All 15 Chapters Covered
SOLUTION MANUAL
, Table of Contents
Chapter 1……………………………………………………..1 Chapter
2……………………………………………………14 Chapter
3…………………………………………………. 47
Chapter 4…………………………………………………. 72
Chapter 5…...……………………………………………….96 Chapter
6……………………………………………….….128 Chapter
7…………………………………………………..151 Chapter
8………………………………………………..…169 Chapter
9…………………………………………………..183 Chapter
10…………………………………………………203 Chapter
11…………………………………………..……..226 Chapter
12…………………………………………………249 Chapter
13………………………………………………. 269
Chapter 14……………………………………………..…..288 Chapter
15…………………………………………..……..305 Sample Formula
Sheet for Exams………………………….
viii
, Chapter 1
This chapter presents a review of some topics from classical physics. I have often heard
from instructors using the book that “my students have already studied a year of introductory
classical physics, so they don’t need the review.” This review chapter gives the opportunity to
present a number of concepts that I have found to cause difficulty for students and to collect those
concepts where they are available for easy reference. For
2
example, all students should know that kinetic energy is 2 mv , but few are readily
1
familiar with kinetic energy as pm , which is used more often in the text. The
expression connecting potential energy difference with potential difference for an electric charge q,
ΔU qΔV , zips by in the blink of an eye in the introductory course and is
rarely used there, while it is of fundamental importance to many experimental set-ups in modern
physics and is used implicitly in almost every chapter. Many introductory courses do not cover
thermodynamics or statistical mechanics, so it is useful to “review” them in this introductory
chapter.
I have observed students in my modern course occasionally struggling with problems
involving linear momentum conservation, another of those classical concepts that resides in the
introductory course. Although we physicists regard momentum conservation as a fundamental law
on the same plane as energy conservation, the latter is frequently invoked throughout the
introductory course while former appears and virtually disappears after a brief analysis of 2-body
collisions. Moreover, some introductory texts present the equations for the final velocities in a one-
dimensional elastic collision, leaving the student with little to do except plus numbers into the
equations. That is, students in the introductory course are rarely called upon to begin momentum
conservation problems with pinitial pfinal . This puts them at a disadvantage in the
application of momentum conservation to problems in modern physics, where many different
forms of momentum may need to be treated in a single situation (for example, classical particles,
relativistic particles, and photons). Chapter 1 therefore contains a brief review of momentum
conservation, including worked sample problems and end-of- chapter exercises.
Placing classical statistical mechanics in Chapter 1 (as compared to its location in Chapter
10 in the 2nd edition) offers a number of advantages. It permits the useful
expression K av 3 2kT to be used throughout the text without additional explanation. The
failure of classical statistical mechanics to account for the heat capacities of diatomic gases
(hydrogen in particular) lays the groundwork for quantum physics. It is especially helpful to
introduce the Maxwell-Boltzmann distribution function early in the text, thus permitting applications
such as the population of molecular rotational states in Chapter 9 and clarifying references to
“population inversion” in the discussion of the laser in Chapter 8. Distribution functions in general
are new topics for most students. They may look like ordinary mathematical functions, but they are
handled and interpreted quite differently. Absent this introduction to a classical distribution function
in Chapter 1, the students’ first exposure to a distribution function will be | |2, which layers an
additional level of confusion on top of the mathematical complications. It is better to have a chance
to cover some of the mathematical details at an earlier stage with a distribution function that is
easier to interpret.
1
, Suggestions for Additional Reading
Some descriptive, historical, philosophical, and nonmathematical texts which give good
background material and are great fun to read:
A. Baker, Modern Physics and Anti-Physics (Addison-Wesley, 1970).
F. Capra, The Tao of Physics (Shambhala Publications, 1975).
K. Ford, Quantum Physics for Everyone (Harvard University Press, 2005).
G. Gamow, Thirty Years that Shook Physics (Doubleday, 1966).
R. March, Physics for Poets (McGraw-Hill, 1978).
E. Segre, From X-Rays to Quarks: Modern Physicists and their Discoveries (Freeman, 1980).
G. L. Trigg, Landmark Experiments in Twentieth Century Physics (Crane, Russak, 1975).
F. A. Wolf, Taking the Quantum Leap (Harper & Row, 1989).
G. Zukav, The Dancing Wu Li Masters, An Overview of the New Physics (Morrow, 1979).
Gamow, Segre, and Trigg contributed directly to the development of modern physics and their
books are written from a perspective that only those who were part of that development can offer.
