f − Ti ) ( 4190 J/kg ⋅ K )( 0.100 kg )(100° C − 23°C )
=
=
= 160s.
P
P
200 J/s 41. (a) We work in Celsius temperature, which poses no difficulty for the J/kg·K values
of specific heat capacity (see Table 183) since a change of Kelvin temperature is
numerically equal to the corresponding change on the Celsius scale. There are three
possibilities:
• None of the ice melts and the waterice system reaches thermal equilibrium at a
temperature that is at or below the melting point of ice.
• The system reaches thermal equilibrium at the melting point of ice, with some of the ice
melted.
• All of the ice melts and the system reaches thermal equilibrium at a temperature at or
above the melting point of ice.
First, suppose that no ice melts. The temperature of the water decreases from TWi = 25°C
to some final temperature Tf and the temperature of the ice increases from TIi = –15°C to
Tf. If mW is the mass of the water and cW is its specific heat then the water rejects heat
 Q  = cW mW (TWi − T f ). If mI is the mass of the ice and cI is its specific heat then the ice absorbs heat
Q = cI mI (T f − TIi ). Since no energy is lost to the environment, these two heats (in absolute value) must be
the same. Consequently,
cW mW (TWi − T f ) = cI mI (T f − TIi ). The solution for the equilibrium temperature is
Tf = cW mW TWi + cI mI TIi
cW mW + cI mI (4190 J / kg ⋅ K)(0.200 kg)(25°C) + (2220 J/kg ⋅ K)(0.100 kg)( −15°C)
(4190 J/kg ⋅ K)(0.200 kg) + (2220 J/kg ⋅ K)(0.100 kg)
= 16.6°C.
= This is above the melting point of ice, which invalidates our assumption that no ice has
melted. That is, the calculation just completed does not take into account the melting of
the ice and is in error. Consequently, we start with a new assumption: that the water and
ice reach thermal equilibrium at Tf = 0°C, with mass m (< mI) of the ice melted. The
magnitude of the heat rejected by the water is
 Q  = cW mW TWi ,
and the heat absorbed by the ice is
Q = cI mI (0 − TIi ) + mLF ,
where LF is the heat of fusion for water. The first term is the energy required to warm all
the ice from its initial temperature to 0°C and the second term is the energy required to
melt mass m of the ice. The two heats are equal, so
cW mW TWi = −cI mI TIi + mLF .
This equation can be solved for the mass m of ice melted:
m=
= cW mW TWi + cI mI TIi
LF
(4190 J / kg ⋅ K)(0.200 kg)(25°C) + (2220 J / kg ⋅ K)(0.100 kg)( −15°C )
333 × 103 J / kg = 5.3 × 10−2 kg = 53g. Since the total mass of ice present initially was 100 g, there is enough ice to bring the
water temperature down to 0°C. This is then the solution: the ice and water reach thermal
equilibrium at a temperature of 0°C with 53 g of ice melted.
(b) Now there is less than 53 g of ice present initially. All the ice melts and the final
temperature is above the melting point of ice. The heat rejected by the water is Q = cW mW (TW i − T f )
and the heat absorbed by the ice and the water it becomes when it melts is
Q = cI mI (0 − TIi ) + cW mI (T f − 0) + mI LF . The first term is the energy required to raise the temperature of the ice to 0°C, the second
term is the energy required to raise the temperature of the melted ice from 0°C to Tf, and
the third term is the energy required to melt all the ice. Since the two heats are equal, cW mW (TW i − T f ) = cI mI (−TI i ) + cW mI T f + mI LF . The solution for Tf is
Tf = cW mW TW i + cI mI TIi − mI LF
cW ( mW + mI ) . Inserting the given values, we obtain Tf = 2.5°C.
43. Over a cycle, the internal energy is the same at the beginning and end, so the heat Q
absorbed equals the work done: Q = W. Over the portion of the cycle from A to B the
pressure p is a linear function of the volume V and we may write
p= 10
⎛ 20
⎞
Pa + ⎜
Pa/m 3 ⎟ V ,
3
⎝ 3
⎠ where the coefficients were chosen so that p = 10 Pa when V = 1.0 m3 and p = 30 Pa
when V = 4.0 m3. The work done by the gas during this portion of the cycle is
4 WAB = ∫ 4 1 pdV = ∫ 4 1 10 2 ⎞
⎛ 10 20 ⎞
⎛ 10
⎜ + V ⎟ dV = ⎜ V + V ⎟
3
⎝ 3 3 ⎠
⎝ 3
⎠1 ⎛ 40 160 10 10 ⎞
− − ⎟ J = 60 J.
= ⎜ +
3
3 3⎠
⎝ 3
The BC portion of the cycle is at constant pressure and the work done by the gas is
WBC = pΔV = (30 Pa)(1.0 m3 – 4.0 m3) = –90 J.
The CA portion of the cycle is at constant volume, so no work is done. The total work
done by the gas is W = WAB + WBC + WCA = 60 J – 90 J + 0 = –30 J and the total heat
absorbed is Q = W = –30 J. This means the gas loses 30 J of energy in the form of heat.
49. (a) The change in internal energy ΔEint is the same for path iaf and path ibf.
According to the first law of thermodynamics, ΔEint = Q – W, where Q is the heat
absorbed and W is the work done by the system. Along iaf
ΔEint = Q – W = 50 cal – 20 cal = 30 cal.
Along ibf ,
W = Q – ΔEint = 36 cal – 30 cal = 6.0 cal.
(b) Since the curved path is traversed from f to i the change in internal energy is –30 cal
and Q = ΔEint + W = –30 cal – 13 cal = – 43 cal. (c) Let ΔEint = Eint, f – Eint, i. Then, Eint, f = ΔEint + Eint, i = 30 cal + 10 cal = 40 cal.
(d) The work Wbf for the path bf is zero, so Qbf = Eint, f – Eint, b = 40 cal – 22 cal = 18 cal.
(e) For the path ibf, Q = 36 cal so Qib = Q – Qbf = 36 cal – 18 cal = 18 cal.
51. The rate of heat flow is given by
Pcond = kA TH − TC
,
L where k is the thermal conductivity of copper (401 W/m·K), A is the crosssectional area
(in a plane perpendicular to the flow), L is the distance along the direction of flow
between the points where the temperature is TH and TC. Thus,
Pcond = ( 401W/m ⋅ K ) ( 90.0 × 10−4 m2 ) (125°C − 10.0°C )
0.250 m = 1.66 × 103 J/s. The thermal conductivity is found in Table 186 of the text. Recall that a change in
Kelvin temperature is numerically equivalent to a change on the Celsius scale.
65. Let h be the thickness of the slab and A be its area. Then, the rate of heat flow through
the slab is
kA ( TH − TC )
Pcond =
h
where k is the thermal conductivity of ice, TH is the temperature of the water (0°C), and
TC is the temperature of the air above the ice (–10°C). The heat leaving the water freezes
it, the heat required to freeze mass m of water being Q = LFm, where LF is the heat of
fusion for water. Differentiate with respect to time and recognize that dQ/dt = Pcond to
obtain
dm
.
Pcond = LF
dt
Now, the mass of the ice is given by m = ρAh, where ρ is the density of ice and h is the
thickness of the ice slab, so dm/dt = ρA(dh/dt) and
Pcond = LF ρ A dh
.
dt We equate the two expressions for Pcond and solve for dh/dt: dh k ( TH − TC )
.
=
dt
LF ρ h
Since 1 cal = 4.186 J and 1 cm = 1 × 10–2 m, the thermal conductivity of ice has the SI
value
k = (0.0040 cal/s·cm·K) (4.186 J/cal)/(1 × 10–2 m/cm) = 1.674 W/m·K.
The density of ice is ρ = 0.92 g/cm3 = 0.92 × 103 kg/m3. Thus, (1.674 W m ⋅ K )( 0°C + 10°C )
dh
=
= 1.1× 10−6 m s = 0.40 cm h .
3
3
3
dt
333 ×10 J kg 0.92 × 10 kg m ( 0.050 m ) ( )( ) 73. The work (the “area under the curve”) for process 1 is 4piVi, so that
Ub – Ua = Q1 – W1 = 6piVi
by the First Law of Thermodynamics.
(a) Path 2 involves more work than path 1 (note the triangle in the figure of area
1
2 (4Vi)(pi/2) = piVi). With W2 = 4piVi + piVi = 5piVi, we obtain
Q2 = W2 + U b − U a = 5 piVi + 6 piVi = 11 piVi .
(b) Path 3 starts at a and ends at b so that ΔU = Ub – Ua = 6piVi.
75. The volume of the disk (thought of as a short cylinder) is πr²L where L = 0.50 cm is
its thickness and r = 8.0 cm is its radius. Eq. 1810, Eq. 1811 and Table 182 (which
gives α = 3.2 × 10−6/C°) lead to
ΔV = (πr²L)(3α)(60°C – 10°C) = 4.83 × 10−2 cm3 .
77. We have W = ∫ p dV (Eq. 1824). Therefore,
W = a ∫ V 2 dV = a 3
(V f − Vi 3 ) = 23 J.
3 81. Following the method of Sample Problem 184 (particularly its third Key Idea), we
have
J J (900 kg·C° )(2.50 kg)(Tf – 92.0°C) + (4190 kg·C° )(8.00 kg)(Tf – 5.0°C) = 0
where Table 183 has been used. Thus we find Tf = 10.5°C. 82. We use Q = –λFmice = W + ΔEint. In this case ΔEint = 0. Since ΔT = 0 for the ideal
gas, then the work done on the gas is
W ' = −W = λ F mi = (333J/g)(100 g) = 33.3kJ.
83. This is similar to Sample Problem 183. An important difference with part (b) of that
sample problem is that, in this case, the final state of the H2O is all liquid at Tf > 0. As
discussed in part (a) of that sample problem, there are three steps to the total process:
Q = m ( cice (0 C° – (–150 C°)) + LF + cliquid ( Tf – 0 C°))
Thus,
Tf = Q/m − (cice(150°) + LF ) cliquid = 79.5°C . Chapter 19 – Student Solutions Manual
7. (a) In solving pV = nRT for n, we first convert the temperature to the Kelvin scale:
T = (40.0 + 273.15) K = 313.15 K . And we convert the volume to SI units: 1000 cm3 =
1000 × 10–6 m3. Now, according to the ideal gas law, ( )( ) 1.01 × 105 Pa 1000 × 10−6 m3
pV
=
= 3.88 × 10−2 mol.
n=
RT
(8.31J/mol ⋅ K )( 313.15 K )
(b) The ideal gas law pV = nRT leads to (
( )( ) 1.06 × 105 Pa 1500 × 10−6 m3
pV
=
= 493K.
T=
nR
3.88 × 10−2 mol (8.31J/mol ⋅ K ) ) We note that the final temperature may be expressed in degrees Celsius as 220°C.
13. Suppose the gas expands from volume Vi to volume Vf during the isothermal portion
of the process. The work it does is
W = ∫ Vf p dV = nRT Vi ∫ Vf Vi V
dV
= nRT ln f ,
V
Vi where the ideal gas law pV = nRT was used to replace p with nRT/V. Now Vi = nRT/pi
and Vf = nRT/pf, so Vf/Vi = pi/pf. Also replace nRT with piVi to obtain W = piVi ln pi
.
pf Since the initial gauge pressure is 1.03 × 105 Pa,
pi = 1.03 × 105 Pa + 1.013 × 105 Pa = 2.04 × 105 Pa.
The final pressure is atmospheric pressure: pf = 1.013 × 105 Pa. Thus ⎛ 2.04 × 105 Pa ⎞
4
W = 2.04 × 105 Pa 0.14 m3 ln ⎜
⎟ = 2.00 × 10 J.
5
⎝ 1.013 × 10 Pa ⎠ ( )( ) During the constant pressure portion of the process the work done by the gas is W = pf(Vi
– Vf). The gas starts in a state with pressure pf, so this is the pressure throughout this
portion of the process. We also note that the volume decreases from Vf to Vi. Now Vf =
piVi/pf, so ⎛
pV ⎞
⎟
⎜
W = p f ⎜Vi − i i ⎟ = ( p f − pi )Vi = (1.013×105 Pa − 2.04×105 Pa )(0.14 m3 )
⎟
⎜
⎟
⎟
⎜
pf ⎠
⎝
= −1.44×104 J.
The total work done by the gas over the entire process is
W = 2.00 × 104 J – 1.44 × 104 J = 5.60 × 103 J.
19. According to kinetic theory, the rms speed is vrms = 3RT
M where T is the temperature and M is the molar mass. See Eq. 1934. According to Table
191, the molar mass of molecular hydrogen is 2.02 g/mol = 2.02 × 10–3 kg/mol, so
vrms = 3 ( 8.31J/mol ⋅ K )( 2.7 K ) = 1.8 × 102 m/s. −3 2.02 × 10 kg/mol 21. Table 191 gives M = 28.0 g/mol for nitrogen. This value can be used in Eq. 1922
with T in Kelvins to obtain the results. A variation on this approach is to set up ratios,
using the fact that Table 191 also gives the rms speed for nitrogen gas at 300 K (the
value is 517 m/s). Here we illustrate the latter approach, using v for vrms:
3RT2 / M
v2
T
=
= 2.
v1
T1
3RT1 / M (a) With T2 = (20.0 + 273.15) K ≈ 293 K, we obtain v2 = ( 517 m/s ) 293K
= 511m/s.
300 K (b) In this case, we set v3 = 1 v2 and solve v3 / v2 = T3 / T2 for T3:
2
2 ⎛v ⎞
⎛1⎞
T3 = T2 ⎜ 3 ⎟ = ( 293K ) ⎜ ⎟ = 73.0 K
⎝2⎠
⎝ v2 ⎠ which we write as 73.0 – 273 = – 200°C.
(c) Now we have v4 = 2v2 and obtain 2 2 ⎛v ⎞
T4 = T2 ⎜ 4 ⎟ = ( 293K )( 4 ) = 1.17 × 103 K
⎝ v2 ⎠ which is equivalent to 899°C.
∑v
, where the sum is over the speeds of the particles
N
and N is the number of particles. Thus 35. (a) The average speed is v = v = (2.0 + 3.0 + 4.0 + 5.0 + 6.0 + 7.0 + 8.0 + 9.0 + 10.0 + 11.0 ) km/s
= 6.5 km/s.
10 (b) The rms speed is given by vrms = ∑v 2 ∑v 2 . Now N = [(2.0) 2 + (3.0) 2 + (4.0) 2 + (5.0) 2 + (6.0) 2
+ (7.0) 2 + (8.0) 2 + (9.0) 2 + (10.0) 2 + (11.0) 2 ] km 2 / s 2 = 505 km 2 / s 2 so
505 km 2 / s 2
=
= 7.1 km/s.
10 vrms 41. (a) The distribution function gives the fraction of particles with speeds between v and
v + dv, so its integral over all speeds is unity: ∫ P(v) dv = 1. Evaluate the integral by
calculating the area under the curve in Fig. 1924. The area of the triangular portion is
half the product of the base and altitude, or 1 av0 . The area of the rectangular portion is
2
the product of the sides, or av0. Thus,
1 ∫ P(v)dv = 2 av 0 + av0 = 3
av0 ,
2 3
so 2 av0 = 1 and av0 = 2/3=0.67. (b) The average speed is given by vavg = ∫ vP ( v ) dv. For the triangular portion of the
distribution P(v) = av/v0, and the contribution of this portion is
a
v0 ∫ v0 0 v 2 dv = 2
a 3 av0 2
v0 =
= v0 ,
3v0
3
9 where 2/3v0 was substituted for a. P(v) = a in the rectangular portion, and the
contribution of this portion is
a∫ 2 v0 v0 v dv = ( ) a
3a 2
2
2
4v0 − v0 =
v0 = v0 .
2
2 Therefore,
vavg = vavg
2
= 1.22 .
v0 + v0 = 1.22v0 ⇒
9
v0 2
(c) The meansquare speed is given by vrms = ∫ v 2 P ( v ) dv. The contribution of the triangular section is
a
v0 ∫ v0 v 3 dv = 0 a 4 1 2
v0 = v0 .
4v0
6 The contribution of the rectangular portion is
a∫ 2 v0 v0 v 2 dv = ( ) a
7 a 3 14 2
3
3
8v0 − v0 =
v0 = v0 .
3
3
9 Thus,
vrms = v
1 2 14 2
v0 + v0 = 1.31v0 ⇒ rms = 1.31 .
6
9
v0 (d) The number of particles with speeds between 1.5v0 and 2v0 is given by N ∫ 2 v0 1.5 v0 P ( v ) dv . The integral is easy to evaluate since P(v) = a throughout the range of integration. Thus
the number of particles with speeds in the given range is N a(2.0v0 – 1.5v0) = 0.5N av0 =
N/3, where 2/3v0 was substituted for a. In other words, the fraction of particles in this
range is 1/3 or 0.33.
45. When the temperature changes by ΔT the internal energy of the first gas changes by
n1C1 ΔT, the internal energy of the second gas changes by n2C2 ΔT, and the internal
energy of the third gas changes by n3C3 ΔT. The change in the internal energy of the
composite gas is
ΔEint = (n1 C1 + n2 C2 + n3 C3) ΔT.
This must be (n1 + n2 + n3) CV ΔT, where CV is the molar specific heat of the mixture.
Thus
CV = n1C1 + n2C2 + n3C3
.
n1 + n2 + n3 With n1=2.40 mol, CV1=12.0 J/mol·K for gas 1, n2=1.50 mol, CV2=12.8 J/mol·K for gas 2,
and n3=3.20 mol, CV3=20.0 J/mol·K for gas 3, we obtain CV =15.8 J/mol·K for the
mixture.
46. Two formulas (other than the first law of thermodynamics) will be of use to us. It is
straightforward to show, from Eq. 1911, that for any process that is depicted as a straight
line on the pV diagram — the work is ⎛ pi + p f ⎞
Wstraight = ⎜
⎟ ΔV
⎝ 2 ⎠
which includes, as special cases, W = pΔV for constantpressure processes and W = 0 for
constantvolume processes. Further, Eq. 1944 with Eq. 1951 gives
⎛ f ⎞
⎛ f ⎞
Eint = n ⎜ ⎟ RT = ⎜ ⎟ pV
⎝2⎠
⎝2⎠ where we have used the ideal gas law in the last step. We emphasize that, in order to
obtain work and energy in Joules, pressure should be in Pascals (N / m2) and volume
should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5.
(a) The internal energy change is 5
5
( pcVc − paVa ) = ( ( 2.0 ×103 Pa )( 4.0 m3 ) − ( 5.0 ×103 Pa )( 2.0 m3 ) )
2
2
3
= −5.0 × 10 J. Eint c − Eint a = (b) The work done during the process represented by the diagonal path is
⎛ p + pc ⎞
Wdiag = ⎜ a
⎟ (Vc − Va ) =
⎝ 2 ⎠ ( 3.5 ×103 Pa )( 2.0 m3 ) which yields Wdiag = 7.0×103 J. Consequently, the first law of thermodynamics gives Qdiag = ΔEint + Wdiag = (−5.0 ×103 + 7.0 × 103 ) J = 2.0 ×103 J.
(c) The fact that ΔEint only depends on the initial and final states, and not on the details of
the “path” between them, means we can write ΔEint = Eint c − Eint a = −5.0 × 103 J for the
indirect path, too. In this case, the work done consists of that done during the constant
pressure part (the horizontal line in the graph) plus that done during the constant volume
part (the vertical line): Windirect = ( 5.0 ×103 Pa )( 2.0 m3 ) + 0 = 1.0 ×104 J. Now, the first law of thermodynamics leads to
Qindirect = ΔEint + Windirect = (−5.0 × 103 + 1.0 × 10 4 ) J = 5.0 × 103 J. 53. (a) Since the process is at constant pressure, energy transferred as heat to the gas is
given by Q = nCp ΔT, where n is the number of moles in the gas, Cp is the molar specific
heat at constant pressure, and ΔT is the increase in temperature. For a diatomic ideal gas
C p = 7 R. Thus,
2
Q= 7
7
nRΔT = ( 4.00 mol )( 8.31J/mol ⋅ K )( 60.0 K ) = 6.98 × 103 J.
2
2 (b) The change in the internal energy is given by ΔEint = nCV ΔT, where CV is the specific
heat at constant volume. For a diatomic ideal gas CV = 5 R , so
2
ΔEint = 5
5
nRΔT = ( 4.00 mol )( 8.31J/mol.K )( 60.0 K ) = 4.99 × 103 J.
2
2 (c) According to the first law of thermodynamics, ΔEint = Q – W, so
W = Q − ΔEint = 6.98 × 103 J − 4.99 × 103 J = 1.99 × 103 J. (d) The change in the total translational kinetic energy is
ΔK = 3
3
nRΔT = ( 4.00 mol )( 8.31J/mol ⋅ K )( 60.0 K ) = 2.99 × 103 J.
2
2 67. In this solution we will use nonstandard notation: writing ρ for weightdensity
(instead of massdensity), where ρc refers to the cool air and ρh refers to the hot air. Then
the condition required by the problem is
Fnet = Fbuoyant – hotairweight – balloonweight
2.67 × 103 N = ρcV – ρhV – 2.45 × 103 N
where V = 2.18 × 103 m3 and ρc = 11.9 N/m3. This condition leads to ρh = 9.55 N/m3.
Using the ideal gas law to write ρh as PMg/RT where P = 101000 Pascals and M = 0.028
kg/m3 (as suggested in the problem), we conclude that the temperature of the enclosed air
should be 349 K.
69. (a) By Eq. 1928, W = –374 J (since the process is an adiabatic compression).
(b) Q = 0 since the process is adiabatic. (c) By first law of thermodynamics, the change in internal energy is ΔEint= Q – W = +374
J.
(d) The change in the average kinetic energy per atom is ΔKavg = ΔEint/N = +3.11 × 10−22
J.
71. This is very similar to Sample Problem 194 (and we use similar notation here) except
for the use of Eq. 1931 for vavg (whereas in that Sample Problem, its value was just
assumed). Thus,
f= speed
distance = 2 p d ⎛16πR⎞
= k ⎜ MT ⎟ .
⎝
⎠ vavg λ Therefore, with p = 2.02 × 103 Pa, d = 290 × 10−12 m and M = 0.032 kg/mol (see Table
191), we obtain f = 7.03 × 109 s−1.
77. (a) The final pressure is
pf = pVi ( 32 atm )(1.0 L )
i
=
= 8.0atm,
4.0 L
Vf (b) For the isothermal process the final temperature of the gas is Tf = Ti = 300 K.
(c) The work done is
⎛ Vf ⎞
⎛ Vf ⎞
⎛ 4.0 L ⎞
W = nRTi ln ⎜ ⎟ = piVi ln ⎜ ⎟ = ( 32 atm ) 1.01× 105 Pa atm 1.0 × 10−3 m3 ln ⎜
⎟
⎝ 1.0 L ⎠
⎝ Vi ⎠
⎝ Vi ⎠
= 4.4 × 103 J.
( )( ) For the adiabatic process piVi γ = p f V fγ . Thus,
(d) The final pressure is
⎛V
p f = pi ⎜ i
⎜ Vf
⎝ γ ⎞
⎛ 1.0 L ⎞
⎟ = ( 32 atm ) ⎜
⎟
⎟
⎝ 4.0 L ⎠
⎠ 53 = 3.2 atm. (e) The final temperature is
Tf = (f) The work done is p f V f Ti
piVi = ( 3.2 atm )( 4.0 L )( 300 K ) = 120 K
( 32 atm )(1.0 L ) . 3
3
W = Q − ΔEint = −ΔEint = − nRΔT = − ( p f V f − piVi )
2
2
3
= − ⎡( 3.2 atm )( 4.0 L ) − ( 32 atm )(1.0 L ) ⎤ 1.01× 105 Pa atm 10−3 m3 L
⎦
2⎣
= 2.9 × 103 J . ( )( ) If the gas is diatomic, then γ = 1.4.
(g) The final pressure is
γ ⎛V
p f = pi ⎜ i
⎜ Vf
⎝ ⎞
⎛ 1.0 L ⎞
⎟ = ( 32 atm ) ⎜
⎟
⎟
⎝ 4.0 L ⎠
⎠ 1.4 = 4.6 atm . (h) The final temperature is
Tf = p f V f Ti = piVi ( 4.6 atm )( 4.0 L )( 300 K ) = 170 K .
( 32 atm )(1.0 L ) (i) The work done is
5
5
W = Q − ΔEint = − nRΔT = − ( p f V f − piVi )
2
2
5
= − ⎡( 4.6 atm )( 4.0 L ) − ( 32 atm )(1.0 L ) ⎤ 1.01× 105 Pa atm 10−3 m3 L
⎦
2⎣
= 3.4 ×103 J. ( )( 81. It is recommended to look over §197 before doing this problem.
(a) We normalize the distribution function as follows: ∫ vo 0 P ( v ) dv = 1 ⇒ C = 3
.
3
vo (b) The average speed is ∫ vo 0 vP ( v ) dv = (c) The rms speed is the square root of ∫ vo 0 ⎛ 3v 2 ⎞
3
v ⎜ 3 ⎟ dv = vo .
4
⎝ vo ⎠ ) ∫ vo 0 vo
⎛ 3v 2 ⎞
3 2
v 2 P ( v ) dv = ∫ v 2 ⎜ 3 ⎟ dv = vo .
0
5
⎝ vo ⎠ Therefore, vrms = 3 5vo ≈ 0.775vo .
83. (a) The temperature is 10.0°C → T = 283 K. Then, with n = 3.50 mol and Vf/V0 = 3/4,
we use Eq. 1914:
⎛ Vf ⎞
W = nRT ln ⎜ ⎟ = − 2.37 kJ.
⎝ V0 ⎠
(b) The internal energy change ΔEint vanishes (for an ideal gas) when ΔT = 0 so that the
First Law of Thermodynamics leads to Q = W = –2.37 kJ. The negative value implies
that the heat transfer is from the sample to its environment. Chapter 20 – Student Solutions Manual
5. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the
volume V, and the temperature T by p = nRT/V. The work done by the gas during the
isothermal expansion is
V2 V2 V1 V1 W = ∫ p dV = n RT ∫ dV
V
= n RT ln 2 .
V
V1 We substitute V2 = 2.00V1 to obtain W = n RT ln2.00 = ( 4.00 mol )( 8.31 J/mol ⋅ K )( 400 K ) ln2.00 = 9.22 ×103 J.
(b) Since the expansion is isothermal, the change in entropy is given by
ΔS = ∫ (1 T ) dQ = Q T , where Q is the heat absorbed. According to the first law of thermodynamics, ΔEint = Q −
W. Now the internal energy of an ideal gas depends only on the temperature and not on
the pressure and volume. Since the expansion is isothermal, ΔEint = 0 and Q = W. Thus,
ΔS = W 9.22 × 103 J
=
= 23.1 J/K.
T
400 K (c) ΔS = 0 for all reversible adiabatic processes.
7. (a) The energy that leaves the aluminum as heat has magnitude Q = maca(Tai − Tf),
where ma is the mass of the aluminum, ca is the specific heat of aluminum, Tai is the
initial temperature of the aluminum, and Tf is the final temperature of the aluminumwater system. The energy that enters the water as heat has magnitude Q = mwcw(Tf − Twi),
where mw is the mass of the water, cw is the specific heat of water, and Twi is the initial
temperature of the water. The two energies are the same in magnitude since no energy is
lost. Thus,
m c T + mw cwTwi
ma ca (Tai − T f ) = mw cw ( T f − Twi ) ⇒ T f = a a ai
.
ma ca + mwcw
The specific heat of aluminum is 900 J/kg⋅K and the specific heat of water is 4190
J/kg⋅K. Thus,
Tf = ( 0.200 kg )( 900 J/kg ⋅ K )(100°C ) + ( 0.0500 kg )( 4190 J/kg ⋅ K )( 20°C ) = 57.0°C = 330 K.
( 0.200 kg )( 900 J/kg ⋅ K ) + ( 0.0500 kg )( 4190 J/kg ⋅ K ) (b) Now temperatures must be given in Kelvins: Tai = 393 K, Twi = 293 K, and Tf = 330
K. For the aluminum, dQ = macadT and the change in entropy is
ΔS a = ∫ Tf
T f dT
dQ
⎛ 330 K ⎞
= ma ca ∫
= ma ca ln
= ( 0.200 kg )( 900 J/kg ⋅ K ) ln ⎜
⎟ = −22.1 J/K.
Tai T
T
Tai
⎝ 373 K ⎠ (c) The entropy change for the water is ΔS w = ∫ T f dT
T
⎛ 330 K ⎞
dQ
= mwcw ∫
= mwcw ln f = (0.0500 kg) (4190 J kg.K) ln ⎜
⎟ = +24.9 J K .
Twi T
T
Twi
⎝ 293K ⎠ (d) The change in the total entropy of the aluminumwater system is
ΔS = ΔSa + ΔSw = −22.1 J/K + 24.9 J/K = +2.8 J/K.
25. (a) The efficiency is ε= TH − TL (235 − 115) K
=
= 0.236 = 23.6% .
TH
(235+273) K We note that a temperature difference has the same value on the Kelvin and Celsius
scales. Since the temperatures in the equation must be in Kelvins, the temperature in the
denominator is converted to the Kelvin scale.
(b) Since the efficiency is given by ε = W/QH, the work done is given by W = ε QH = 0.236 (6.30 ×104 J) = 1.49 ×104 J .
29. (a) Energy is added as heat during the portion of the process from a to b. This portion
occurs at constant volume (Vb), so Qin = nCV ΔT. The gas is a monatomic ideal gas, so
CV = 3R / 2 and the ideal gas law gives
ΔT = (1/nR)(pb Vb – pa Va) = (1/nR)(pb – pa) Vb.
Thus, Qin = 3
2 ( pb − pa ) Vb . Vb and pb are given. We need to find pa. Now pa is the same as pc and points c and b are connected by an adiabatic process. Thus, pcVcγ = pbVbγ and
γ ⎛V ⎞
⎛ 1 ⎞
pa = pc = ⎜ b ⎟ pb = ⎜
⎟
⎝ 8.00 ⎠
⎝ Vc ⎠ The energy added as heat is 53 (1.013 ×10 6 ) Pa = 3.167 × 104 Pa. Qin = ( )( ) 3
1.013 × 106 Pa − 3.167 × 104 Pa 1.00 × 10−3 m3 = 1.47 × 103 J.
2 (b) Energy leaves the gas as heat during the portion of the process from c to a. This is a
constant pressure process, so
Qout = nC p ΔT =
= ( 5
5
( paVa − pcVc ) = pa (Va − Vc )
2
2 ) ( ) 5
3.167 × 104 Pa ( −7.00 ) 1.00 × 10−3 m3 = −5.54 × 102 J,
2 or  Qout = 5.54 × 10 2 J . The substitutions Va – Vc = Va – 8.00 Va = – 7.00 Va and C p = 5 R
2
were made.
(c) For a complete cycle, the change in the internal energy is zero and
W = Q = 1.47 × 103 J – 5.54 × 102 J = 9.18 × 102 J.
(d) The efficiency is ε = W/Qin = (9.18 × 102 J)/(1.47 × 103 J) = 0.624 = 62.4%.
37. A Carnot refrigerator working between a hot reservoir at temperature TH and a cold
reservoir at temperature TL has a coefficient of performance K that is given by
K= TL
.
TH − TL For the refrigerator of this problem, TH = 96° F = 309 K and TL = 70° F = 294 K, so K =
(294 K)/(309 K – 294 K) = 19.6. The coefficient of performance is the energy QL drawn
from the cold reservoir as heat divided by the work done: K = QL/W. Thus,
QL = KW = (19.6)(1.0 J) = 20 J.
39. The coefficient of performance for a refrigerator is given by K = QL/W, where QL is
the energy absorbed from the cold reservoir as heat and W is the work done during the
refrigeration cycle, a negative value. The first law of thermodynamics yields QH + QL –
W = 0 for an integer number of cycles. Here QH is the energy ejected to the hot reservoir
as heat. Thus, QL = W – QH. QH is negative and greater in magnitude than W, so QL =
QH – W. Thus,
Q −W
K= H
.
W
The solution for W is W = QH/(K + 1). In one hour, W = 7.54 MJ
= 1.57 MJ.
3.8 + 1 The rate at which work is done is (1.57 × 106 J)/(3600 s) = 440 W.
47. (a) Suppose there are nL molecules in the left third of the box, nC molecules in the
center third, and nR molecules in the right third. There are N! arrangements of the N
molecules, but nL! are simply rearrangements of the nL molecules in the right third, nC!
are rearrangements of the nC molecules in the center third, and nR! are rearrangements of
the nR molecules in the right third. These rearrangements do not produce a new
configuration. Thus, the multiplicity is
W= N!
.
nL !nC !nR ! (b) If half the molecules are in the right half of the box and the other half are in the left
half of the box, then the multiplicity is WB = N!
.
( N 2 )!( N 2 ) ! If onethird of the molecules are in each third of the box, then the multiplicity is WA =
The ratio is N!
.
( N 3)!( N 3) !( N 3)! ( N 2 ) !( N 2 ) ! .
WA
=
WB ( N 3) !( N 3) !( N 3) ! (c) For N = 100,
WA
50!50!
=
= 4.2 × 1016.
WB 33!33!34! 49. Using Eq. 1934 and Eq. 1935, we arrive at
Δv = ( 3 − 2 ) RT/M
(a) We find, with M = 28 g/mol = 0.028 kg/mol (see Table 191), Δvi=87 m/s at 250 K,
(b) and Δvf =122 ≈1.2×102 m/s at 500 K.
(c) The expression above for Δv implies T = M
2
2 (Δv)
R( 3 − 2 ) which we can plug into Eq. 204 to yield
ΔS = nR ln(Vf /Vi) + nCV ln(Tf /Ti) = 0 + nCV ln[(Δvf)2/(Δvi)2] = 2nCV ln(Δvf /Δvi).
Using Table 193 to get CV = 5R/2 (see also Table 192) we then find, for n = 1.5 mol, ΔS
= 22 J/K.
55. Except for the phase change (which just uses Eq. 202), this has some similarities
with Sample Problem 202. Using constants available in the Chapter 19 tables, we
compute
Lf
ΔS = m[cice ln(273/253) + 273 + cwater ln(313/273)] = 1.18 × 103 J/K.
1 63. (a) Eq. 2015 can be written as QH = QL(1 + 1/KC ) = (35)(1 + 4.6 ) = 42.6 kJ.
(b) Similarly, Eq. 2014 leads to W = QL/K = 35/4.6 = 7.61 kJ.
67. We adapt the discussion of §207 to 3 and 5 particles (as opposed to the 6 particle
situation treated in that section).
(a) The least multiplicity configuration is when all the particles are in the same half of the
box. In this case, using Eq. 2020, we have
W= 3!
= 1.
3!0! (b) Similarly for box B, W = 5!/(5!0!) = 1 in the “least” case.
(c) The most likely configuration in the 3 particle case is to have 2 on one side and 1 on
the other. Thus,
3!
= 3.
W=
2!1!
(d) The most likely configuration in the 5 particle case is to have 3 on one side and 2 on
the other. Thus,
5!
= 10.
W=
3!2!
(e) We use Eq. 2021 with our result in part (c) to obtain S = k ln W = (1.38 × 10−23 ) ln3 = 1.5 × 10−23 J/K. (f) Similarly for the 5 particle case (using the result from part (d)), we find
S = k ln 10 = 3.2 × 10−23 J/K. Chapter 21 1
The magnitude of the force that either charge exerts on the other is given by
F = 1 jq1 jjq2 j
;
4¼²0 r2 where r is the distance between them. Thus
s
jq1 jjq2 j
r=
4¼²0 F
s
(8:99 £ 109 N ¢ m2 =C2 )(26:0 £ 10¡6 C)(47:0 £ 10¡6 C)
= 1:38 m :
=
5:70 N
5
The magnitude of the force of either of the charges on the other is given by
F = 1 q(Q ¡ q)
;
4¼²0
r2 where r is the distance between the charges. You want the value of q that maximizes the function
f (q) = q(Q ¡ q). Set the derivative df =dq equal to zero. This yields Q ¡ 2q = 0, or q = Q=2.
7
Assume the spheres are far apart. Then the charge distribution on each of them is spherically
symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges and choose the
coordinate system so the force on q2 is positive if it is repelled by q1 . Take the distance between
the charges to be r. Then the force on q2 is
Fa = ¡ 1 q1 q2
:
4¼²0 r2 The negative sign indicates that the spheres attract each other.
