Instructor Solutions Manual for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley Beloit College
Volume 1: Chapters 1-24
A Note To The Instructor...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server access must be restricted to your students.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it will sometimes obfuscate the approach
if viewed by a novice.
There are some traditional formula, such as
2 2
v =v + 2axx,
x 0x
which are not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercises and problems in the text
which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in partic-
ular, I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual
with considerably more detail and, when appropriate, include discussion on any physical implications
of the answer. These student solutions carefully discuss the steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented. You are
encouraged to refer students to the Student’s Solution Manual for these exercises and problems.
However, the material from the Student’s Solution Manual must not be copied.
Paul Stanley
Beloit College
,E1-1 (a) Megaphones; (b) Microphones; (c) Decacards (Deck of Cards); (d) Gigalows (Gigolos);
(e) Terabulls (Terribles); (f) Decimates; (g) Centipedes; (h) Nanonanettes (?); (i) Picoboos (Peek-a-
Boo); (j) Attoboys (’atta boy); (k) Two Hectowithits (To Heck With It); (l) Two Kilomockingbirds
(To Kill A Mockingbird, or Tequila Mockingbird).
E1-2 (a) $36, 000/52 week = $692/week. (b) $10, 000, 000/(20 × 12 month) = $41, 700/month. (c)
30 × 109/8 = 3.75 × 109.
Multiply out the factors which make up a century.
365 days 24 hours 60 minutes
1 century = 100 years
1 year 1 day 1 hour
This gives 5.256 × 107 minutes in a century, so a microcentury is 52.56 minutes.
The percentage difference from Fermi’s approximation is (2.56 min)/(50 min) × 100% or 5.12%.
E1-4 (3000 mi)/(3 hr) = 1000 mi/timezone-hour. There are 24 time-zones, so the circumference
is approximately 24 × 1000 mi = 24, 000 miles.
E1-5 Actual number of seconds in a year is
24 hr 60 min 60 s 7
(365.25 days) = 3.1558 × 10 s.
1 day 1 hr 1 min
The percentage error of the approximation is then
3.1416 × 107 s − 3.1558 × 107 s
= —0.45 % .
3.1558 × 107 s
E1-6 (a) 10−8 seconds per shake means 108 shakes per second. There are
365 days 24 hr 60 min 60 s
= 3.1536 × 107 s/year.
1 year 1 day 1 hr 1 min
This means there are more shakes in a second.
(b) Humans have existed for a fraction of
6 10 4
10 years/10 years = 10− .
That fraction of a day is
4 60 min 60 s
10− (24 hr) = 8.64 s.
1 hr 1 min
We’ll assume, for convenience only, that the runner with the longer time ran exactly one
mile. Let the speed of the runner with the shorter time be given by v1, and call the distance actually
ran by this runner d1. Then v1 = d1/t1. Similarly, v2 = d2/t2 for the other runner, and d2 = 1 mile.
We want to know when v1 > v2. Substitute our expressions for speed, and get d1/t1 > d2/t2.
Rearrange, and d1/d2 > t1/t2 or d1/d2 > 0.99937. Then d1 > 0.99937 mile (5280 feet/1 mile) or
d1 > 5276.7 feet is the condition that the first runner was indeed faster. The first track can be no
more than 3.3 feet too short to guarantee that the first runner was faster.
,E1-8 We will wait until a day’s worth of minutes have been gained. That would be
60 min
(24 hr) = 1440 min
1 hr
The clock gains one minute per day, so we need to wait 1,440 days, or almost four years. Of course,
if it is an older clock with hands that only read 12 hours (instead of 24), then after only 720 days
the clock would be correct.
First find the “logarithmic average” by
1
17 15
log tav = log(5 × 10 ) + log(6 × 10− ) ,
2
17 15
= log 5 × 10 × 6 × 10− ,
2
1 √
= log 3000 = log 3000 .
Solve, and tav = 54.8 seconds.
E1-10 After 20 centuries the day would have increased in length by a total of 20 0.001 s = 0.02 s.
The cumulative effect would by the product of the average increase and the number of days; that
1
average is half of the maximum, so the cumulative effect is (2000)(365)(0.02 s) = 7300 s. That’s
about 2 hours.