The books by Capra, Wolf, and Zukav offer controversial interpretations of quantum mechanics as
connected to eastern mysticism, spiritualism, or consciousness.
Materials for Active Engagement in the Classroom
A. Reading Quizzes
1. In an ideal gas at temperature T, the average speed of the molecules:
(1) increases as the square of the temperature.
(2) increases linearly with the temperature.
(3) increases as the square root of the temperature.
(4) is independent of the temperature.
2. The heat capacity of molecular hydrogen gas can take values of 3R/2, 5R/2, and 7R/2 at
different temperatures. Which value is correct at low temperatures?
(1) 3R/2 (2) 5R/2 (3) 7R/2
Answers 1. 3 2. 1
B. Conceptual and Discussion Questions
1. Equal numbers of molecules of hydrogen gas (molecular mass = 2 u) and helium gas
(molecular mass = 4 u) are in equilibrium in a container.
(a) What is the ratio of the average kinetic energy of a hydrogen molecule to the average
kinetic energy of a helium molecule?
K H / K He (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2
, (b) What is the ratio of the average speed of a hydrogen molecule to the average speed
of a helium molecule?
vH / vHe (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4 (C)
(c) What is the ratio of the pressure exerted on the walls of the container by the
hydrogen gas to the pressure exerted on the walls by the helium gas?
P /P (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
H He
2. Containers 1 and 3 have volumes of 1 m3 and container 2 has a volume of 2 m3. Containers
1 and 2 contain helium gas, and container 3 contains neon gas. All three containers have a
temperature of 300 K and a pressure of 1 atm.
He He Ne
1 2 3
(a) Rank the average speeds of the molecules in the containers in order from largest to
smallest.
(1) 1 > 2 > 3 (2) 1 = 2 > 3 (3) 1 = 2 = 3
(4) 3 > 1 > 2 (5) 3 > 1 = 2 (6) 2 > 1 > 3
(b) In which container is the average kinetic energy per molecule the largest?
(1) 1 (2) 2 (3) 3
(4) 1 and 2 (5) 1 and 3 (6) All the same
3. (a) Consider diatomic nitrogen gas at room temperature, in which only the translational
and rotational motions are possible. Suppose that 100 J of energy is transferred to the
gas at constant volume. How much of this energy goes into the translational kinetic
energy of the molecules?
(1) 20 J (2) 40 J (3) 50 J
(4) 60 J (5) 80 J (6) 100 J
(b) Now suppose that the gas is at a higher temperature, so that vibrational motion is also
possible. Compared with the situation at room temperature, is the fraction of the added energy
that goes into translational kinetic energy:
(1) smaller? (2) the same? (3) greater?
Answers 1. (a) 4 (b) 3 (c) 4 2. (a) 2 (b) 6 3. (a) 4 (b) 1
3
, Sample Exam Questions
A. Multiple Choice
1. A container holds gas molecules of mass m at a temperature T. A small probe inserted into
the container measures the value of the x component of the velocity of
2
the molecules. What is the average value of 2 mv x for these molecules?
1
(a) kT
3
(b) kT
1
(c) kT (d) 3kT
2 2
2. A container holds N molecules of a diatomic gas at temperature T. At this temperature,
rotational and vibrational motions of the gas molecules are allowed. A quantity of energy E
is transferred to the gas. What fraction of this added energy is responsible for increasing the
temperature of the gas?
(a) All of the added energy (b) 3/5 (c) 2/5 (d) 2/7 (e) 3/7
3. Two identical containers with fixed volumes hold equal amounts of Ne gas and N2 gas at the
same temperature of 1000 K. Equal amounts of heat energy are then transferred to the two
gases. How do the final temperatures of the two gases compare?
(a) T(Ne) = T(N2) (b) T(Ne) > T(N2) (c) T(Ne) < T(N2)
Answers 1. b 2. e 3. b
B. Conceptual
1. A container of volume V holds an equilibrium mixture of N molecules of oxygen gas O2
(molecular mass = 32.0 u) and also 2N molecules of He gas (mass = 4.00 u). Is the average
molecular energy of O2 greater than, equal to, or less than the average molecular energy of
He? EXPLAIN YOUR ANSWER.