After the wire is connected, the spheres, being identical, have the same charge. Since charge
is conserved, the total charge is the same as it was originally. This means the charge on each
sphere is (q1 + q2 )=2. The force is now one of repulsion and is given by
Fb = 1 (q1 + q2 )2
:
4¼²0
4r2
Chapter 21 131 Solve the two force equations simultaneously for q1 and q2 . The first gives
q1 q2 = ¡4¼²0 r 2 Fa = ¡
and the second gives
q1 + q2 = 2r
Thus p (0:500 m)2 (0:108 N)
8:99 £ 109 s 4¼²0 Fb = 2(0:500 m) N¢ m2 =C2 = ¡3:00 £ 10¡12 C2 0:0360 N
8:99 £ 109 N ¢ m2 =C2 = 2:00 £ 10¡6 C : ¡(3:00 £ 10¡12 C2 )
q1
and substitution into the second equation gives
q2 = ¡3:00 £ 10¡12 C2
= 2:00 £ 10¡6 C :
q1
Multiply by q1 to obtain the quadratic equation
q1 + The solutions are 2
q1 ¡ (2:00 £ 10¡6 C)q1 ¡ 3:00 £ 10¡12 C2 = 0 : p
(¡2:00 £ 10¡6 C)2 + 4(3:00 £ 10¡12 C2 )
:
q1 =
2
If the positive sign is used, q1 = 3:00 £ 10¡6 C and if the negative sign is used, q1 = ¡1:00 £
10¡6 C. Use q2 = (¡3:00 £ 10¡12 )=q1 to calculate q2 . If q1 = 3:00 £ 10¡6 C, then q2 =
¡1:00 £ 10¡6 C and if q1 = ¡1:00 £ 10¡6 C, then q2 = 3:00 £ 10¡6 C. Since the spheres are
identical, the solutions are essentially the same: one sphere originally had charge ¡1:00 £ 10¡6 C
and the other had charge +3:00 £ 10¡6 C.
2:00 £ 10¡6 C § 19
If the system of three particles is to be in equilibrium, the force
on each particle must be zero. Let the charge on the third particle
be q0 . The third particle must lie on the x axis since otherwise the
two forces on it would not be along the same line and could not
Ã x ¡ ¡ L¡x ¡
¡ !Ã
!
sum to zero. Thus the y coordinate of the particle must be zero.
²
²
²
The third particle must lie between the other two since otherwise
q
q0
4:00q
the forces acting on it would be in the same direction and would
not sum to zero. Suppose the third particle is a distance x from
the particle with charge q, as shown on the diagram to the right.
The force acting on it is then given by
∙
¸
1
4:00qq0
qq0
= 0;
F0 =
¡
4¼²0 x2
(L ¡ x)2
where the positive direction was taken to be toward the right. Solve this equation for x. Canceling
common factors yields 1=x2 = 4:00=(L¡x)2 and taking the square root yields 1=x = 2:00=(L¡x).
The solution is x = 0:333L.
132 Chapter 21 The force on q is ∙
¸
1
qq0 4:00q 2
= 0:
+
Fq =
4¼²0 x2
L2 Solve for q0 : q0 = ¡4:00qx2 =L2 = ¡0:444q, where x = 0:333L was used.
The force on the particle with charge 4:00q is
∙
¸
∙
¸
1
1
4:00q2 4:00qq0
4:00q 2 4:00(0:444)q2
F4q =
=
+
+
4¼²0
(L ¡ x)2
4¼²0
(0:444)L2
L2
L2
∙
¸
1
4:00q2 4:00q2
= 0:
=
¡
4¼²0
L2
L2
With q0 = ¡0:444q and x = 0:333L, all three charges are in equilibrium. 25
(a) The magnitude of the force between the ions is given by
q2
;
4¼²0 r2
where q is the charge on either of them and r is the distance between them. Solve for the charge:
s
p
3:7 £ 10¡9 N
= 3:2 £ 10¡19 C :
q = r 4¼²0 F = (5:0 £ 10¡10 m)
2
8:99 £ 109 N ¢ m2 =C
F = (b) Let N be the number of electrons missing from each ion. Then N e = q and
N= q
3:2 £ 10¡19 C
= 2:
=
e 1:60 £ 10¡19 C 35
(a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine
ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion
that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another,
diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair
exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero
and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero.
(b) Rather than remove a cesium ion, superpose charge ¡e at the position of one cesium ion.
This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it is
equivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum to
zero, so the only force on the chlorine ion is the force of the added charge.
p
The length of a body diagonal of a cube is 3a, where p is the length of a cube edge. Thus the
a
distance from the center of the cube to a corner is d = ( 3=2)a. The force has magnitude
F =
= 1 e2
1
e2
=
4¼²0 d2 4¼²0 (3=4)a2
(8:99 £ 109 N ¢ m2 =C2 )(1:60 £ 10¡19 C)2
= 1:9 £ 10¡9 N :
(3=4)(0:40 £ 10¡9 m)2
Chapter 21 133 Since both the added charge and the chlorine ion are negative, the force is one of repulsion. The
chlorine ion is pulled away from the site of the missing cesium ion.
37
None of the reactions given include a beta decay, so the number of protons, the number of
neutrons, and the number of electrons are each conserved. Atomic numbers (numbers of protons
and numbers of electrons) and molar masses (combined numbers of protons and neutrons) can
be found in Appendix F of the text.
(a) 1 H has 1 proton, 1 electron, and 0 neutrons and 9 Be has 4 protons, 4 electrons, and 9 ¡ 4 = 5
neutrons, so X has 1 + 4 = 5 protons, 1 + 4 = 5 electrons, and 0 + 5 ¡ 1 = 4 neutrons. One of
the neutrons is freed in the reaction. X must be boron with a molar mass of 5 g=mol + 4 g=mol =
9 g=mol: 9 B.
(b) 12 C has 6 protons, 6 electrons, and 12 ¡ 6 = 6 neutrons and 1 H has 1 proton, 1 electron, and
0 neutrons, so X has 6 + 1 = 7 protons, 6 + 1 = 7 electrons, and 6 + 0 = 6 neutrons. It must be
nitrogen with a molar mass of 7 g=mol + 6 g=mol = 13 g=mol: 13 N.
(c) 15 N has 7 protons, 7 electrons, and 15 ¡ 7 = 8 neutrons; 1 H has 1 proton, 1 electron, and 0
neutrons; and 4 He has 2 protons, 2 electrons, and 4 ¡ 2 = 2 neutrons; so X has 7 + 1 ¡ 2 = 6
protons, 6 electrons, and 8 + 0 ¡ 2 = 6 neutrons. It must be carbon with a molar mass of
6 g=mol + 6 g=mol = 12 g=mol: 12 C.
39
The magnitude of the force of particle 1 on particle 2 is
F = 1 jq1 jjq2 j
:
4¼²0 d2 + d2
1
2 The signs of the charges are the same, so the particles repel each other along the line that runs
q
through them. This line makes an angle µ with the x axis such that cos µ = d2 = d2 + d2 , so the
1
2
x component of the force is
1 jq1 jjq2 j
1 jq1 jjq2 jd2
cos µ =
2 + d2
4¼²0 d1
4¼²0 (d2 + d2 )3=2
2
1
2
24(1:60 £ 10¡19 C)2 (6:00 £ 10¡3 m)
= (8:99 £ 109 C2 =N ¢ m2 )
= 1:31 £ 10¡22 N :
[(2:00 £ 10¡3 m)2 + (6:00 £ 10¡3 m)2 ]3=2 Fx = 50
The magnitude of the gravitational force on a proton near Earth’s surface is mg, where m is
the mass of the proton (1:67 £ 10¡27 kg from Appendix B). The electrostatic force between two
protons is F = (1=4¼²0 )(e2 =d2 ), where d is their separation. Equate these forces to each other
and solve for d. The result is
s
s
(1:60 £ 10¡19 C)2
1 e2
2
= (8:99 £ 10¡9 N ¢ m2 =C )
= 0:119 m :
d=
2
4¼²0 mg
(1:67 £ 10¡27 kg)(9:8 m=s )
134 Chapter 21 60
The magnitude of the force of particle 1 on particle 4 is
F1 = ¡19
1 jq1 jjq4 j
C)(3:20 £ 10¡19 C)
2 (3:20 £ 10
= (8:99 £ 109 N ¢ m2 =C )
= 1:02 £ 10¡24 N :
4¼²0 d2
(0:0300 m)2
1 The charges have opposite signs, so the particles attract each other and the vector force is
~
F1 = ¡(1:02¡24 N)(cos 35:0± ) ˆ ¡ (1:02 £ 10¡24 N)(sin 35:0± ) ˆ
i
j
i
j
= ¡(8:36 £ 10¡25 N) ˆ ¡ (5:85 £ 10¡25 N) ˆ :
Particles 2 and 3 repel each other. The force of particle 2 on particle 4 is
1 jq2 jjq4 j ˆ
~
j
F2 = ¡
4¼²0 d2
2
2 = ¡(8:99 £ 109 N ¢ m2 =C ) (3:20 £ 10¡19 C)(3:20 £ 10¡19 C)
ˆ = ¡(2:30 £ 10¡24 N) ˆ :
j
j
(0:0200 m)2 Particles 3 and 4 repel each other and the force of particle 3 on particle 4 is
1 j2q3 jjq4 j ˆ
~
i
F3 = ¡
4¼²0
d2
3
= ¡(8:99 £ 109 N ¢ m2 =C2 ) (6:40 £ 10¡19 C)(3:20 £ 10¡19 C)
= ¡(4:60 £ 10¡24 C) ˆ :
i
(0:0200 m)2 The net force is the vector sum of the three forces. The x component is Fx = ¡18:36£10¡25 N¡
4:60 N = ¡5:44 £ 10¡24 N and the y component is Fy = ¡5:85 £ 10¡25 N ¡ 2:30 £ 10¡24 N =
¡2:89 £ 10¡24 N. The magnitude of the force is
q
p
2
2
F = Fx + Fy = (¡5:44 £ 10¡24 N)2 + (¡2:89 £ 10¡24 N)2 = 6:16 £ 1024 N : The tangent of the angle µ between the net force and the positive x axis is tanµ = Fy =Fx =
(¡2:89 £ 10¡24 N)=(¡5:44 £ 10¡24 N) = 0:531 and the angle is either 28± or 208± . The later
angle is associated with a vector that has negative x and y components and so is the correct
angle.
69
The net force on particle 3 is the vector sum of the forces of particles 1 and 2 and for this to
be zero the two forces must be along the same line. Since electrostatic forces are along the lines
that join the particles, particle 3 must be on the x axis. Its y coordinate is zero.
Particle 3 is repelled by one of the other charges and attracted by the other. As a result, particle
3 cannot be between the other two particles and must be either to the left of particle 1 or to the
right of particle 2. Since the magnitude of q1 is greater than the magnitude of q2 , particle 3 must
Chapter 21 135 be closer to particle 2 than to particle 1 and so must be to the right of particle 2. Let x be the
coordinate of particle 3. The the x component of the force on it is
Fx = 1 h q1 q3
q2 q3 i
:
+
4¼²0 x2
(x ¡ L)2 If Fx = 0 the solution for x is
p
p
¡q1 =q2
¡(¡5:00q)=(2:00q)
L= p
L = 2:72L :
x= p
¡q1 =q2 ¡ 1
¡(¡5:00q)=(2:00q) ¡ 1 136 Chapter 21 Chapter 22 3
Since the magnitude of the electric field produced by a point particle with charge q is given
by E = jqj=4¼²0 r2 , where r is the distance from the particle to the point where the field has
magnitude E, the magnitude of the charge is
jqj = 4¼0 r2 E = (0:50 m)2 (2:0 N=C)
8:99 £ 109 N ¢ m2 =C2 = 5:6 £ 10¡11 C : 5
Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is
exactly the same as it would be if the charge were all at the center. That is, the magnitude of the
field is
q
E=
;
4¼²0 R2
where q is the magnitude of the total charge and R is the sphere radius. The magnitude of the
total charge is Ze, so
2 E= Ze
(8:99 £ 109 N ¢ m2 =C )(94)(1:60 £ 10¡19 C)
=
= 3:07 £ 1021 N=C :
4¼²0 R2
(6:64 £ 10¡15 m)2 The field is normal to the surface and since the charge is positive it points outward from the
surface.
9
Choose the coordinate axes as shown on the diagram
to the right. At the center of the square, the electric
fields produced by the particles at the lower left and
upper right corners are both along the x axis and each
points away from the center and toward the particle
that
p produces it. Since each particle is a distance d =
p
2a=2 = a= 2 away from the center, the net field due
to these two particles is
∙ q
1
2q
Ex =
¡ 2
2 =2
4¼²0 a
a =2
= ¸ ..
..
.
. y
.
..
.. a ..
..
. ..
.
q ².. . ..
..
. ..
..
. ..
..
. ..
..
. d
..
.
.
. ..
..
. a .
..
¡q.. ².
.
..
.
.
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. .
..
.. ..
..
.
.
..
.. .
..
..
..
..
. .
..
.. ..
..
. .
..
.. ..
..
. .
..
.. ..
..
. .
..
.. ..
..
. .
..
.. .
.
..
.. .
..
.. .
..
.. .
.
.. x ² ¡2q ..
... d
..
..
. ..
..
.
. .
..
.. ².2q
..
.. ..
..
. ..
..
. ..
.. (8:99 £ 109 N ¢ m2 =C2 )(1:0 £ 10¡8 C)
q
1
=
= 7:19 £ 104 N=C.
4¼²0 a2 =2
(0:050 m)2 =2
Chapter 22 137 At the center of the square, the field produced by the particles at the upper left and lower right
corners are both along the y axis and each points away from the particle that produces it. The
net field produced at the center by these particles is
∙
¸
q
1
1
q
2q
¡ 2
=
= 7:19 £ 104 N=C :
Ey =
2 =2
2 =2
4¼²0 a
4¼²0 a
a =2
The magnitude of the net field is
q
q
2 + E2 =
E = Ex
2(7:19 £ 104 N=C)2 = 1:02 £ 105 N=C
y
and the angle it makes with the x axis is µ = tan¡1 Ey
= tan¡1 (1) = 45± :
Ex It is upward in the diagram, from the center of the square toward the center of the upper side.
21
Think of the quadrupole as composed of two dipoles, each with dipole moment of magnitude
p = qd. The moments point in opposite directions and produce fields in opposite directions at
points on the quadrupole axis. Consider the point P on the axis, a distance z to the right of the
quadrupole center and take a rightward pointing field to be positive. Then the field produced
by the right dipole of the pair is qd=2¼²0 (z ¡ d=2)3 and the field produced by the left dipole
is ¡qd=2¼²0 (z + d=2)3 . Use the binomial expansions (z ¡ d=2)¡3 ¼ z ¡3 ¡ 3z ¡4 (¡d=2) and
(z + d=2)¡3 ¼ z ¡3 ¡ 3z ¡4 (d=2) to obtain
∙
¸
6qd2
3d
3d
1
qd
1
+ 4¡ 3+ 4 =
:
E=
2¼²0 z 3 2z
2z
4¼²0 z 4
z
Let Q = 2qd2 . Then
E= 3Q
:
4¼²0 z 4 27
(a) The linear charge density ¸ is the charge per unit length of rod. Since the charge is uniformly
distributed on the rod, ¸ = ¡q=L = ¡(4:23 £ 10¡15 C)=(0:0815 m) = ¡5:19 £ 10¡14 C=m.
(b) and (c) Position the origin at the left end of
dx
P
the rod, as shown in the diagram. Let dx be an
²
infinitesimal length of rod at x. The charge in
x
this segment is dq = ¸ dx. Since the segment
L
L+a
0
may be taken to be a point particle, the electric field it produces at point P has only an x
component and this component is given by
dEx =
138 Chapter 22 1
¸ dx
:
4¼²0 (L + a ¡ x)2 The total electric field produced at P by the whole rod is the integral
¯L
Z L
¯
¸
¸
dx
1
¯
=
Ex =
4¼²0 0 (L + a ¡ x)2 4¼²0 L + a ¡ x ¯0
∙
¸
1
L
1
¸
¸
¡
=
:
=
4¼²0 a L + a
4¼²0 a(L + a)
When ¡q=L is substituted for ¸ the result is
Ex = ¡ 2 (8:99 £ 109 N ¢ m2 =C )(4:23 £ 10¡15 C)
1
q
= ¡1:57 £ 10¡3 N=C :
=¡
4¼²0 a(L + a)
(0:120 m)(0:0815 m + 0:120 m) The negative sign indicates that the field is toward the rod and makes an angle of 180± with the
positive x direction.
(d) Now
2 (8:99 £ 109 N ¢ m2 =C )(4:23 £ 10¡15 C)
1
q
= ¡1:52 £ 10¡8 N=C :
Ex = ¡
=¡
4¼²0 a(L + a)
(50 m)(0:0815 m + 50 m)
(e) The field of a point particle at the origin is
q
(8:99 £ 109 N ¢ m2 =C2 )(4:23 £ 10¡15 C)
=¡
= ¡1:52 £ 10¡8 N=C :
Ex = ¡
4¼²0 a2
(50 m)2
35
At a point on the axis of a uniformly charged disk a distance z above the center of the disk, the
magnitude of the electric field is
∙
¸
¾
z
E=
1¡ p
;
2²0
z 2 + R2 where R is the radius of the disk and ¾ is the surface charge density on the disk. See Eq. 22–26.
The magnitude of the field at the center of the disk (z = 0) is Ec = ¾=2²0 . You want to solve
for the value of z such that E=Ec = 1=2. This means
z
1
E
=1¡ p
=
2 + R2
2
Ec
z
or 1
z
p
= :
z 2 + R2 2
Square bothp
sides, then multiply them by z 2 + R2 to obtain z 2 = (z 2 =4) + (R2 =4). Thus z 2 = R2 =3
p
and z = R= 3 = (0:600 m)= 3 = 0:346 m.
39
The magnitude of the force acting on the electron is F = eE, where E is the magnitude of the
electric field at its location. The acceleration of the electron is given by Newton’s second law:
a= F eE (1:60 £ 10¡19 C)(2:00 £ 104 N=C)
= 3:51 £ 1015 m=s2 :
=
=
9:11 £ 10¡31 kg
m
m
Chapter 22 139 43
(a) The magnitude of the force on the particle is given by F = qE, where q is the magnitude of
the charge carried by the particle and E is the magnitude of the electric field at the location of
the particle. Thus
F 3:0 £ 10¡6 N
= 1:5 £ 103 N=C :
E=
=
2:0 £ 10¡9 C
q
The force points downward and the charge is negative, so the field points upward.
(b) The magnitude of the electrostatic force on a proton is
Fe = eE = (1:60 £ 10¡19 C)(1:5 £ 103 N=C) = 2:4 £ 10¡16 N :
(c) A proton is positively charged, so the force is in the same direction as the field, upward.
(d) The magnitude of the gravitational force on the proton is
2 Fg = mg = (1:67 £ 10¡27 kg)(9:8 m=s ) = 1:64 £ 10¡26 N :
The force is downward.
(e) The ratio of the force magnitudes is
2:4 £ 10¡16 N
Fe
= 1:5 £ 1010 :
=
¡26 N
Fg 1:64 £ 10
45
(a) The magnitude of the force acting on the proton is F = eE, where E is the magnitude of the
electric field. According to Newton’s second law, the acceleration of the proton is a = F=m =
eE=m, where m is the mass of the proton. Thus
a= (1:60 £ 10¡19 C)(2:00 £ 104 N=C)
= 1:92 £ 1012 m=s2 :
1:67 £ 10¡27 kg 2
(b) Assume the proton starts from rest and use the kinematic equation v 2 = v0 + 2ax (or else
x = 1 at2 and v = at) to show that
2
q
p
v = 2ax = 2(1:92 £ 1012 m=s2 )(0:0100 m) = 1:96 £ 105 m=s : 57
(a) If q is the positive charge in the dipole and d is the separation of the charged particles, the
magnitude of the dipole moment is p = qd = (1:50£ 10¡9 C)(6:20£10¡6 m) = 9:30£ 10¡15 C¢m.
(b) If the initial angle between the dipole moment and the electric field is µ0 and the final angle
is µ, then the change in the potential energy as the dipole swings from µ = 0 to µ = 180± is
¢U = ¡pE(cos µ ¡ cos µ0 ) = ¡(9:30 £ 10¡15 C ¢ m)(1100 N=C)(cos 180± ¡ cos 0)
= 2:05 £ 10¡11 J : 140 Chapter 22 79
(a) and (b) Since the field at the point on the x axis with coordinate x = 2:0 cm is in the positive
x direction you know that the charged particle is on the x axis. The line through the point
with coordinates x = 3:0 cm and y = 3:0 cm and parallel to the field at that point must pass
through the position of the particle. Such a line has slope (3:0)=(4:0) = 0:75 and its equation is
y = 0:57 + (0:75)x. The solution for y = 0 is x = ¡1:0 cm, so the particle is located at the point
with coordinates x = ¡1:0 cm and y = 0.
(c) The magnitude of the field at the point on the x axis with coordinate x = 2:0 cm is given by
E = (1=4¼²0 )q=(2:0 cm ¡ x)2 , so
q = 4¼²0 x2 E = (0:020 m + 0:010 m)2 (100 N=C)
8:99 £ 109 N ¢ 2
m2 =C = 1:0 £ 10¡11 C : 81
~
(a) The potential energy of an electric dipole with dipole moment p in an electric field E is
~
U = ¡~ ¢ E = (1:24 £ 10¡30 C ¢ m)(3:00 ˆ + 4:00 ˆ) ¢ (4000 N=C) ˆ
p ~
i
j
i = ¡(1:24 £ 10¡30 C ¢ m)(3:00)(4000 N=C) = ¡1:49 £ 10¡26 J : Here we used ~ ¢ ~ = ax bx + ay by + az bz to evaluate the scalar product.
a b
(b) The torque is
ˆ
i
j
i
~ = p £ E = (px ˆ + py ˆ) £ (Ex ˆ) = ¡py Ex k
¿ ~ ~ ˆ
= ¡(4:00)(1:24 £ 10¡30 C ¢ m)(4000 N=C) = ¡(1:98 £ 10¡26 N ¢ m) k : (c) The work done by the agent is equal to the change in the potential energy of the dipole. The
initial potential energy is Ui = ¡1:49 £ 10¡26 J, as computed in part (a). The final potential
energy is
Uf = (1:24 £ 10¡30 C ¢ m)(¡4:00 ˆ + 3:00 ˆ) ¢ (4000 N=C) ˆ
i
j
i
= ¡(1:24 £ 10¡30 C ¢ m)(¡4:00)(4000 N=C) = +1:98 £ 10¡26 J : The work done by the agent is W = (1:98 £ 10¡26 J) ¡ (¡1:49 £ 10¡26 J) = 3:47 £ 10¡26 J. Chapter 22 141 Chapter 23 1
~
~
The vector area A and the electric field E are shown on the diagram to the right. The angle µ between them is 180± ¡ 35± = 145± ,
~ ~
so the electric flux through the area is © = E ¢ A = EA cos µ =
¡3
±
2
(1800 N=C)(3:2 £ 10 m) cos 145 = ¡1:5 £ 10¡2 N ¢ m2 =C. .
.
. A .................... ± .
.
.
. 35 ..
.
.
..
.
.
..
......
.......
..
...
... .
.
.
.
..
. ...
.
. ...
. ..
.
. ..
. ..
.
.
..
.
...
.
.
.
.
.
..
.
.
..
.
..
.
.
.
..
..
..
..
..
.. ...
.. ..
...
..
.
..
..
..
..
..
..
..
..
..
. ..
. ..
.. .
....
....
...
... µ ............. E 9
Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the upper
face, and E` be the magnitude of the field at the lower face. Since the field is downward, the
flux through the upper face is negative and the flux through the lower face is positive. The flux
through the other faces is zero, so the total flux through the cube surface is © = A(E` ¡ Eu ).
The net charge inside the cube is given by Gauss’ law:
q = ²0 © = ²0 A(E` ¡ Eu ) = (8:85 £ 10¡12 C2 =N ¢ m2 )(100 m)2 (100 N=C ¡ 60:0 N=C)
= 3:54 £ 10¡6 C = 3:54 ¹C : 19
(a) The charge on the surface of the sphere is the product of the surface charge density ¾ and
the surface area of the sphere (4¼r2 , where r is the radius). Thus
2 q = 4¼r ¾ = 4¼ µ 1:2 m
2 ¶2 (8:1 £ 10¡6 C=m2 ) = 3:7 £ 10¡5 C : (b) Choose a Gaussian surface in the form a sphere, concentric with the conducting sphere and
with a slightly larger radius. The flux through the surface is given by Gauss’ law:
©= 3:7 £ 10¡5 C
q
=
= 4:1 £ 106 N ¢ m2 =C :
²0 8:85 £ 10¡12 C2 =N ¢ m2 23
The magnitude of the electric field produced by a uniformly charged infinite line is E = ¸=2¼²0 r,
where ¸ is the linear charge density and r is the distance from the line to the point where the
field is measured. See Eq. 23–12. Thus
¸ = 2¼²0 Er = 2¼(8:85 £ 10¡12 C2 =N ¢ m2 )(4:5 £ 104 N=C)(2:0 m) = 5:0 £ 10¡6 C=m :
142 Chapter 23 27
Assume the charge density of both the conducting rod and the shell are uniform. Neglect fringing.
Symmetry can be used to show that the electric field is radial, both between the rod and the shell
and outside the shell. It is zero, of course, inside the rod and inside the shell since they are
conductors.
(a) and (b) Take the Gaussian surface to be a cylinder of length L and radius r, concentric
with the conducting rod and shell and with its curved surface outside the shell. The area of the
curved surface is 2¼rL. The field is normal to the curved portion of the surface and has uniform
magnitude over it, so the flux through this portion of the surface is © = 2¼rLE, where E is the
magnitude of the field at the Gaussian surface. The flux through the ends is zero. The charge
enclosed by the Gaussian surface is Q1 ¡ 2:00Q1 = ¡Q1 . Gauss’ law yields 2¼r²0 LE = ¡Q1 ,
so
2
(8:99 £ 109 N ¢ m2 =C )(3:40 £ 10¡12 C)
Q1
= ¡0:214 N=C :
=¡
E=¡
2¼²0 Lr
(11:00 m)(26:0 £ 10¡3 m) The magnitude of the field is 0:214 N=C. The negative sign indicates that the field points inward.
(c) and (d) Take the Gaussian surface to be a cylinder of length L and radius r, concentric with
the conducting rod and shell and with its curved surface between the conducting rod and the
shell. As in (a), the flux through the curved portion of the surface is © = 2¼rLE, where E is
the magnitude of the field at the Gaussian surface, and the flux through the ends is zero. The
charge enclosed by the Gaussian surface is only the charge Q1 on the conducting rod. Gauss’
law yields 2¼²0 rLE = Q1 , so
2 E= 2(8:99 £ 109 N ¢ m2 =C )(3:40 £ 10¡12 C)
Q1
= +0:855 N=C :
=
2¼²0 Lr
(11:00 m)(6:50 £ 10¡3 m) The positive sign indicates that the field points outward.
(e) Consider a Gaussian surface in the form of a cylinder of length L with the curved portion
of its surface completely within the shell. The electric field is zero at all points on the curved
surface and is parallel to the ends, so the total electric flux through the Gaussian surface is zero
and the net charge within it is zero. Since the conducting rod, which is inside the Gaussian
cylinder, has charge Q1 , the inner surface of the shell must have charge ¡Q1 = ¡3:40 £ 10¡12 C.
(f) Since the shell has total charge ¡2:00Q1 and has charge ¡Q1 on its inner surface, it must
have charge ¡Q1 = ¡3:40 £ 10¡12 C on its outer surface.
35
(a) To calculate the electric field at a point very close to the center of a large, uniformly charged
conducting plate, we may replace the finite plate with an infinite plate with the same area charge
density and take the magnitude of the field to be E = ¾=²0 , where ¾ is the area charge density
for the surface just under the point. The charge is distributed uniformly over both sides of the
original plate, with half being on the side near the field point. Thus
¾= 6:0 £ 10¡6 C
q
=
= 4:69 £ 10¡4 C=m2 :
2A
2(0:080 m)2
Chapter 23 143 The magnitude of the field is
¾
4:69 £ 10¡4 C=m2
E=
=
= 5:3 £ 107 N=C :
²0 8:85 £ 10¡12 C2 =N ¢ m2
The field is normal to the plate and since the charge on the plate is positive, it points away from
the plate.
(b) At a point far away from the plate, the electric field is nearly that of a point particle with
charge equal to the total charge on the plate. The magnitude of the field is E = q=4¼²0 r 2 , where
r is the distance from the plate. Thus
(8:99 £ 109 N ¢ m2 =C2 )(6:0 £ 10¡6 C)
E=
= 60 N=C :
(30 m)2
41
The forces on the ball are shown in the diagram to the right. The gravitational force has magnitude mg, where m is the mass of the ball; the
electrical force has magnitude qE, where q is the charge on the ball and
E is the electric field at the position of the ball; and the tension in the
thread is denoted by T . The electric field produced by the plate is normal
to the plate and points to the right. Since the ball is positively charged,
the electric force on it also points to the right. The tension in the thread
makes the angle µ (= 30± ) with the vertical. T ...
...
.
....
....
...
..
..
. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...
..
.
.
..
.
................
. ...............
..
.
.. ..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
... .
. ...
..
.
..
.
..
.
. µ ² qE mg Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal components
yields qE ¡ T sin µ = 0 and the sum of the vertical components yields T cos µ ¡ mg = 0.
The expression T = qE= sin µ, from the first equation, is substituted into the second to obtain
qE = mg tan µ.
The electric field produced by a large uniform plane of charge is given by E = ¾=2²0 , where ¾
is the surface charge density. Thus
q¾
= mg tan µ
2²0
and
¾= 2²0 mg tan µ
q
2 2(8:85 £ 10¡12 C2 =N ¢ m2 )(1:0 £ 10¡6 kg)(9:8 m=s ) tan 30±
=
2:0 £ 10¡8 C
2
¡9
= 5:0 £ 10 C=m :
45
Charge is distributed uniformly over the surface of the sphere and the electric field it produces at
points outside the sphere is like the field of a point particle with charge equal to the net charge
on the sphere. That is, the magnitude of the field is given by E = q=4¼²0 r2 , where q is the
144 Chapter 23 magnitude of the charge on the sphere and r is the distance from the center of the sphere to the
point where the field is measured. Thus
q = 4¼²0 r2 E = (0:15 m)2 (3:0 £ 103 N=C)
2 8:99 £ 109 N ¢ m2 =C = 7:5 £ 10¡9 C : The field points inward, toward the sphere center, so the charge is negative: ¡7:5 £ 10¡9 C.
49
To find an expression for the electric field inside the shell in terms of A and the distance from
the center of the shell, select A so the field does not depend on the distance.
Use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical
shell and within it (a < rg < b). Gauss’ law is used to find the magnitude of the electric field a
distance rg from the shell center.
The charge that is both in the shell and within the Gaussian sphere is given by the integral
R
qenc = ½ dV over the portion of the shell within the Gaussian surface. Since the charge
distribution has spherical symmetry, we may take dV to be the volume of a spherical shell with
radius r and infinitesimal thickness dr: dV = 4¼r 2 dr. Thus
Z rg
Z rg
Z rg
A 2
2
2
qenc = 4¼
r dr = 4¼A
½r dr = 4¼
r dr = 2¼A(rg ¡ a2 ) :
r
a
a
a
2
The total charge inside the Gaussian surface is q + qenc = q + 2¼A(rg ¡ a2 ).
2
The electric field is radial, so the flux through the Gaussian surface is © = 4¼rg E, where E is
the magnitude of the field. Gauss’ law yields
2
2
4¼²0 Erg = q + 2¼A(rg ¡ a2 ) : Solve for E: ∙
¸
2¼Aa2
1
q
:
+ 2¼A ¡
E=
2
2
4¼²0 rg
rg For the field to be uniform, the first and last terms in the brackets must cancel. They do if
2
q ¡ 2¼Aa2 = 0 or A = q=2¼a2 = (45:0 £ 10¡15 C)=2¼(2:00 £ 10¡2 m)2 = 1:79 £ 10¡11 C=m .
59
(a) The magnitude E1 of the electric field produced by the charge q on the spherical shell is
2
E1 = q=4¼²0 Ro , where Ro is the radius of the outer surface of the shell. Thus
2
q = 4¼²0 E1 Ro = (450 N=C)(0:20 m)2
2 8:99 £ 109 N ¢ m2 =C = 2:0 £ 10¡9 C : (b) Since the field at P is outward and is reduced in magnitude the field of Q must be inward. Q
is a negative charge and the magnitude of its field at P is E2 = 450 N=C ¡ 180 N=C = 270 N=C.
The value of Q is
2
Q = 4¼²0 E2 Ro = ¡ (270 N=C)(0:20 m)2
= ¡1:2 £ 10¡9 C :
8:99 £ 109 N ¢ m2 =C)2
Chapter 23 145 (c) Gauss’ law tells us that since the electric field is zero inside a conductor the net charge inside
a spherical surface with a radius that is slightly larger than the inside radius of the shell must be
zero. Thus the charge on the inside surface of the shell is +1:2 £ 10¡9 C.
(d) The remaining charge on the shell must be on its outer surface and this is 2:0 £ 10¡9 C ¡
1:2 £ 10¡9 C = +0:8 £ 10¡9 C.
69
(a) Draw a spherical Gaussian surface with radius r, concentric with the shells. The electric
field, if it exists, is radial and so is normal to the surface. The integral in Gauss’ law is
H
~ ~
E ¢ dA = 4¼r2 E, where E is the radial component of the field. For r < a then charge enclosed
is zero. Gauss’ law gives 4¼r2 E = 0, so E = 0.
(b) For a < r < b the charge enclosed by the Gaussian surface is qa , so the law gives 4¼r 2 E =
qa =²0 and E = qa =4¼²0 r 2 .
(c) For r > b the charge enclosed by the Gaussian surface is qa + qb , so 4¼²0 E = (qa + qb )=²0
and E = (qa + qb )=4¼²0 r2 .
(d) Consider first a spherical Gaussian with radius just slightly greater than a. The electric field
is zero everywhere on this surface, so according to Gauss’ law it encloses zero net charge. Since
there is no charge in the cavity the charge on the inner surface of the smaller shell is zero. The
total charge on the smaller shell is qa and this must reside on the outer surface. Now consider a
spherical Gaussian surface with radius slight larger than the inner radius of the larger shell. This
surface also encloses zero net charge, which is the sum of the charge on the outer surface of the
smaller shell and the charge on the inner surface of the larger shell. Thus the charge on the inner
surface of the larger shell is ¡qa . The net charge on the larger shell is qb , with ¡qa on its inner
surface, so the charge on its outer surface must be qb ¡ (¡qa ) = qb + qa .
76
(a) The magnitude of the electric field due to a large uniformly charged plate is given by ¾=2²0 ,
where ¾ is the surface charge density. In the region between the oppositely charged plates the
fields of the plates are in the same direction, so the net field has magnitude E = ¾=²0 . The
electrical force on an electron has magnitude eE = e¾=²0 and the gravitational force on it is
mg, where m is it mass. If these forces are to balance, they must have the same magnitude, so
mg = e¾=²0 and
2 ¾= mg²0 (9:11 £ 10¡31 kg)(9:8 m=s )(8:85 £ 10¡12 C2 =N ¢ m2 )
2
= 4:9 £ 10¡22 C=m :
=
1:60 £ 10¡19 C
e (b) The gravitational force is downward, so the electrical force must be upward. Since an electron
is negatively charged the electrical force on it is opposite to the electric field, so the electric field
must be downward.
79
(a) Let Q be the net charge on the shell, qi be the charge on its inner surface and qo be the charge
on its outer surface. Then Q = qi + qo and qi = Q ¡ qo = (¡10 ¹C) ¡ (¡14 ¹C) = +4 ¹C.
146 Chapter 23 (b) Let q be the charge on the particle. Gauss’ law tells us that since the electric field is zero
inside the conducting shell the net charge inside any spherical surface that entirely within the
shell is zero. Thus the sum of the charge on the particle and on the inner surface of the shell is
zero, so q + qi = 0 and q = ¡qi = ¡4 ¹C. Chapter 23 147 Chapter 24 3
(a) An ampere is a coulomb per second, so
¶
µ
s´
C¢h ³
3600
= 3:0 £ 105 C :
84 A ¢ h = 84
s
h (b) The change in potential energy is ¢U = q ¢V = (3:0 £ 105 C)(12 V) = 3:6 £ 106 J. 5
The electric field produced by an infinite sheet of charge has magnitude E = ¾=2²0 , where ¾ is
the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a
coordinate system at the sheet and take the x axis to be parallel to the field and positive in the
direction of the field. Then the electric potential is
Z x
V = Vs ¡
E dx = Vs ¡ Ex ;
0 where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x; that
is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ¢x
then their potentials differ in magnitude by ¢V = E¢x = (¾=2²0 )¢x. Thus
¢x = 2(8:85 £ 10¡12 C2 =N ¢ m2 )(50 V)
2²0 ¢V
=
= 8:8 £ 10¡3 m :
¡6 C=m2
¾
0:10 £ 10 19
(a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius
R of the drop are related by V = q=4¼²0 R. Thus
(8:99 £ 109 N ¢ m2 =C2 )(30 £ 10¡12 C)
q
= 5:4 £ 10¡4 m :
=
R=
4¼²0 V
500 V
(b) After the drops combine the total volume is twice the volume of an original drop, so the
radius R0 of the combined drop is given by (R0 )3 = 2R3 and R0 = 21=3 R. The charge is twice
the charge of original drop: q0 = 2q. Thus
V0 = 1 q0
1
2q
= 22=3 V = 22=3 (500 V) = 790 V :
=
0
4¼²0 R
4¼²0 21=3 R 29
The disk is uniformly charged. This means that when the full disk is present each quadrant
contributes equally to the electric potential at P , so the potential at P due to a single quadrant
is onefourth the potential due to the entire disk. First find an expression for the potential at P
due to the entire disk.