E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
the Moon has farther to go to complete a phase. In 27.3 days the Moon may have orbited through
360◦, but since the Earth moved through (27.3/365) 360◦ = 27◦ the Moon needs to move 27◦
farther to catch up. That will take (27◦/360◦) 27.3 days = 2.05 days, but in that time the Earth
would have moved on yet farther, and the moon will need to catch up again. How much farther?
(2.05/365) 360◦ = 2.02◦ which means (2.02◦/360◦) 27.3 days = 0.153 days. The total so far is
2.2 days longer; we could go farther, but at our accuracy level, it isn’t worth it.
E1-12 (1.9 m)(3.281 ft/1.000 m) = 6.2 ft, or just under 6 feet, 3 inches.
E1-13 (a) 100 meters = 328.1 feet (Appendix G), or 328.1/3 = 10.9 yards. This is 28 feet longer
than 100 yards, or (28 ft)(0.3048 m/ft) = 8.5 m. (b) A metric mile is (1500 m)(6.214 10−4 mi/m) =
0.932 mi. I’d rather run the metric mile.
E1-14 There are
365.25 days 24 hr 60 min 60 s
300, 000 years = 9.5 × 1012s
1 year 1 day 1 hr 1 min
that will elapse before the cesium clock is in error by 1 s. This is almost 1 part in 1013. This kind
of accuracy with respect to 2572 miles is
13 1609 m
10− (2572 mi) = 413 nm
1 mi
, The volume of Antarctica is approximated by the area of the base time the height; the
area of the base is the area of a semicircle. Then
1 2
V = Ah = πr h.
The volume is
1 2 16 3
V = (3.14)(2000 × 1000 m) (3000 m) = 1.88 × 10 m
3
16 3 100 cm 22 3
= 1.88 × 10 m × = 1.88 × 10 cm .
1m
E1-16 The volume is (77×104 m2)(26 m) = 2.00×107 m3. This is equivalent to
7 3 3 3 3
(2.00×10 m )(10− km/m) = 0.02 km .
2
E1-17 (a) C = 2πr = 2π(6.37 × 103 km) = 4.00 × 104 km. (b) A = 4πr = 4π(6.37 × 103 km)2 =
4 3
5.10 × 108 km. (c) V = π(6.37 × 103 km)3 = 1.08 × 1012 km .
E1-18 The conversions: squirrel, 19 km/hr(1000 m/km)/(3600 s/hr) = 5.3 m/s;
rabbit, 30 knots(1.688ft/s/knot)(0.3048 m/ft) = 15 m/s;
snail, 0.030 mi/hr(1609 m/mi)/(3600 s/hr) = 0.013 m/s;
spider, 1.8 ft/s(0.3048 m/ft) = 0.55 m/s;
cheetah, 1.9 km/min(1000 m/km)/(60 s/min) = 32 m/s;
human, 1000 cm/s/(100 cm/m) = 10 m/s;
fox, 1100 m/min/(60 s/min) = 18 m/s;
lion, 1900 km/day(1000 m/km)/(86, 400 s/day) = 22 m/s.
The order is snail, spider, squirrel, human, rabbit, fox, lion, cheetah.
One light-year is the distance traveled by light in one year, or (3 × 108 m/s) × (1 year).
Then
mi light-year 1609 m 1 hr 100 year
19, 200 ,
hr (3 × 108 m/s) × (1 year) 1 mi 3600 s 1 century
which is equal to 0.00286 light-year/century.
E1-20 Start with the British units inverted,
3
gal 231 in 1.639 × 10−2 L mi 2
= 7 84 10− L/km
30.0 mi gal 1.609 km
E1-21 (b) A light-year is
5 3600 s 24 hr 12
(3.00 × 10 km/s) (365 days) = 9.46 × 10 km.
1 hr 1 day
A parsec is
1.50 × 108 km 360◦ 1.50 × 108 km 360◦
= = 3 09 1013 km
2π rad 2π rad
(a) (1.5 × 108 km)/(3.09 × 1013 km/pc = 4.85 × 10−6 pc. (1.5 × 108 km)/(9.46 × 1012 km/ly) =
1.59 × 10−5 ly.
