Complete All Chapters
,Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 1: Chapters 1-24
Complete All Chapters
, 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
vx2 = v0x
2
+ 2ax x,
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
1
Complete All Chapters
, 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 .
E1-3 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
(365.25 days) = 3.1558 × 107 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
106 years/1010 years = 10−4 .
That fraction of a day is
60 min 60 s
10−4 (24 hr) = 8.64 s.
1 hr 1 min
E1-7 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.
2
Complete All Chapters
,Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 1: Chapters 1-24
Complete All Chapters
, 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
vx2 = v0x
2
+ 2ax x,
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
1
Complete All Chapters
, 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 .
E1-3 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
(365.25 days) = 3.1558 × 107 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
106 years/1010 years = 10−4 .
That fraction of a day is
60 min 60 s
10−4 (24 hr) = 8.64 s.
1 hr 1 min
E1-7 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.
2
Complete All Chapters