148 Chapter 24 Consider a ring of charge with radius r and width dr. Its area is 2¼r dr and it contains charge
p
dq = 2¼¾r dr. All the charge in it is a distance r2 + D2 from P , so the potential it produces
at P is
¾r dr
1 2¼¾r dr
p
p
=
:
dV =
2 + D2
4¼²0 r
2²0 r2 + D2
The total potential at P is
Z R
¯R
i
¾ p 2
¾ hp 2
¾
r dr
¯
p
=
r + D2 ¯ =
R + D2 ¡ D :
V =
2²0 0
2²0
0
r2 + D2 2²0 The potential Vsq at P due to a single quadrant is
i
¾ hp 2
V
=
R + D2 ¡ D
Vsq =
4 8²0
2
hp
i
7:73 £ 10¡15 C=m
(0:640 m)2 + (0:259 m)2 ¡ 0:259 m
=
8(8:85 £ 10¡12 C2 =N ¢ m2 )
= 4:71 £ 10¡5 V : 39
Take the negatives of the partial derivatives of the electric potential with respect to the coordinates
and evaluate the results for x = 3:00 m, y = ¡2:00 m, and z = 4:00 m. This yields
@V
4
4
= ¡(2:00 V=m )yz 2 = ¡(2:00 V=m )((¡2:00 m)(4:00 m)2 = 64:0 V=m ;
@x
@V
4
4
Ey = ¡
= ¡(2:00 V=m )xz 2 = ¡(2:00 V=m )(3:00 m)(4:00 m)2 = ¡96:0 V=m ;
@y
@V
4
4
= 2(2:00 V=m )xyz = ¡2(2:00 V=m )(3:00 m)(¡2:00 m)(4:00 m) = 96:0 V=m :
Ez = ¡
@z Ex = ¡ The magnitude of the electric field is
q
q
2 + E2 + E2 =
E = Ex
(64:0 V=m)2 + (¡96:0 V=m)2 + (96:0 V=m)2 = 1:50 £ 102 V=m :
y
z
41
The work required is equal to the potential energy of the system, relative to a potential energy of
zero for infinite separation. Number the particles 1, 2, 3, and 4, in clockwise order starting with
the particle in the upper left corner of the arrangement. The potential energy of the interaction
of particles 1 and 2 is
2 (8:99 £ 109 N ¢ m2 =C )(2:30 £ 10¡12 C)(¡2:30 £ 10¡12 C)
q1 q2
U12 =
=
4¼²0 a
0:640 m
¡14
= ¡7:43 £ 10
J:
p
The distance between particles 1 and 3 is 2a and both these particles are positively charged,
p
so the potential energy of the interaction between particles 1 and 3 is U13 = ¡U12 = 2 = +5:25 £
Chapter 24 149 10¡14 J. The potential energy of the interaction between particles 1 and 4 is U14 = U12
¡7:43 £ 10¡14 J. The potential energy of the interaction between particles 2 and 3 is U23
U12 = ¡7:43 £ 10¡14 J. The potential energy of the interaction between particles 2 and 4
U24 = U13 = 5:25 £ 10¡14 J. The potential energy of the interaction between particles 3 and 4
U34 = U12 = ¡7:43 £ 10¡14 J. =
=
is
is The total potential energy of the system is U = U12 + U13 + U14 + U23 + U24 + U34
= ¡7:43 £ 10¡14 J + 5:25 £ 10¡14 J ¡ 7:43 £ 10¡14 J ¡ 7:43 £ 10¡14 J
¡ 7:43 £ 10¡14 J + 5:25 £ 10¡14 J = ¡1:92 £ 10¡13 J : This is equal to the work that must be done to assemble the system from infinite separation.
59
(a) Use conservation of mechanical energy. The potential energy when the moving particle is at
any coordinate y is qV , where V is the electric potential produced at that place by the two fixed
particles. That is,
2Q
p
U =q
;
4¼²0 x2 + y 2
where x is the coordinate and Q is the charge of either one of the fixed particles. The factor
2 appears since the two fixed particles produce the same potential at points on the y axis.
Conservation of mechanical energy yields
0
1
2Q
2qQ @
1
2Q
1
A;
p
q
p
Kf = Ki + q
= Ki +
¡q
¡q
2
2
4¼²0
4¼²0 x2 + yi
x2 + yi
4¼² x2 + y 2
x2 + y 2
0 f f where K is the kinetic energy of the moving particle, the subscript i refers to the initial position
of the moving particle, and the subscript f refers to the final position. Numerically
#
"
2(¡15 £ 10¡6 C)(50 £ 10¡6 C)
1
1
p
Kf = 1:2 J +
¡p
= 3:0 J :
4¼(8:85 £ 10¡12 C2 =N ¢ m2 )
(3:0 m)2 + (4:0 m)2
(3:0 m)2 (b) Now Kf = 0 and we solve the energy conservation equation for yf . Conservation of energy
first yields Uf = Ki + Ui . The initial potential energy is
Ui = 2qQ
2(¡15 £ 10¡6 C)(50 £ 10¡6 C)
p
p
=
= ¡2:7 J :
2
4¼²0 x2 + yi 4¼(8:85 £ 10¡12 C2 =N ¢ m2 ) (3:0 m)2 + (4:0 m)2 Thus Kf = 1:2 J ¡ 2:7 J = ¡1:5 J.
Now Uf =
150 Chapter 24 2qQ
q
2
4¼²0 x2 + yf so
y=¡ sµ 2qQ
4¼²0 Uf ¶2 ¡ x2 = s
µ 2(¡15 £ 10¡6 C)(50 £ 10¡6 C)
4¼(8:85 £ 10¡12 C2 =N ¢ m2 )(¡1:5 J) ¶2 ¡ (3:0 m)2 = ¡8:5 m : 63
If the electric potential is zero at infinity, then the electric potential at the surface of the sphere
is given by V = q=4¼²0 r, where q is the charge on the sphere and r is its radius. Thus
q = 4¼²0 rV = (0:15 m)(1500 V)
2 8:99 £ 109 N ¢ m2 =C = 2:5 £ 10¡8 C : 65
(a) The electric potential is the sum of the contributions of the individual spheres. Let q1 be
the charge on one, q2 be the charge on the other, and d be their separation. The point halfway
between them is the same distance d=2 (= 1:0 m) from the center of each sphere, so the potential
at the halfway point is
2 V = q1 + q2
(8:99 £ 109 N ¢ m2 =C )(1:0 £ 10¡8 C ¡ 3:0 £ 10¡8 C)
= ¡1:80 £ 102 V :
=
4¼²0 d=2
1:0 m (b) The distance from the center of one sphere to the surface of the other is d ¡ R, where R is
the radius of either sphere. The potential of either one of the spheres is due to the charge on that
sphere and the charge on the other sphere. The potential at the surface of sphere 1 is
∙
¸
q2
1
q1
+
V1 =
4¼²0 R d ¡ R
¸
∙
3:0 £ 10¡8 C
1:0 £ 10¡8 C
2
9
2
¡
= (8:99 £ 10 N ¢ m =C )
0:030 m
2:0 m ¡ 0:030 m
3
= 2:9 £ 10 V :
(c) The potential at the surface of sphere 2 is
∙
¸
q2
1
q1
+
V2 =
4¼²0 d ¡ R R
¸
∙
3:0 £ 10¡8 C
1:0 £ 10¡8 C
2
9
2
¡
= (8:99 £ 10 N ¢ m =C )
2:0 m ¡ 0:030 m
0:030 m
3
= ¡8:9 £ 10 V :
75
The initial potential energy of the threeparticle system is Ui = 2(q2 =4¼²0 L) + Ufixed , where q is
the charge on each particle, L is the length of a triangle side, and Ufixed is the potential energy
associated with the interaction of the two fixed particles. The factor 2 appears since the potential
Chapter 24 151 energy is the same for the interaction of the movable particle and each of the fixed particles. The
final potential energy is Uf = 2[q 2 =4¼²0 (L=2)] + Ufixed and the change in the potential energy is
µ
¶
1
2q 2
2q 2 2
¡
=
:
¢U = Uf ¡ Ui =
4¼²0 L L
4¼²0 L
This is the work that is done by the external agent. If P is the rate with energy is supplied by
the agent and t is the time for the move, then P t = ¢U , and
t=
This is 2:1 d. ¢U
2q2
2(8:99 £ 109 N ¢ m2 =C2 )(0:12 C)2
= 1:83 £ 105 s :
=
=
4¼²0 LP
(1:7 m)(0:83 £ 103 W)
P 77
(a) Use Gauss’ law to find an expression for the electric field. The Gaussian surface is a
cylindrical surface that is concentric with the cylinder and has a radius r that is greater than
the radius of the cylinder. The electric field is normal to the Gaussian surface and has uniform
H
~
~
magnitude on it, so the integral in Gauss’ law is E ¢ dA = 2¼rEL, where L is the length of
the Gaussian surface. The charge enclosed is ¸L, where ¸ is the charge per unit length on the
cylinder. Thus 2¼rRLE = ¸L=²0 and E = ¸=2¼²0 r.
Let EB be the magnitude of the field at B and rB be the distance from the central axis to B. Let
EC be the magnitude of the field at C and rC be the distance from the central axis to C. Since
E is inversely proportional to the distance from the central axis,
EC = rB
2:0 cm
(160 N=C) = 64 N=C :
EB =
5:0 cm
rC (b) The magnitude of the field a distance r from the central axis is E = (rB =r)EB , so the
potential difference of points B and C is
µ ¶
Z rB
rB
rB
EB dr = ¡rB EB ln
VB ¡ VC = ¡
r
rC
rC
¶
µ
0:020 m
= 2:9 V :
= ¡(0:020 m)(160 N=C) ln
0:050 m
(c) The cylinder is conducting, so all points inside have the same potential, namely VB , so
VA ¡ VB = 0.
85
Consider a point on the z axis that has coordinate z. All points on the ring are the same distance
p
from the point. The distance is r = R2 + z 2 , where R is the radius of the ring. If the electric
potential is taken to be zero at points that are infinitely far from the ring, then the potential at
the point is
Q
p
V =
;
4¼²0 R2 + z 2
152 Chapter 24 where Q is the charge on the ring. Thus
∙
¸
Q
1
1
p
¡
VB ¡ VA =
4¼²0
R2 + z 2 R " 2 1 = (8:99 £ 109 N ¢ m2 =C )(16:0 £ 10¡6 C) p (0:0300 m)2 + (0:0400 m)2 # ¡ 0:300 m = ¡1:92 £ 106 V : 93
(a) For r > r2 the field is like that of a point charge and
V = 1 Q
;
4¼²0 r where the zero of potential was taken to be at infinity.
(b) To find the potential in the region r1 < r < r2 , first use Gauss’s law to find an expression
for the electric field, then integrate along a radial path from r2 to r. The Gaussian surface is
a sphere of radius r, concentric with the shell. The field is radial and therefore normal to the
surface. Its magnitude is uniform over the surface, so the flux through the surface is © = 4¼r2 E.
3
3
The volume of the shell is (4¼=3)(r2 ¡ r1 ), so the charge density is
½= 3Q
3
3
4¼(r2 ¡ r1 ) and the charge enclosed by the Gaussian surface is
¶
µ 3
µ ¶
3
r ¡ r1
4¼
3
3
:
(r ¡ r1 )½ = Q 3
q=
3
3
r2 ¡ r1
Gauss’ law yields µ 3
r 3 ¡ r1
4¼²0 r E = Q 3
3
r2 ¡ r1
2 and the magnitude of the electric field is
E= ¶ 3
r3 ¡ r1
Q
:
3
3
4¼²0 r 2 (r2 ¡ r1 ) If Vs is the electric potential at the outer surface of the shell (r = r2 ) then the potential a distance
r from the center is given by
¶
Z rµ
Z r
3
r1
Q
1
r ¡ 2 dr
V = Vs ¡
E dr = Vs ¡
3
3
4¼²0 r2 ¡ r1 r2
r
r
2
µ 2
¶
r2 r 3 r 3
Q
1
r
¡ 2+ 1¡ 1 :
= Vs ¡
3
3
4¼²0 r2 ¡ r1
2
2
r
r2
Chapter 24 153 The potential at the outer surface is found by placing r = r2 in the expression found in part (a).
It is Vs = Q=4¼²0 r2 . Make this substitution and collect like terms to find
µ 2
¶
3
1
Q
3r2 r 2 r1
¡
¡
:
V =
3
3
4¼²0 r2 ¡ r1
2
2
r
3
3
Since ½ = 3Q=4¼(r2 ¡ r1 ) this can also be written
µ 2
¶
3
½ 3r2 r 2 r1
¡
¡
:
V =
3²0
2
2
r (c) The electric field vanishes in the cavity, so the potential is everywhere the same inside and
has the same value as at a point on the inside surface of the shell. Put r = r1 in the result of part
(b). After collecting terms the result is
V = or in terms of the charge density 2
2
Q 3(r2 ¡ r1 )
;
3
3
4¼²0 2(r2 ¡ r1 ) ½ 2
2
(r ¡ r1 ) :
2²0 2
(d) The solutions agree at r = r1 and at r = r2 .
V = 95
The electric potential of a dipole at a point a distance r away is given by Eq. 24–30:
V = 1 p cos µ
;
4¼²0 r2 where p is the magnitude of the dipole moment and µ is the angle between the dipole moment
and the position vector of the point. The potential at infinity was taken to be zero. Take the z
axis to be the dipole axis and consider a point with z positive (on the positive side of the dipole).
For this point r = z and µ = 0. The z component of the electric field is
¶
µ
p
@
@V
p
=
=¡
Ez = ¡
:
2¼²0 z 3
@x
@z 4¼²0 z 2 This is the only nonvanishing component at a point on the dipole axis.
For a point with a negative value of z, r = ¡z and cos µ = ¡1, so
¶
µ
p
@
¡p
=¡
Ez = ¡
:
2
2¼²0 z 3
@z 4¼²0 z 103
(a) The electric potential at the surface of the sphere is given by V = q=4¼²0 R, where q is the
charge on the sphere and R is the sphere radius. The charge on the sphere when the potential
reaches 1000 V is
(0:010 m)(1000 V)
q = 4¼²0 rV =
= 1:11 £ 10¡9 C :
9 N ¢ m2 =C2
8:99 £ 10
154 Chapter 24 The number of electrons that enter the sphere is N = q=e = (1:11 £ 10¡9 C)=(1:60 £ 10¡19 C) =
6:95£109 . Let R be the decay rate and t be the time for the potential to reach it final value. Since
half the resulting electrons enter the sphere N = (P=2)t and t = 2N=P = 2(6:95 £ 109 )=(3:70 £
108 s¡1 ) = 38 s.
(b) The increase in temperature is ¢T = N ¢E=C, where E is the energy deposited by a single
electron and C is the heat capacity of the sphere. Since N = (P=2)t, this is ¢T = (P=2)t¢E=C
and
t= 2C ¢T
2(14 J=K)(5:0 K)
= 2:4 £ 107 s :
=
8 s¡1 )(100 £ 103 eV)(1:60 £ 10¡19 J=eV)
(3:70 £ 10
P ¢E This is about 280 d. Chapter 24 155 Chapter 25 5
(a) The capacitance of a parallelplate capacitor is given by C = ²0 A=d, where A is the area of
each plate and d is the plate separation. Since the plates are circular, the plate area is A = ¼R2 ,
where R is the radius of a plate. Thus
²0 ¼R2 (8:85 £ 10¡12 F=m)¼(8:20 £ 10¡2 m)2
= 1:44 £ 10¡10 F = 144 pF :
=
¡3 m
1:30 £ 10
d
(b) The charge on the positive plate is given by q = CV , where V is the potential difference
across the plates. Thus q = (1:44 £ 10¡10 F)(120 V) = 1:73 £ 10¡8 C = 17:3 nC.
C= 15
The charge initially on the charged capacitor is given by q = C1 V0 , where C1 (= 100 pF) is the
capacitance and V0 (= 50 V) is the initial potential difference. After the battery is disconnected
and the second capacitor wired in parallel to the first, the charge on the first capacitor is q1 = C1 V ,
where v (= 35 V) is the new potential difference. Since charge is conserved in the process, the
charge on the second capacitor is q2 = q ¡q1 , where C2 is the capacitance of the second capacitor.
Substitute C1 V0 for q and C1 V for q1 to obtain q2 = C1 (V0 ¡ V ). The potential difference across
the second capacitor is also V , so the capacitance is
C2 = q2 V0 ¡ V
50 V ¡ 35 V
(100 pF) = 43 pF :
=
C1 =
35 V
V
V 19
(a) After the switches are closed, the potential differences across the capacitors are the same and
the two capacitors are in parallel. The potential difference from a to b is given by Vab = Q=Ceq ,
where Q is the net charge on the combination and Ceq is the equivalent capacitance.
The equivalent capacitance is Ceq = C1 + C2 = 4:0 £ 10¡6 F. The total charge on the combination
is the net charge on either pair of connected plates. The charge on capacitor 1 is
q1 = C1 V = (1:0 £ 10¡6 F)(100 V) = 1:0 £ 10¡4 C
and the charge on capacitor 2 is
q2 = C2 V = (3:0 £ 10¡6 F)(100 V) = 3:0 £ 10¡4 C ;
so the net charge on the combination is 3:0£10¡4 C¡1:0£10¡4 C = 2:0£10¡4 C. The potential
difference is
2:0 £ 10¡4 C
= 50 V :
Vab =
4:0 £ 10¡6 F
(b) The charge on capacitor 1 is now q1 = C1 Vab = (1:0 £ 10¡6 F)(50 V) = 5:0 £ 10¡5 C.
156 Chapter 25 (c) The charge on capacitor 2 is now q2 = C2 Vab = (3:0 £ 10¡6 F)(50 V) = 1:5 £ 10¡4 C.
29
The total energy is the sum of the energies stored in the individual capacitors. Since they are
connected in parallel, the potential difference V across the capacitors is the same and the total
energy is U = 1 (C1 + C2 )V 2 = 1 (2:0 £ 10¡6 F + 4:0 £ 10¡6 F)(300 V)2 = 0:27 J.
2
2
35
(a) Let q be the charge on the positive plate. Since the capacitance of a parallelplate capacitor
is given by ²0 A=d, the charge is q = CV = ²0 AV =d. After the plates are pulled apart, their
separation is d0 and the potential difference is V 0 . Then q = ²0 AV 0 =d0 and
d0 ²0 A
d0
8:00 mm
d0
(6:00 V) = 16:0 V :
q=
V = V =
V =
3:00 mm
²0 A
²0 A d
d
0 (b) The initial energy stored in the capacitor is
1
²0 AV 2 (8:85 £ 10¡12 F=m)(8:50 £ 10¡4 m2 )(6:00 V)
=
= 4:51 £ 10¡11 J
Ui = CV 2 =
2
2d
2(3:00 £ 10¡3 mm)
and the final energy stored is
1
1 ²0 A 0 2 (8:85 £ 10¡12 F=m)(8:50 £ 10¡4 m2 )(16:0 V)
= 1:20£10¡10 J :
(V ) =
Uf = C 0 (V 0 )2 =
2
2 d0
2(8:00 £ 10¡3 mm)
(c) The work done to pull the plates apart is the difference in the energy: W = Uf ¡ Ui =
1:20 £ 10¡10 J ¡ 4:51 £ 10¡11 J = 7:49 £ 10¡11 J.
43
The capacitance of a cylindrical capacitor is given by
C = ∙C0 = 2¼∙²0 L
;
ln(b=a) where C0 is the capacitance without the dielectric, ∙ is the dielectric constant, L is the length, a
is the inner radius, and b is the outer radius. See Eq. 25–14. The capacitance per unit length of
the cable is 45 2¼(2:6)(8:85 £ 10¡12 F=m)
2¼∙²0
C
£
¤ = 8:1 £ 10¡11 F=m = 81 pF=m :
=
=
L ln(b=a)
ln (0:60 mm)=(0:10 mm) The capacitance is given by C = ∙C0 = ∙²0 A=d, where C0 is the capacitance without the
dielectric, ∙ is the dielectric constant, A is the plate area, and d is the plate separation. The
Chapter 25 157 electric field between the plates is given by E = V =d, where V is the potential difference between
the plates. Thus d = V =E and C = ∙²0 AE=V . Solve for A:
A= CV
:
∙²0 E For the area to be a minimum, the electric field must be the greatest it can be without breakdown
occurring. That is,
A= (7:0 £ 10¡8 F)(4:0 £ 103 V)
= 0:63 m2 :
¡12 F=m)(18 £ 106 V=m)
2:8(8:85 £ 10 51
(a) The electric field in the region between the plates is given by E = V =d, where V is the
potential difference between the plates and d is the plate separation. The capacitance is given by
C = ∙²0 A=d, where A is the plate area and ∙ is the dielectric constant, so d = ∙²0 A=C and
E= (50 V)(100 £ 10¡12 F)
VC
= 1:0 £ 104 V=m :
=
∙²0 A 5:4(8:85 £ 10¡12 F=m)(100 £ 10¡4 m2 ) (b) The free charge on the plates is qf = CV = (100 £ 10¡12 F)(50 V) = 5:0 £ 10¡9 C.
(c) The electric field is produced by both the free and induced charge. Since the field of a large
uniform layer of charge is q=2²0 A, the field between the plates is
E= qf
qf
qi
qi
+
¡
¡
;
2²0 A 2²0 A 2²0 A 2²0 A where the first term is due to the positive free charge on one plate, the second is due to the
negative free charge on the other plate, the third is due to the positive induced charge on one
dielectric surface, and the fourth is due to the negative induced charge on the other dielectric
surface. Note that the field due to the induced charge is opposite the field due to the free charge,
so the fields tend to cancel. The induced charge is therefore
qi = qf ¡ ²0 AE = 5:0 £ 10¡9 C ¡ (8:85 £ 10¡12 F=m)(100 £ 10¡4 m2 )(1:0 £ 104 V=m)
= 4:1 £ 10¡9 C = 4:1 nC : 61
Capacitors 3 and 4 are in parallel and may be replaced by a capacitor with capacitance C34 =
C3 + C4 = 30 ¹F. Capacitors 1, 2, and the equivalent capacitor that replaced 3 and 4 are all
in series, so the sum of their potential differences must equal the potential difference across the
battery. Since all of these capacitors have the same capacitance the potential difference across
each of them is onethird the battery potential difference or 3:0 V. The potential difference across
capacitor 4 is the same as the potential difference across the equivalent capacitor that replaced 3
and 4, so the charge on capacitor 4 is q4 = C4 V4 = (15 £ 10¡6 F)(3:0 V) = 45 £ 10¡6 C.
158 Chapter 25 69
(a) and (b) The capacitors have the same plate separation d and the same potential difference V
across their plates, so the electric field are the same within them. The magnitude of the field in
either one is E = V =d = (600 V)=(3:00 £ 10¡3 m) = 2:00 £ 105 V=m.
(c) Let A be the area of a plate. Then the surface charge density on the positive plate is ¾A =
2
qA =A = CA V =A = (²0 A=d)V =A = ²0 V =d = ²0 E = (8:85 £ 10¡12 N ¢ m2 =C )(2:00 £ 105 V=m) =
1:77 £ 10¡6 C=m2 , where CV was substituted for q and the expression ²0 A=d for the capacitance
of a parallelplate capacitor was substituted for C.
(d) Now the capacitance is ∙²0 A=d, where ∙ is the dielectric constant. The surface charge density
2
2
on the positive plate is ¾B = ∙²0 E = ∙¾A = (2:60)(1:77 £ 10¡6 C=m ) = 4:60 £ 10¡6 C=m .
(e) The electric field in B is produced by the charge on the plates and the induced charge
together while the field in A is produced by the charge on the plates alone. since the fields are
the same ¾B + ¾induced = ¾A , so ¾induced = ¾A ¡ ¾B = 1:77 £ 10¡6 C=m2 ¡ 4:60 £ 10¡6 C=m2 =
2
¡2:83 £ 10¡6 C=m .
73
The electric field in the lower region is due to the charge on both plates and the charge induced
on the upper and lower surfaces of the dielectric in the region. The charge induced on the
dielectric surfaces of the upper region has the same magnitude but opposite sign on the two
surfaces and so produces a net field of zero in the lower region. Similarly, the electric field in
the upper region is due to the charge on the plates and the charge induced on the upper and lower
surfaces of dielectric in that region. Thus the electric field in the upper region has magnitude
Eupper = q∙upper ²0 A and the potential difference across that region is Vupper = Eupper d, where d is
the thickness of the region. The electric field in the lower region is Elower = q∙lower ²0 A and the
potential difference across that region is Vlower = Elower d. The sum of the potential differences
must equal the potential difference V across the entire capacitor, so
¸
∙
1
qd
1
:
+
V = Eupper d + Elower d =
²0 A ∙upper ∙lower
The solution for q is
∙upper ∙lower ²0 A
(3:00)(4:00) (8:85 £ 10¡12 N ¢ m2 =C2 )(2:00 £ 10¡2 m2 )
(7:00 V)
V =
q=
3:00 + 4:00
2:00 £ 10¡3 m
∙upper + ∙lower d
= 1:06 £ 10¡9 C : Chapter 25 159 Chapter 26 7
(a) The magnitude of the current density is given by J = nqvd , where n is the number of
particles per unit volume, q is the charge on each particle, and vd is the drift speed of the
particles. The particle concentration is n = 2:0 £ 108 cm¡3 = 2:0 £ 1014 m¡3 , the charge is
q = 2e = 2(1:60 £ 10¡19 C) = 3:20 £ 10¡19 C, and the drift speed is 1:0 £ 105 m=s. Thus
2 J = (2 £ 1014 m¡3 )(3:2 £ 10¡19 C)(1:0 £ 105 m=s) = 6:4 A=m : (b) Since the particles are positively charged, the current density is in the same direction as their
motion, to the north.
(c) The current cannot be calculated unless the crosssectional area of the beam is known. Then
i = JA can be used.
17
The resistance of the wire is given by R = ½L=A, where ½ is the resistivity of the material, L
is the length of the wire, and A is the crosssectional area of the wire. The crosssectional area
is A = ¼r2 = ¼(0:50 £ 10¡3 m)2 = 7:85 £ 10¡7 m2 . Here r = 0:50 mm = 0:50 £ 10¡3 m is the
radius of the wire. Thus
½= RA (50 £ 10¡3 )(7:85 £ 10¡7 m2 )
= 2:0 £ 10¡8 ¢ m :
=
2:0 m
L 19
The resistance of the coil is given by R = ½L=A, where L is the length of the wire, ½ is the
resistivity of copper, and A is the crosssectional area of the wire. Since each turn of wire has
length 2¼r, where r is the radius of the coil, L = (250)2¼r = (250)(2¼)(0:12 m) = 188:5 m. If rw
2
is the radius of the wire, its crosssectional area is A = ¼rw = ¼(0:65£10¡3 m)2 = 1:33£10¡6 m2 .
According to Table 26–1, the resistivity of copper is 1:69 £ 10¡8 ¢ m. Thus
R= ½L (1:69 £ 10¡8 ¢ m)(188:5 m)
=
= 2:4 :
1:33 £ 10¡6 m2
A 21
Since the mass and density of the material do not change, the volume remains the same. If L0
is the original length, L is the new length, A0 is the original crosssectional area, and A is the
new crosssectional area, then L0 A0 = LA and A = L0 A0 =L = L0 A0 =3L0 = A0 =3. The new
resistance is
½L0
½L ½3L0
=
=9
= 9R0 ;
R=
A
A0 =3
A0
160 Chapter 26 where R0 is the original resistance. Thus R = 9(6:0 ) = 54 .
23
The resistance of conductor A is given by
RA = ½L
;
2
¼rA where rA is the radius of the conductor. If ro is the outside radius of conductor B and ri is its
2
2
inside radius, then its crosssectional area is ¼(ro ¡ ri ) and its resistance is
RB =
The ratio is ½L
:
2
¡ ri ) 2
¼(ro 2
2
RA ro ¡ ri (1:0 mm)2 ¡ (0:50 mm)2
=
=
= 3:0 :
2
(0:50 mm)2
RB
rA 39
(a) Electrical energy is transferred to internal energy at a rate given by
P = V2
;
R where V is the potential difference across the heater and R is the resistance of the heater. Thus
(120 V)2
= 1:0 £ 103 W = 1:0 kW :
P =
14
(b) The cost is given by
C = (1:0 kW)(5:0 h)($0:050 =kW ¢ h) = $0:25 :
43
(a) Let P be the rate of energy dissipation, i be the current in the heater, and V be the potential
difference across the heater. They are related by P = iV . Solve for i:
i= P 1250 W
= 10:9 A :
=
115 V
V (b) According to the definition of resistance V = iR, where R is the resistance of the heater.
Solve for R:
V
115 V
R=
= 10:6 :
=
10:9 A
i
(c) The thermal energy E produced by the heater in time t (= 1:0 h = 3600 s) is
E = P t = (1250 W)(3600 s) = 4:5 £ 106 J :
Chapter 26 161 53
(a) and (b) Calculate the electrical resistances of the wires. Let ½C be the resistivity of wire C,
rC be its radius, and LC be its length. Then the resistance of this wire is
RC = ½C 1:0 m
LC
= (2:0 £ 10¡6 ¢ m)
= 2:54 :
2
¼(0:50 £ 10¡3 m)2
¼rC Let ½D be the resistivity of wire D, rD be its radius, and LD be its length. Then the resistance
of this wire is
RD = ½D 1:0 m
LD
= (1:0 £ 10¡6 ¢ m)
= 5:09 :
2
¼(0:25 £ 10¡3 m)2
¼rD If i is the current in the wire, the potential difference between points 1 and 2 is
¢V12 = iRC = (2:0 A)(2:54 ) = 5:1 V
and the potential difference between points 2 and 3 is
¢V23 = iRD = (2:0 A)(5:09 ) = 10 V :
(c) and (d) The rate of energy dissipation between points 1 and 2 is
P12 = i2 RC = (2:0 A)2 (2:54 ) = 10 W
and the rate of energy dissipation between points 2 and 3 is
P23 = i2 RD = (2:0 A)2 (5:09 ) = 20 W : 55
(a) The charge that strikes the surface in time ¢t is given by ¢q = i ¢t, where i is the current.
Since each particle carries charge 2e, the number of particles that strike the surface is
N= ¢q i ¢t (0:25 £ 10¡6 A)(3:0 s)
=
=
= 2:3 £ 1012 :
¡19 C)
2e
2e
2(1:6 £ 10 (b) Now let N be the number of particles in a length L of the beam. They will all pass through
the beam cross section at one end in time t = L=v, where v is the particle speed. The current is
the charge that moves through the cross section per unit time. That is, i = 2eN=t = 2eN v=L.
Thus, N = iL=2ev.
Now find the particle speed. The kinetic energy of a particle is
K = 20 MeV = (20 £ 106 eV)(1:60 £ 10¡19 J=eV) = 3:2 £ 10¡12 J :
p
Since K = 1 mv 2 , v = 2K=m. The mass of an alpha particle is four times the mass of a proton
2
or m = 4(1:67 £ 10¡27 kg) = 6:68 £ 10¡27 kg, so
s
2(3:2 £ 10¡12 J)
= 3:1 £ 107 m=s
v=
¡27 kg
6:68 £ 10
162 Chapter 26 and (0:25 £ 10¡6 A)(20 £ 10¡2 m)
iL
=
= 5:0 £ 103 :
N=
¡19 C)(3:1 £ 107 m=s)
2ev 2(1:60 £ 10 (c) Use conservation of energy. The initial kinetic energy is zero, the final kinetic energy is
20 MeV = 3:2 £ 10¡12 J, the initial potential energy is qV = 2eV , and the final potential energy
is zero. Here V is the electric potential through which the particles are accelerated. Conservation
of energy leads to Kf = Ui = 2eV , so
V = 3:2 £ 10¡12 J
Kf
=
= 10 £ 106 V :
2e
2(1:60 £ 10¡19 C) 59
Let RH be the resistance at the higher temperature (800± C) and let RL be the resistance at the
lower temperature (200± C). Since the potential difference is the same for the two temperatures,
the rate of energy dissipation at the lower temperature is PL = V 2 =RL , and the rate of energy
dissipation at the higher temperature is PH = V 2 =RH , so PL = (RH =RL )PH . Now RL =
RH + ®RH ¢T , where ¢T is the temperature difference TL ¡ TH = ¡600± C. Thus,
PL = 500 W
RH
PH
=
= 660 W :
PH =
1 + ® ¢T 1 + (4:0 £ 10¡4 =± C)(¡600± C)
RH + ®RH ¢T 75
If the resistivity is ½0 at temperature T0 , then the resistivity at temperature T is ½ = ½0 +®½0 (T ¡T0 ),
where ® is the temperature coefficient of resistivity. The solution for T is
T = ½ ¡ ½0 + ®½0 T0
:
®½0 Substitute ½ = 2½0 to obtain
T = T0 + 1
1
= 20:0± C +
= 250± C :
4:3 £ 10¡3 K¡1
® The value of ® was obtained from Table 26–1. Chapter 26 163 Chapter 27 7
(a) Let i be the current in the circuit and take it to be positive if it is to the left in R1 . Use
Kirchhoff’s loop rule: E1 ¡ iR2 ¡ iR1 ¡ E2 = 0. Solve for i:
i= 12 V ¡ 6:0 V
E1 ¡ E2
= 0:50 A :
=
R1 + R2 4:0 + 8:0 A positive value was obtained, so the current is counterclockwise around the circuit.
(b) and (c) If i is the current in a resistor with resistance R, then the power dissipated by that
resistor is given by P = i2 R. For R1 the power dissipated is
P1 = (0:50 A)2 (4:0 ) = 1:0 W
and for R2 the power dissipated is
P2 = (0:50 A)2 (8:0 ) = 2:0 W :
(d) and (e) If i is the current in a battery with emf E, then the battery supplies energy at the rate
P = iE provided the current and emf are in the same direction. The battery absorbs energy at
the rate P = iE if the current and emf are in opposite directions. For battery 1 the power is
P1 = (0:50 A)(12 V) = 6:0 W
and for battery 2 it is
P2 = (0:50 A)(6:0 V) = 3:0 W :
(f) and (g) In battery 1, the current is in the same direction as the emf so this battery supplies
energy to the circuit. The battery is discharging. The current in battery 2 is opposite the direction
of the emf, so this battery absorbs energy from the circuit. It is charging.
13
(a) If i is the current and ¢V is the potential difference, then the power absorbed is given by
P = i ¢V . Thus
P 50 W
= 50 V :
=
¢V =
1:0 A
i
Since energy is absorbed, point A is at a higher potential than point B; that is, VA ¡ VB = 50 V. (b) The endtoend potential difference is given by VA ¡ VB = +iR + E, where E is the emf of
element C and is taken to be positive if it is to the left in the diagram. Thus E = VA ¡ VB ¡ iR =
50 V ¡ (1:0 A)(2:0 ) = 48 V.
164 Chapter 27 (c) A positive value was obtained for E, so it is toward the left. The negative terminal is at B.
21
R... ...
(a) and (b) The circuit is shown in the diagram to
.
.
.
.
..
..
.
.
..
..
..
. ..
..
..
..
..
. .. . .. . ..
..
. .. . . . . . .
. .. . .. . .. . .. .
. .. .. .. .. .. .. .. .
the right. The current is taken to be positive if it is
. . .. . .. . .. .
. . . . .
. .. . .
..
..
..
..
..
..
..
..
..
..
...
...
.
.
.
.
.
.
.
.
.
clockwise. The potential difference across battery 1
¡! i
is given by V1 = E ¡ ir1 and for this to be zero, the
current must be i = E=r1 . Kirchhoff’s loop rule gives
2E ¡ ir1 ¡ ir2 ¡ iR = 0. Substitute i = E=r1 and
.
.
.
.
.
..
.
..
..
..
..
..
.
.
..
..
..
..
. .. .. ..
. .. .. ..
. .. .. ..
. .. .. ..
. .. . .. .
. .. . .. .
. .
. .
. .
. .
solve for R. You should get R = r1 ¡ r2 = 0:016 ¡
. .
. .
. .
. .
. ..
. ..
. ..
. ..
..
..
..
..
.
.
.
.
.
.
.
.
r2
r1
0:012 = 0:004 .
E
E
Now assume that the potential difference across battery 2 is zero and carry out the same analysis. You
should find R = r2 ¡ r1 . Since r1 > r2 and R must
be positive, this situation is not possible. Only the potential difference across the battery with
the larger internal resistance can be made to vanish with the proper choice of R.
29
Let r be the resistance of each of the thin wires. Since they are in parallel, the resistance R of
the composite can be determined from
1 9
= ;
R r
or R = r=9. Now
4½`
r=
¼d2
and
4½`
;
R=
¼D 2
where ½ is the resistivity of copper. Here ¼d2 =4 was used for the crosssectional area of any one
of the original wires and ¼D2 =4 was used for the crosssectional area of the replacement wire.
Here d and D are diameters. Since the replacement wire is to have the same resistance as the
composite,
4½`
4½`
=
:
2
9¼d2
¼D
Solve for D and obtain D = 3d.