Physics
by
Halliday, Resnick, and Krane
Paul Stanley Beloit College
Volume 1: Chapters 1-24
A Note To The Instructor...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server access must be restricted to your students.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it will sometimes obfuscate the approach
if viewed by a novice.
There are some traditional formula, such as
2 2
v =v + 2axx,
x 0x
which are not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercises and problems in the text
which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in partic-
ular, I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual
with considerably more detail and, when appropriate, include discussion on any physical implications
of the answer. These student solutions carefully discuss the steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented. You are
encouraged to refer students to the Student’s Solution Manual for these exercises and problems.
However, the material from the Student’s Solution Manual must not be copied.
Paul Stanley
Beloit College
,E1-1 (a) Megaphones; (b) Microphones; (c) Decacards (Deck of Cards); (d) Gigalows (Gigolos);
(e) Terabulls (Terribles); (f) Decimates; (g) Centipedes; (h) Nanonanettes (?); (i) Picoboos (Peek-a-
Boo); (j) Attoboys (’atta boy); (k) Two Hectowithits (To Heck With It); (l) Two Kilomockingbirds
(To Kill A Mockingbird, or Tequila Mockingbird).
E1-2 (a) $36, 000/52 week = $692/week. (b) $10, 000, 000/(20 × 12 month) = $41, 700/month. (c)
30 × 109/8 = 3.75 × 109.
Multiply out the factors which make up a century.
365 days 24 hours 60 minutes
1 century = 100 years
1 year 1 day 1 hour
This gives 5.256 × 107 minutes in a century, so a microcentury is 52.56 minutes.
The percentage difference from Fermi’s approximation is (2.56 min)/(50 min) × 100% or 5.12%.
E1-4 (3000 mi)/(3 hr) = 1000 mi/timezone-hour. There are 24 time-zones, so the circumference
is approximately 24 × 1000 mi = 24, 000 miles.
E1-5 Actual number of seconds in a year is
24 hr 60 min 60 s 7
(365.25 days) = 3.1558 × 10 s.
1 day 1 hr 1 min
The percentage error of the approximation is then
3.1416 × 107 s − 3.1558 × 107 s
= —0.45 % .
3.1558 × 107 s
E1-6 (a) 10−8 seconds per shake means 108 shakes per second. There are
365 days 24 hr 60 min 60 s
= 3.1536 × 107 s/year.
1 year 1 day 1 hr 1 min
This means there are more shakes in a second.
(b) Humans have existed for a fraction of
6 10 4
10 years/10 years = 10− .
That fraction of a day is
4 60 min 60 s
10− (24 hr) = 8.64 s.
1 hr 1 min
We’ll assume, for convenience only, that the runner with the longer time ran exactly one
mile. Let the speed of the runner with the shorter time be given by v1, and call the distance actually
ran by this runner d1. Then v1 = d1/t1. Similarly, v2 = d2/t2 for the other runner, and d2 = 1 mile.
We want to know when v1 > v2. Substitute our expressions for speed, and get d1/t1 > d2/t2.
Rearrange, and d1/d2 > t1/t2 or d1/d2 > 0.99937. Then d1 > 0.99937 mile (5280 feet/1 mile) or
d1 > 5276.7 feet is the condition that the first runner was indeed faster. The first track can be no
more than 3.3 feet too short to guarantee that the first runner was faster.
,E1-8 We will wait until a day’s worth of minutes have been gained. That would be
60 min
(24 hr) = 1440 min
1 hr
The clock gains one minute per day, so we need to wait 1,440 days, or almost four years. Of course,
if it is an older clock with hands that only read 12 hours (instead of 24), then after only 720 days
the clock would be correct.
First find the “logarithmic average” by
1
17 15
log tav = log(5 × 10 ) + log(6 × 10− ) ,
2
17 15
= log 5 × 10 × 6 × 10− ,
2
1 √
= log 3000 = log 3000 .
Solve, and tav = 54.8 seconds.
E1-10 After 20 centuries the day would have increased in length by a total of 20 0.001 s = 0.02 s.
The cumulative effect would by the product of the average increase and the number of days; that
1
average is half of the maximum, so the cumulative effect is (2000)(365)(0.02 s) = 7300 s. That’s
about 2 hours.