2. Consider two containers of identical volumes. Container 1 holds N molecules of He at
temperature T. Container 2 holds the same number N molecules of H2 at the same
temperature T. Is the average energy per molecule of He greater than, less than, or the
same as the average energy per molecule of H2? EXPLAIN YOUR ANSWER.
Answers 1. equal to 2. the same as
C. Problems
1. N molecules of a gas are confined in a container at temperature T. A measuring device in the
container can determine the number of molecules in a range of 0.002v at any speed v, that is,
the number of molecules with speeds between 0.999v and 1.001v. When the device is set for
molecules at the speed vrms, the result is N1. When it is set for molecules at the speed 2vrms,
the result is N2. Find the value of N1/N2. Your answer should be a pure number, involving no
symbols or variables.
2. A container holds 2.5 moles of helium (a gas with one atom per molecule; atomic mass =
4.00 u; molar mass = 0.00400 kg) at a temperature of 342 K. What fraction of the gas
molecules has translational kinetic energies between 0.01480 eV and 0.01520 eV?
4
,3. One mole of N2 gas (molecular mass = 28 u) is confined to a container at a temperature of
387 K. At this temperature, you may assume that the molecules are free to both rotate and
vibrate.
(a) What fraction of the molecules has translational kinetic energies within 1% of the
average translational kinetic energy?
(b) Find the total internal energy of the gas.
Answers: 1. 11.25 2. 0.0066 3. (a) 0.0093 (b) 11.2 kJ
5
, Problem Solutions
1. (a) Conservation of momentum gives px,initial px,final , or
mHvH,initial mHevHe,initial mHvH,final mHevHe,final
Solving for vHe,final with vHe,initial 0 , we obtain
vHe,final m (v v )
H H,initial H,final
mHe
(1.674 10 27 kg)[1.1250 107 m/s ( 6.724 106 m/s)]
4.527 106 m/s
27
6.646 10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic
collision. Conservation of energy then gives Kinitial Kfinal , or
1 2 1 2 1 2 1 2
2
mHvH,initial 2
mHevHe,initial mv
2 H H,final 2
mHevHe,final
Solving for vHe,final with vHe,initial 0 , we obtain
2
mH (vH,initial vH,final
2
)
vHe,final
mHe
(1.674 1027 kg)[(1.1250 107 m/s) 2 (6.724 106 m/s) 2 ]
4.527 106 m/s
6.646 1027 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum gives:
px,initial px,final : mHevHe,initial mHevHe,final cos He mOvO,final cos O
py,initial py,final : 0 mHevHe,final sin He mOvO,final sin O
From the second equation,
m Hev He,final sin He
(6.6465 10 27 kg)(6.636 106 m/s)(sin 84.7 )
v 2.551 106 m/s
O,final m sin (2.6560 10 26 kg)[sin( 40.4 )]
O O
(b) From the first momentum equation,
6
, m v cos m v cos
vHe,initial He He,final
mHe
He O O,final O
(6.6465 10 27 kg)(6.636 106 m/s)(cos 84.7 ) (2.6560 10 26 kg)(2.551 106 m/s)[cos( 40.4 )]
6.6465 10 27 kg
6
8.376 10 m/s
3. (a) Using conservation of momentum for this one-dimensional situation, we have
px,initial px,final , or
mHevHe mN vN mDvD mOvO
Solving for vO with vN 0 , we obtain
m Hev He m Dv (3.016 u)(6.346 106 m/s) (2.014 u)(1.531 107 m/s)
vO D
7.79 105 m/s
mO 15.003 u
(b) The kinetic energies are:
K 1
m v 2 1 m v 2 1 (3.016 u)(1.6605 10 27 kg/u)(6.346 106 m/s)2 1.008 10 13 J
initial He He
2 2 N N 2
K 1
m v 2 1 m v 2 1 (2.014 u)(1.6605 10 27 kg/u)(1.531 107 m/s)2
final 2 D D 2 O O 2
1
(15.003 u)(1.6605 10 27 kg/u)(7.79 105 m/s)2 3.995 10 13 J
2
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic energy.