33
Replace the two resistors on the left with their equivalent resistor. They are in parallel, so the
equivalent resistance is Req = 1:0 . The circuit now consists of the two emf devices and four
resistors. Take the current to be upward in the righthand emf device. Then the loop rule gives
E2 ¡ iReq ¡ 3iR ¡ E2 , where R = 2:0 . The current is
i= 12 V ¡ 5:0 V
E2 ¡ E1
=
= 1:0 A :
Req + 3R 1:0 + 3(2:0 ) Chapter 27 165 To find the potential at point 1 take a path from ground, through the equivalent resistor and E2 ,
to the point. The result is V1 = iReq ¡ E1 = (1:0 A)(1:0 ) ¡ 12 V = ¡11 V. To find the potential
at point 2 continue the path through the lowest resistor on the digram. It is V2 = V1 + iR =
¡11 V + (1:0 A)(2:0 ) = ¡9:0 V.
47
(a) and (b) The copper wire and the aluminum jacket are connected in parallel, so the potential
difference is the same for them. Since the potential difference is the product of the current and
the resistance, iC RC = iA RA , where iC is the current in the copper, iA is the current in the
aluminum, RC is the resistance of the copper, and RA is the resistance of the aluminum. The
resistance of either component is given by R = ½L=A, where ½ is the resistivity, L is the length,
and A is the crosssectional area. The resistance of the copper wire is
RC = ½C L
¼a2 and the resistance of the aluminum jacket is
RA = ½A L
:
¼(b2 ¡ a2 ) Substitute these expressions into iC RC = iA RA and cancel the common factors L and ¼ to obtain
iA ½A
iC ½C
= 2
:
2
a
b ¡ a2 Solve this equation simultaneously with i = iC + iA , where i is the total current. You should get
iC = a2 ½C i
(b2 ¡ a2 )½C + a2 ½A iA = (b2 ¡ a2 )½C i
:
(b2 ¡ a2 )½C + a2 ½A and The denominators are the same and each has the value
£
¤
(b2 ¡ a2 )½C + a2 ½A = (0:380 £ 10¡3 m)2 ¡ (0:250 £ 10¡3 m)2 (1:69 £ 10¡8 ¢ m)
+ (0:250 £ 10¡3 m)2 (2:75 £ 10¡8 ¢ m) = 3:10 £ 10¡15 ¢ m3 :
Thus
iC =
and 166 (0:250 £ 10¡3 m)2 (2:75 £ 10¡8 ¢ m)(2:00 A)
= 1:11 A
3:10 £ 10¡15 ¢ m3 ¤
£
(0:380 £ 10¡3 m)2 ¡ (0:250 £ 10¡3 m)2 (1:69 £ 10¡8 ¢ m)(2:00 A)
iA =
3:10 £ 10¡15 ¢ m3
= 0:893 A :
Chapter 27 (c) Consider the copper wire. If V is the potential difference, then the current is given by
V = iC RC = iC ½C L=¼a2 , so
L= ¼(0:250 £ 10¡3 m)2 (12:0 V)
¼a2 V
= 126 m :
=
(1:11 A)(1:69 £ 10¡8 ¢ m)
iC ½C 57
During charging the charge on the positive plate of the capacitor is given by Eq. 27–33, with
RC = ¿ . That is,
i
h
q = CE 1 ¡ e¡t=¿ ;
where C is the capacitance, E is applied emf, and ¿ is the time constant. You want the time for
which q = 0:990CE, so
0:990 = 1 ¡ e¡t=¿ :
Thus e¡t=¿ = 0:010 :
Take the natural logarithm of both sides to obtain t=¿ = ¡ ln 0:010 = 4:61 and t = 4:61¿ .
65
(a), (b), and (c) At t = 0, the capacitor is completely uncharged and the current in the capacitor
branch is as it would be if the capacitor were replaced by a wire. Let i1 be the current in R1 and
take it to be positive if it is to the right. Let i2 be the current in R2 and take it to be positive
if it is downward. Let i3 be the current in R3 and take it to be positive if it is downward. The
junction rule produces i1 = i2 + i3 , the loop rule applied to the lefthand loop produces
E ¡ i1 R1 ¡ i2 R2 = 0 ;
and the loop rule applied to the righthand loop produces
i2 R2 ¡ i3 R3 = 0 :
Since the resistances are all the same, you can simplify the mathematics by replacing R1 , R2 ,
and R3 with R. The solution to the three simultaneous equations is
2(1:2 £ 103 V)
2E
=
= 1:1 £ 10¡3 A
i1 =
6 )
3R 3(0:73 £ 10
and
i2 = i3 = 1:2 £ 103 V
E
=
= 5:5 £ 10¡4 A :
3R 3(0:73 £ 106 ) (d), (e), and (f) At t = 1, the capacitor is fully charged and the current in the capacitor branch
is zero. Then i1 = i2 and the loop rule yields
E ¡ i1 R1 ¡ i1 R2 = 0 :
Chapter 27 167 The solution is 1:2 £ 103 V
E
=
= 8:2 £ 10¡4 A :
6 )
2R 2(0:73 £ 10
(g) and (h) The potential difference across resistor 2 is V2 = i2 R2 . At t = 0 it is
i1 = i2 = and at t = 1 it is V2 = (5:5 £ 10¡4 A)(0:73 £ 106 ) = 4:0 £ 102 V V2 = (8:2 £ 10¡4 A)(0:73 £ 106 ) = 6:0 £ 102 V :
V2
(i) The graph of V2 versus t is shown to the right.
E=2
E=3 ...............
.................
.....................
....................
............
............
.........
.........
.......
.......
......
......
...
....
.....
.....
....
....
....
....
. ..
....
...
...
.
.
...
... E=6
t
73
As the capacitor discharges the potential difference across its plates at time t is given by V =
V0 e¡t=¿ , where V0 is the potential difference at time t = 0 and ¿ is the capacitive time constant.
This equation is solved for the time constant, with result
t
¿ =¡
:
ln(V =V0 )
Since the time constant is ¿ = RC, where RR is the resistance and C is the capacitance,
t
R=¡
:
C ln(V =V0 )
For the smaller time interval
R=¡ 10:0 £ 10¡6 s
¶ = 24:8 :
µ
0:800 V
¡6 F) ln
(0:220 £ 10
5:00 V and for the larger time interval
R=¡ 6:00 £ 10¡3 s
¶ = 1:49 £ 104 :
µ
0:800 V
(0:220 £ 10¡6 F) ln
5:00 V 75
(a) Let i be the current, which is the same in both wires, and E be the applied potential difference.
Then the loop equation gives E ¡ iRA ¡ iRB = 0 and the current is
i=
168 Chapter 27 E
60:0 V
= 70:1 A :
=
RA + RB 0:127 + 0:729 The current density in wire A is
JA = 70:1 A
i
=
= 1:32 £ 107 A=m2 :
2
¡3 m)2
¼rA ¼(1:30 £ 10 (b) The potential difference across wire A is VA = iRA = (70:1 A)(0:127 ) = 8:90 V.
(c) The resistance is RA = ½A L=A, where ½ is the resistivity, A is the crosssectional area, and
L is the length. The resistivity of wire A is
RA A (0:127 )¼(1:30 £ 10¡3 m)2
= 1:69 £ 10¡8 ¢ m :
=
40:0 m
L
According to Table 26–1 the material is copper.
(d) Since wire B has the same diameter and length as wire A and carries the same current, the
2
current density in it is the same, 1:32 £ 107 A=m .
(e) The potential difference across wire B is VB = iRB = (70:1 A)(0:729 ) = 51:1 V.
(f) The resistivity of wire B is
½A = RB A (0:729 )¼(1:30 £ 10¡3 m)2
= 9:68 £ 10¡8 ¢ m :
=
40:0 m
L
According to Table 26–1 the material is iron.
½B = 77
The three circuit elements are in series, so the current is the same in all of them. Since the battery
is discharging, the potential difference across its terminals is Vbatt = E ¡ ir, where E is its emf
and r is its internal resistance. Thus
E ¡V
12 V ¡ 11:4 V
= 0:012 :
r=
=
50 A
i
This is less than 0:0200 , so the battery is not defective.
The resistance of the cable is Rcable = Vcable =i = (3:0 V)=(50 A = 0:060 , which is greater than
0:040 . The cable is defective.
The potential difference across the motor is Vmotor = 11:4 V ¡ 3:0 V = 8:4 V and its resistance
is Rmotor = Vmotor =i = (8:4 V)=(50 A) = 0:17 , which is less than 0:200 . The motor is not
defective.
85
Let RS0 be the resistance of the silicon resistor at 20± and RI0 be the resistance of the iron resistor
at that temperature. At some other temperature T the resistance of the silicon resistor is RS =
RS0 + ®S RS0 (T ¡ 20± C) and the resistance of the iron resistor is RI = RI0 + ®I RI0 (T ¡ 20± C).
Here ®S and ®I are the temperature coefficients of resistivity. The resistors are series so the
resistance of the combination is
R = RS0 + RI0 + (®S RS0 + ®I RI0 )(T ¡ 20± C) : We want RS0 + RI0 to be 1000 and ®S RS0 + ®I RI0 to be zero. Then the resistance of the
combination will be independent of the temperature.
Chapter 27 169 The second equation gives RI0 = ¡(®S =®I )RS0 and when this is used to substitute for RI0 in
the first equation the result is RS0 ¡ (®S =®I )RS0 = 1000 . The solution for RS0 is 1000
1000
RS0 = ®S
=
= 85 ;
¡3 ¡1
¡ 1 ¡70 £ 10 K ¡ 1
®I
6:5 £ 10¡3 K¡1
where values for the temperature coefficients of resistivity were obtained from Table 26–1. The
resistance of the iron resistor is RI0 = 1000 ¡ 85 = 915 . 95
When the capacitor is fully charged the potential difference across its plates is E and the energy
stored in it is U = 1 CE 2 .
2
(a) The current is given as a function of time by i = (E=R)e¡t=¿ , where ¿ (= RC) is the capacitive
time constant. The rate with which the emf device supplies energy is PE = iE and the energy
supplied in fully charging the capacitor is
Z
Z 1
E 2 1 ¡t=¿
E 2 ¿ E 2 RC
=
= CE 2 :
PE dt =
e
dt =
EE =
R 0
R
R
0
This is twice the energy stored in the capacitor.
(b) The rate with which energy is dissipated in the resistor is PR = i2 R and the energy dissipated
as the capacitor is fully charged is
Z
Z 1
E 2 1 ¡2t=¿
E 2 ¿ E 2 RC 1
=
= 2 CE 2 :
PR dt =
e
dt =
ER =
2R
2R
R 0
0 97
(a) Immediately after the switch is closed the capacitor is uncharged and since the charge on
the capacitor is given by q = CVC , the potential difference across its plates is zero. Apply
the loop rule to the righthand loop to find that the potential difference across R2 must also be
zero. Now apply the loop rule to the lefthand loop to find that E ¡ i1 R1 = 0 and i1 = E=R1 =
(30 V)=(20 £ 103 ) = 1:5 £ 10¡3 A.
(b) Since the potential difference across R2 is zero and this potential difference is given by
VR2 = i2 R2 , i2 = 0.
(c) A long time later, when the capacitor is fully charged, the current is zero in the capacitor
branch and the current is the same in the two resistors. The loop rule applied to the lefthand loop
gives E ¡ iR1 ¡ iR2 = 0, so i = E=(R1 + R2 ) = (30 V)=(20 £ 103 + 10 £ 103 ) = 1:0 £ 10¡3 A.
99
(a) R2 and R3 are in parallel, with an equivalent resistance of R2 R3 =(R2 + R3 ), and this combination is in series with R1 , so the circuit can be reduced to a single loop with an emf E and a
resistance Req = R1 + R2 R3 =(R2 + R3 ) = (R1 R2 + R1 R3 + R2 R3 )=(R1 + R2 ). The current is
i=
170 Chapter 27 (R2 + R3 )E
E
=
:
Req R1 R2 + R1 R3 + R2 R3 The rate with which the battery supplies energy is
P = iE = (R2 + R3 )E 2
:
R1 R2 + R1 R3 + R2 R3 The derivative with respect to R3 is
2
E2
E 2 R2
(R2 + R3 )(R1 + R2 )E 2
dP
=
=¡
¡
;
(R1 R2 + R1 R3 + R2 R3 )2
dR3 R1 R2 + R1 R3 + R2 R3 (R1 R2 + R1 R3 + R2 R3 )2 where the last form was obtained with a little algebra. The derivative is negative for all (positive)
values of the resistances, so P has its maximum value for R3 = 0.
(b) Substitute R3 = 0 in the expression for P to obtain
R1 E 2 E 2 12:0 V
= 14:4 W :
=
=
P =
R1 R2 R1 10:0
101
If the batteries are connected in series the total emf in the circuit is N E and the equivalent
resistance is R + nr, so the current is i = N E=(R + N r). If R = r, then i = N E=(N + 1)r.
If the batteries are connected in parallel then the emf in the circuit is E and the equivalent
resistance is R + r=N , so the current is i = E=(R + r=N ) = N E=(N R + r). If R = r, i =
N E=(N + 1)r, the same as when they are connected in series. Chapter 27 171 Chapter 28 3
(a) The magnitude of the magnetic force on the proton is given by FB = evB sin Á, where v is
the speed of the proton, B is the magnitude of the magnetic field, and Á is the angle between
the particle velocity and the field when they are drawn with their tails at the same point. Thus
FB
6:50 £ 10¡17 N
=
= 4:00 £ 105 m=s :
eB sin Á (1:60 £ 10¡19 C)(2:60 £ 10¡3 T) sin 23:0±
(b) The kinetic energy of the proton is
v= K = 1 mv 2 = 1 (1:67 £ 10¡27 kg)(4:00 £ 105 m=s)2 = 1:34 £ 10¡16 J :
2
2 This is (1:34 £ 10¡16 J)=(1:60 £ 10¡19 J=eV) = 835 eV.
17 (a) Since the kinetic energy is given by K = 1 mv 2 , where m is the mass of the electron and v
2
is its speed,
s
r
2K
2(1:20 £ 103 eV)(1:60 £ 10¡19 J=eV)
= 2:05 £ 107 m=s :
=
v=
9:11 £ 10¡31 kg
m (b) The magnitude of the magnetic force is given by evB and the acceleration of the electron is
given by v 2 =r, where r is the radius of the orbit. Newton’s second law is evB = mv 2 =r, so mv (9:11 £ 10¡31 kg)(2:05 £ 107 m=s)
= 4:68 £ 10¡4 T = 468 ¹T :
=
¡19 C)(25:0 £ 10¡2 m)
(1:60 £ 10
er
(c) The frequency f is the number of times the electron goes around per unit time, so
B= v
2:05 £ 107 m=s
=
= 1:31 £ 107 Hz = 13:1 MHz :
2¼r 2¼(25:0 £ 10¡2 m)
(d) The period is the reciprocal of the frequency:
1
1
= 7:63 £ 10¡8 s = 76:3 ns :
T = =
7 Hz
f 1:31 £ 10
f= 29
(a) If v is the speed of the positron, then v sin Á is the component of its velocity in the plane
that is perpendicular to the magnetic field. Here Á is the angle between the velocity and the field
(89± ). Newton’s second law yields eBv sin Á = m(v sin Á)2 =r, where r is the radius of the orbit.
Thus r = (mv=eB) sin Á. The period is given by
2¼r
2¼m
2¼(9:11 £ 10¡31 kg)
= 3:58 £ 10¡10 s :
=
=
¡19 C)(0:100 T)
(1:60 £ 10
v sin Á
eB
The expression for r was substituted to obtain the second expression for T .
T = 172 Chapter 28 (b) The pitch p is the distance traveled along the line of the magnetic field in a time interval of
one period. Thus p = vT cos Á. Use the kinetic energy to find the speed: K = 1 mv 2 yields
2
s
r
2K
2(2:0 £ 103 eV)(1:60 £ 10¡19 J=eV)
= 2:651 £ 107 m=s :
v=
=
9:11 £ 10¡31 kg
m
Thus
p = (2:651 £ 107 m=s)(3:58 £ 10¡10 s) cos 89:0± = 1:66 £ 10¡4 m : (c) The orbit radius is
r= mv sin Á (9:11 £ 10¡31 kg)(2:651 £ 107 m=s) sin 89:0±
= 1:51 £ 10¡3 m :
=
(1:60 £ 10¡19 C)(0:100 T)
eB 41
(a) The magnitude of the magnetic force on the wire is given by FB = iLB sin Á, where i is the
current in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and Á
is the angle between the current and the field. In this case Á = 70± . Thus
FB = (5000 A)(100 m)(60:0 £ 10¡6 T) sin 70± = 28:2 N :
~
~ ~
(b) Apply the righthand rule to the vector product FB = iL £ B to show that the force is to the
west.
47 z. y ....
.
.
.
.
............... a
. .......
.
............
.
.
.......
.
.
...
.
.
.
..
.
.
.
.
.
.
.
.
...
.
....
.
..
.
.
..
.
.. i
.
.
.
.
....
.
.
.
...
.
..
.
.
.
..
.
........
.
.
b.
.
.
.
.
.
.
.
.
.
.
.
.
.
...........................................
.
.
.
.
.
~
...
...
B
.
~ ....
F .......................... .
.
.
.
.
........ ....... ....... ....... ....... ....... ........ ....... ....
.
... .
.
.. ...
.. ....
.
........... µ . x
.
..
.
..
............ .
.
..
..
.
......
.
.
..
.. ........
........ µ
........
........
.
a ...............
.......
.
.
.
.
.
.
z ~
F x .
.
.
.
.
.
.
.
.
.
.
... .
.....
.
.
..
.
..
..
. The situation is shown in the left diagram above. The y axis is along the hinge and the magnetic
field is in the positive x direction. A torque around the hinge is associated with the wire opposite
the hinge and not with the other wires. The force on this wire is in the positive z direction and
has magnitude F = N ibB, where N is the number of turns.
The right diagram shows the view from above. The magnitude of the torque is given by
¿ = F a cos µ = N ibBa cos µ
= 20(0:10 A)(0:10 m)(0:50 £ 10¡3 T)(0:050 m) cos 30±
= 4:3 £ 10¡3 N ¢ m : Chapter 28 173 Use the righthand rule to show that the torque is directed downward, in the negative y direction.
Thus ~ = ¡(4:3 £ 10¡3 N ¢ m) ˆ.
¿
j
55
(a) The magnitude of the magnetic dipole moment is given by ¹ = N iA, where N is the number
of turns, i is the current in each turn, and A is the area of a loop. In this case the loops are
circular, so A = ¼r 2 , where r is the radius of a turn. Thus
i= ¹
2:30 A ¢ m2
=
= 12:7 A :
N ¼r2 (160)(¼)(0:0190 m)2 (b) The maximum torque occurs when the dipole moment is perpendicular to the field (or the
plane of the loop is parallel to the field). It is given by ¿ = ¹B = (2:30 A ¢ m2 )(35:0 £ 10¡3 T) =
8:05 £ 10¡2 N ¢ m.
59
The magnitude of a magnetic dipole moment of a current loop is given by ¹ = iA, where i is
the current in the loop and A is the area of the loop. Each of these loops is a circle and its area
is given by A = ¼R2 , where R is the radius. Thus the dipole moment of the inner loop has a
2
magnitude of ¹i = i¼r1 = (7:00 A)¼(0:200 m)2 = 0:880 A ¢ m2 and the dipole moment of the outer
2
loop has a magnitude of ¹o = i¼r2 = (7:00 A¼(0:300 m)2 = 1:979 A ¢ m2 .
(a) Both currents are clockwise in Fig. 28–51 so, according to the righthand rule, both dipole
moments are directed into the page. The magnitude of the net dipole moment is the sum of the
magnitudes of the individual moments: ¹net = ¹i + ¹o = 0:880 A ¢ m2 + 1:979 A ¢ m2 = 2:86 A ¢ m2 .
The net dipole moment is directed into the page.
(b) Now the dipole moment of the inner loop is directed out of the page. The moments are in
opposite directions, so the magnitude of the net moment is ¹net = ¹o ¡ ¹i = 1:979 A ¢ m2 ¡
0:880 A ¢ m2 = 1:10 A ¢ m2 . The net dipole moment is again into the page.
63
If N closed loops are formed from the wire of length L, the circumference of each loop is L=N ,
the radius of each loop is R = L=2¼N , and the area of each loop is A = ¼R2 = ¼(L=2¼N )2 =
L2 =4¼N 2 . For maximum torque, orient the plane of the loops parallel to the magnetic field, so
the dipole moment is perpendicular to the field. The magnitude of the torque is then
µ 2 ¶
iL2 B
L
B=
:
¿ = N iAB = (N i)
4¼N 2
4¼N To maximize the torque, take N to have the smallest possible value, 1. Then
¿= iL2 B (4:51 £ 10¡3 A)(0:250 m)2 (5:71 £ 10¡3 T)
=
= 1:28 £ 10¡7 N ¢ m :
4¼
4¼ 65
(a) the magnetic potential energy is given by U = ¡~ ¢ B, where ~ is the magnetic dipole moment
¹ ~
¹
~ is the magnetic field. The magnitude of the magnetic moment is ¹ = N iA,
of the coil and B
174 Chapter 28 where i is the current in the coil, A is the area of the coil, and N is the number of turns. The
moment is in the negative y direction, as you can tell by wrapping the fingers of your right hand
around the coil in the direction of the current. Your thumb is then in the negative y direction.
Thus ~ = ¡(3:00)(2:00 A)(4:00 £ 10¡3 m2 ) ˆ = ¡(2:40 £ 10¡2 A ¢ m2 ) ˆ. The magnetic potential
¹
j
j
energy is
ˆ
U = ¡(¹y ˆ) ¢ (Bx ˆ + By ˆ + Bz k) = ¡¹y By
j
i
j
= ¡(¡2:40 £ 10¡2 A ¢ m2 )(¡3:00 £ 10¡3 T) = ¡7:20 £ 10¡5 J ; where ˆ ¢ ˆ = 0, ˆ ¢ ˆ = 1, and ˆ ¢ k = 0 were used.
j i
j j
j ˆ
(b) The magnetic torque on the coil is ˆ
ˆ
~ = ~ £ B = (¹y ˆ) £ (Bx ˆ + By ˆ + Bz k) = ¹y Bz ˆ ¡ ¹y Bx k
¿ ¹ ~
j
i
j
i
ˆ
i
= (¡2:40 £ 10¡2 A ¢ m2 )(¡4:00 £ 10¡3 T) ˆ ¡ (¡2:40 £ 10¡2 A ¢ m2 )(2:00 £ 10¡3 T) k
ˆ
i
= (9:6 £ 10¡5 N ¢ m) ˆ + (4:80 £ 10¡5 N ¢ m)k ; ˆ j j
where ˆ £ ˆ = ¡k, ˆ £ ˆ = 0, and ˆ £ k = ˆ were used.
j i
j ˆ i
73
~
~ v ~
~
~
The net force on the electron is given by F = ¡e(E + ~ £ B), where E is the electric field, B
is the magnetic field, and ~ is the electron’s velocity. Since the electron moves with constant
v
velocity you know that the net force must vanish. Thus
~
ˆ
E = ¡~ £ B = ¡(v ˆ) £ (B k) = ¡vB ˆ = ¡(100 m=s)(5:00 T) ˆ = (500 V=m) ˆ :
v ~
i
j
j
j
75
(a) and (b) Suppose the particles are accelerated from rest through an electric potential difference
V . Since energy is conserved the kinetic energy of a particle is K = qV , where q is the
particle’s charge. The ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy is
Kp =K® = e=2e = 0:50. The ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic
energy is Kd =K® = e=2e = 0:50.
(c) The magnitude of the magnetic force on a particle is qvB and, according to Newton’s second
law,p must equal mv 2 =R, where v is its speed and R is the radius of its orbit. Since
this
p
v = 2K=m = 2qV =m,
s
r
r
m 2K
m 2qV
mv
2m
=
=
=
R=
:
qB qB
m
qB
m
qB 2
The ratio of the radius of the deuteron’s path to the radius of the proton’s path is
r
r
2:0 u e
Rd
= 1:4 :
=
1:0 u e
Rp
Since the radius of the proton’s path is 10 cm, the radius of the deuteron’s path is (1:4)(10 cm) =
14 cm.
Chapter 28 175 (d) The ratio of the radius of the alpha particle’s path to the radius of the proton’s path is
r
r
R®
4:0 u e
= 1:4 :
=
1:0 u 2e
Rp
Since the radius of the proton’s path is 10 cm, the radius of the deuteron’s path is (1:4)(10 cm) =
14 cm.
77
i
j
i
Take the velocity of the particle to be ~ = vx ˆ + vy ˆ and the magnetic field to be B ˆ. The
v
magnetic force on the particle is then
~
ˆ
F = q~ £ B = q(vx ˆ + vy ˆ) £ (B ˆ) = ¡qvy B k ;
v ~
i
j
i
ˆ
where q is the charge of the particle. We used ˆ £ ˆ = 0 and ˆ £ ˆ = ¡k. The charge is
i i
j i
q= 0:48 N
F
= ¡4:0 £ 10¡2 C :
=
3 m=s)(sin 37± )(5:0 £ 10¡3 T)
¡vy B ¡(4:0 £ 10 81
(a) If K is the kinetic energy of the electron and m is its mass, then its speed is
s
r
2K
2(12 £ 103 eV)(1:60 £ 10¡19 J=eV)
=
v=
= 6:49 £ 107 m=s :
9:11 £ 10¡31
m
Since the electron is traveling along a line that is parallel to the horizontal component of Earth’s
magnetic field, that component does not enter into the calculation of the magnetic force on the
electron. The magnitude of the force on the electron is evB and since F = ma, where a is the
magnitude of its acceleration, evB = ma and
evB (1:60 £ 10¡19 C)(6:49 £ 107 m=s)(55:0 £ 10¡6 T)
= 6:3 £ 1014 m=s2 :
=
a=
¡31 kg
9:11 £ 10
m (b) If the electron does not get far from the x axis we may neglect the influence of the horizontal
component of Earth’s field and assume the electron follows a circular path. Its acceleration is
given by a = v 2 =R, where R is the radius of the path. Thus
y
v 2 (6:49 £ 107 m=s)2
...
...
...
...
=
R=
= 6:72 m :
...
...
...
...
a
...
...
6:27 £ 1014 m=s2
..
.
µ .......................
...R
...
...
...
The solid curve on the diagram is the path. Suppose it subtends
.
the angle µ at its center. d (= 0:200 m) is the distance traveled
along the x axis and ` is the deflection. Thep
right triangle
yields d = R sin µ, so sin µ = d=R and cos µ = 1 ¡ sin2 µ =
p
1 ¡ (x=R)2 . The triangle also gives ` = R ¡ R cos µ, so
p
` = R ¡ R 1 ¡ (x=R)2 . Substitute R = 6:72 m and d = 0:2 m
to obtain ` = 0:0030 m.
176 Chapter 28 ...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
. ....
......
...
....... `
.........
....
.........................
d x Chapter 29 7
(a) If the currents are parallel, the two magnetic fields are in opposite directions in the region
between the wires. Since the currents are the same, the net field is zero along the line that runs
halfway between the wires. There is no possible current for which the field does not vanish. If
there is to be a field on the bisecting line the currents must be in opposite directions. Then the
fields are in the same direction in the region between the wires.
(b) At a point halfway between the wires, the fields have the same magnitude, ¹0 i=2¼r. Thus
the net field at the midpoint has magnitude B = ¹0 i=¼r and
¼rB ¼(0:040 m)(300 £ 10¡6 T)
= 30 A :
=
i=
4¼ £ 10¡7 T ¢ m=A
¹0
15
Sum the fields of the two straight wires and the circular arc. Look at the derivation of the
expression for the field of a long straight wire, leading to Eq. 29–6. Since the wires we are
considering are infinite in only one direction, the field of either of them is half the field of an
infinite wire. That is, the magnitude is ¹0 i=4¼R, where R is the distance from the end of the
wire to the center of the arc. It is the radius of the arc. The fields of both wires are out of the
page at the center of the arc.
Now find an expression for the field of the arc at its center. Divide the arc into infinitesimal
segments. Each segment produces a field in the same direction. If ds is the length of a segment,
the magnitude of the field it produces at the arc center is (¹0 i=4¼R2 ) ds. If µ is the angle
subtended by the arc in radians, then Rµ is the length of the arc and the net field of the arc is
¹0 iµ=4¼R. For the arc of the diagram, the field is into the page. The net field at the center, due
to the wires and arc together, is
B= ¹0 i
¹0 iµ
¹0 i
¹0 i
+
¡
=
(2 ¡ µ) :
4¼R 4¼R 4¼R 4¼R For this to vanish, µ must be exactly 2 radians.
19
Each wire produces a field with magnitude given by B = ¹0 i=2¼r, where r is the distance from
the corner of the square to the center. According to the Pythagorean theorem, the diagonal of
p
p
p
the square has length 2a, so r = a= 2 and B = ¹0 i= 2¼a. The fields due to the wires at the
upper left and lower right corners both point toward the upper right corner of the square. The
Chapter 29 177 fields due to the wires at the upper right and lower left corners both point toward the upper left
corner. The horizontal components cancel and the vertical components sum to
¹0 i
2¹0 i
cos 45± =
Bnet = 4 p
¼a
2¼a
¡7
2(4¼ £ 10 T ¢ m=A)(20 A)
= 8:0 £ 10¡5 T :
=
¼(0:20 m)
p
~
In the calculation cos 45± was replaced with 1= 2. In unit vector notation B = (8:0 £ 10¡5 T) ˆ.
j
21
Follow the same steps as in the solution of Problem 17 above but change the lower limit of
integration to ¡L, and the upper limit to 0. The magnitude of the net field is
¯0
¯
¹0 iR 1
dx
x
¯ = ¹0 i p L
=
2 (x2 + R2 )1=2 ¯
2
2 3=2
4¼ R
4¼R L2 + R2
¡L (x + R )
¡L
4¼ £ 10¡7 T ¢ m=A)(0:693 A)
0:136 m
p
=
= 1:32 £ 10¡7 T :
4¼(0:251 m)
(0:136 m)2 + (0:251 m)2 ¹0 iR
B=
4¼ 31 Z 0 The current per unit width of the strip is i=w and the current through a width dx is (i=w) dx.
Treat this as a long straight wire. The magnitude of the field it produces at a point that is a
distance d from the edge of the strip is dB = (¹0 =2¼)(i=w) dx=x and the net field is
Z d+w
d+w
¹0
¹0 i
dx
ln
=
B=
2¼w d
2¼w
x
d
¡7
(4¼ £ 10 T ¢ m=A)(4:61 £ 10¡6 A) 0:0216 m + 0:0491 m
ln
=
2¼(0:0491 m)
0:0216 m
¡11
= 2:23 £ 10
T:
35
The magnitude of the force of wire 1 on wire 2 is given by
¹0 i1 i2 =2¼r, where i1 is the current in wire 1, i2 is the current
in wire 2, and r is the separation of the wires. The distance
q
between the wires is r = d2 + d2 . Since the currents are
1
2
in opposite directions the wires repel each other so the force
on wire 2 is along the line that joins the wires and is away
from wire 1. ...
...
¯..............
" .....................
q
j
...
...
...
...
...
...
...
...
d2 + d2
...
...
1
2
...
...
d1
...
...
...
...
...
...
...
...
...
...
..
.
j
# ¡ ¡ d ¡ ¡.................
Ã ¡ 2 ¡ !....................... µ ...
...
...
...
...
...
...
...
...
... .
... .
... .
.......
......
...
... x
~
F To find the x component of the force, multiply the magnitude q the force by the cosine of the
of
angle µ that the force makes with the x axis. This is cos µ = d2 = d2 + d2 . Thus the x component
1
2 178 Chapter 29 of the force is
¹0 i1 i2 d2
Fx =
2¼ d2 + d2
1
2
¡7
(2¼ £ 10 T ¢ m=A)(4:00 £ 10¡3 A)(6:80 £ 10¡3 A)
0:0500 m
=
2¼
(0:024 m)2 + (5:00 m)2
= 8:84 £ 10¡11 T :
43
(a) Two of the currents are out of the page and one is into the page, so the net current enclosed
by the path is 2:0 A, out of the page. Since the path is traversed in the clockwise sense, a current
into the page is positive and a current out of the page is negative, as indicated by the righthand
rule associated with Ampere’s law. Thus ienc = ¡i and
I
~ s
B ¢ d~ = ¡¹0 i = ¡(4¼ £ 10¡7 T ¢ m=A)(2:0 A) = ¡2:5 £ 10¡6 T ¢ m : (b) The net current enclosed by the path is zero (two currents are out of the page and two are
H
~ s
into the page), so B ¢ d~ = ¹0 ienc = 0. 53
(a) Assume that the point is inside the solenoid. The field of the solenoid at the point is parallel
to the solenoid axis and the field of the wire is perpendicular to the solenoid axis. The net field
makes an angle of 45± with the axis if these two fields have equal magnitudes.
The magnitude of the magnetic field produced by a solenoid at a point inside is given by Bsol =
¹0 isol n, where n is the number of turns per unit length and isol is the current in the solenoid.
The magnitude of the magnetic field produced by a long straight wire at a point a distance r
away is given by Bwire = ¹0 iwire =2¼r, where iwire is the current in the wire. We want ¹0 nisol =
¹0 iwire =2¼r. The solution for r is
r= iwire
6:00 A
= 4:77 £ 10¡2 m = 4:77 cm :
=
2 m¡1 )(20:0 £ 10¡3 A)
2¼nisol 2¼(10:0 £ 10 This distance is less than the radius of the solenoid, so the point is indeed inside as we assumed.
(b) The magnitude of the either field at the point is
Bsol = Bwire = ¹0 nisol = (4¼ £ 10¡7 T ¢ m=A)(10:0 £ 102 m¡1 )(20:0 £ 10¡3 A) = 2:51 £ 10¡5 T :
Each of the two fields is a vector component of the net field, so the magnitude of the net field
p
is the square root of the sum of the squares of the individual fields: B = 2(2:51 £ 10¡5 T)2 =
3:55 £ 10¡5 T.
57
The magnitude of the dipole moment is given by ¹ = N iA, where N is the number of turns, i
is the current, and A is the area. Use A = ¼R2 , where R is the radius. Thus
¹ = N i¼R2 = (200)(0:30 A)¼(0:050 m)2 = 0:47 A ¢ m2 :
Chapter 29 179 65
(a) Take the magnetic field at a point within the hole to be the sum of the fields due to two
current distributions. The first is the solid cylinder obtained by filling the hole and has a current
density that is the same as that in the original cylinder with the hole. The second is the solid
cylinder that fills the hole. It has a current density with the same magnitude as that of the original
cylinder but it is in the opposite direction. Notice that if these two situations are superposed, the
total current in the region of the hole is zero.
Recall that a solid cylinder carrying current i, uniformly distributed over a cross section, produces
a magnetic field with magnitude B = ¹0 ir=2¼R2 a distance r from its axis, inside the cylinder.
Here R is the radius of the cylinder.
For the cylinder of this problem, the current density is
J= i
i
;
=
2 ¡ b2 )
A ¼(a where A (= ¼(a2 ¡ b2 )) is the crosssectional area of the cylinder with the hole. The current in
the cylinder without the hole is
ia2
a2 ¡ b2
and the magnetic field it produces at a point inside, a distance r1 from its axis, has magnitude
i1 = JA1 = ¼Ja2 = ¹0 ir1
¹0 i1 r1
¹0 ir1 a2
=
:
=
B1 =
2
2 (a2 ¡ b2 )
2¼a
2¼a
2¼(a2 ¡ b2 ) The current in the cylinder that fills the hole is ib2
a2 ¡ b2
and the field it produces at a point inside, a distance r2 from the its axis, has magnitude
i2 = ¼Jb2 = B2 = ¹0 ir2
¹0 i2 r2
¹0 ir2 b2
=
:
=
2
2 (a2 ¡ b2 )
2¼b
2¼b
2¼(a2 ¡ b2 ) At the center of the hole, this field is zero and the field there is exactly the same as it would be
if the hole were filled. Place r1 = d in the expression for B1 and obtain
B= (4¼ £ 10¡7 T ¢ m=A)(5:25 A)(0:0200 m)
¹0 id
=
= 1:53 £ 10¡5 T
2¼(a2 ¡ b2 )
2¼[(0:0400 m)2 ¡ (0:0150 m)2 ] for the field at the center of the hole. The field points upward in the diagram if the current is
out of the page.
(b) If b = 0, the formula for the field becomes
¹0 id
:
2¼a2
This correctly gives the field of a solid cylinder carrying a uniform current i, at a point inside
the cylinder a distance d from the axis. If d = 0, the formula gives B = 0. This is correct for the
field on the axis of a cylindrical shell carrying a uniform current.
B= 180 Chapter 29 (c) The diagram shows the situation in a crosssectional
plane of the cylinder. P is a point within the hole, A is
on the axis of the cylinder, and C is on the axis of the
hole. The magnetic field due to the cylinder without the
hole, carrying a uniform current out of the page, is labeled
~
B1 and the magnetic field of the cylinder that fills the hole,
~
carrying a uniform current into the page, is labeled B2 . The
line from A to P makes the angle µ1 with the line that joins
the centers of the cylinders and the line from C to P makes
~
the angle µ2 with that line, as shown. B1 is perpendicular
to the line from A to P and so makes the angle µ1 with the
~
vertical. Similarly, B2 is perpendicular to the line from C
to P and so makes the angle µ2 with the vertical. y ~
B1 ~
B2 ......
.....
.
....
....