E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
the Moon has farther to go to complete a phase. In 27.3 days the Moon may have orbited through
360◦, but since the Earth moved through (27.3/365) 360◦ = 27◦ the Moon needs to move 27◦
farther to catch up. That will take (27◦/360◦) 27.3 days = 2.05 days, but in that time the Earth
would have moved on yet farther, and the moon will need to catch up again. How much farther?
(2.05/365) 360◦ = 2.02◦ which means (2.02◦/360◦) 27.3 days = 0.153 days. The total so far is
2.2 days longer; we could go farther, but at our accuracy level, it isn’t worth it.
E1-12 (1.9 m)(3.281 ft/1.000 m) = 6.2 ft, or just under 6 feet, 3 inches.
E1-13 (a) 100 meters = 328.1 feet (Appendix G), or 328.1/3 = 10.9 yards. This is 28 feet longer
than 100 yards, or (28 ft)(0.3048 m/ft) = 8.5 m. (b) A metric mile is (1500 m)(6.214 10−4 mi/m) =
0.932 mi. I’d rather run the metric mile.
E1-14 There are
365.25 days 24 hr 60 min 60 s
300, 000 years = 9.5 × 1012s
1 year 1 day 1 hr 1 min
that will elapse before the cesium clock is in error by 1 s. This is almost 1 part in 1013. This kind
of accuracy with respect to 2572 miles is
13 1609 m
10− (2572 mi) = 413 nm
1 mi
, The volume of Antarctica is approximated by the area of the base time the height; the
area of the base is the area of a semicircle. Then
1 2
V = Ah = πr h.
The volume is
1 2 16 3
V = (3.14)(2000 × 1000 m) (3000 m) = 1.88 × 10 m
3
16 3 100 cm 22 3
= 1.88 × 10 m × = 1.88 × 10 cm .
1m
E1-16 The volume is (77×104 m2)(26 m) = 2.00×107 m3. This is equivalent to
7 3 3 3 3
(2.00×10 m )(10− km/m) = 0.02 km .
2
E1-17 (a) C = 2πr = 2π(6.37 × 103 km) = 4.00 × 104 km. (b) A = 4πr = 4π(6.37 × 103 km)2 =
4 3
5.10 × 108 km. (c) V = π(6.37 × 103 km)3 = 1.08 × 1012 km .
E1-18 The conversions: squirrel, 19 km/hr(1000 m/km)/(3600 s/hr) = 5.3 m/s;
rabbit, 30 knots(1.688ft/s/knot)(0.3048 m/ft) = 15 m/s;
snail, 0.030 mi/hr(1609 m/mi)/(3600 s/hr) = 0.013 m/s;
spider, 1.8 ft/s(0.3048 m/ft) = 0.55 m/s;
cheetah, 1.9 km/min(1000 m/km)/(60 s/min) = 32 m/s;
human, 1000 cm/s/(100 cm/m) = 10 m/s;
fox, 1100 m/min/(60 s/min) = 18 m/s;
lion, 1900 km/day(1000 m/km)/(86, 400 s/day) = 22 m/s.
The order is snail, spider, squirrel, human, rabbit, fox, lion, cheetah.
One light-year is the distance traveled by light in one year, or (3 × 108 m/s) × (1 year).
Then
mi light-year 1609 m 1 hr 100 year
19, 200 ,
hr (3 × 108 m/s) × (1 year) 1 mi 3600 s 1 century
which is equal to 0.00286 light-year/century.
E1-20 Start with the British units inverted,
3
gal 231 in 1.639 × 10−2 L mi 2
= 7 84 10− L/km
30.0 mi gal 1.609 km
E1-21 (b) A light-year is
5 3600 s 24 hr 12
(3.00 × 10 km/s) (365 days) = 9.46 × 10 km.
1 hr 1 day
A parsec is
1.50 × 108 km 360◦ 1.50 × 108 km 360◦
= = 3 09 1013 km
2π rad 2π rad
(a) (1.5 × 108 km)/(3.09 × 1013 km/pc = 4.85 × 10−6 pc. (1.5 × 108 km)/(9.46 × 1012 km/ly) =
1.59 × 10−5 ly.