The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds
v1 and v2 . Conservation of momentum along the x direction ( px,initial px,final ) gives
0 m1v1 m2v2 or v1 v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 K2 92.2 keV
Because v1 v2 , it follows that K1 K 2 46.1 keV , so
v 2K1 2(46.1 103 eV)(1.602 10 19 J/eV) 1.49 106 m/s
m1 (4.00 u)(1.66051027 kg/u)
6
v v 1.49 10 m/s
2 1
7
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8
SOLUTION MANUAL
, Table of Contents
Chapter 1……………………………………………………..1 Chapter
2……………………………………………………14 Chapter
3…………………………………………………. 47
Chapter 4…………………………………………………. 72
Chapter 5…...……………………………………………….96 Chapter
6……………………………………………….….128 Chapter
7…………………………………………………..151 Chapter
8………………………………………………..…169 Chapter
9…………………………………………………..183 Chapter
10…………………………………………………203 Chapter
11…………………………………………..……..226 Chapter
12…………………………………………………249 Chapter
13………………………………………………. 269
Chapter 14……………………………………………..…..288 Chapter
15…………………………………………..……..305 Sample Formula
Sheet for Exams………………………….
viii
, Chapter 1
This chapter presents a review of some topics from classical physics. I have often heard
from instructors using the book that “my students have already studied a year of introductory
classical physics, so they don’t need the review.” This review chapter gives the opportunity to
present a number of concepts that I have found to cause difficulty for students and to collect those
concepts where they are available for easy reference. For
2
example, all students should know that kinetic energy is 2 mv , but few are readily
1
familiar with kinetic energy as pm , which is used more often in the text. The
expression connecting potential energy difference with potential difference for an electric charge q,
ΔU qΔV , zips by in the blink of an eye in the introductory course and is
rarely used there, while it is of fundamental importance to many experimental set-ups in modern
physics and is used implicitly in almost every chapter. Many introductory courses do not cover
thermodynamics or statistical mechanics, so it is useful to “review” them in this introductory
chapter.
I have observed students in my modern course occasionally struggling with problems
involving linear momentum conservation, another of those classical concepts that resides in the
introductory course. Although we physicists regard momentum conservation as a fundamental law
on the same plane as energy conservation, the latter is frequently invoked throughout the
introductory course while former appears and virtually disappears after a brief analysis of 2-body
collisions. Moreover, some introductory texts present the equations for the final velocities in a one-
dimensional elastic collision, leaving the student with little to do except plus numbers into the
equations. That is, students in the introductory course are rarely called upon to begin momentum
conservation problems with pinitial pfinal . This puts them at a disadvantage in the
application of momentum conservation to problems in modern physics, where many different
forms of momentum may need to be treated in a single situation (for example, classical particles,
relativistic particles, and photons). Chapter 1 therefore contains a brief review of momentum
conservation, including worked sample problems and end-of- chapter exercises.
Placing classical statistical mechanics in Chapter 1 (as compared to its location in Chapter
10 in the 2nd edition) offers a number of advantages. It permits the useful
expression K av 3 2kT to be used throughout the text without additional explanation. The
failure of classical statistical mechanics to account for the heat capacities of diatomic gases
(hydrogen in particular) lays the groundwork for quantum physics. It is especially helpful to
introduce the Maxwell-Boltzmann distribution function early in the text, thus permitting applications
such as the population of molecular rotational states in Chapter 9 and clarifying references to
“population inversion” in the discussion of the laser in Chapter 8. Distribution functions in general
are new topics for most students. They may look like ordinary mathematical functions, but they are
handled and interpreted quite differently. Absent this introduction to a classical distribution function
in Chapter 1, the students’ first exposure to a distribution function will be | |2, which layers an
additional level of confusion on top of the mathematical complications. It is better to have a chance
to cover some of the mathematical details at an earlier stage with a distribution function that is
easier to interpret.