. ....
. .....
...
...
......
.....
...
...
...
. ...
.
...
.......
...
..... .
...
.....
...
...
.....
...
.....
....
...
...
.....
....
...
..
..
...
...
....
....
.
...
...
...
... ........
..... .....
....
..
. .
.
. ..
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..
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..
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..
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..
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..
.
..
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.
.
.
.
.
.
.. µ1 µ2
P x r1 r2 µ2 µ1 A d C The x component of the total field is
Bx = B2 sin µ2 ¡ B1 sin µ1 =
= ¹0 ir2
¹0 ir1
sin µ2 ¡
sin µ1
2 ¡ b2 )
2¼(a
2¼(a2 ¡ b2 ) ¹0 i
[r2 sin µ2 ¡ r1 sin µ1 ] :
2¼(a2 ¡ b2 ) As the diagram shows, r2 sin µ2 = r1 sin µ1 , so Bx = 0. The y component is given by
By = B2 cos µ2 + B1 cos µ1 =
= ¹0 ir2
¹0 ir1
cos µ2 +
cos µ1
2¼(a2 ¡ b2 )
2¼(a2 ¡ b2 ) ¹0 i
[r2 cos µ2 + r1 cos µ1 ] :
2¼(a2 ¡ b2 ) The diagram shows that r2 cos µ2 + r1 cos µ1 = d, so
By = ¹0 id
:
2¼(a2 ¡ b2 ) This is identical to the result found in part (a) for the field on the axis of the hole. It is independent
of r1 , r2 , µ1 , and µ2 , showing that the field is uniform in the hole.
71
Use the BiotSavart law in the form
s r
~ ¹0 i¢~ £ ~ :
B=
4¼
r3
ˆ
ˆ
Take ¢~ to be ¢s ˆ, and ~ to be x ˆ + y ˆ +z k. Then ¢~ £~ = ¢s ˆ £(x ˆ +y ˆ +pk) = ¢s(z ˆ ¡x k),
s
j
r
i j
s r
j
i j zˆ
i
2 + y 2 + z 2 . The
ˆ j j
where ˆ £ ˆ = ¡k, ˆ £ ˆ = 0, and ˆ £ k = ˆ were used. In addition, r = x
j i
j ˆ i
BiotSavart equation becomes
ˆ
ˆ
~ ¹0 i ¢s(z i ¡ z k) :
B=
2 + y 2 + z 2 )3=2
4¼ (x
Chapter 29 181 (a) For x = 0, y = 0, and z = 5:0 m,
¡7
ˆ
~ 4¼ £ 10 T ¢ m=A (2:0 A)(0:030 m)(5:0 m) i = (2:4 £ 10¡10 T) ˆ :
i
B=
3
4¼
(5:0 m) ~
(b) For x = 0, y = 6:0 m, and z = 0, B = 0.
(c) For x = 7:0 m, y = 7:0 m, and z = 0,
¡7
ˆ
~ 4¼ £ 10 T ¢ m=A (2:0 A)(0:030 m)(¡7:0 m) k = (4:3 £ 10¡11 T) k :
ˆ
B=
4¼
[(7:0 m)2 + (7:0 m)2 ]3=2 (d) For x = ¡3:0 m, y = ¡4:0 m, and z = 0,
¡7
ˆ
~ 4¼ £ 10 T ¢ m=A (2:0 A)(0:030 m)(3:0 m) k = (1:4 £ 10¡10 T) k :
ˆ
B=
4¼
[(¡3:0 m)2 + (¡4:0 m)2 ]3=2 77
First consider the finite wire segment shown on the right.
It extends from y = ¡d to y = a¡d, where a is the length
of the segment, and it carries current i in the positive y
direction. Let dy be an infinitesimal length of wire at
coordinate y. According to the BiotSavart law the magnitude of the magnetic field at P due to this infinitesimal
length is dB = (¹0 =4¼)(i sin µ=r 2 ) dy. Now r 2 = y 2 + R2
p
and sin µ = R=r = R= y 2 + R2 , so
¹0
iR
dy
dB =
2 + R2 )3=2
4¼ (y y i .
.
.
.
..
..
.....
.
... .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. R x ...
....
.......
.......
.......
......
......
......
......
.......
.....
......
.......
.......
......
......
..... .
.......
......
...... .µ
.
dy . P r and the field of the entire segment is
¹0
iR
B=
4¼ Z a¡d
¡d i
y
dy = ¹0 =4¼
R
(y 2 + R2 )3=2 " a¡d d p
+p
R2 + (a ¡ d)2
R2 + d2 # ; where integral 19 of Appendix E was used.
All four sides of the square produce magnetic fields that are into the page at P, so we sum their
magnitudes. To calculate the field of the left side of the square put d = 3a=4 and R = a=4. The
result is
¸
∙
¹O 4i
3
¹0 4i 1
p +p
Bleft =
=
(1:66) :
4¼ a
3¼ a
10
2
The field of the upper side of the square is the same. To calculate the field of the right side of
the square put d = a=4 and R = 3a=4. The result is
∙
¸
1
¹O 4i
¹0 4i
3
p +p
Bright =
(0:341) :
=
4¼ 3a
3¼ a
18
10
182 Chapter 29 The field of the bottom side is the same. The total field at P is
B = Bleft + Bupper + Bright + Blower =
= ¹0 4i
(1:66 + 1:66 + 0:341 + 0:341)
4¼ a 4¼ £ 10¡7 T ¢ m=A 4(10 A)
(4:00) = 2:0 £ 10¡4 T :
4¼
0:080 m 79
.......
......
....
..
(a) Suppose the field is not parallel to the sheet, as shown
........
.........
.....
.....
....
..
in the upper diagram. Reverse the direction of the current.
According to the BiotSavart law, the field reverses, so it will
be as in the second diagram. Now rotate the sheet by 180±
about a line that is perpendicular to the sheet. The field,
of course, will rotate with it and end up in the direction
.....
.....
..... .
.....
..... ..
..... ..
shown in the third diagram. The current distribution is now
..
.. ..
.....
.......
exactly as it was originally, so the field must also be as it
was originally. But it is not. Only if the field is parallel
to the sheet will be final direction of the field be the same
as the original direction. If the current is out of the page,
. ..
.....
.....
.....
.......
.. .....
any infinitesimal portion of the sheet in the form of a long
...
......
....
.......
straight wire produces a field that is to the left above the
sheet and to the right below the sheet. The field must be as
drawn in Fig. 29–85.
(b) Integrate the tangential component of the magnetic field
~
B Ã¡
around the rectangular loop shown with dotted lines. The
upper and lower edges are the same distance from the current
sheet and each has length L. This means the field has the
~
¡! B
same magnitude along these edges. It points to the left along
Ã¡ L ¡¡
¡¡
¡!
the upper edge and to the right along the lower.
If the integration is carried out in the counterclockwise sense, the contribution of the upper edge
is BL, the contribution of the lower edge is also BL, and the contribution of each of the sides
H
~ s
is zero because the field is perpendicular to the sides. Thus B ¢ d~ = 2BL. The total current
through the loop is ¸L. Ampere’s law yields 2BL = ¹0 ¸L, so B = ¹0 ¸=2.
81
(a) Use a circular Amperian path that has radius r and is concentric with the cylindrical shell
as shown by the dotted circle on Fig. 29–86. The magnetic field is tangent to the path and
has uniform magnitude on it, so the integral on the left side of the Ampere’s law equation is
H
~ s
B ¢d~ = 2¼rB. The current through the Amperian path is the current through the region outside
the circle of radius b and inside the circle of radius r. Since the current is uniformly distributed
through a cross section of the shell, the enclosed current is i(r2 ¡ b2 )=(a2 ¡ b2 ). Thus
2¼rB = r2 ¡ b2
i
a2 ¡ b2
Chapter 29 183 and
B= ¹0 i
r 2 ¡ b2
:
2¼(a2 ¡ b2 )
r (b) When r = a this expression reduce to B = ¹0 i=2¼r, which is the correct expression for the
field of a long straight wire. When r = b it reduces to B = 0, which is correct since there is no
field inside the shell. When b = 0 it reduces to B = ¹0 ir=2¼a2 , which is correct for the field
inside a cylindrical conductor.
(c) The graph is shown below.
B (T)
10 £ 10¡4 .
.
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..
..
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..
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.
. 8 £ 10¡4
6 £ 10¡4
4 £ 10¡4
2 £ 10¡4
0 0:01 0:02 0:03
r (m) 0:04 0:05 0:06 89
The result of Problem 11 is used four times, once for each of
the sides of the square loop. A point on the axis of the loop
is also on a perpendicular bisector of each of the loop sides.
The diagram shows the field due to one of the loop sides,
the one on the left. In the expression found in Problem 11,
p
p
replace L with a and R with x2 + a2 =4 = 1 4x2 + a2 .
2
The field due to the side is therefore P .....
........
. ...
....
....
.
..
....
....
. .
.
. .
.... .
.... .
.
.....
....
.
......
......
.
. .
.. .
.
.
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..
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..
.
.
.
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..
..
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..
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..
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.
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..
.
..
.
.
..
. ......................................................................
.
. .......................................................................
. .
.
.
.
.
. .
.
..
..
. ..
.
.
. ...
. ..
.
..
.
..
. ..
..
..
. ..
.
..
.
..
...
..
.
..
.
..
..
..
..
..
.... ... ... ... ... ...
.
.
..
.... .... .... .... .... ...
..
..
..
..
..
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.
..
...........................................................................
...........................................................................
.
.
..
....
....
... µ ~
B x R µ ¹0 ia
p
:
B= p
¼ 4x2 + a2 4x2 + 2a2 a=2 The field is in the plane of the dotted triangle shown and
is perpendicular to the line from the midpoint of the loop
side to the point P. Therefore it makes the angle µ with the
vertical. i When the fields of the four sides are summed vectorially the horizontal components add to zero.
The vertical components are all the same, so the total field is given by
Btotal = 4B cos µ =
184 Chapter 29 4Ba
4Ba
=p
:
2R
4x2 + a2 Thus
Btotal = 4¹0 ia2
p
:
¼(4x2 + a2 ) 4x2 + 2a2 For x = 0, the expression reduces to
Btotal p
2 2¹0 i
4¹0 ia2
p =
;
=
¼a
¼a2 2a in agreement with the result of Problem 12.
91 H
~ s
Use Ampere’s law: B ¢d~ = ¹0 ienc , where the integral is around a closed loop and ienc is the net
current through the loop. For the dashed loop shown on the diagram i = 0. Assume the integral
R
~ s
B ¢ d~ is zero along the bottom, right, and top sides of the loop as it would be if the field lines
are as shown on the diagram. Along the right side the field is zero and along the top and bottom
sides the field is perpendicular to d~. If ` is the length of the left edge, then direct integration
s
H
~ ¢ d~ = B`, where B is the magnitude of the field at the left side of the loop. Since
yields B s
neither B nor ` is zero, Ampere’s law is contradicted. We conclude that the geometry shown for
the magnetic field lines is in error. The lines actually bulge outward and their density decreases
gradually, not precipitously as shown. Chapter 29 185 Chapter 30 5
The magnitude of the magnetic field inside the solenoid is B = ¹0 nis , where n is the number
of turns per unit length and is is the current. The field is parallel to the solenoid axis, so the
2
2
flux through a cross section of the solenoid is ©B = As B = ¹0 ¼rs nis , where As (= ¼rs ) is the
crosssectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this is
also the flux through the coil. The emf in the coil has magnitude
E= N d©B
dis
2
= ¹0 ¼rs N n
dt
dt and the current in the coil is 2
E ¹0 ¼rs N n dis
=
;
R
R
dt
where N is the number of turns in the coil and R is the resistance of the coil. The current
changes linearly by 3:0 A in 50 ms, so dis =dt = (3:0 A)=(50 £ 10¡3 s) = 60 A=s. Thus ic = ic = (4¼ £ 10¡7 T ¢ m=A)¼(0:016 m)2 (120)(220 £ 102 m¡1 )
(60 A=s) = 3:0 £ 10¡2 A :
5:3 21
(a) In the region of the smaller loop, the magnetic field produced by the larger loop may be taken
to be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 29–26,
with z = x and much greater than R, gives
B= ¹0 iR2
2x3 for the magnitude. The field is upward in the diagram. The magnetic flux through the smaller
loop is the product of this field and the area (¼r2 ) of the smaller loop:
©B = ¼¹0 ir2 R2
:
2x3 (b) The emf is given by Faraday’s law:
µ
¶
¶µ
µ
µ ¶
¶
3 dx
3¼¹0 ir2 R2 v
d©B
¼¹0 ir2 R2
d
¼¹0 ir 2 R2
1
=¡
¡ 4
=¡
=
E=¡
:
2
2
2x4
dt
dt x3
x dt
(c) The field of the larger loop is upward and decreases with distance away from the loop. As
the smaller loop moves away, the flux through it decreases. The induced current is directed so
as to produce a magnetic field that is upward through the smaller loop, in the same direction as
186 Chapter 30 the field of the larger loop. It is counterclockwise as viewed from above, in the same direction
as the current in the larger loop.
29
Thermal energy is generated at the rate E 2 =R, where E is the emf in the wire and R is the
resistance of the wire. The resistance is given by R = ½L=A, where ½ is the resistivity of copper,
L is the length of the wire, and A is the crosssectional area of the wire. The resistivity can be
found in Table 26–1. Thus
R= ½L (1:69 £ 10¡8 ¢ m)(0:500 m)
=
= 1:076 £ 10¡2 :
A
¼(0:500 £ 10¡3 m)2 Faraday’s law is used to find the emf. If B is the magnitude of the magnetic field through the
loop, then E = A dB=dt, where A is the area of the loop. The radius r of the loop is r = L=2¼
and its area is ¼r 2 = ¼L2 =4¼ 2 = L2 =4¼. Thus
E= L2 dB (0:500 m)2
(10:0 £ 10¡3 T=s) = 1:989 £ 10¡4 V :
=
4¼ dt
4¼ The rate of thermal energy generation is
P = E 2 (1:989 £ 10¡4 V)2
=
= 3:68 £ 10¡6 W :
¡2
1:076 £ 10
R 37
(a) The field point is inside the solenoid, so Eq. 30–25 applies. The magnitude of the induced
electric field is
E= 1 dB
1
r = (6:5 £ 10¡3 T=s)(0:0220 m) = 7:15 £ 10¡5 V=m :
2 dt
2 (b) Now the field point is outside the solenoid and Eq. 30–27 applies. The magnitude of the
induced field is
E= (0:0600 m)2
1 dB R2 1
= 1:43 £ 10¡4 V=m :
= (6:5 £ 10¡3 T=s)
2 dt r
2
(0:0820 m) 51
Starting with zero current when the switch is closed, at time t = 0, the current in an RL series
circuit at a later time t is given by
´
E ³
1 ¡ e¡t=¿L ;
i=
R where ¿L is the inductive time constant, E is the emf, and R is the resistance. You want to
calculate the time t for which i = 0:9990E=R. This means
´
E ³
E
1 ¡ e¡t=¿L ;
0:9990 =
R R
Chapter 30 187 so
0:9990 = 1 ¡ e¡t=¿L or e¡t=¿L = 0:0010 :
Take the natural logarithm of both sides to obtain ¡(t=¿L ) = ln(0:0010) = ¡6:91. That is, 6:91
inductive time constants must elapse.
55
(a) If the battery is switched into the circuit at time t = 0, then the current at a later time t is
given by
´
E ³
1 ¡ e¡t=¿L ;
i=
R
where ¿L = L=R. You want to find the time for which i = 0:800E=R. This means
0:800 = 1 ¡ e¡t=¿L
or
e¡t=¿L = 0:200 :
Take the natural logarithm of both sides to obtain ¡(t=¿L ) = ln(0:200) = ¡1:609. Thus
t = 1:609¿L = 1:609L 1:609(6:30 £ 10¡6 H)
=
= 8:45 £ 10¡9 s :
1:20 £ 103
R (b) At t = 1:0¿L the current in the circuit is
¶
µ
¢
¢
¡
E ¡
14:0 V
¡1:0
1¡e
=
i=
1 ¡ e¡1:0 = 7:37 £ 10¡3 A :
3
1:20 £ 10
R
59
(a) Assume i is from left to right through the closed switch. Let i1 be the current in the resistor
and take it to be downward. Let i2 be the current in the inductor and also take it to be downward.
The junction rule gives i = i1 + i2 and the loop rule gives i1 R ¡ L(di2 =dt) = 0. Since di=dt = 0,
the junction rule yields (di1 =dt) = ¡(di2 =dt). Substitute into the loop equation to obtain
L di1
+ i1 R = 0 :
dt This equation is similar to Eq. 30–44, and its solution is the function given as Eq. 30–45:
i1 = i0 e¡Rt=L ;
where i0 is the current through the resistor at t = 0, just after the switch is closed. Now, just
after the switch is closed, the inductor prevents the rapid buildup of current in its branch, so at
that time, i2 = 0 and i1 = i. Thus i0 = i, so
i1 = ie¡Rt=L
188 Chapter 30 and h ¡Rt=L i2 = i ¡ i1 = i 1 ¡ e
(b) When i2 = i1 , i : e¡Rt=L = 1 ¡ e¡Rt=L ;
so 1
:
2
Take the natural logarithm of both sides and use ln(1=2) = ¡ ln 2 to obtain (Rt=L) = ln 2 or
e¡Rt=L = t= L
ln 2 :
R 63
(a) If the battery is applied at time t = 0, the current is given by
i= ´
E ³
1 ¡ e¡t=¿L ;
R where E is the emf of the battery, R is the resistance, and ¿L is the inductive time constant. In
terms of R and the inductance L, ¿L = L=R. Solve the current equation for the time constant.
First obtain
iR
;
e¡t=¿L = 1 ¡
E
then take the natural logarithm of both sides to obtain
¸
∙
t
iR
:
¡ = ln 1 ¡
E
¿L
Since ¸
¸
∙
(2:00 £ 10¡3 A)(10:0 £ 103 )
iR
= ¡0:5108 ;
= ln 1 ¡
ln 1 ¡
50:0 V
E
∙ the inductive time constant is ¿L = t=0:5108 = (5:00 £ 10¡3 s)=(0:5108) = 9:79 £ 10¡3 s and the
inductance is
L = ¿L R = (9:79 £ 10¡3 s)(10:0 £ 103 ) = 97:9 H :
(b) The energy stored in the coil is
1
1
UB = Li2 = (97:9 H)(2:00 £ 10¡3 A)2 = 1:96 £ 10¡4 J :
2
2
69
(a) At any point, the magnetic energy density is given by uB = B 2 =2¹0 , where B is the magnitude
of the magnetic field at that point. Inside a solenoid, B = ¹0 ni, where n is the number of turns
Chapter 30 189 per unit length and i is the current. For the solenoid of this problem, n = (950)=(0:850 m) =
1:118 £ 103 m¡1 . The magnetic energy density is
1
1
uB = ¹0 n2 i2 = (4¼ £ 10¡7 T ¢ m=A)(1:118 £ 103 m¡1 )2 (6:60 A)2 = 34:2 J=m3 :
2
2 (b) Since the magnetic field is uniform inside an ideal solenoid, the total energy stored in the
field is UB = uB V , where V is the volume of the solenoid. V is calculated as the product of the
crosssectional area and the length. Thus
UB = (34:2 J=m3 )(17:0 £ 10¡4 m2 )(0:850 m) = 4:94 £ 10¡2 J :
73
(a) The mutual inductance M is given by
E1 = M di2
;
dt where E1 is the emf in coil 1 due to the changing current i2 in coil 2. Thus
M= 25:0 £ 10¡3 V
E1
= 1:67 £ 10¡3 H :
=
15:0 A=s
di2 =dt (b) The flux linkage in coil 2 is
N2 ©21 = M i1 = (1:67 £ 10¡3 H)(3:60 A) = 6:01 £ 10¡3 Wb :
75
(a) Assume the current is changing at the rate di=dt and calculate the total emf across both coils.
First consider the lefthand coil. The magnetic field due to the current in that coil points to the
left. So does the magnetic field due to the current in coil 2. When the current increases, both
fields increase and both changes in flux contribute emf’s in the same direction. Thus the emf in
coil 1 is
di
E1 = ¡ (L1 + M )
:
dt
The magnetic field in coil 2 due to the current in that coil points to the left, as does the field in
coil 2 due to the current in coil 1. The two sources of emf are again in the same direction and
the emf in coil 2 is
di
:
E2 = ¡ (L2 + M )
dt
The total emf across both coils is
E = E1 + E2 = ¡ (L1 + L2 + 2M ) di
:
dt This is exactly the emf that would be produced if the coils were replaced by a single coil with
inductance Leq = L1 + L2 + 2M .
190 Chapter 30 (b) Reverse the leads of coil 2 so the current enters at the back of the coil rather than the front
as pictured in the diagram. Then the field produced by coil 2 at the site of coil 1 is opposite the
field produced by coil 1 itself. The fluxes have opposite signs. An increasing current in coil 1
tends to increase the flux in that coil but an increasing current in coil 2 tends to decrease it. The
emf across coil 1 is
di
E1 = ¡ (L1 ¡ M )
:
dt
Similarly the emf across coil 2 is
E2 = ¡ (L2 ¡ M ) di
:
dt The total emf across both coils is
di
:
dt
This the same as the emf that would be produced by a single coil with inductance Leq = L1 +
L2 ¡ 2M .
E = ¡ (L1 + L2 ¡ 2M ) 79
(a) The electric field lines are circles that are concentric with the cylindrical region and the
magnitude of the field is uniform around any circle. Thus the emf around a circle of radius r
H
~ s
is E = E ¢ d~ = 2¼rE. Here r is inside the cylindrical region so the magnetic flux is ¼r2 B.
According to Faraday’s law 2¼rE = ¡¼r2 (dB=dt) and
dB
= ¡ 1 (0:050 m)(¡10 £ 10¡3 T=s) = 2:5 £ 10¡4 V=m :
2
dt
Since the normal used to compute the flux was taken to be into the page, in the direction of the
magnetic field, the positive direction for the electric is clockwise. The calculated value of E is
~
positive, so the electric field at point a is toward the left and E = ¡(2:5 £ 10¡4 V=m) ˆ.
i
~
~
The force on the electron is F = ¡eE and, according to Newton’s second law, its acceleration is
E = ¡1r
2 ~=
a ~
~
eE
(1:60 £ 10¡19 C)(¡2:5 £ 10¡4 V=m) ˆ
F
i
2
= (4:4 £ 107 m=s ) ˆ :
=¡
=¡
i
9:11 £ 10¡31 kg
m
m The mass and charge of an electron can be found in Appendix B.
(b) The electric field at r = 0 is zero, so the force and acceleration of an election placed at point
b are zero.
(c) The electric field at point c has the same magnitude as the field at point a but now the field
2
~
is to the right. That is E = (2:5 £ 10¡4 V=m) ˆ and ~ = ¡(4:4 £ 107 m=s ) ˆ.
i
a
i
81
(a) The magnetic flux through the loop is ©B = BA, where B is the magnitude of the magnetic
field and A is the area of the loop. The magnitude of the average emf is given by Faraday’s law
: Eavg = B¢A=¢t, where ¢A is the change in the area in time ¢t. Since the final area is zero,
the change in area is the initial area and Eavg = BA=¢t = (2:0 T)(0:20 m)2 =(0:20 s) = 0:40 V.
Chapter 30 191 (b) The average current in the loop is the emf divided by the resistance of the loop: iavg =
Eavg =R = (0:40 V)=(20 £ 10¡3 ) = 20 A.
85
(a), (b), (c), (d), and (e) Just after the switch is closed the current i2 through the inductor is zero.
The loop rule applied to the left loop gives E ¡ I1 R1 = 0, so i1 = E=R1 = (10 V)=(5:0 ) = 2:0 A.
The junction rule gives is = i1 = 2:0 A. Since i2 = 0, the potential difference across R2 is
V2 = i2 R2 = 0. The potential differences across the inductor and resistor must sum to E and, since
V2 = 0, VL = E = 10 V. The rate of change of i2 is di2 =dt = VL =L = (10 V)=(5:0 H) = 2:0 A=s.
(g), (h), (i), (j), (k), and (l) After the switch has been closed for a long time the current i2
reaches a constant value. Since its derivative is zero the potential difference across the inductor
is VL = 0. The potential differences across both R1 and R2 are equal to the emf of the battery,
so i1 = E=R1 = (10 V)=(5:0 ) = 2:0 A and i2 = E=R2 = (10 V)=(10 ) = 1:0 A. The junction
rule gives is = i1 + i2 = 3:0 A.
95
(a) Because the inductor is in series with the battery the current in the circuit builds slowly and
just after the switch is closed it is zero.
(b) Since all currents are zero just after the switch is closed the emf of the inductor must match the
emf of the battery in magnitude. Thus L(dibat =dt) = E and dibat = E=L = (40 V)=(50 £ 10¡3 H) =
8:0 £ 102 A=s.
(c) Replace the two resistors in parallel with their equivalent resistor. The equivalent resistance
is
R1 R2
(20 k)(20 k)
= 10 k :
Req =
=
20 k + 20 k
R1 + R2
The current as a function of time is given by
i
E h
ibat =
1 ¡ e¡t=¿L ;
Req
where ¿L is the inductive time constant. Its value is ¿L = L=Req = (50 £ 10¡3 H)=(10 £ 103 ) =
5:0 £ 10¡6 s. At t = 3:0 £ 10¡6 s, t=¿L = (3:0)=(5:0) = 0:60 and ¤
£
40 V
1 ¡ e¡0:60 = 1:8 £ 10¡3 A :
3
10 £ 10
(d) Differentiate the expression for ibat to obtain
ibat = E 1 ¡t=¿L E ¡t=¿L
dibat
=
= e
e
;
dt
Req ¿L
L
where ¿L = L=Req was used to obtain the last form. At t = 3:0 £ 10¡6 s 40 V
dibat
e¡0:60 = 4:4 £ 102 A=s :
=
50 £ 10¡3 H
dt
(e) A long time after the switch is closed the currents are constant and the emf of the inductor is
zero. The current in the battery is ibat = E=Req = (40 V)=(10 £ 103 ) = 4:0 £ 10¡3 A. 192 Chapter 30 (f) The currents are constant and dibat =dt = 0.
97
(a) and (b) Take clockwise current to be positive and counterclockwise current to be negative.
Then according to the righthand rule we must take the normal to the loop to be into the page,
so the flux is negative if the magnetic field is out of the page and positive if it is into the page.
Assume the field in region 1 is out of the page. We will obtain a negative result for the field
if the assumption is incorrect. Let x be the distance that the front edge of the loop is into
region 1. Then while the loop is entering this region flux is ¡B1 Hx and, according to Faraday’s
law, the emf induced around the loop is E = B1 H(dx=dt) = B1 Hv. The current in the loop is
i = E=R = B1 Hv=R, so
B1 = iR (3:0 £ 10¡6 A)(0:020 )
= 1:0 £ 10¡5 T :
=
(0:0150 m)(0:40 m=s)
Hv The field is positive and therefore out of the page.
(c) and (d) Assume that the field B2 of region 2 is out of the page. Let x now be the distance the
front end of the loop is into region 2 as the loop enters that region. The flux is ¡B1 H(D ¡ x) ¡
B2 Hx, the emf is E = ¡B1 Hv + B2 Hv = (B2 ¡ B1 )Hv, and the current is i = (B2 ¡ B1 )Hv=R.
The field of region 2 is
B2 = B1 + iR
(¡2:0 £ 10¡6 A(0:020 )
= 3:3 £ 10¡6 T :
= 1:0 £ 10¡5 T +
(0:015 m)(0:40 m=s)
Hv The field is positive, indicating that it is out of the page. Chapter 30 193 Chapter 31 7
(a) The mass m corresponds to the inductance, so m = 1:25 kg.
(b) The spring constant k corresponds to the reciprocal of the capacitance. Since the total energy is
given by U = Q2 =2C, where Q is the maximum charge on the capacitor and C is the capacitance,
¢2
¡
175 £ 10¡6 C
Q2
=
= 2:69 £ 10¡3 F
C=
2U 2(5:70 £ 10¡6 J)
and
k= 1
= 372 N=m :
2:69 £ 10¡3 m=N (c) The maximum displacement xm corresponds to the maximum charge, so
xm = 1:75 £ 10¡4 m :
(d) The maximum speed vm corresponds to the maximum current. The maximum current is
175 £ 10¡6 C
Q
= 3:02 £ 10¡3 A :
=p
I = Q! = p
LC
(1:25 H)(2:69 £ 10¡3 F) Thus vm = 3:02 £ 10¡3 m=s.
15 (a) Since p frequency of oscillation f is related to the inductance L and capacitance C by
the
p
f = 1=2¼ LC, the smaller value of C gives the larger value of f . Hence, fmax = 1=2¼ LCmin ,
p
fmin = 1=2¼ LCmax, and
p
p
fmax
365 pF
Cmax
= 6:0 :
= p
= p
fmin
10 pF
Cmin
(b) You want to choose the additional capacitance C so the ratio of the frequencies is
r= 1:60 MHz
= 2:96 :
0:54 MHz Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that
of the tuning capacitor. If C is in picofarads, then
p
C + 365 pF
p
= 2:96 :
C + 10 pF
194 Chapter 31 The solution for C is
C= (365 pF) ¡ (2:96)2 (10 pF)
= 36 pF :
(2:96)2 ¡ 1 p
(c) Solve f = 1=2¼ LC for L. For the minimum frequency, C = 365 pF + 36 pF = 401 pF and
f = 0:54 MHz. Thus
1
1
L=
=
= 2:2 £ 10¡4 H :
2 Cf 2
2 (401 £ 10¡12 F)(0:54 £ 106 Hz)2
(2¼)
(2¼)
27
Let t be a time at which the capacitor is fully charged in some cycle and let qmax 1 be the charge
on the capacitor then. The energy in the capacitor at that time is
U (t) = 2
qmax 1 Q2 ¡Rt=L
=
e
;
2C
2C where
qmax 1 = Q e¡Rt=2L
was used. Here Q is the charge at t = 0. One cycle later, the maximum charge is
qmax 2 = Q e¡R(t+T )=2L
and the energy is 2
qmax 2 Q2 ¡R(t+T )=L
=
e
;
2C
2C
where T is the period of oscillation. The fractional loss in energy is U (t + T ) = ¢U U (t) ¡ U (t + T ) e¡Rt=L ¡ e¡R(t+T )=L
=
=
= 1 ¡ e¡RT =L :
U
U (t)
e¡Rt=L
Assume that RT =L is small compared to 1 (the resistance is small) and use the Maclaurin series
to expand the exponential. The first two terms are:
RT
e¡RT =L ¼ 1 ¡
:
L
Replace T with 2¼=!, where ! is the angular frequency of oscillation. Thus
µ
¶
RT
RT 2¼R
¢U
¼1¡ 1¡
=
=
:
U
L
L
!L
33
(a) The generator emf is a maximum when sin(!d t¡¼=4) = 1 or !d t¡¼=4 = (¼=2)§2n¼, where
n is an integer, including zero. The first time this occurs after t = 0 is when !d t ¡ ¼=4 = ¼=2 or
3¼
3¼
= 6:73 £ 10¡3 s :
t=
=
4!d 4(350 s¡1 )
(b) The current is a maximum when sin(!d t ¡ 3¼=4) = 1, or !d t ¡ 3¼=4 = ¼=2 § 2n¼. The first
time this occurs after t = 0 is when
5¼
5¼
= 1:12 £ 10¡2 s :
=
t=
¡1 )
4!d 4(350 s
(c) The current lags the inductor by ¼=2 rad, so the circuit element must be an inductor.
Chapter 31 195 (d) The current amplitude I is related to the voltage amplitude VL by VL = IXL , where XL is
the inductive reactance, given by XL = !d L. Furthermore, since there is only one element in
the circuit, the amplitude of the potential difference across the element must be the same as the
amplitude of the generator emf: VL = Em . Thus Em = I!d L and
L= 30:0 V
Em
= 0:138 H :
=
¡3 A)(350 rad=s)
I!d (620 £ 10 39
(a) The capacitive reactance is
XC = 1
1
1
= 37:9 :
=
=
!d C 2¼fd C 2¼(60:0 Hz)(70:0 £ 10¡6 F) The inductive reactance is XL = !d L = 1¼fd L = 2¼(60:0 Hz)(230 £ 10¡3 H) = 86:7 :
The impedance is
Z= p R2 + (XL ¡ XC )2 = (b) The phase angle is
Á = tan ¡1 µ XL ¡ XC
R (c) The current amplitude is
I= p
(200 )2 + (37:9 ¡ 86:7 )2 = 206 : ¶ = tan ¡1 µ 86:7 ¡ 37:9
200 ¶ = 13:7± : Em 36:0 V
= 0:175 A :
=
206
Z (d) The voltage amplitudes are
VR = IR = (0:175 A)(200 ) = 35:0 V,
VL = IXL = (0:i75 A)(86:7 ) = 15:2 V, Em ...
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.
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.
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. VL VR VL ¡ VC and
VC = IXC = (0:175 A)(37:9 ) = 6; 63 V.
Note that XL > XC , so that Em leads I. The phasor
diagram is drawn to scale on the right. Á VC 45
(a) For a given amplitude Em of the generator emf, the current amplitude is given by
I=
196 Chapter 31 Em
Em
=p
;
2 + (! L ¡ 1=! C)2
Z
R
d
d where R is the resistance, L is the inductance, C is the capacitance, and !d is the angular
frequency. To find the maximum, set the derivative with respect to !d equal to zero and solve
for !d . The derivative is
∙
¸
¸∙
£ 2
¤
1
1
dI
2 ¡3=2
!d L ¡
:
L+ 2
= ¡Em R + (!d L ¡ 1=!d C)
d!d
!d C
!d C
p
The only factor that can equal zero is !d L ¡ (1=!d C) and it does for !d = 1= LC. For the
given circuit,
1
1
= 224 rad=s :
=p
!d = p
LC
(1:00 H)(20:0 £ 10¡6 F)
(b) For this value of the angular frequency, the impedance is Z = R and the current amplitude is
I= Em 30:0 V
= 6:00 A :
=
5:00
R (c) and (d) You want to find the values of !d for which I = Em =2R. This means
Em p R2 + (!d L ¡ 1=!d C)2 = Em
:
2R Cancel the factors Em that appear on both sides, square both sides, and set the reciprocals of the
two sides equal to each other to obtain
¶2
µ
1
2
= 4R2 :
R + !d L ¡
!d C
Thus µ ¶2
1
!d L ¡
= 3R2 :
!d C
Now take the square root of both sides and multiply by !d C to obtain
´
³p
2
3CR ¡ 1 = 0 ;
!d (LC) § !d where the symbol § indicates the two possible signs for the square root. The last equation is a
quadratic equation for !d . Its solutions are
p
p
§ 3CR § 3C 2 R2 + 4LC
:
!d =
2LC
You want the two positive solutions. The smaller of these is
p
p
¡ 3CR + 3C 2 R2 + 4LC
!2 =
2LC
p
¡ 3(20:0 £ 10¡6 F)(5:00 )
=
2(1:00 H)(20:0 £ 10¡6 F)
p
3(20:0 £ 10¡6 F)2 (5:00 )2 + 4(1:00 H)(20:0 £ 10¡6 F)
+
2(1:00 H)(20:0 £ 10¡6 F)
= 219 rad=s
Chapter 31 197 and the larger is
p
p
+ 3CR + 3C 2 R2 + 4LC
!1 =
2LC
p
+ 3(20:0 £ 10¡6 F)(5:00 )
=
2(1:00 H)(20:0 £ 10¡6 F)
p
3(20:0 £ 10¡6 F)2 (5:00 )2 + 4(1:00 H)(20:0 £ 10¡6 F)
+
2(1:00 H)(20:0 £ 10¡6 F)
= 228 rad=s :
(e) The fractional width is
!1 ¡ !2 228 rad=s ¡ 219 rad=s
= 0:04 :
=
224 rad=s
!0
49
Use the expressions found in Problem 31–45:
p
p
+ 3CR + 3C 2 R2 + 4LC
!1 =
2LC
and p
p
¡ 3CR + 3C 2 R2 + 4LC
:
!2 =
2LC Also use Thus 1
:
!=p
LC
r
p
p
¢!d !1 ¡ !2 2 3CR LC
3C
=R
=
=
:
2LC
!
!
L 55
(a) The impedance is given by
Z= p
R2 + (XL ¡ XC )2 ; where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.
Thus
p
Z = (12:0 )2 + (1:30 ¡ 0)2 = 12:1 :
(b) The average rate at which energy is supplied to the air conditioner is given by
Pavg =
198 Chapter 31 2
Erms
cos Á ;
Z where cos Á is the power factor. Now
cos Á =
so
Pavg 12
R
= 0:992 ;
=
Z 12:1 ¸
(120 V)2
(0:992) = 1:18 £ 103 W :
=
12:1
∙ 57
(a) The power factor is cos Á, where Á is the phase angle when the current is written i =
I sin(!d t ¡ Á). Thus Á = ¡42:0± and cos Á = cos(¡42:0± ) = 0:743.
(b) Since Á < 0, !d t ¡ Á > !d t and the current leads the emf.
(c) The phase angle is given by tan Á = (XL ¡ XC )=R, where XL is the inductive reactance,
XC is the capacitive reactance, and R is the resistance. Now tan Á = tan(¡42:0± ) = ¡0:900, a
negative number. This means XL ¡ XC is negative, or XC > XL . The circuit in the box is
predominantly capacitive.
(d) If the circuit is in resonance, XL is the same as XC , tan Á is zero, and Á would be zero.
Since Á is not zero, we conclude the circuit is not in resonance.
(e), (f), and (g) Since tan Á is negative and finite, neither the capacitive reactance nor the resistance
is zero. This means the box must contain a capacitor and a resistor. The inductive reactance may
be zero, so there need not be an inductor. If there is an inductor, its reactance must be less than
that of the capacitor at the operating frequency.
(h) The average power is
1
1
Pavg = Em I cos Á = (75:0 V)(1:20 A)(0:743) = 33:4 W :
2
2
(i) The answers above depend on the frequency only through the phase angle Á, which is given.
If values are given for R, L, and C, then the value of the frequency would also be needed to
compute the power factor.
63
(a) If Np is the number of primary turns and Ns is the number of secondary turns, then
¶
µ
Ns
10
(120 V) = 2:4 V :
Vp =
Vs =
500
Np
(b) and (c) The current in the secondary is given by Ohm’s law:
Is = Vs 2:4 V
= 0:16 A :
=
15
Rs The current in the primary is
Ns
Ip =
Is =
Np µ ¶
10
(0:16 A) = 3:2 £ 10¡3 A :
500
Chapter 31 199 67
Use the trigonometric identity, found in Appendix E,
¶
µ
¶
µ
®+¯
®¡¯
cos
;
sin ® ¡ sin ¯ = 2 sin
2
2
where ® and ¯ are any two angles. Thus
V1 ¡ V2 = A sin(!d t) ¡ A sin(!d t ¡ 120± ) = 2A sin(120± ) cos(!d t ¡ 60± ) =
p
where sin(120± ) = 3=2 was used. Similarly, p
3A cos(!d t ¡ 60± ) ; p
V1 ¡V3 = A sin(!d t)¡A sin(!d t¡240± ) = 2A sin(240± ) cos(!d t¡120± ) = ¡ 3A cos(!d t¡120± ) ;
p
where sin(240± ) = ¡ 3=2 was used, and
V2 ¡ V3 = A sin(!d t ¡ 120± ) ¡ A sin(!d t ¡ 240± ) = 2A sin(120± ) cos(!d t ¡ 180± )
p
= 3A cos(!d t ¡ 180± ) :
All of these are sinusoidal functions of !d and all have amplitudes of p 3A. 71
(a) Let VC be the maximum potential difference across the capacitor, VL be the maximum potential
difference across the inductor, and VR be the maximum potential difference across the resistor.
Then the phase constant Á is
µ
¶
µ
¶
¡1 VL ¡ VC
¡1 2:00VR ¡ VR
= tan
= tan¡1 (1:00) = 45:0± :
tan
VR
VR
(b) Since the maximum emf is related to the current amplitude by Em = IZ, where Z is the
impedance and R = Z cos Á,
R= Em cos Á (30:0 V) cos 45±
= 70:7 :
=
300 £ 10¡3 A
I 73 p
(a) The frequency of oscillation of an LC circuit is f = 1=2¼ LC, where L is the inductance
and C is the capacitance. Thus
L= 1
1
= 6:89 £ 10¡7 H :
= 2
4¼ 2 f 2 C 4¼ (10:4 £ 103 Hz)2 (340 £ 10¡6 F) (b) The total energy is U = 1 LI 2 , where I is the current amplitude. Thus U =
2
10¡7 H)(7:20 £ 10¡3 A)2 = 1:79 £ 10¡11 J.
200 Chapter 31 1
2 (6:89 £ (c) The total energy is also given by U = Q2 =2C, where Q is the charge amplitude. Thus
p
p
Q = 2U C = 2(1:79 £ 10¡11 J)(340 £ 10¡6 F) = 1:10 £ 10¡7 C. 83
(a) The total energy U of the circuit is the sum of the energy UE stored in the capacitor and
the energy UB stored in the inductor at the same time. Since UB = 2:00UE , the total energy is
U = 3:00UE . Now U = Q2 =2C and UE = q 2 =2C, where Q is the maximum charge, q is the
charge when the magnetic energyp twice the electrical energy, and C is the capacitance. Thus
is
2
2
Q =2C = 3:00q =2C and q = Q= 3:00 = 0:577Q.
(b) If the capacitor has maximum charge at time t = 0, then q = Q cos(!t), where ! is the angular
frequency of oscillation. This means !t = cos¡1 (0:577) = 0:964 rad. Since ! = 2¼=T , where T
is the period,
0:964
T = 0:153T :
t=
2¼
85
(a) The energy stored in a capacitor is given by UE = q 2 =2C, where q is the charge and C is the
capacitance. Now q 2 is periodic with a period of T =2, where T is the period of the driving emf,
so UE has the same value at the beginning and end of each cycle. Actually UE has the same
value at the beginning and end of each half cycle.
(b) The energy stored in an inductor is given by Li2 =2, where i is the current and L is the
inductance. The square of the current is periodic with a period of T =2, so it has the same value
at the beginning and end of each cycle.
(c) The rate with which the driving emf device supplies energy is
PE = iE = IEm sin(!d t) sin(!d t ¡ Á) ; where I is the current amplitude, Em is the emf amplitude, ! is the angular frequency, and Á is
a phase constant. The energy supplied over a cycle is
Z T
Z T
EE =
sin(!d t) sin(!d t ¡ Á) dt
PE dt = IEm
0
0
Z T
sin(!d t) [sin(!d t)cos(Á) ¡ cos(!d t) sin(Á)] dt ;
= IEm
0 where the trigonometric identity sin(® ¡ ¯) = sin ® cos ¯ ¡ cos ® sin ¯ was used. Now the integral
of sin2 (!d t) over a cycle is T =2 and the integral of sin(!d t) cos(!d t) over a cycle is zero, so
EE = 1 IEm cos Á.
2
(d) The rate of energy dissipation in a resistor is given by
PR = i2 R = I 2 sin2 (!d t ¡ Á) and the energy dissipated over a cycle is
Z T
2
sin2 (!d t ¡ Á) dt = 1 I 2 RT :
ER = I
2
0 (e) Now Em = IZ, where Z is the impedance, and R = Z cos Á, so EE =
1 2
2 I RT = ER . 1 2
2I TZ Chapter 31 cos Á =
201 Chapter 32 3 H
H
~
~
~
~
(a) Use Gauss’ law for magnetism: B ¢ dA = 0. Write B ¢ dA = ©1 + ©2 + ©C , where ©1
is the magnetic flux through the first end mentioned, ©2 is the magnetic flux through the second
end mentioned, and ©C is the magnetic flux through the curved surface. Over the first end, the
magnetic field is inward, so the flux is ©1 = ¡25:0 ¹Wb. Over the second end, the magnetic
field is uniform, normal to the surface, and outward, so the flux is ©2 = AB = ¼r2 B, where A
is the area of the end and r is the radius of the cylinder. Its value is
©2 = ¼(0:120 m)2 (1:60 £ 10¡3 T) = +7:24 £ 10¡5 Wb = +72:4 ¹Wb :
Since the three fluxes must sum to zero,
©C = ¡©1 ¡ ©2 = 25:0 ¹Wb ¡ 72:4 ¹Wb = ¡47:4 ¹Wb :
(b) The minus sign indicates that the flux is inward through the curved surface.
5
Consider a circle of radius r (= 6:0 mm), between the plates and with its center on the axis of
the capacitor. The current through this circle is zero, so the AmpereMaxwell law becomes
I
d©E
~ s
B ¢ d~ = ¹0 ²0
;
dt
~
where B is the magnetic field at points on the circle and ©E is the electric flux through the
H
~ s
circle. The magnetic field is tangent to the circle at all points on it, so B ¢ d~ = 2¼rB. The
2
electric flux through the circle is ©E = ¼R E, where R (= 3:0 mm) is the radius of a capacitor
plate. When these substitutions are made, the AmpereMaxwell law becomes
2¼rB = ¹0 ²0 ¼R2 dE
:
dt Thus
2rB
dE
2(6:0 £ 10¡3 m)(2:0 £ 10¡7 T)
=
=
= 2:4 £ 1013 V=m ¢ s :
dt
¹0 ²0 R2 (4¼ £ 10¡7 H=m)(8:85 £ 10¡12 Fm)(3:0 £ 10¡3 m)2
13
The displacement current is given by
id = ²0 A
202 Chapter 32 dE
;
dt where A is the area of a plate and E is the magnitude of the electric field between the plates.
The field between the plates is uniform, so E = V =d, where V is the potential difference across
the plates and d is the plate separation. Thus
id = ²0 A dV
:
d dt Now ²0 A=d is the capacitance C of a parallelplate capacitor without a dielectric, so
id = C dV
:
dt 21
(a) For a parallelplate capacitor, the charge q on the positive plate is given by q = (²0 A=d)V ,
where A is the plate area, d is the plate separation, and V is the potential difference between the
plates. In terms of the electric field E between the plates, V = Ed, so q = ²0 AE = ²0 ©E , where
©E is the total electric flux through the region between the plates. The true current into the
positive plate is i = dq=dt = ²0 d©E =dt = id total , where id total is the total displacement current
between the plates. Thus id total = 2:0 A.
(b) Since id total = ²0 d©E =dt = ²0 A dE=dt,
2:0 A
dE id total
=
=
= 2:3 £ 1011 V=m ¢ s :
¡12 F=m)(1:0 m)2
(8:85 £ 10
dt
²0 A (c) The displacement current is uniformly distributed over the area. If a is the area enclosed by
the dashed lines and A is the area of a plate, then the displacement current through the dashed
path is
a
(0:50 m)2
id enc = id total =
(2:0 A) = 0:50 A :
(1:0 m)2
A
(d) According to Maxwell’s law of induction,
I
~ s
B ¢ d~ = ¹0 id enc = (4¼ £ 10¡7 H=m)(0:50 A) = 6:3 £ 10¡7 T ¢ m :
Notice that the integral is around the dashed path and the displacement current on the right
side of the Maxwell’s law equation is the displacement current through that path, not the total
displacement current. 35
(a) The z component of the orbital angular momentum is given by Lorb, z = m` h=2¼, where h is
the Planck constant. Since m` = 0, Lorb, z = 0.
(b) The z component of the orbital contribution to the magnetic dipole moment is given by
¹orb, z = ¡m` ¹B , where ¹B is the Bohr magneton. Since m` = 0, ¹orb, z = 0.
(c) The potential energy associated with the orbital contribution to the magnetic dipole moment
is given by U = ¡¹orb, z Bext , where Bext is the z component of the external magnetic field. Since
¹orb, z = 0, U = 0.
Chapter 32 203 (d) The z component of the spin magnetic dipole moment is either +¹B or ¡¹B , so the potential
energy is either
U = ¡¹B Bext = ¡(9:27 £ 10¡24 J=T)(35 £ 10¡3 T) = ¡3:2 £ 10¡25 J :
or U = +3:2 £ 10¡25 J. (e) Substitute m` into the equations given above. The z component of the orbital angular
momentum is
Lorb, z = m` h (¡3)(6:626 £ 10¡34 J ¢ s)
=
= ¡3:2 £ 10¡34 J ¢ s :
2¼
2¼ (f) The z component of the orbital contribution to the magnetic dipole moment is
¹orb, z = ¡m` ¹B = ¡(¡3)(9:27 £ 10¡24 J=T) = 2:8 £ 10¡23 J=T :
(g) The potential energy associated with the orbital contribution to the magnetic dipole moment
is
U = ¡¹orb, z Bext = ¡(2:78 £ 10¡23 J=T)(35 £ 10¡3 T) = ¡9:7 £ 10¡25 J :
(h) The potential energy associated with spin does not depend on m` . It is §3:2 £ 10¡25 J.
39
The magnetization is the dipole moment per unit volume, so the dipole moment is given by
¹ = M V , where M is the magnetization and V is the volume of the cylinder. Use V = ¼r 2 L,
where r is the radius of the cylinder and L is its length. Thus
¹ = M ¼r 2 L = (5:30 £ 103 A=m)¼(0:500 £ 10¡2 m)2 (5:00 £ 10¡2 m) = 2:08 £ 10¡2 J=T :
45
(a) The number of atoms per unit volume in states with the dipole moment aligned with the
magnetic field is N+ = Ae¹B=kT and the number per unit volume in states with the dipole
moment antialigned is N¡ = Ae¡¹B=kT , where A is a constant of proportionality. The total
¡
¢
number of atoms per unit volume is N = N+ + N¡ = A e¹B=kT + e¡¹B=kT . Thus
A= e¹B=kT N
:
+ e¡¹B=kT The magnetization is the net dipole moment per unit volume. Subtract the magnitude of the total
dipole moment per unit volume of the antialigned moments from the total dipole moment per unit
volume of the aligned moments. The result is
¡
¢
N ¹e¹B=kT ¡ N ¹e¡¹B=kT N ¹ e¹B=kT ¡ e¡¹B=kT
M=
=
= N ¹ tanh(¹B=kT ) :
e¹B=kT + e¡¹B=kT
e¹B=kT + e¡¹B=kT
204 Chapter 32 (b) If ¹B ¿ kT , then e¹B=kT ¼ 1 + ¹B=kT and e¡¹B=kT ¼ 1 ¡ ¹B=kT . (See Appendix E for
the power series expansion of the exponential function.) The expression for the magnetization
becomes
£
¤
N ¹ (1 + ¹B=kT ) ¡ (1 ¡ ¹B=kT )
N ¹2 B
=
:
M¼
(1 + ¹B=kT ) + (1 ¡ ¹B=kT )
kT (c) If ¹B À kT , then e¡¹B=kT is negligible compared to e¹B=kT in both the numerator and
denominator of the expression for M . Thus
M¼ N ¹e¹B=kT
= N¹ :
e¹B=kT (d) The expression for M predicts that it is linear in B=kT for ¹B=kT small and independent
of B=kT for ¹B=kT large. The figure agrees with these predictions.
47
(a) The field of a dipole along its axis is given by Eq. 29–27:
¹
~ ¹0 ~ ;
B=
2¼ z 3
where ¹ is the dipole moment and z is the distance from the dipole. Thus the magnitude of the
magnetic field is
B= (4¼ £ 10¡7 T ¢ m=A)(1:5 £ 10¡23 J=T)
= 3:0 £ 10¡6 T :
2¼(10 £ 10¡9 m)3 ~
(b) The energy of a magnetic dipole with dipole moment ¹ in a magnetic field B is given by
~
U = ¡~ ¢ B = ¡¹B cos Á, where Á is the angle between the dipole moment and the field. The
¹ ~
energy required to turn it end for end (from Á = 0± to Á = 180± ) is
¢U = ¡¹B(cos 180± ¡ cos 0± ) = 2¹B = 2(1:5 £ 10¡23 J=T)(3:0 £ 10¡6 T)
= 9:0 £ 10¡29 J = 5:6 £ 10¡10 eV : The mean kinetic energy of translation at room temperature is about 0:04 eV (see Eq. 19–24 or
Sample Problem 32–3). Thus if dipoledipole interactions were responsible for aligning dipoles,
collisions would easily randomize the directions of the moments and they would not remain
aligned.
53
(a) If the magnetization of the sphere is saturated, the total dipole moment is ¹total = N ¹, where
N is the number of iron atoms in the sphere and ¹ is the dipole moment of an iron atom. We
wish to find the radius of an iron sphere with N iron atoms. The mass of such a sphere is N m,
where m is the mass of an iron atom. It is also given by 4¼½R3 =3, where ½ is the density of
iron and R is the radius of the sphere. Thus N m = 4¼½R3 =3 and
N= 4¼½R3
:
3m
Chapter 32 205 Substitute this into ¹total = N ¹ to obtain
¹total =
Solve for R and obtain ∙ 4¼½R3 ¹
:
3m 3m¹total
R=
4¼½¹
The mass of an iron atom is ¸1=3 : m = 56 u = (56 u)(1:66 £ 10¡27 kg=u) = 9:30 £ 10¡26 kg : So R= " 3(9:30 £ 10¡26 kg)(8:0 £ 1022 J=T)
3 4¼(14 £ 103 kg=m )(2:1 £ 10¡23 J=T) #1=3 = 1:8 £ 105 m : (b) The volume of the sphere is
4¼ 3 4¼
Vs =
R =
(1:82 £ 105 m)3 = 2:53 £ 1016 m3
3
3
and the volume of Earth is
4¼
(6:37 £ 106 m)3 = 1:08 £ 1021 m3 ;
Ve =
3
so the fraction of Earth’s volume that is occupied by the sphere is
2:53 £ 1016 m3
= 2:3 £ 10¡5 :
1:08 £ 1021 m3
The radius of Earth was obtained from Appendix C.
55
(a) The horizontal and vertical directions are perpendicular to each other, so the magnitude of
the field is
q
q
q
¹0 ¹
2 + B 2 = ¹0 ¹
2 ¸ + 4 sin2 ¸ =
B = Bh
cos m
1 ¡ sin2 ¸m + 4 sin2 ¸m
m
v
3
3
4¼r
4¼r
q
¹0 ¹
=
1 + 3 sin2 ¸m ;
4¼r3 where the trigonometric identity cos2 ¸m = 1 ¡ sin2 ¸m was used.
(b) The tangent of the inclination angle is
¶µ
¶
µ
Bv
2 sin ¸m
4¼r3
¹0 ¹
tan Ái =
= 2 tan ¸m ;
=
=
2¼r3 sin ¸m
cos ¸m
Bh
¹0 ¹ cos ¸m where tan ¸m = (sin ¸m )=(cos ¸m ) was used. 61
(a) The z component of the orbital angular momentum can have the values Lorb,z = m` h=2¼,
where m` can take on any integer value from ¡3 to +3, inclusive. There are seven such values
(¡3, ¡2, ¡1, 0, +1, +2, and +3).
206 Chapter 32 (b) The z component of the orbital magnetic moment is given by ¹orb,z = ¡m` eh=4¼m, where
m is the electron mass. Since there is a different value for each possible value of m` , there are
seven different values in all.
(c) The greatest possible value of Lorb,z occurs if m` = +3 is 3h=2¼.
(d) The greatest value of ¹orb, z is 3eh=4¼m.
(e) Add the orbital and spin angular momenta: Lnet,z = Lorb,z + Ls;z = (m` h=2¼) + (ms h=2¼). To
obtain the maximum value, set m` equal to +3 and ms equal to + 1 . The result is Lnet,z = 3:5h=2¼.
2
(f) Write Lnet,z = M h=2¼, where M is half an odd integer. M can take on all such values from
¡3:5 to +3:5. There are eight of these: ¡3:5, ¡2:5, ¡1:5, ¡0:5, +0:5, +1:5, +2:5, and +3:5. Chapter 32 207 Chapter 33 5
If f is the frequency and ¸ is the wavelength of an electromagnetic wave, then f ¸ = c. The
frequency is the same as the frequency of oscillation of the current in the LC circuit of the
p
generator. That is, f = 1=2¼ LC, where C is the capacitance and L is the inductance. Thus
¸
p
= c:
2¼ LC
The solution for L is
L= ¸2
(550 £ 10¡9 m)2
= 2
= 5:00 £ 10¡21 H :
4¼ 2 Cc2 4¼ (17 £ 10¡12 F)(3:00 £ 108 m=s)2 This is exceedingly small. 21
The plasma completely reflects all the energy incident on it, so the radiation pressure is given by
pr = 2I=c, where I is the intensity. The intensity is I = P=A, where P is the power and A is
the area intercepted by the radiation. Thus
pr = 2(1:5 £ 109 W)
2P
= 1:0 £ 107 Pa = 10 MPa :
=
Ac (1:00 £ 10¡6 m2 )(3:00 £ 108 m=s) 23
Let f be the fraction of the incident beam intensity that is reflected. The fraction absorbed is
1 ¡ f. The reflected portion exerts a radiation pressure of pr = (2f I0 )=c and the absorbed portion
exerts a radiation pressure of pa = (1 ¡ f)I0 =c, where I0 is the incident intensity. The factor 2
enters the first expression because the momentum of the reflected portion is reversed. The total
radiation pressure is the sum of the two contributions:
2fI0 + (1 ¡ f )I0 (1 + f)I0
=
:
c
c
To relate the intensity and energy density, consider a tube with length ` and crosssectional
area A, lying with its axis along the propagation direction of an electromagnetic wave. The
electromagnetic energy inside is U = uA`, where u is the energy density. All this energy will
pass through the end in time t = `=c so the intensity is
ptotal = pr + pa = U
uA`c
=
= uc :
At
A`
Thus u = I=c. The intensity and energy density are inherently positive, regardless of the propagation direction.
I= 208 Chapter 33 For the partially reflected and partially absorbed wave, the intensity just outside the surface is
I = I0 + f I0 = (1 + f )I0 , where the first term is associated with the incident beam and the second
is associated with the reflected beam. The energy density is, therefore,
u= I (1 + f )I0
=
;
c
c the same as radiation pressure.
25
(a) Since c = ¸f , where ¸ is the wavelength and f is the frequency of the wave,
f= c 3:00 £ 108 m=s
= 1:0 £ 108 Hz :
=
3:0 m
¸ (b) The angular frequency is
! = 2¼f = 2¼(1:0 £ 108 Hz) = 6:3 £ 108 rad=s :
(c) The angular wave number is
k= 2¼
2¼
= 2:1 rad=m :
=
3:0 m
¸ (d) The magnetic field amplitude is
Bm = 300 V=m
Em
= 1:00 £ 10¡6 T :
=
3:00 £ 108 m=s
c ~
~
~ ~
(e) B must be in the positive z direction when E is in the positive y direction in order for E £ B
to be in the positive x direction (the direction of propagation).
(f) The timeaveraged rate of energy flow or intensity of the wave is
I= 2
(300 V=m)2
Em
= 1:2 £ 102 W=m2 :
=
¡7 H=m)(3:00 £ 108 m=s)
2¹0 c 2(4¼ £ 10 (g) Since the sheet is perfectly absorbing, the rate per unit area with which momentum is delivered
to it is I=c, so
2
dp IA (119 W=m )(2:0 m2 )
= 8:0 £ 10¡7 N :
=
=
8 m=s
3:00 £ 10
dt
c
(h) The radiation pressure is pr = dp=dt 8:0 £ 10¡7 N
=
= 4:0 £ 10¡7 Pa :
2:0 m2
A 27
If the beam carries energy U away from the spaceship, then it also carries momentum p = U=c
away. Since the total momentum of the spaceship and light is conserved, this is the magnitude of
Chapter 33 209 the momentum acquired by the spaceship. If P is the power of the laser, then the energy carried
away in time t is U = P t. Thus p = P t=c and, if m is mass of the spaceship, its speed is
v= P t (10 £ 103 W)(1 d)(8:64 £ 104 s=d)
p
= 1:9 £ 10¡3 m=s = 1:9 mm=s :
=
=
(1:5 £ 103 kg)(3:00 £ 108 m=s)
m mc 35
Let I0 be in the intensity of the unpolarized light that is incident on the first polarizing sheet.
Then the transmitted intensity is I1 = 1 I0 and the direction of polarization of the transmitted light
2
is µ1 (= 40± ) counterclockwise from the y axis in the diagram.
The polarizing direction of the second sheet is µ2 (= 20± ) clockwise from the y axis so the angle
between the direction of polarization of the light that is incident on that sheet and the polarizing
direction of the of the sheet is 40± + 20± = 60± . The transmitted intensity is
1
I2 = I1 cos2 60± = I0 cos2 60±
2
and the direction of polarization of the transmitted light is 20± clockwise from the y axis.
The polarizing direction of the third sheet is µ3 (= 40± ) counterclockwise from the y axis so the
angle between the direction of polarization of the light incident on that sheet and the polarizing
direction of the sheet is 20± + 40± = 60± . The transmitted intensity is
1
I3 = I2 cos2 60± = I0 cos4 60± = 3:1 £ 10¡2 :
2
3:1% of the light’s initial intensity is transmitted.
43
(a) The rotation cannot be done with a single sheet. If a sheet is placed with its polarizing
direction at an angle of 90± to the direction of polarization of the incident radiation, no radiation
is transmitted.
It can be done with two sheets. Place the first sheet with its polarizing direction at some angle
µ, between 0 and 90± , to the direction of polarization of the incident radiation. Place the second
sheet with its polarizing direction at 90± to the polarization direction of the incident radiation.
The transmitted radiation is then polarized at 90± to the incident polarization direction. The
intensity is I0 cos2 µ cos2 (90± ¡ µ) = I0 cos2 µ sin2 µ, where I0 is the incident radiation. If µ is not
0 or 90± , the transmitted intensity is not zero.
(b) Consider n sheets, with the polarizing direction of the first sheet making an angle of µ = 90± =n
with the direction of polarization of the incident radiation and with the polarizing direction of
each successive sheet rotated 90± =n in the same direction from the polarizing direction of the
previous sheet. The transmitted radiation is polarized with its direction of polarization making
an angle of 90± with the direction of polarization of the incident radiation. The intensity is
I = I0 cos2n (90± =n). You want the smallest integer value of n for which this is greater than
0:60I0 .
210 Chapter 33 Start with n = 2 and calculate cos2n (90± =n). If the result is greater than 0:60, you have obtained
the solution. If it is less, increase n by 1 and try again. Repeat this process, increasing n by 1
each time, until you have a value for which cos2n (90± =n) is greater than 0:60. The first one will
be n = 5.
55
Look at the diagram on the right. The two angles labeled ® have the same value. µ2 is the angle of refraction. Because the dotted lines are perpendicular to
the prism surface µ2 + ® = 90± and ® = 90± ¡ µ2 .
Because the interior angles of a triangle sum to 180± ,
180± ¡ 2µ2 + Á = 180± and µ2 = Á=2.
Now look at the next diagram and consider the triangle
formed by the two normals and the ray in the interior.
The two equal interior angles each have the value µ¡µ2 .
Because the exterior angle of a triangle is equal to the
sum of the two opposite interior angles, Ã = 2(µ ¡ µ2 )
and µ = µ2 +Ã=2. Upon substitution for µ2 this becomes
µ = (Á + Ã)=2. .
...
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.........................................................................
. Á µ ® ® µ2 µ2 µ Á Ã µ µ2 µ2 µ According to the law of refraction the index of refraction of the prism material is
n= sin(Á + Ã)=2
sin µ
:
=
sin µ2
sin Á=2 65
(a) No refraction occurs at the surface ab, so the angle of incidence at surface ac is 90± ¡ Á.
For total internal reflection at the second surface, ng sin(90± ¡ Á) must be greater than na .
Here ng is the index of refraction for the glass and na is the index of refraction for air. Since
sin(90± ¡ Á) = cos Á, you want the largest value of Á for which ng cos Á ¸ na . Recall that
cos Á decreases as Á increases from zero. When Á has the largest value for which total internal
reflection occurs, then ng cos Á = na , or
¶
µ ¶
µ
1
¡1 na
¡1
= 48:9± :
= cos
Á = cos
1:52
ng
The index of refraction for air was taken to be unity.
(b) Replace the air with water. If nw (= 1:33) is the index of refraction for water, then the largest
value of Á for which total internal reflection occurs is
¶
µ ¶
µ
¡1 nw
¡1 1:33
= 29:0± :
= cos
Á = cos
1:52
ng
Chapter 33 211 69
The angle of incidence µB for which reflected light is fully polarized is given by Eq. 33–49 of
the text. If n1 is the index of refraction for the medium¢of incidence and n2 is the index of
¡
¡
¢
refraction for the second medium, then µB = tan¡1 n2 =n1 = tan¡1 1:53=1:33 = 63:8± .
73 Let µ1 (= 45± ) be the angle of incidence at the first surface and µ2 be the angle of refraction
there. Let µ3 be the angle of incidence at the second surface. The condition for total internal
reflection at the second surface is n sin µ3 ¸ 1. You want to find the smallest value of the index
of refraction n for which this inequality holds.
The law of refraction, applied to the first surface, yields n sin µ2 = sin µ1 . Consideration of the
triangle formed by the surface of the slab and the ray in the slab tells us that µ3 = 90± ¡µ2 . Thus the
condition for total internal reflection becomes 1 ∙ n sin(90± ¡µ2 ) = n cos µ2 . Square this equation
and use sin2 µ2 + cos2 µ2 = 1 to obtain 1 ∙ n2 (1 ¡ sin2 µ2 ). Now substitute sin µ2 = (1=n) sin µ1
to obtain
Ã
!
sin2 µ1
2
1∙n 1¡
= n2 ¡ sin2 µ1 :
2
n
The largest value of n for which this equation is true is the value for which 1 = n2 ¡ sin2 µ1 .
Solve for n:
q
p
n = 1 + sin2 µ1 = 1 + sin2 45± = 1:22 :
75
Let µ be the angle of incidence and µ2 be the angle
of refraction at the left face of the plate. Let n
be the index of refraction of the glass. Then, the
law of refraction yields sin µ = n sin µ2 . The angle
of incidence at the right face is also µ2 . If µ3 is
the angle of emergence there, then n sin µ2 = sin µ3 .
Thus sin µ3 = sin µ and µ3 = µ. The emerging ray is
parallel to the incident ray.
You wish to derive an expression for x in terms
of µ. If D is the length of the ray in the glass,
then D cos µ2 = t and D = t= cos µ2 . The angle ®
in the diagram equals µ ¡ µ2 and x = D sin ® =
D sin(µ ¡ µ2 ). Thus
x= Ã¡¡ t ¡¡!
¡¡
¡¡ ..
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.
. µ ® µ2 µ2 x µ3 t sin(µ ¡ µ2 )
:
cos µ2 If all the angles µ, µ2 , µ3 , and µ¡µ2 are small and measured in radians, then sin µ ¼ µ, sin µ2 ¼ µ2 ,
sin(µ ¡ µ2 ) ¼ µ ¡ µ2 , and cos µ2 ¼ 1. Thus x ¼ t(µ ¡ µ2 ). The law of refraction applied to the
212 Chapter 33 point of incidence at the left face of the plate is now µ ¼ nµ2 , so µ2 ¼ µ=n and
¶
µ
(n ¡ 1)tµ
µ
=
:
x¼t µ¡
n
n
77
The time for light to travel a distance d in free space is t = d=c, where c is the speed of light
(3:00 £ 108 m=s).
(a) Take d to be 150 km = 150 £ 103 m. Then,
t= 150 £ 103 m
d
= 5:00 £ 10¡4 s :
=
c 3:00 £ 108 m=s (b) At full moon, the Moon and Sun are on opposite sides of Earth, so the distance traveled by
the light is d = (1:5 £ 108 km) + 2(3:8 £ 105 km) = 1:51 £ 108 km = 1:51 £ 1011 m. The time
taken by light to travel this distance is
t= 1:51 £ 1011 m
d
= 500 s = 8:4 min :
=
c 3:00 £ 108 m=s The distances are given in the problem.
(c) Take d to be 2(1:3 £ 109 km) = 2:6 £ 1012 m. Then,
t= 2:6 £ 1012 m
d
= 8:7 £ 103 s = 2:4 h :
=
c 3:00 £ 108 m=s (d) Take d to be 6500 ly and the speed of light to be 1:00 ly=y. Then,
t= 6500 ly
d
= 6500 y :
=
c 1:00 ly=y The explosion took place in the year 1054 ¡ 6500 = ¡5446 or B.C. 5446.
79
(a) The amplitude of the magnetic field is B = E=c = (5:00 V=m)=(3:00 £ 108 m=s) = 1:67 £
10¡8 T. According to the argument of the trigonometric function in the expression for the electric
field, the wave is moving in the negative z direction and the electric field is parallel to the y
~ ~
~
axis. In order for E £ B to be in the negative z direction, B must be in the positive x direction
~
when E is in the positive y direction. Thus
Bx = (1:67 £ 10¡8 T) sin[(1:00 £ 106 m¡1 )z + !t]
is the only nonvanishing component of the magnetic field.
The angular wave number is k = 1:00 £ 106 m¡1 so the angular frequency is ! = kc = (1:00 £
106 m¡1 )(3:00 £ 108 m=s) = 3:00 £ 1014 s¡1 and
Bx = (1:67 £ 10¡8 T) sin[(1:00 £ 106 m¡1 )z + (3:00 £ 1014 s¡1 )t] : (b) The wavelength is ¸ = 2¼=k = 2¼=(1:00 £ 106 m¡1 ) = 6:28 £ 10¡6 m. Chapter 33 213 (c) The period is T = 2¼=! = 2¼=(3:00 £ 1014 s¡1 ) = 2:09 £ 10¡14 s.
2
(d) The intensity of this wave is I = Em =2¹0 c = (5:00 V=m)2 =2(4¼£10¡7 H=m)(3:00£108 m=s =
2
0:0332 W=m . (f) A wavelength of 6:28 £ 10¡6 m places this wave in the infrared portion of the
electromagnetic spectrum. See Fig. 331.
83
(a) The power is the same through any hemisphere centered at the source. The area of a hemisphere of radius r is A = 2¼r2 . In this case r is the distance from the source to the aircraft.
Thus the intensity at the aircraft is I = P=A = P=2¼r2 = (180 £ 103 W)=2¼(90 £ 103 m)2 =
2
3:5 £ 10¡6 W=m .
(b) The power of the reflection is the product of the intensity at the aircraft and the cross section
of the aircraft: Pr = (3:5 £ 10¡6 W=m2 )(0:22 m2 ) = 7:8 £ 10¡7 W.
(c) The intensity at the detector is Pr =2¼r 2 = (7:8 £ 10¡7 W)=2¼(90 £ 103 m)2 = 1:5 £
2
10¡17 W=m .
2
(d) Since the intensity is given by I = Em =2¹0 c,
q
p
2
Em = 2¹0 cI = 2(4¼ £ 10¡7 H=m)(3:00 £ 108 m=s)(1:5 £ 10¡17 W=m ) = 1:1 £ 10¡7 V=m :
p
p
(e) The rms value of the magnetic field is Brms = Em = 2c = (1:1 £ 10¡7 V=m)=( 2)(3:00 £
108 m=s) = 2:5 £ 10¡16 T.
91
The critical angle for total internal reflection is given by µc = sin¡1 (1=n). For n = 1:456 this
angle is µc = 43:38± and for n = 1:470 it is µc = 42:86± .
(a) An incidence angle of 42:00± is less than the critical angle for both red and blue light. The
refracted light is white.
(b) An incidence angle of 43:10± is less than the critical angle for red light and greater than the
critical angle for blue light. Red light is refracted but blue light is not. The refracted light is
reddish.
(c) An incidence angle of 44:00± is greater than the critical angle for both red and blue light.
Neither is refracted.
103
(a) Take the derivative of the functions given for E and B, then substitute them into
@ 2E
@2E
= c2 2
@t2
@x and @2B
@ 2B
= c2 2 :
@t2
@x The derivatives of E are @ 2 E=@t2 = ¡! 2 Em sin(kx ¡ !t) and @ 2 E=@x2 = ¡k 2 Em sin(kx ¡ !t),
so the wave equation for the electric field yields ! 2 = c2 k 2 . Since ! = ck the function satisfies
the wave equation. Similarly, the derivatives of B are @ 2 B=@t2 = ¡! 2 Bm sin(kx ¡ !t) and
@ 2 B=@x2 = ¡k2 Bm sin(kx ¡ !t) and the wave equation for the magnetic field yields ! 2 = c2 k 2 .
Since ! = ck the function satisfies the wave equation.
214 Chapter 33 (b) Let u = kx § !t and consider f to be a function of u, which in turn is a function of x and
t. Then the chain rule of the calculus gives
@ 2 E d2 f
= 2
@t2
du
and µ @ 2 E d2 f
=
@x2 du2 µ ¶2 = d2 f 2
!
du2 ¶2 = d2 f 2
k :
du2 @u
@t @u
@x Substitution into the wave equation again yields !2 = c2 k2 , so the function obeys the wave
equation. A similar analysis shows that the function for B also satisfies the wave equation. Chapter 33 215 Chapter 34 5
..
.
.
.
..
.
..
The light bulb is labeled O and its image is la.
.
..
.
..
" O ²............
.
.
..
.
..
.
.
.
..
..
.
..
..
.
.
j
beled I on the digram to the right. Consider the
.
..
.
..
..
.
.
..
. µ ...
..
..
.
.
.
..
..
..
.
..
.
.
..
..
.
..
..
.
.
two rays shown on the diagram to the right. One
..
.
..
..
..
d1
.
.
.
..
..
..
.
..
.
.
..
..
.
..
..
.
.
.. µ
.
..
..
.
..
µ....
enters the water at A and is reflected from the
.
..
.
..
j
.
.
.
.
..
..
.
.. C
..
E ............ water line
# A ..........
..
..
mirror at B. This ray is perpendicular to the wa.
.
.
..
.
.
..
.
.
.
.
.
.
.
.
.
.
.
"
.
.
.
.
.
.. .
.
.. ..
.
.
.
.
ter line and mirror. The second ray leaves the
.
.
.
.
.
.
.
.
.
.
.
.. .
j
.
.
.. ..
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.. .
.
.
.. .
.
.
lightbulb at the angle µ, enters the water at C,
.
.
.
.
.
µ0 ...........µ..0...µ0......... µ0
.
d2
.
.
.
.
..
.
.
..
.
.