1
, Suggestions for Additional Reading
Some descriptive, historical, philosophical, and nonmathematical texts which give good
background material and are great fun to read:
A. Baker, Modern Physics and Anti-Physics (Addison-Wesley, 1970).
F. Capra, The Tao of Physics (Shambhala Publications, 1975).
K. Ford, Quantum Physics for Everyone (Harvard University Press, 2005).
G. Gamow, Thirty Years that Shook Physics (Doubleday, 1966).
R. March, Physics for Poets (McGraw-Hill, 1978).
E. Segre, From X-Rays to Quarks: Modern Physicists and their Discoveries (Freeman, 1980).
G. L. Trigg, Landmark Experiments in Twentieth Century Physics (Crane, Russak, 1975).
F. A. Wolf, Taking the Quantum Leap (Harper & Row, 1989).
G. Zukav, The Dancing Wu Li Masters, An Overview of the New Physics (Morrow, 1979).
Gamow, Segre, and Trigg contributed directly to the development of modern physics and their
books are written from a perspective that only those who were part of that development can offer.
The books by Capra, Wolf, and Zukav offer controversial interpretations of quantum mechanics as
connected to eastern mysticism, spiritualism, or consciousness.
Materials for Active Engagement in the Classroom
A. Reading Quizzes
1. In an ideal gas at temperature T, the average speed of the molecules:
(1) increases as the square of the temperature.
(2) increases linearly with the temperature.
(3) increases as the square root of the temperature.
(4) is independent of the temperature.
2. The heat capacity of molecular hydrogen gas can take values of 3R/2, 5R/2, and 7R/2 at
different temperatures. Which value is correct at low temperatures?
(1) 3R/2 (2) 5R/2 (3) 7R/2
Answers 1. 3 2. 1
B. Conceptual and Discussion Questions
1. Equal numbers of molecules of hydrogen gas (molecular mass = 2 u) and helium gas
(molecular mass = 4 u) are in equilibrium in a container.
(a) What is the ratio of the average kinetic energy of a hydrogen molecule to the average
kinetic energy of a helium molecule?
K H / K He (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2
, (b) What is the ratio of the average speed of a hydrogen molecule to the average speed
of a helium molecule?
vH / vHe (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4 (C)
(c) What is the ratio of the pressure exerted on the walls of the container by the
hydrogen gas to the pressure exerted on the walls by the helium gas?
P /P (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
H He
2. Containers 1 and 3 have volumes of 1 m3 and container 2 has a volume of 2 m3. Containers
1 and 2 contain helium gas, and container 3 contains neon gas. All three containers have a
temperature of 300 K and a pressure of 1 atm.
He He Ne
1 2 3
(a) Rank the average speeds of the molecules in the containers in order from largest to
smallest.
(1) 1 > 2 > 3 (2) 1 = 2 > 3 (3) 1 = 2 = 3
(4) 3 > 1 > 2 (5) 3 > 1 = 2 (6) 2 > 1 > 3
(b) In which container is the average kinetic energy per molecule the largest?
(1) 1 (2) 2 (3) 3
(4) 1 and 2 (5) 1 and 3 (6) All the same
3. (a) Consider diatomic nitrogen gas at room temperature, in which only the translational
and rotational motions are possible. Suppose that 100 J of energy is transferred to the
gas at constant volume. How much of this energy goes into the translational kinetic
energy of the molecules?
(1) 20 J (2) 40 J (3) 50 J
(4) 60 J (5) 80 J (6) 100 J
(b) Now suppose that the gas is at a higher temperature, so that vibrational motion is also
possible. Compared with the situation at room temperature, is the fraction of the added energy
that goes into translational kinetic energy:
(1) smaller? (2) the same? (3) greater?
Answers 1. (a) 4 (b) 3 (c) 4 2. (a) 2 (b) 6 3. (a) 4 (b) 1
3
, Sample Exam Questions
A. Multiple Choice
1. A container holds gas molecules of mass m at a temperature T. A small probe inserted into
the container measures the value of the x component of the velocity of
2
the molecules. What is the average value of 2 mv x for these molecules?
1
(a) kT
3
(b) kT
1
(c) kT (d) 3kT
2 2
2. A container holds N molecules of a diatomic gas at temperature T. At this temperature,
rotational and vibrational motions of the gas molecules are allowed. A quantity of energy E
is transferred to the gas. What fraction of this added energy is responsible for increasing the
temperature of the gas?
(a) All of the added energy (b) 3/5 (c) 2/5 (d) 2/7 (e) 3/7
3. Two identical containers with fixed volumes hold equal amounts of Ne gas and N2 gas at the
same temperature of 1000 K. Equal amounts of heat energy are then transferred to the two
gases. How do the final temperatures of the two gases compare?