.
where it is refracted. It is reflected from the mir..
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.. .. .
.
.
j
.. .. ..
.
.
. .
.
.
. ..
.
. .
. . D
ror at D and leaves the water at E. At C the angle
..
..
mirror
..
..
..
# B .........
.
.
.
.
..
..
.
.
.
"
of incidence is µ and the angle of refraction is
.
.
..
.
.
.
.
.
j
.
..
.
.
.
.
µ0 . At D the angles of incidence and reflection
.
.
.
..
.
.
.
.
d3
.
.
.
..
are both µ0 . At E the angle of incidence is µ0 and
.
.
.
.
.
..
.
.
.
j
. ..
. .
.
..
the angle of refraction is µ. The dotted lines that
..
#
.
.
.
.
.
I²
.
.
meet at I represent extensions of the emerging
.
.
.
.
.
.
.
.
rays. Light appears to come from I. We want to
.
compute d3 .
Consideration of the triangle OBE tells us that the distance d2 + d3 is L tan(90± ¡ µ) = L= tan µ,
where L is the distance between A and E. Consideration of the triangle OBC tells us that
the distance between A and C is d1 tan µ and consideration of the triangle CDE tells us that
the distance between C and E is 2d2 tan µ0 , so L = d1 tan µ + 2d2 tan µ0 , d2 + d3 = (d1 tan µ +
2d2 tan µ0 )= tan µ, and
d1 tan µ + 2d2 tan µ0
¡ d2 :
d3 =
tan µ Apply the law of refraction at point C: sin µ = n sin µ0 , where n is the index of refraction of
water. Since the angles µ and µ0 are small we may approximate their sines by their tangents and
write tan µ = n tan µ0 . Us this to substitute for tan µ in the expression for d3 to obtain
d3 = nd1 + 2d2
(1:33)(250 cm) + 2(200 cm)
¡ 200 cm = 350 cm ;
¡ d2 =
1:33
n where the index of refraction of water was taken to be 1:33.
11
(a) The radius of curvature r and focal length f are positive for a concave mirror and are related
by f = r=2, so r = 2(+12 cm) = +24 cm.
216 Chapter 34 (b) Since (1=p) + (1=i) = 1=f , where i is the image distance,
i= (12 cm)(18 cm)
fp
= 36 cm :
=
p ¡ f 18 cm ¡ 12 cm (c) The magnification is m = ¡i=p = ¡(36 cm)=(18 cm = ¡2:0.
(d) The value obtained for i is positive, so the image is real.
(e) The value obtained for the magnification is negative, so the image is inverted.
(f) Real images are formed by mirrors on the same side as the object. Since the image here is
real it is on the same side of the mirror as the object.
9
(a) The radius of curvature r and focal length f are positive for a concave mirror and are related
by f = r=2, so r = 2(+18 cm) = +36 cm.
(b) Since (1=p) + (1=i) = 1=f , where i is the image distance,
i= fp
(18 cm)(12 cm)
= ¡36 cm :
=
p ¡ f 12 cm ¡ 18 cm (c) The magnification is m = ¡i=p = ¡(¡36 cm)=(12 cm = 3:0.
(d) The value obtained for i is negative, so the image is virtual.
(e) The value obtained for the magnification is positive, so the image is not inverted.
(f) Real images are formed by mirrors on the same side as the object and virtual images are
formed on the opposite side. Since the image here is virtual it is on the opposite side of the
mirror from the object.
15
(a) The radius of curvature r and focal length f are negative for a convex mirror and are related
by f = r=2, so r = 2(¡10 cm) = ¡20 cm.
(b) Since (1=p) + (1=i) = 1=f , where i is the image distance,
i= (¡10 cm)(8 cm)
fp
= ¡4:44 cm :
=
p ¡ f (8 cm) ¡ (¡10 cm) (c) The magnification is m = ¡i=p = ¡(¡4:44 cm)=(8 cm = +0:56.
(d) The value obtained for i is negative, so the image is virtual.
(e) The value obtained for the magnification is positive, so the image is not inverted.
(f) Real images are formed by mirrors on the same side as the object and virtual images are
formed on the opposite side. Since the image here is virtual it is on the opposite side of the
mirror from the object
27
Since the mirror is convex the radius of curvature is negative. The focal length is f = r=2 =
(¡40 cm)=2 = ¡20 cm.
Chapter 34 217 Since (1=p) + (1=i) = (1=f ),
p= if
:
i¡f This yields p = +5:0 cm if i = ¡4:0 cm and p = ¡3:3 cm if i = ¡4:0 cm. Since p must be positive
we select i = ¡4:0 cm and take p to be +5:0 cm. The magnification is m = ¡i=p = ¡(¡4:0 cm)=(5:0 cm) = +0:80. Since the image distance is
negative the image is virtual and on the opposite side of the mirror from the object. Since the
magnification is positive the image is not inverted.
29 Since the magnification m is m = ¡i=p, where p is the object distance and i is the image
distance, i = ¡mp. Use this to substitute for i in (1=p) + (1=i) = (1=f ), where f is the focal
length. The solve for p. The result is
¶
µ
¶
µ
1
1
= §120 cm :
p= f 1¡
= (§30 cm) 1 ¡
0:20
m
Since p must be positive we must use the lower sign. Thus the focal length is ¡30 cm and
the radius of curvature is r = 2f == 60 cm. Since the focal length and radius of curvature are
negative the mirror is convex.
The object distance is 1:2 m and the image distance is i = ¡mp = ¡(0:20)(120 cm) = ¡24 cm. Since the image distance is negative the image is virtual and on the opposite side of the mirror
from the object. Since the magnification is positive the image is not inverted.
35 Solve n1 n2 n2 ¡ n1
+
=
p
i
r for r. the result is
r= (¡13 cm)(+10 cm)
ip(n2 ¡ n1 )
= ¡33 cm :
=
((1:0)(¡13 cm) + (1:5)(+10 cm)
n1 i + n2 p Since the image distance is negative the image is virtual and appears on the same side of the
surface as the object.
37
Solve n1 n2 n2 ¡ n1
+
=
p
i
r for r. the result is
i=
218 (1:0)(+30 cm)(+70 cm)
n2 rp
= ¡26 cm :
=
(n2 ¡ n1 )p ¡ n1 r (1:0 ¡ 1:5)(+70 cm) ¡ (1:5)(+30 cm) Chapter 34 Since the image distance is negative the image is virtual and appears on the same side of the
surface as the object.
41
Use the lens maker’s equation, Eq. 34–10:
¶
µ
1
1
1
;
= (n ¡ 1)
¡
f
r1 r2
where f is the focal length, n is the index of refraction, r1 is the radius of curvature of the first
surface encountered by the light and r2 is the radius of curvature of the second surface. Since
one surface has twice the radius of the other and since one surface is convex to the incoming
light while the other is concave, set r2 = ¡2r1 to obtain
¶
µ
3(n ¡ 1)
1
1
1
=
= (n ¡ 1)
+
:
2r1
f
r1 2r1
Solve for r1 : 3(n ¡ 1)f 3(1:5 ¡ 1)(60 mm)
=
= 45 mm :
2
2
The radii are 45 mm and 90 mm.
r1 = 47
The object distance p and image distance i obey (1=p) + (1=i) = (1=f ), where f is the focal
length. In addition, p + i = L, where L (= 44 cm) is the distance from the slide to the screen. Use
i = L ¡ p to substitute for i in the first equation and obtain p2 ¡ pL + Lf = 0. The solution is
p
p
L § L2 ¡ 4Lf (44 cm) § 4(44 cm)(11 cm)
=
= 22 cm :
p=
2
2
51
The lens is diverging, so the focal length is negative. Solve (1=p) + (1=i) = (1=f ) for i. The
result is
(+8:0 cm)(¡12 cm)
pf
= ¡4:8 cm :
=
i=
p ¡ f (8:0 cm) ¡ (¡12 cm) The magnification is m = ¡i=p = ¡(¡4:8 cm)=(+8:0 cm) = 0:60. Since the image distance is
negative the image is virtual and appears on the same side of the lens as the object. Since the
magnification is positive the image is not inverted.
55
The lens is converging, so the focal length is positive. Solve (1=p) + (1=i) = (1=f ) for i. The
result is
(+45 cm)(+20 cm)
pf
= +36 cm :
=
i=
p ¡ f (45 cm) ¡ (+20 cm)
Chapter 34 219 The magnification is m = ¡i=p = ¡(36 cm)=(45 cm) = ¡0:80. Since the image distance is
positive the image is real and appears on the opposite side of the lens from the object. Since the
magnification is negative the image is inverted.
61
The focal length is
f= (+30 cm)(¡42 cm)
r1 r2
=
= +31:8 cm :
(n ¡ 1)(r2 ¡ r1 ) (1:55 ¡ 1)[(¡42 cm) ¡ (+30 cm)] Solve (1=p) + (1=i) = (1=f ) for i. the result is
i= (+75 cm)(+31:8 cm)
pf
= 55 cm :
=
p ¡ f (+75 cm) ¡ (+31:8 cm) The magnification is m = ¡i=p = ¡(55 cm)=(75 cm) = ¡0:73. Since the image distance is positive the image is real and on the opposite side of the lends from
the object. Since the magnification is negative the image is inverted.
75 Since m = ¡i=p, i = ¡mp = ¡(+1:25)(+16 cm) = ¡20 cm. Solve (1=p) + (1=i) = (1=f ) for f .
The result is
(+16 cm)(¡20 cm)
pi
= +80 cm :
=
f=
p + i (+16 cm) + (¡20 cm)
Since f is positive the lens is a converging lens. Since the image distance is negative the image
is virtual and appears on the same side of the lens as the object. Since the magnification is
positive the image is not inverted.
79
The image is on the same side of the lens as the object. This means that the image is virtual and
the image distance is negative. Solve (1=p) + (1=i) = (1=f ) for i. The result is
i= pf
:
p¡f and the magnification is
f
i
:
m=¡ =¡
p
p¡f Since the magnification is less than 1:0, f must be negative and the lens must be a diverging
lens. The image distance is
i= (+5:0 cm)(¡10 cm)
= ¡3:3 cm :
(5:0 cm) ¡ (¡10 cm) and the magnification is m = ¡i=p = ¡(¡3:3 cm)=(5:0 cm) = 0:66 cm.
220 Chapter 34 Since the magnification is positive the image is not inverted.
81
Lens 1 is converging and so has a positive focal length. Solve (1=p1 ) + (1=i1 ) = (1=f1 ) for the
image distance i1 associated with the image produced by this lens. The result is
i1 = p1 f1
(20 cm)(+9:0 cm)
= 16:4 cm :
=
p1 ¡ f1 (20 cm) ¡ (9:0 cm) This image is the object for lens 2. The object distance is d¡p2 = (8:0 cm)¡(16:4 cm) = ¡8:4 cm.
The negative sign indicates that the image is behind the second lens. The lens equation is still
valid. The second lens has a positive focal length and the image distance for the image it forms
is
p2 f 2
(¡8:4 cm)(5:0 cm)
= +3:1 cm :
i2 =
=
p2 ¡ f2 (¡8:4 cm) ¡ (5:0 cm)
The overall magnification is the product of the individual magnifications:
¶µ
¶µ
¶ µ
¶
µ
3:1 cm
i2
16:4 cm
i1
¡
¡
= ¡
= ¡0:30 :
m = m1 m2 = ¡
20 cm
¡8:4 cm
p1
p2
Since the final image distance is positive the final image is real and on the opposite side of lens
2 from the object. Since the magnification is negative the image is inverted. 89
(a) If L is the distance between the lenses, then according to Fig. 34–20, the tube length is
s = L ¡ fob ¡ fey = 25:0 cm ¡ 4:00 cm ¡ 8:00 cm = 13:0 cm.
(b) Solve (1=p) + (1=i) = (1=fob ) for p. The image distance is i = fob + s = 4:00 cm + 13:0 cm =
17:0 cm, so
(17:0 cm)(4:00 cm)
ifob
= 5:23 cm :
=
p=
i ¡ fob 17:0 cm ¡ 4:00 cm
(c) The magnification of the objective is
17:0 cm
i
= ¡3:25 :
m=¡ =¡
5:23 cm
p
(d) The angular magnification of the eyepiece is
mµ = 25 cm
25 cm
= 3:13 :
=
8:00 cm
fey (e) The overall magnification of the microscope is
M = mmµ = (¡3:25)(3:13) = ¡10:2 :
93
(a) When the eye is relaxed, its lens focuses faraway objects on the retina, a distance i behind
the lens. Set p = 1 in the thin lens equation to obtain 1=i = 1=f , where f is the focal length
Chapter 34 221 of the relaxed effective lens. Thus i = f = 2:50 cm. When the eye focuses on closer objects, the
image distance i remains the same but the object distance and focal length change. If p is the
new object distance and f 0 is the new focal length, then
1
1 1
+ = 0:
p i f
Substitute i = f and solve for f 0 . You should obtain
f0 = (40:0 cm)(2:50 cm)
pf
= 2:35 cm :
=
f + p 40:0 cm + 2:50 cm (b) Consider the lensmaker’s equation
¶
µ
1
1
1
;
= (n ¡ 1)
¡
f
r1 r2
where r1 and r2 are the radii of curvature of the two surfaces of the lens and n is the index of
refraction of the lens material. For the lens pictured in Fig. 34–46, r1 and r2 have about the same
magnitude, r1 is positive, and r2 is negative. Since the focal length decreases, the combination
(1=r1 ) ¡ (1=r2 ) must increase. This can be accomplished by decreasing the magnitudes of either
or both radii.
103
For a thin lens, (1=p) + (1=i) = (1=f ), where p is the object distance, i is the image distance, and
f is the focal length. Solve for i:
fp
:
i=
p¡f Let p = f + x, where x is positive if the object is outside the focal point and negative if it is
inside. Then
f (f + x)
i=
:
x
Now let i = f + x0 , where x0 is positive if the image is outside the focal point and negative if it
is inside. Then
f2
f (f + x)
¡f =
x0 = i ¡ f =
x
x
and xx0 = f 2 .
105
Place an object far away from the composite lens and find the image distance i. Since the image
is at a focal point, i = f , the effective focal length of the composite. The final image is produced
by two lenses, with the image of the first lens being the object for the second. For the first lens,
(1=p1 ) + (1=i1 ) = (1=f1 ), where f1 is the focal length of this lens and i1 is the image distance for
the image it forms. Since p1 = 1, i1 = f1 .
222 Chapter 34 The thin lens equation, applied to the second lens, is (1=p2 ) + (1=i2 ) = (1=f2 ), where p2 is the
object distance, i2 is the image distance, and f2 is the focal length. If the thicknesses of the
lenses can be ignored, the object distance for the second lens is p2 = ¡i1 . The negative sign
must be used since the image formed by the first lens is beyond the second lens if i1 is positive.
This means the object for the second lens is virtual and the object distance is negative. If i1 is
negative, the image formed by the first lens is in front of the second lens and p2 is positive. In
the thin lens equation, replace p2 with ¡f1 and i2 with f to obtain
¡ 1
1 1
+ =
:
f1 f f2 The solution for f is
f= f1 f2
:
f1 + f2 107
(a) and (b) Since the height of the image is twice the height of the fly and since the fly and its
image have the same orientation the magnification of the lens is m = +2:0. Since m = ¡i=p,
where p is the object distance and i is the image distance, i = ¡2p. Now jp + ij = d, so j ¡ pj = d
and p = d = 20 cm. The image distance is ¡40 cm.
Solve (1=p) + (1=i) = (1=f ) for f . the result is
f= pi
(20 cm)(¡40 cm)
= +40 cm :
=
p + i (20 cm) + (¡40 cm) (c) and (d) Now m = +0:5 and i = ¡0:5p. Since jp + ij = d, 0:5p = d and p = 2d = 40 cm. The
image distance is ¡20 cm and the focal length is
f= (40 cm)(¡20 cm)
pi
= ¡40 cm :
=
p + i (40 cm) + (¡20 cm) Chapter 34 223 Chapter 35 5
(a) Take the phases of both waves to be zero at the front surfaces of the layers. The phase of
the first wave at the back surface of the glass is given by Á1 = k1 L ¡ !t, where k1 (= 2¼=¸1 ) is
the angular wave number and ¸1 is the wavelength in glass. Similarly, the phase of the second
wave at the back surface of the plastic is given by Á2 = k2 L ¡ !t, where k2 (= 2¼=¸2 ) is the
angular wave number and ¸2 is the wavelength in plastic. The angular frequencies are the same
since the waves have the same wavelength in air and the frequency of a wave does not change
when the wave enters another medium. The phase difference is
¶
µ
1
1
L:
Á1 ¡ Á2 = (k1 ¡ k2 )L = 2¼
¡
¸1 ¸2
Now ¸1 = ¸air =n1 , where ¸air is the wavelength in air and n1 is the index of refraction of the
glass. Similarly, ¸2 = ¸air =n2 , where n2 is the index of refraction of the plastic. This means that
the phase difference is Á1 ¡ Á2 = (2¼=¸air )(n1 ¡ n2 )L. The value of L that makes this 5:65 rad is
L= (Á1 ¡ Á2 )¸air 5:65(400 £ 10¡9 m)
=
= 3:60 £ 10¡6 m :
2¼(n1 ¡ n2 )
2¼(1:60 ¡ 1:50) (b) 5:65 rad is less than 2¼ rad (= 6:28 rad), the phase difference for completely constructive
interference, and greater than ¼ rad (= 3:14 rad), the phase difference for completely destructive interference. The interference is therefore intermediate, neither completely constructive nor
completely destructive. It is, however, closer to completely constructive than to completely destructive.
15
Interference maxima occur at angles µ such that d sin µ = m¸, where d is the separation of the
sources, ¸ is the wavelength, and m is an integer. Since d = 2:0 m and ¸ = 0:50 m, this means
that sin µ = 0:25m. You want all values of m (positive and negative) for which j0:25mj ∙ 1.
These are ¡4, ¡3, ¡2, ¡1, 0, +1, +2, +3, and +4. For each of these except ¡4 and +4, there are
two different values for µ. A single value of µ (¡90± ) is associated with m = ¡4 and a single
value (¡90± ) is associated with m = +4. There are sixteen different angles in all and therefore
sixteen maxima.
17
The angular positions of the maxima of a twoslit interference pattern are given by d sin µ = m¸,
where d is the slit separation, ¸ is the wavelength, and m is an integer. If µ is small, sin µ may be
approximated by µ in radians. Then dµ = m¸. The angular separation of two adjacent maxima
224 Chapter 35 is ¢µ = ¸=d. Let ¸0 be the wavelength for which the angular separation is 10:0% greater. Then
1:10¸=d = ¸0 =d or ¸0 = 1:10¸ = 1:10(589 nm) = 648 nm.
19
The condition for a maximum in the twoslit interference pattern is d sin µ = m¸, where d is the
slit separation, ¸ is the wavelength, m is an integer, and µ is the angle made by the interfering
rays with the forward direction. If µ is small, sin µ may be approximated by µ in radians. Then
dµ = m¸ and the angular separation of adjacent maxima, one associated with the integer m and
the other associated with the integer m + 1, is given by ¢µ = ¸=d. The separation on a screen a
distance D away is given by ¢y = D ¢µ = ¸D=d. Thus
¢y = (500 £ 10¡9 m)(5:40 m)
= 2:25 £ 10¡3 m = 2:25 mm :
1:20 £ 10¡3 m 21
The maxima of a twoslit interference pattern are at angles µ that are given by d sin µ = m¸,
where d is the slit separation, ¸ is the wavelength, and m is an integer. If µ is small, sin µ may
be replaced by µ in radians. Then dµ = m¸. The angular separation of two maxima associated
with different wavelengths but the same value of m is ¢µ = (m=d)(¸2 ¡ ¸1 ) and the separation
on a screen a distance D away is
¸
∙
mD
(¸2 ¡ ¸1 )
¢y = D tan ¢µ ¼ D ¢µ =
d
¸
∙
3(1:0 m)
(600 £ 10¡9 m ¡ 480 £ 10¡9 m) = 7:2 £ 10¡5 m :
=
5:0 £ 10¡3 m
The small angle approximation tan ¢µ ¼ ¢µ was made. ¢µ must be in radians.
29
The phasor diagram is shown to the right. Here E1 = 1:00, E2 = 2:00,
and Á = 60± . The resultant amplitude Em is given by the trigonometric
law of cosines:
2
2
2
Em = E1 + E2 ¡ 2E1 E2 cos(180± ¡ Á) ; so
Em = ....
..
......
... ..
.
....
.. ..
..
. .. ...
. .. ...
.. ..
.. ..
.. ...
.
. ...
.
.. ..
.
.. ..
.
..
..
..
..
.
..
..
.
..
.
..
.
..
..
..
..
..
..
.
..
.
..
..
.
..
..
.
..
..
.
.. .
.. .
..
..
.. .
.
...
..
.
..
..
.
..
..
.
..
..
....
.....
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.. ..
.
.
.. .
.. .
.
.. .
.
.. .
..
.
.. E2 Em Á E1 p
(1:00)2 + (2:00)2 ¡ 2(1:00)(2:00) cos 120± = 2:65 : 39
For complete destructive interference, you want the waves reflected from the front and back of
the coating to differ in phase by an odd multiple of ¼ rad. Each wave is incident on a medium of
higher index of refraction from a medium of lower index, so both suffer phase changes of ¼ rad
Chapter 35 225 on reflection. If L is the thickness of the coating, the wave reflected from the back surface travels
a distance 2L farther than the wave reflected from the front. The phase difference is 2L(2¼=¸c ),
where ¸c is the wavelength in the coating. If n is the index of refraction of the coating, ¸c = ¸=n,
where ¸ is the wavelength in vacuum, and the phase difference is 2nL(2¼=¸). Solve
µ ¶
2¼
2nL
= (2m + 1)¼
¸
for L. Here m is an integer. The result is L= (2m + 1)¸
:
4n To find the least thickness for which destructive interference occurs, take m = 0. Then
L= 600 £ 10¡9 m
¸
=
= 1:2 £ 10¡7 m :
4n
4(1:25) 41
Since n1 is greater than n2 there is no change in phase on reflection from the first surface. Since
n2 is less than n3 there is a change in phase of ¼ rad on reflection from the second surface. One
wave travels a distance 2L further than the other, so the difference in the phases of the two waves
is 4¼L=¸2 + ¼, where ¸2 is the wavelength in medium 2. Since interference produces a minimum
the phase difference must be an odd multiple of ¼. Thus 4¼L=¸2 + ¼ = (2m + 1)¼, where m is
an integer or zero. Replace ¸2 with ¸=n2 , where ¸ is the wavelength in air, and solve for ¸. The
result is
4Ln2 2(380 nm)(1:1:34) 1018 nm
=
=
:
¸=
2m
m
m
For m = 1, ¸ = 1018 nm and for m = 2, ¸ = (1018 nm)=2 = 509 nm. Other wavelengths are
shorter. Only ¸ = 509 nm is in the visible range.
47
There is a phase shift on reflection of ¼ for both waves and one wave travels a distance 2L
further than the other, so the phase difference of the reflected waves is 4¼L=¸2 , where ¸2 is the
wavelength in medium 2. Since the result of the interference is a minimum of intensity the phase
difference must be an odd multiple of ¼. Thus 4¼L=¸2 = (2m + 1)¼, where m is an integer or
zero. Replace ¸2 with ¸=n2 , where ¸ is the wavelength in air, and solve for ¸. The result is
¸= 4(210 nm)(1:46) 1226 nm
4Ln2
=
=
:
2m + 1
2m + 1
2m + 1 For m = 1, ¸ = (1226 nm)=3 = 409 nm. This is in the visible range. Other values of m are
associated with wavelengths that are not in the visible range.
53
(a) Oil has a greater index of refraction than air and water has a still greater index of refraction.
There is a change of phase of ¼ rad at each reflection. One wave travels a distance 2L further
226 Chapter 35 than the other, where L is the thickness of the oil. The phase difference of the two reflected
waves is 4¼L=¸o , where ¸ is the wavelength in oil, and this must be equal to a multiple of 2¼
for a bright reflection. Thus 4¼L=¸o = 2m¼, where m is an integer. Use ¸ = no ¸o , where no is
the index of refraction for oil, to find the wavelength in air. The result is
2no L 2(1:20)(460 nm) 1104 nm
=
=
:
m
m
m
For m = 1, ¸ = 1104 nm; for m = 2, ¸ = (1104 nm)=2 = 552 nm; and for m = 3, ¸ =
(1104 nm)=3 = 368 nm. Other wavelengths are shorter. Only ¸ = 552 nm is in the visible range.
(b) A maximum in transmission occurs for wavelengths for which the reflection is a minimum.
The phases of the two reflected waves then differ by an odd multiple of ¼ rad. This means
4¼L=¸o = (2m + 1)¼ and
¸= 4(1:20)(460 nm) 2208 nm
4no L
=
=
:
2m + 1
2m + 1
2m + 1
For m = 0, ¸ = 2208 nm; for m = 1, ¸ = (2208 nm)=3 = 736 nm; and for m = 3, ¸ =
(2208 nm)=5 = 442 nm. Other wavelengths are shorter. Only ¸ = 442 nm falls in the visible
range.
¸= 63
One wave travels a distance 2L further than the other. This wave is reflected twice, once from
the back surface and once from the front surface. Since n2 is greater than n3 there is no change
in phase at the backsurface reflection. Since n1 is greater than n2 there is a phase change of ¼
at the frontsurface reflection. Thus the phase difference of the two waves as they exit material 2
is 4¼L=¸2 + ¼, where ¸2 is the wavelength in material 2. For a maximum in intensity the phase
difference is a multiple of 2¼. Thus 4¼L=¸2 + ¼ = 2m¼, where m is an integer. The solution
for ¸2 is
4(415 nm) 1660 nm
4L
=
=
:
¸2 =
2m ¡ 1
2m ¡ 1
2m ¡ 1
The wavelength in air is
(1:59)(1660 nm) 2639 nm
=
:
2m ¡ 1
2m ¡ 1
For m = 1, ¸ = 2639 nm; for m = 2, ¸ = 880 nm; for m = 3, ¸ = 528 nm; and for m = 4,
¸ = 377 nm. Other wavelengths are shorter. Only ¸ = 528 nm is in the visible range.
¸ = n 2 ¸2 = 71
Consider the interference of waves reflected from the top and bottom surfaces of the air film. The
wave reflected from the upper surface does not change phase on reflection but the wave reflected
from the bottom surface changes phase by ¼ rad. At a place where the thickness of the air film
is L, the condition for fully constructive interference is 2L = (m + 1 )¸, where ¸ (= 683 nm) is
2
the wavelength and m is an integer. This is satisfied for m = 140:
L= (m + 1 )¸ (140:5)(683 £ 10¡9 m)
2
=
= 4:80 £ 10¡5 m = 0:048 mm :
2
2
Chapter 35 227 At the thin end of the air film, there is a bright fringe. It is associated with m = 0. There are,
therefore, 140 bright fringes in all.
75
Consider the interference pattern formed by waves reflected from the upper and lower surfaces
of the air wedge. The wave reflected from the lower surface undergoes a ¼rad phase change
while the wave reflected from the upper surface does not. At a place where the thickness of
the wedge is d, the condition for a maximum in intensity is 2d = (m + 1 )¸, where ¸ is the
2
wavelength in air and m is an integer. Thus d = (2m + 1)¸=4. As the geometry of Fig. 35–47
p
shows, d = R ¡ R2 ¡ r2 , where R is the radius of curvature of the lens and r is the radius of
p
a Newton’s ring. Thus (2m + 1)¸=4 = R ¡ R2 ¡ r2 . Solve for r. First rearrange the terms so
the equation becomes
p
(2m + 1)¸
:
R2 ¡ r 2 = R ¡
4
Now square both sides and solve for r2 . When you take the square root, you should get
r
(2m + 1)R¸ (2m + 1)2 ¸2
¡
:
r=
2
16 If R is much larger than a wavelength, the first term dominates the second and
r
(2m + 1)R¸
r=
:
2
81
Let Á1 be the phase difference of the waves in the two arms when the tube has air in it and let Á2
be the phase difference when the tube is evacuated. These are different because the wavelength
in air is different from the wavelength in vacuum. If ¸ is the wavelength in vacuum, then the
wavelength in air is ¸=n, where n is the index of refraction of air. This means
¸
∙
4¼(n ¡ 1)L
2¼n 2¼
Á1 ¡ Á2 = 2L
¡
=
;
¸
¸
¸
where L is the length of the tube. The factor 2 arises because the light traverses the tube twice,
once on the way to a mirror and once after reflection from the mirror.
Each shift by one fringe corresponds to a change in phase of 2¼ rad, so if the interference pattern
shifts by N fringes as the tube is evacuated,
4¼(n ¡ 1)L
= 2N ¼
¸
and
n= 1+
228 Chapter 35 60(500 £ 10¡9 m)
N¸
= 1+
= 1:00030 :
2L
2(5:0 £ 10¡2 m) 87
Suppose the wave that goes directly to the receiver travels a distance L1 and the reflected wave
travels a distance L2 . Since the index of refraction of water is greater than that of air this last wave
suffers a phase change on reflection of half a wavelength. To obtain constructive interference at
the receiver the difference L2 ¡ L2 in the distances traveled must be an odd multiple of a half
wavelength.
T
Look at the diagram on the right. The right triangle
.....
....
.
" ²...........................................................
on the left, formed by the vertical line from the water
.....
.....
..
..
.....
.....
..
j
..
..... L1
.....
..
.....
.....
..
.....
.....
to the transmitter T, the ray incident on the water,
..
..
.....
.....
..
..
.....
.....
j
..
.....
.....
..
.....
.....
..
..
.....
.....
..
and the water line, gives Da = a= tan µ and the right
..
.....
.....
j
..
.....
.....
.. L
..... R
.....
..
..
2a
....
...
.
..
..
.
.²
..
..
..
triangle on the right, formed by the vertical line from
a
..
..
..
..
"
..
..
..
..
.
..
..
..
..
j
..
..
the water to the receiver R, the reflected ray, and the
j
L2b .............
..
..
..
..
.
.
..
..
..
..
x
j
..
..
.
..
water line gives Db = x= tan µ. Since Da + Db = D,
..
..
..
..
..
..
..
..
..
j
..
..
..
j
..
..
..
..
µ ..................... µ
#
#
a+x
:
tan µ =
Ã ¡ Da ¡ ¡ ¡ Db ¡
¡¡
¡ !Ã
!
D
p
Use the identity sin2 µ = tan2 µ=(1 + tan2 µ) to show that sin µ = (a + x)= D 2 + (a + x)2 . This
means
p
a D 2 + (a + x)2
a
=
L2a =
sin µ
a+x
and
p
x D 2 + (a + x)2
x
=
;
L2b =
sin µ
a+x
so
p
(a + x) D2 + (a + x)2 p 2
= D + (a + x)2 :
L2 = L2a + L2b =
a+x Use the binomial theorem, with D 2 large and a2 + x2 small, to approximate this expression:
L2 ¼ D + (a + x)2 =2D.
p
The distance traveled by the direct wave is L1 = D 2 + (a ¡ x)2 . Use the binomial theorem to
approximate this expression: L1 ¼ D + (a ¡ x)2 =2D. Thus
L2 ¡ L1 ¼ D + a2 ¡ 2ax + x2 2ax
a2 + 2ax + x2
¡D¡
=
:
2D
2D
D Set this equal to (m + 1 )¸, where m is zero or a positive integer. Solve for x. The result is
2
x = (m + 1 )(D=2a)¸.
2
89
Bright fringes occur at an angle µ such that d sin µ = m¸, where d is the slit separation, ¸ is the
wavelength in the medium of propagation, and m is an integer. Near the center of the pattern the
angles are small and sin µ can be approximated by µ in radians. Thus µ = m¸=d and the angular
separation of two adjacent bright fringes is ¢µ = ¸=d. When the arrangement is immersed in
water the angular separation of the fringes becomes ¢µ0 = ¸w =d, where ¸w is the wavelength in
Chapter 35 229 water. Since ¸w = ¸=nw , where nw is the index of refraction of water, ¢µ0 = ¸=nw d = (¢µ)=nw .
Since the units of the angles cancel from this equation we may substitute the angles in degrees
and obtain ¢ 0 = 0:30± =1:33 = 0:23± .
93
(a) For wavelength ¸ dark bands occur where the path difference is an odd multiple of ¸=2. That
is, where the path difference is (2m + 1)¸=2, where m is an integer. The fourth dark band from
the central bright fringe is associated with m = 3 and is 7¸=2 = 7(500 nm)=2 = 1750 nm.
(b) The angular position µ of the first bright band on either side of the central band is given by
sin µ = ¸=d, where d is the slit separation. The distance on the screen is given by ¢y = D tan µ,
where D is the distance from the slits to the screen. Because µ is small its sine and tangent are
very nearly equal and ¢y = D sin µ = D¸=d.
Dark bands have angular positions that are given by sin µ = (m + 1 )¸=d and, for the fourth dark
2
band, m = 3 and sin µ4 = (7=2)¸=d. Its distance on the screen from the central fringe is ¢y4 =
D tan µ4 = D sin µ4 = 7D¸=2d. This means that D¸=d = 2¢y4 =7 = 2(1:68 cm)=7 = 0:48 cm.
Note that this is ¢y.
99
Minima occur at angles µ for which sin µ = (m + 1 )¸=d, where ¸ is the wavelength, d is the slit
2
separation, and m is an integer. For the first minimum, m = 0 and sin µ1 = ¸=2d. For the tenth
minimum, m = 9 and sin µ10 = 19¸=2d.
The distance on the screen from the central fringe to a minimum is y = D tan µ, where D is the
distance from the slits to the screen. Since the angle is small we may approximate its tangent
with its sine and write y = D sin µ = D(m + 1 )¸=d. Thus the separation of the first and tenth
2
minima is
¶
µ
9D¸
D 19¸ ¸
¡
=
¢y =
2
2
d
d
and d ¢y (0:150 £ 10¡3 m)(18:0 £ 10¡3 m)
=
= 6:00 £ 10¡7 m :
¸=
¡2 m)
9D
9(50:0 £ 10 103
The difference in the path lengths of the two beams is 2x, so their difference in phase when
they reach the detector is Á = 4¼x=¸, where ¸ is the wavelength. Assume their amplitudes
are the same. According to Eq. 35–22 the intensity associated with the addition of two waves
is proportional to the square of the cosine of half their phase difference. Thus the intensity of
the light observed in the interferometer is proportional to cos2 (2¼x=¸). Since the intensity is
maximum when x = 0 (and the arms have equal lengths), the constant of proportionality is the
maximum intensity Im and I = Im cos2 (2¼x=¸). 230 Chapter 35 Chapter 36 9
The condition for a minimum of intensity in a singleslit diffraction pattern is a sin µ = m¸,
where a is the slit width, ¸ is the wavelength, and m is an integer. To find the angular position
of the first minimum to one side of the central maximum, set m = 1:
¶
µ ¶
µ
589 £ 10¡9 m
¡1 ¸
¡1
= 5:89 £ 10¡4 rad :
= sin
µ1 = sin
1:00 £ 10¡3 m
a
If D is the distance from the slit to the screen, the distance on the screen from the center of the
pattern to the minimum is y1 = D tan µ1 = (3:00 m) tan(5:89 £ 10¡4 rad) = 1:767 £ 10¡3 m.
To find the second minimum, set m = 2:
¸
∙
¡9
m)
¡1 2(589 £ 10
= 1:178 £ 10¡3 rad :
µ2 = sin
1:00 £ 10¡3 m
The distance from the pattern center to the minimum is y2 = D tan µ2 = (3:00 m) tan(1:178 £
10¡3 rad) = 3:534 £ 10¡3 m. The separation of the two minima is ¢y = y2 ¡ y1 = 3:534 mm ¡
1:767 mm = 1:77 mm.
17
(a) The intensity for a singleslit diffraction pattern is given by
I = Im sin2 ®
;
®2 where ® = (¼a=¸) sin µ, a is the slit width and ¸ is the wavelength. The angle µ is measured
from the forward direction. You want I = Im =2, so
1
sin2 ® = ®2 :
2
(b) Evaluate sin2 ® and ®2 =2 for ® = 1:39 rad and compare the results. To be sure that 1:39 rad
is closer to the correct value for ® than any other value with three significant digits, you should
also try 1:385 rad and 1:395 rad.
(c) Since ® = (¼a=¸) sin µ,
µ ¶
¡1 ®¸
µ = sin
:
¼a
Now ®=¼ = 1:39=¼ = 0:442, so ¡1 µ = sin µ 0:442¸
a ¶ :
Chapter 36 231 The angular separation of the two points of half intensity, one on either side of the center of the
diffraction pattern, is
µ
¶
¡1 0:442¸
:
¢µ = 2µ = 2 sin
a
(d) For a=¸ = 1:0,
¢µ = 2 sin¡1 (0:442=1:0) = 0:916 rad = 52:5± :
(e) For a=¸ = 5:0,
¢µ = 2 sin¡1 (0:442=5:0) = 0:177 rad = 10:1± :
(f) For a=¸ = 10,
¢µ = 2 sin¡1 (0:442=10) = 0:0884 rad = 5:06± :
21
(a) Use the Rayleigh criteria. To resolve two point sources, the central maximum of the diffraction
pattern of one must lie at or beyond the first minimum of the diffraction pattern of the other.
This means the angular separation of the sources must be at least µR = 1:22¸=d, where ¸ is the
wavelength and d is the diameter of the aperture. For the headlights of this problem,
µR = 1:22(550 £ 10¡9 m)
= 1:3 £ 10¡4 rad :
5:0 £ 10¡3 m (b) If D is the distance from the headlights to the eye when the headlights are just resolvable
and ` is the separation of the headlights, then ` = D tan µR ¼ DµR , where the small angle
approximation tan µR ¼ µR was made. This is valid if µR is measured in radians. Thus
D = `=µR = (1:4 m)=(1:34 £ 10¡4 rad) = 1:0 £ 104 m = 10 km.
25
(a) Use the Rayleigh criteria: two objects can be resolved if their angular separation at the
observer is greater than µR = 1:22¸=d, where ¸ is the wavelength of the light and d is the
diameter of the aperture (the eye or mirror). If D is the distance from the observer to the objects,
then the smallest separation ` they can have and still be resolvable is ` = D tan µR ¼ DµR , where
µR is measured in radians. The small angle approximation tan µR ¼ µR was made. Thus
`= 1:22D¸ 1:22(8:0 £ 1010 m)(550 £ 10¡9 m)
= 1:1 £ 107 m = 1:1 £ 104 km :
=
5:0 £ 10¡3 m
d This distance is greater than the diameter of Mars. One part of the planet’s surface cannot be
resolved from another part.
(b) Now d = 5:1 m and
`=
232 Chapter 36 1:22(8:0 £ 1010 m)(550 £ 10¡9 m)
= 1:1 £ 104 m = 11 km :
5:1 m 29
(a) The first minimum in the diffraction pattern is at an angular position µ, measured from the
center of the pattern, such that sin µ = 1:22¸=d, where ¸ is the wavelength and d is the diameter
of the antenna. If f is the frequency, then the wavelength is
¸=
Thus
¡1 µ = sin µ 1:22¸
d c 3:00 £ 108 m=s
= 1:36 £ 10¡3 m :
=
220 £ 109 Hz
f ¶ = sin ¡1 µ 1:22(1:36 £ 10¡3 m)
55:0 £ 10¡2 m ¶ = 3:02 £ 10¡3 rad : The angular width of the central maximum is twice this, or 6:04 £ 10¡3 rad (0:346± ).
(b) Now ¸ = 1:6 cm and d = 2:3 m, so
µ
¶
¡2
m)
¡1 1:22(1:6 £ 10
= 8:5 £ 10¡3 rad :
µ = sin
2:3 m
The angular width of the central maximum is 1:7 £ 10¡2 rad (0:97± ).
39
(a) The angular positions µ of the bright interference fringes are given by d sin µ = m¸, where d is
the slit separation, ¸ is the wavelength, and m is an integer. The first diffraction minimum occurs
at the angle µ1 given by a sin µ1 = ¸, where a is the slit width. The diffraction peak extends
from ¡µ1 to +µ1 , so you want to count the number of values of m for which ¡µ1 < µ < +µ1 , or
what is the same, the number of values of m for which ¡ sin µ1 < sin µ < + sin µ1 . This means
¡1=a < m=d < 1=a or ¡d=a < m < +d=a. Now d=a = (0:150 £ 10¡3 m)=(30:0 £ 10¡6 m) =
5:00, so the values of m are m = ¡4, ¡3, ¡2, ¡1, 0, +1, +2, +3, and +4. There are nine fringes.
(b) The intensity at the screen is given by
I = Im ¡ 2 ¢
cos ¯ µ sin ®
® ¶2 ; where ® = (¼a=¸) sin µ, ¯ = (¼d=¸) sin µ, and Im is the intensity at the center of the pattern.
For the third bright interference fringe, d sin µ = 3¸, so ¯ = 3¼ rad and cos2 ¯ = 1. Similarly,
® = 3¼a=d = 3¼=5:00 = 0:600¼ rad and
µ sin ®
® ¶2 = µ sin 0:600¼
0:600¼ ¶2 = 0:255 : The intensity ratio is I=Im = 0:255.
45
The ruling separation is d = 1=(400 mm¡1 ) = 2:5 £ 10¡3 mm. Diffraction lines occur at angles µ
such that d sin µ = m¸, where ¸ is the wavelength and m is an integer. Notice that for a given
Chapter 36 233 order, the line associated with a long wavelength is produced at a greater angle than the line
associated with a shorter wavelength. Take ¸ to be the longest wavelength in the visible spectrum
(700 nm) and find the greatest integer value of m such that µ is less than 90± . That is, find the
greatest integer value of m for which m¸ < d. Since d=¸ = (2:5£10¡6 m)=(700£10¡9 m) = 3:57,
that value is m = 3. There are three complete orders on each side of the m = 0 order. The
second and third orders overlap.
51
(a) Maxima of a diffraction grating pattern occur at angles µ given by d sin µ = m¸, where d is the
slit separation, ¸ is the wavelength, and m is an integer. The two lines are adjacent, so their order
numbers differ by unity. Let m be the order number for the line with sin µ = 0:2 and m + 1 be the
order number for the line with sin µ = 0:3. Then 0:2d = m¸ and 0:3d = (m+1)¸. Subtract the first
equation from the second to obtain 0:1d = ¸, or d = ¸=0:1 = (600 £ 10¡9 m)=0:1 = 6:0 £ 10¡6 m.
(b) Minima of the singleslit diffraction pattern occur at angles µ given by a sin µ = m¸, where a
is the slit width. Since the fourthorder interference maximum is missing, it must fall at one of
these angles. If a is the smallest slit width for which this order is missing, the angle must be given
by a sin µ = ¸. It is also given by d sin µ = 4¸, so a = d=4 = (6:0 £ 10¡6 m)=4 = 1:5 £ 10¡6 m. (c) First, set µ = 90± and find the largest value of m for which m¸ < d sin µ. This is the highest
order that is diffracted toward the screen. The condition is the same as m < d=¸ and since
d=¸ = (6:0 £ 10¡6 m)=(600 £ 10¡9 m) = 10:0, the highest order seen is the m = 9 order. (d) and (e) The fourth and eighth orders are missing so the observable orders are m = 0, 1, 2,
3, 5, 6, 7, and 9. The second highest order is the m = 7 order and the third highest order is the
m = 6 order.
61
If a grating just resolves two wavelengths whose average is ¸avg and whose separation is ¢¸,
then its resolving power is defined by R = ¸avg =¢¸. The text shows this is N m, where N is
the number of rulings in the grating and m is the order of the lines. Thus ¸avg =¢¸ = N m and
N= ¸avg
656:3 nm
= 3:65 £ 103 rulings :
=
m ¢¸ (1)(0:18 nm) 73
We want the reflections to obey the Bragg condition 2d sin µ = m¸, where µ is the angle between
the incoming rays and the reflecting planes, ¸ is the wavelength, and m is an integer. Solve for
µ:
¸
¸
∙
∙
¡9
m)m
¡1 m¸
¡1 (0:125 £ 10
= sin
= sin¡1 (0:2480m) :
µ = sin
¡9 m)
2d
2(0:252 £ 10
For m = 1 this gives µ = 14:4± . The crystal should be turned 45± ¡ 14:4± = 30:6± clockwise. For m = 2 it gives µ = 29:7± . The crystal should be turned 45± ¡ 29:7± = 15:3± clockwise. For m = 3 it gives µ = 48:1± . The crystal should be turned 48:1± ¡ 45± = 3:1± counterclockwise.
234 Chapter 36 For m = 4 it gives µ = 82:8± . The crystal should be turned 82:8± ¡45± = 37:8± counterclockwise.
There are no intensity maxima for m > 4 as you can verify by noting that m¸=2d is greater than
1 for m greater than 4. For clockwise turns the smaller value is 15:3± and the larger value is
30:6± . For counterclockwise turns the smaller value is 3:1± and the larger value is 37:8± .
77
Intensity maxima occur at angles µ such that d sin µ = m¸, where d is the separation of adjacent
rulings and ¸ is the wavelength. Here the ruling separation is 1=(200 mm¡1 ) = 5:00£10¡3 mm =
5:00 £ 10¡6 m. Thus
¸= d sin µ (5:00 £ 10¡6 m) sin 30:0± 2:50 £ 10¡6 m
=
=
:
m
m
m For m = 1, ¸ = 2500 nm; for m = 2, ¸ = 1250 nm; for m ¡ 3, ¸ = 833 nm; for m = 4,
¸ = 625 nm; for m = 5, ¸ = 500 nm, and for m = 6, ¸ = 417 nm. Only the last three are in
the visible range, so the longest wavelength in the visible range is 625 nm, the next longest is
500 nm, and the third longest is 417 nm.
79
Suppose mo is the order of the minimum for orange light, with wavelength ¸o , and mbg is
the order of the minimum for bluegreen light, with wavelength ¸bg . Then a sin µ = mo ¸o and
a sin µ = mbg ¸bg . Thus mo ¸o = mbg ¸bg and mbg =mo = ¸o =¸bg = (600 nm)=(500 nm) = 6=5.
The smallest two integers with this ratio are mbg = 6 and mo = 5. The slit width is
a= 5(600 £ 10¡9 m
mo ¸o
=
= 3:0 £ 10¡3 m :
sin µ
sin(1:00 £ 10¡3 rad) Other values for mo and mbg are possible but these are associated with a wider slit.
81
(a) Since the first minimum of the diffraction pattern occurs at the angle µ such that sin µ = ¸=a,
where ¸ is the wavelength and a is the slit width, the central maximum extends from µ1 =
¡ sin¡1 (¸=a) to µ2 = + sin¡1 (¸=a). Maxima of the twoslit interference pattern are at angles µ
such that sin µ = m¸=d, where d is the slit separation and m is an integer. We wish to know
the number of values of m such that sin¡1 (m¸=d) lies between ¡ sin¡1 (¸=a) and + sin(¸=a) or,
what is the same, the number of values of m such that m=d lies between ¡1=a and +1=a. The
greatest m can be is the greatest integer that is smaller than d=a = (14 ¹m)=(2:0 ¹m) = 7. (The
m = 7 maximum does not appear since it coincides with a minimum of the diffraction pattern.)
There are 13 such values: 0, §1, §2, §3; §4; §5, and §6. Thus 13 interference maxima appear
in the central diffraction envelope.
(b) The first diffraction envelope extends from µ1 = sin¡1 (¸=a) to µ2 = sin¡1 (2¸=a). Thus we
wish to know the number of values of m such that m=d is greater than 1=a and less than 2=a.
Since d = 7:0a, m can be 8, 9, 10, 11, 12, or 13. That is, there are 6 interference maxima in the
first diffraction envelope.
Chapter 36 235 93
If you divide the original slit into N strips and represent the light from each strip, when it reaches
the screen, by a phasor, then at the central maximum in the diffraction pattern you add N phasors,
all in the same direction and each with the same amplitude. The intensity there is proportional
to N 2 . If you double the slit width, you need 2N phasors if they are each to have the amplitude
of the phasors you used for the narrow slit. The intensity at the central maximum is proportional
to (2N )2 and is, therefore, four times the intensity for the narrow slit. The energy reaching the
screen per unit time, however, is only twice the energy reaching it per unit time when the narrow
slit is in place. The energy is simply redistributed. For example, the central peak is now half as
wide and the integral of the intensity over the peak is only twice the analogous integral for the
narrow slit.
95
(a) Since the resolving power of a grating is given by R = ¸=¢¸ and by N m, the range of
wavelengths that can just be resolved in order m is ¢¸ = ¸=N m. Here N is the number
of rulings in the grating and ¸ is the average wavelength. The frequency f is related to the
wavelength by f ¸ = c, where c is the speed of light. This means f ¢¸ + ¸ ¢f = 0, so
¸
¸2
¢¸ = ¡ ¢f = ¡ ¢f ;
f
c
where f = c=¸ was used. The negative sign means that an increase in frequency corresponds to
a decrease in wavelength. We may interpret ¢f as the range of frequencies that can be resolved
and take it to be positive. Then
¸
¸2
¢f =
c
Nm
and
c
¢f =
:
N m¸
(b) The difference in travel time for waves traveling along the two extreme rays is ¢t = ¢L=c,
where ¢L is the difference in path length. The waves originate at slits that are separated by
(N ¡ 1)d, where d is the slit separation and N is the number of slits, so the path difference is
¢L = (N ¡ 1)d sin µ and the time difference is
¢t = (N ¡ 1)d sin µ
:
c If N is large, this may be approximated by ¢t = (N d=c) sin µ. The lens does not affect the travel
time.
(c) Substitute the expressions you derived for ¢t and ¢f to obtain
³ c ´ µ N d sin µ ¶ d sin µ
¢f ¢t =
=
= 1:
N m¸
c
m¸
The condition d sin µ = m¸ for a diffraction line was used to obtain the last result.
236 Chapter 36 101
The dispersion of a grating is given by D = dµ=d¸, where µ is the angular position of a line
associated with wavelength ¸. The angular position and wavelength are related by ` sin µ = m¸,
where ` is the slit separation and m is an integer. Differentiate this with respect to µ to obtain
(dµ=d¸) ` cos µ = m or
`µ
m
D=
=
:
`¸ ` cos µ
Now m = (`=¸) sin µ, so
tan µ
` sin µ
=
:
D=
`¸ cos µ
¸
The trigonometric identity tan µ = sin µ= cos µ was used. Chapter 36 237 Chapter 37 11
(a) The rest length L0 (= 130 m) of the spaceship and its length L as measured by the timing
p
p
station are related by L = L0 =° = L0 1 ¡ ¯ 2 , where ° = 1= 1 ¡ ¯ 2 and ¯ = v=c. Thus
p
L = (130 m) 1 ¡ (0:740)2 = 87:4 m.
(b) The time interval for the passage of the spaceship is
¢t = L
87:4 m
= 3:94 £ 10¡7 s :
=
v (0:740)(2:9979 £ 108 m=s) 19
The proper time is not measured by clocks in either frame S or frame S 0 since a single clock at rest
in either frame cannot be present at the origin and at the event. The full Lorentz transformation
must be used:
x0 = °[x ¡ vt]
t0 = °[t ¡ ¯x=c] ;
p
p
where ¯ = v=c = 0:950 and ° = 1= 1 ¡ ¯ 2 = 1= 1 ¡ (0:950)2 = 3:2026. Thus
£
¤
x0 = (3:2026) 100 £ 103 m ¡ (0:950)(2:9979 £ 108 m=s)(200 £ 10¡6 s and = 1:38 £ 105 m = 138 km ¸
∙
(0:950)(100 £ 103 m)
¡6
= ¡3:74 £ 10¡4 s = ¡374 ¹s :
t = (3:2026) 200 £ 10 s ¡
8 m=s
2:9979 £ 10
0 21
(a) The Lorentz factor is
°=p 1 1
=p
= 1:25 :
1 ¡ (0:600)2
1 ¡ ¯2 (b) In the unprimed frame, the time for the clock to travel from the origin to x = 180 m is
t= x
180 m
= 1:00 £ 10¡6 s :
=
v (0:600)(2:9979 £ 108 m=s) The proper time interval between the two events (at the origin and at x = 180 m) is measured by
the clock itself. The reading on the clock at the beginning of the interval is zero, so the reading
at the end is
t 1:00 £ 10¡6 s
t0 = =
= 8:00 £ 10¡7 s :
1:25
°
238 Chapter 37 29
Use Eq. 37–29 with u0 = 0:40c and v = 0:60c. Then
u= 0:40c + 0:60c
= 0:81c :
1 + (0:40c)(0:60c)=c2 33
Calculate the speed of the micrometeorite relative to the spaceship. Let S 0 be the reference frame
for which the data is given and attach frame S to the spaceship. Suppose the micrometeorite is
going in the positive x direction and the spaceship is going in the negative x direction, both as
viewed from S 0 . Then, in Eq. 37–29, u0 = 0:82c and v = 0:82c. Notice that v in the equation
is the velocity of S 0 relative to S. Thus the velocity of the micrometeorite in the frame of the
spaceship is
0:82c + 0:82c
u0 + v
=
= 0:9806c :
u=
1 + u0 v=c2 1 + (0:82c)(0:82c)=c2
The time for the micrometeorite to pass the spaceship is
¢t = 350 m
L
= 1:19 £ 10¡6 s :
=
u (0:9806)(2:9979 £ 108 m=s) 37
The spaceship is moving away from Earth, so the frequency received is given by
s
1¡¯
;
f = f0
1+¯
where f0 is the frequency in the frame of the spaceship, ¯ = v=c, and v is the speed of the
spaceship relative to Earth. See Eq. 37–31. Thus
r
1 ¡ 0:9000
= 22:9 MHz :
f = (100 MHz)
1 + 0:9000
39
The spaceship is moving away from Earth, so the frequency received is given by
s
1¡¯
;
f = f0
1+¯
where f0 is the frequency in the frame of the spaceship, ¯ = v=c, and v is the speed of the
spaceship relative to Earth. See Eq. 37–31. The frequency f and wavelength ¸ are related by
f ¸ = c, so if ¸0 is the wavelength of the light as seen on the spaceship and ¸ is the wavelength
detected on Earth, then
s
r
1+¯
1 + 0:20
= (450 nm)
= 550 nm :
¸ = ¸0
1¡¯
1 ¡ 0:20
Chapter 37 239 This is in the yellowgreen portion of the visible spectrum.
43
Use the two expressions for the total energy: E = mc2 + K and E = °mc2 , where m is the mass
p
of an electron, K is the kinetic energy, and ° = 1= 1 ¡ ¯ 2 . Thus mc2 + K = °mc2 and
° =1+ K
(100:000 £ 106 eV)(1:602 176 462 J=eV)
=1+
= 196:695 :
(9:109 381 88 £ 10¡31 kg)(2:997 924 58 £ 108 m=s)2
mc2 Now ° 2 = 1=(1 ¡ ¯ 2 ), so ¯= s 1
1¡ 2 =
° s 1¡ 1
= 0:999 987 :
(196:695)2 53
The energy equivalent of one tablet is mc2 = (320 £ 10¡6 kg)(2:9979 £ 108 m=s)2 = 2:88 £ 1013 J.
This provides the same energy as (2:88 £ 1013 J)=(3:65 £ 107 J=L) = 7:89 £ 105 L of gasoline.
The distance the car can go is d = (7:89 £ 105 L)(12:75 km=L) = 1:01 £ 107 km.
71 p
The energy of the electron is given by E = mc2 = 1 ¡ (v=c)2 , which yields
s
s
¸2
∙ 2 ¸2
∙
mc
(9:11 £ 10¡31 kg)(2:9979 £ 108 m=s)2
= 0:99999994c ¼ c
c= 1¡
v = 1¡
(1533 MeV)(1:602 £ 10¡13 J=MeV)
E for the speed v of the electron. In the rest frame of Earth the trip took time t = 26 y. A clock
traveling with the electron records the proper time of the trip, so the trip in the rest frame of the
electron took time t0 = t=°. Now
°= E
1533 MeV)(1:602 £ 10¡13 J=MeV)
= 3:0 £ 103
=
mc2 (9:11 £ 10¡31 kg)(2:9979 £ 108 m=s) and t0 = (26 y)=(3:0 £ 103 ) = 8:7 £ 10¡3 y. The distance traveled is 8:7 £ 10¡3 ly.
73
Start with (pc)2 = K 2 + 2Kmc2 , where p is the momentum of the particle, K is its kinetic energy,
and m is its mass. For an electron mc2 = 0:511 MeV, so
p
p
pc = K 2 + 2Kmc2 = (2:00 MeV)2 + 2(2:00 MeV)(0:511 MeV) = 2:46 MeV : Thus p = 2:46 MeV=c. 75
The work required is the increased in the energy of the proton. The energy is given by E =
mc2 =[1 ¡ (v=c)2 ]. Let v1 be the initial speed and v2 be the final speed. Then the work is 240 938 MeV
mc2
938 MeV
=p
= 189 MeV ;
¡p
¡p
2
2
2
1 ¡ (v2 =c)
1 ¡ (v=c)
1 ¡ (0:9860)
1 ¡ (0:9850)2 W =p mc2 Chapter 37 where mc2 = 938 MeV was used.
77
(a) Let v be the speed of either satellite, relative to Earth. According to the Galilean velocity
transformation equation the relative speed is vrel = 2v = 2(2:7 £ 104 km=h = 5:4 £ 104 km=h.
(b) The correct relativistic transformation equation is
vrel =
The fractional error is
fract err = 2v
:
2
1 + v2
c 1
2v ¡ vrel
=1¡
:
2
2v
1 + v2
c The speed of light is 1:08 £ 109 km=h, so
fract err = 1
= 6:3 £ 10¡10 :
4
2
(2:7 £ 10 km=h)
1+
(1:08 £ 109 km=h)2 Chapter 37 241 Chapter 38 7
(a) Let R be the rate of photon emission (number of photons emitted per unit time) and let E be
the energy of a single photon. Then the power output of a lamp is given by P = RE if all the
power goes into photon production. Now E = hf = hc=¸, where h is the Planck constant, f is
the frequency of the light emitted, and ¸ is the wavelength. Thus P = Rhc=¸ and R = ¸P=hc.
The lamp emitting light with the longer wavelength (the 700nm lamp) emits more photons per
unit time. The energy of each photon is less so it must emit photons at a greater rate.
(b) Let R be the rate of photon production for the 700 nm lamp Then
R= ¸P
(700 £ 10¡9 m)(400 J=s)
= 1:41 £ 1021 photon=s :
=
hc (6:626 £ 10¡34 J ¢ s)(2:9979 £ 108 m=s) 17
The energy of an incident photon is E = hf = hc=¸, where h is the Planck constant, f is the
frequency of the electromagnetic radiation, and ¸ is its wavelength. The kinetic energy of the
most energetic electron emitted is Km = E ¡ © = (hc=¸) ¡ ©, where © is the work function for
sodium. The stopping potential V0 is related to the maximum kinetic energy by eV0 = Km , so
eV0 = (hc=¸) ¡ © and
¸= (6:626 £ 19¡34 J ¢ s)(2:9979 £ 108 m=s)
hc
=
= 1:7 £ 10¡7 m :
(5:0 eV + 2:2 eV)(1:602 £ 10¡19 J=eV)
eV0 + © Here eV0 = 5:0 eV was used.
21
(a) The kinetic energy Km of the fastest electron emitted is given by Km = hf ¡ © = (hc=¸) ¡ ©,
where © is the work function of aluminum, f is the frequency of the incident radiation, and ¸ is
its wavelength. The relationship f = c=¸ was used to obtain the second form. Thus
Km = (6:626 £ 10¡34 J ¢ s)(2:9979 £ 108 m=s)
¡ 4:20 eV = 2:00 eV :
(200 £ 10¡9 m)(1:602 £ 10¡19 J=eV) (b) The slowest electron just breaks free of the surface and so has zero kinetic energy.
(c) The stopping potential V0 is given by Km = eV0 , so V0 = Km =e = (2:00 eV)=e = 2:00 V.
(d) The value of the cutoff wavelength is such that Km = 0. Thus hc=¸ = © or
¸=
242 Chapter 38 hc (6:626 £ 10¡34 J ¢ s)(2:9979 £ 108 m=s)
= 2:95 £ 10¡7 m :
=
(4:2 eV)(1:602 £ 10¡19 J=eV)
© If the wavelength is longer, the photon energy is less and a photon does not have sufficient energy
to knock even the most energetic electron out of the aluminum sample.
29
(a) When a photon scatters from an electron initially at rest, the change in wavelength is given
by ¢¸ = (h=mc)(1 ¡ cos Á), where m is the mass of an electron and Á is the scattering angle.
Now h=mc = 2:43 £ 10¡12 m = 2:43 pm, so ¢¸ = (2:43 pm)(1 ¡ cos 30± ) = 0:326 pm. The final
wavelength is ¸0 = ¸ + ¢¸ = 2:4 pm + 0:326 pm = 2:73 pm.
(b) Now ¢¸ = (2:43 pm)(1 ¡ cos 120± ) = 3:645 pm and ¸0 = 2:4 pm + 3:645 pm = 6:05 pm.
43
Since the kinetic energy K and momentum p are related by K = p2 =2m, the momentum of the
p
p
electron is p = 2mK and the wavelength of its matter wave is ¸ = h=p = h= 2mK. Replace
K with eV , where V is the accelerating potential and e is the fundamental charge, to obtain
6:626 £ 10¡34 J ¢ s
h
=p
¸= p
2meV
2(9:109 £ 10¡31 kg)(1:602 £ 10¡19 C)(25:0 £ 103 V)
= 7:75 £ 10¡12 m = 7:75 pm :
47
(a) The kinetic energy acquired is K = qV , where q is the charge on an ion and V is the
accelerating potential. Thus K = (1:602 £ 10¡19 C)(300 V) = 4:80 £ 10¡17 J. The mass of
a single sodium atom is, from Appendix F, m = (22:9898 g=mol)=(6:02 £ 1023 atom=mol) =
3:819 £ 10¡23 g = 3:819 £ 10¡26 kg. Thus the momentum of an ion is
p
p
p = 2mK = 2(3:819 £ 10¡26 kg)(4:80 £ 10¡17 J) = 1:91 £ 10¡21 kg ¢ m=s :
(b) The de Broglie wavelength is
¸= 6:63 £ 10¡34 J ¢ s
h
= 3:47 £ 10¡13 m :
=
p 1:91 £ 10¡21 kg ¢ m=s 49
Since the kinetic energy K and momentum p are related by K = p2 =2m, the momentum of the
p
p
electron is p = 2mK and the wavelength of its matter wave is ¸ = h=p = h= 2mK. Thus
µ ¶2
µ
¶2
1
1
h
6:626 £ 10¡34 J ¢ s)
=
K=
2m ¸
2(9:11 £ 10¡31 kg)
590 £ 10¡9 m)
= 6:92 £ 10¡25 J = 4:33 £ 10¡6 eV :
59
The angular wave number k is related to the wavelength ¸ by k = 2¼=¸ and the wavelength is
related to the particle momentum p by ¸ = h=p, so k = 2¼p=h. Now the kinetic energy K and
Chapter 38 243 the momentum are related by K = p2 =2m, where m is the mass of the particle. Thus p =
and
p
2¼ 2mK
:
k=
h p
2mK 61
For U = U0 , Schrodinger’s equation becomes
¨
d2 Ã 8¼2 m
+
[E ¡ U0 ] Ã = 0 :
dx2
h2
Substitute Ã = Ã0 eikx . The second derivative is d2 Ã=dx2 = ¡k 2 Ã0 eikx = ¡k 2 Ã. The result is
¡k2 Ã + 8¼2 m
[E ¡ U0 ] Ã = 0 :
h2 Solve for k and obtain
k= r 2¼ p
8¼ 2 m
[E ¡ U0 ] =
2m [E ¡ U0 ] :
2
h
h 67
(a) If m is the mass of the particle and E is its energy, then the transmission coefficient for a
barrier of height U and width L is given by
T = e¡2kL ;
where r 8¼ 2 m(U ¡ E)
:
h2
If the change ¢U in U is small (as it is), the change in the transmission coefficient is given by
k= ¢T =
Now Thus dk
1
= p
dU 2 U ¡ E r dk
dT
¢U = ¡2LT
¢U :
dU
dU 1
8¼ 2 m
=
2
2(U ¡ E)
h r k
8¼2 m(U ¡ E)
:
=
2
2(U ¡ E)
h ¢U
:
U ¡E
For the data of Sample Problem 38–7, 2kL = 10:0, so kL = 5:0 and
¢T = ¡LT k ¢U
(0:010)(6:8 eV)
¢T
= ¡0:20 :
= ¡kL
= ¡(5:0)
6:8 eV ¡ 5:1 eV
T
U ¡E There is a 20% decrease in the transmission coefficient.
244 Chapter 38 (b) The change in the transmission coefficient is given by
¢T = dT
¢L = ¡2ke¡2kL ¢L = ¡2kT ¢L
dL and ¢T
= ¡2k ¢L = ¡2(6:67 £ 109 m¡1 )(0:010)(750 £ 10¡12 m) = ¡0:10 :
T
There is a 10% decrease in the transmission coefficient.
(c) The change in the transmission coefficient is given by
¢T = dT
dk
dk
¢E = ¡2Le¡2kL
¢E = ¡2LT
¢E :
dE
dE
dE Now dk=dE = ¡dk=dU = ¡k=2(U ¡ E), so
¢T
¢E
(0:010)(5:1 eV)
= 0:15 :
= kL
= (5:0)
6:8 eV ¡ 5:1 eV
T
U ¡E There is a 15% increase in the transmission coefficient. 79
The uncertainty in the momentum is ¢p = m ¢v = (0:50 kg)(1:0 m=s) = 0:50 kg ¢ m=s, where ¢v
is the uncertainty in the velocity. Solve the uncertainty relationship ¢x ¢p ¸ for the minimum
h
uncertainty in the coordinate x: ¢x =
h=¢p = (0:60 J ¢ s)=2¼(0:50 kg ¢ m=s) = 0:19 m. Chapter 38 245 Chapter 39 13
R
The probability that the electron is found in any interval is given by P = jÃj2 dx, where the
integral is over the interval. If the interval width ¢x is small, the probability can be approximated
by P = jÃj2 ¢x, where the wave function is evaluated for the center of the interval, say. For an
electron trapped in an infinite well of width L, the ground state probability density is
2 2 ³ ¼x ´
2
jÃj = sin
;
L
L
so
µ
¶
³ ¼x ´
2 ¢x
P =
sin2
:
L
L
(a) Take L = 100 pm, x = 25 pm, and ¢x = 5:0 pm. Then
·
¸
·
¸
2(5:0 pm)
2 ¼(25 pm)
P =
sin
= 0:050 :
100 pm
100 pm
(b) Take L = 100 pm, x = 50 pm, and ¢x = 5:0 pm. Then
·
¸
·
¸
2(5:0 pm)
2 ¼(50 pm)
P =
sin
= 0:10 :
100 pm
100 pm
(c) Take L = 100 pm, x = 90 pm, and ¢x = 5:0 pm. Then
·
¸
·
¸
2(5:0 pm)
2 ¼(90 pm)
P =
sin
= 0:0095 :
100 pm
100 pm
25
The energy levels are given by
Enx ny "
#
"
#
n2
h2 n2 n2
h2
y
y
x
=
+
=
n2 +
;
8m L2 L2
8mL2 x 4
x
y where the substitutions Lx = L and Ly = 2L were made. In units of h2 =8mL2 , the energy
levels are given by n2 + n2 =4. The lowest five levels are E1;1 = 1:25, E1;2 = 2:00, E1;3 = 3:25,
x
y
E2;1 = 4:25, and E2;2 = E1;4 = 5:00. A little thought should convince you that there are no other
possible values for the energy less than 5.
The frequency of the light emitted or absorbed when the electron goes from an initial state i to a
final state f is f = (Ef ¡ Ei )=h and in units of h=8mL2 is simply the difference in the values of
n2 + n2 =4 for the two states. The possible frequencies are 0:75 (1,2 ¡! 1,1), 2:00 (1,3 ¡ 1,1),
!
x
y
3:00 (2,1 ¡ 1,1), 3:75 (2,2 ¡! 1,1), 1:25 (1,3 ¡! 1,2), 2:25 (2,1 ¡ 1,2), 3:00 (2,2 ¡ 1,2),
!
!
!
2.
1:00 (2,1 ¡! 1,3), 1:75 (2,2 ¡! 1,3), 0:75 (2,2 ¡! 2,1), all in units of h=8mL
246 Chapter 39 There are 8 different frequencies in all. In units of h=8mL2 the lowest is 0:75, the second lowest
is 1:00, and the third lowest is 1:25. The highest is 3:75, the second highest is 3:00, and the
third highest is 2:25.
33
If kinetic energy is not conserved some of the neutron’s initial kinetic energy is used to excite the
hydrogen atom. The least energy that the hydrogen atom can accept is the difference between the
first excited state (n = 2) and the ground state (n = 1). Since the energy of a state with principal
quantum number n is ¡(13:6 eV)=n2 , the smallest excitation energy is 13:6 eV¡(13:6 eV)=(2)2 =
10:2 eV. The neutron does not have sufficient kinetic energy to excite the hydrogen atom, so the
hydrogen atom is left in its ground state and all the initial kinetic energy of the neutron ends up
as the final kinetic energies of the neutron and atom. The collision must be elastic.
37
The energy E of the photon emitted when a hydrogen atom jumps from a state with principal
quantum number u to a state with principal quantum number ` is given by
µ
¶
1
1
E =A 2 ¡ 2 ;
`
u
where A = 13:6 eV. The frequency f of the electromagnetic wave is given by f = E=h and the
wavelength is given by ¸ = c=f. Thus
µ
¶
1 f
E
A 1
1
= =
=
¡
:
¸ c hc hc `2 u2
The shortest wavelength occurs at the series limit, for which u = 1. For the Balmer series,
` = 2 and the shortest wavelength is ¸B = 4hc=A. For the Lyman series, ` = 1 and the shortest
wavelength is ¸L = hc=A. The ratio is ¸ B =¸L = 4.
43
The proposed wave function is
1
Ã = p 3=2 e¡r=a ;
¼a
where a is the Bohr radius. Substitute this into the right side of Schr odinger’s equation and show
¨
that the result is zero. The derivative is
dÃ
1
= ¡ p 5=2 e¡r=a ;
dr
¼a
so
r2
and dÃ
r2
= ¡ p 5=2 e ¡r =a
dr
¼a µ
¶
·
¸
·
¸
1 d
1
2 1 ¡r=a 1
2 1
2 dÃ
r
= p 5=2 ¡ +
e
=
¡ +
Ã:
r 2 dr
dr
¼a
r a
a
r a
Chapter 39 247 Now the energy of the ground state is given by E = ¡me4 =8² 2 h2 and the Bohr radius is given
0
by a = h2 ²0 =¼me2 , so E = ¡e2 =8¼²0 a. The potential energy is given by U = ¡e2 =4¼²0 r, so
·
¸
·
¸
8¼2 m
8¼2 m
e2
e2
8¼ 2 m e 2
1 2
[E ¡ U ] Ã =
¡
+
Ã=
¡ +
Ã
h2
h2
8¼² 0 a 4¼²0 r
h2 8¼²0
a r
·
¸
·
¸
¼me 2
1 2
1
1 2
= 2
¡ +
Ã=
¡ +
Ã:
h ²0
a r
a
a r
The two terms in Schrodinger’s equation obviously cancel and the proposed function Ã satisfies
¨
that equation.
47
The radial probability function for the ground state of hydrogen is P (r) = R 2 =a3 )e¡2r =a , where
(4r
1
a is the Bohr radius. (See Eq. 39–44.) You want to evaluate the integral 0 P (r) dr. Eq. 15 in
the integral table of Appendix E is an integral of this form. Set n = 2 and replace a in the given
formula with 2=a and x with r. Then
Z 1
Z
4 1 2 ¡2r =a
4
2
P (r) dr = 3
r e
dr = 3
= 1:
a 0
a (2=a)3
0
49
(a) Ã210 is real. Simply square it to obtain the probability density:
r2 ¡r =a
e
cos2 µ :
32¼a5
(b) Each of the other functions is multiplied by its complex conjugate, obtained by replacing i
with ¡i in the function. Since eiÁe¡iÁ = e0 = 1, the result is the square of the function without
the exponential factor:
r2 ¡r=a 2
jÃ21+1 j2 =
e
sin µ
64¼a5
r2 ¡r =a 2
jÃ21¡1 j2 =
e
sin µ :
64¼a5
The last two functions lead to the same probability density.
(c) For m` = 0 the radial probability density decreases strongly with distance from the nucleus,
is greatest along the z axis, and for a given distance from the nucleus decreases in proportion to
cos2 µ for points away from the z axis. This is consistent with the dot plot of Fig. 39–24 (a).
For m` = §1 the radial probability density decreases strongly with distance from the nucleus, is
greatest in the x; y plane, and for a given distance from the nucleus decreases in proportion to
sin2 µ for points away from that plane. Thus it is consistent with the dot plot of Fig. 3924(b).
(d) The total probability density for the three states is the sum:
·
¸
r2 ¡r=a
1 2
1 2
2
2
2
2
jÃ 210 j + jÃ21+1 j + jÃ 21¡1 j =
e
cos µ + sin µ + sin µ
32¼a5
2
2
r2 ¡r=a
=
e
:
32¼a5
jÃ 210 j2 = 248 Chapter 39 The trigonometric identity cos2 µ + sin2 µ = 1 was used. The total probability density does not
depend on µ or Á. It is spherically symmetric.
57
The wave function is Ã = p ¡kx
Ce
. Substitute this function into Schr odinger’s equation,
¨
¡ Since d2 Ã=dx2 = p h2 d2 Ã
+ U0 Ã = EÃ :
8¼2 m dx2 Ck 2 e¡kx = k2 Ã, the result is
h2 k2
Ã + U0 Ã = EÃ :
8¼2 m The solution for k is r 8¼2 m
(U0 ¡ E) :
h2
Thus the function given for Ã is a solution to Schrodinger’s equation provided k has the value
¨
calculated from the expression given above.
k= Chapter 39 249 ...
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This note was uploaded on 10/04/2009 for the course PHYS 89882 taught by Professor Koskelo during the Spring '09 term at San Mateo Colleges.
 Spring '09
 Koskelo
 Physics

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