(a) T(Ne) = T(N2) (b) T(Ne) > T(N2) (c) T(Ne) < T(N2)
Answers 1. b 2. e 3. b
B. Conceptual
1. A container of volume V holds an equilibrium mixture of N molecules of oxygen gas O2
(molecular mass = 32.0 u) and also 2N molecules of He gas (mass = 4.00 u). Is the average
molecular energy of O2 greater than, equal to, or less than the average molecular energy of
He? EXPLAIN YOUR ANSWER.
2. Consider two containers of identical volumes. Container 1 holds N molecules of He at
temperature T. Container 2 holds the same number N molecules of H2 at the same
temperature T. Is the average energy per molecule of He greater than, less than, or the
same as the average energy per molecule of H2? EXPLAIN YOUR ANSWER.
Answers 1. equal to 2. the same as
C. Problems
1. N molecules of a gas are confined in a container at temperature T. A measuring device in the
container can determine the number of molecules in a range of 0.002v at any speed v, that is,
the number of molecules with speeds between 0.999v and 1.001v. When the device is set for
molecules at the speed vrms, the result is N1. When it is set for molecules at the speed 2vrms,
the result is N2. Find the value of N1/N2. Your answer should be a pure number, involving no
symbols or variables.
2. A container holds 2.5 moles of helium (a gas with one atom per molecule; atomic mass =
4.00 u; molar mass = 0.00400 kg) at a temperature of 342 K. What fraction of the gas
molecules has translational kinetic energies between 0.01480 eV and 0.01520 eV?
4
,3. One mole of N2 gas (molecular mass = 28 u) is confined to a container at a temperature of
387 K. At this temperature, you may assume that the molecules are free to both rotate and
vibrate.
(a) What fraction of the molecules has translational kinetic energies within 1% of the
average translational kinetic energy?
(b) Find the total internal energy of the gas.
Answers: 1. 11.25 2. 0.0066 3. (a) 0.0093 (b) 11.2 kJ
5
, Problem Solutions
1. (a) Conservation of momentum gives px,initial px,final , or
mHvH,initial mHevHe,initial mHvH,final mHevHe,final
Solving for vHe,final with vHe,initial 0 , we obtain
vHe,final m (v v )
H H,initial H,final
mHe
(1.674 10 27 kg)[1.1250 107 m/s ( 6.724 106 m/s)]
4.527 106 m/s
27
6.646 10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic
collision. Conservation of energy then gives Kinitial Kfinal , or
1 2 1 2 1 2 1 2
2
mHvH,initial 2
mHevHe,initial mv
2 H H,final 2
mHevHe,final
Solving for vHe,final with vHe,initial 0 , we obtain
2
mH (vH,initial vH,final
2
)
vHe,final
mHe
(1.674 1027 kg)[(1.1250 107 m/s) 2 (6.724 106 m/s) 2 ]
4.527 106 m/s
6.646 1027 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum gives:
px,initial px,final : mHevHe,initial mHevHe,final cos He mOvO,final cos O
py,initial py,final : 0 mHevHe,final sin He mOvO,final sin O
From the second equation,
m Hev He,final sin He
(6.6465 10 27 kg)(6.636 106 m/s)(sin 84.7 )
v 2.551 106 m/s
O,final m sin (2.6560 10 26 kg)[sin( 40.4 )]
O O
(b) From the first momentum equation,
6
, m v cos m v cos
vHe,initial He He,final
mHe
He O O,final O
(6.6465 10 27 kg)(6.636 106 m/s)(cos 84.7 ) (2.6560 10 26 kg)(2.551 106 m/s)[cos( 40.4 )]
6.6465 10 27 kg
6
8.376 10 m/s
3. (a) Using conservation of momentum for this one-dimensional situation, we have
px,initial px,final , or
mHevHe mN vN mDvD mOvO
Solving for vO with vN 0 , we obtain
m Hev He m Dv (3.016 u)(6.346 106 m/s) (2.014 u)(1.531 107 m/s)
vO D
7.79 105 m/s
mO 15.003 u
(b) The kinetic energies are:
K 1
m v 2 1 m v 2 1 (3.016 u)(1.6605 10 27 kg/u)(6.346 106 m/s)2 1.008 10 13 J
initial He He
2 2 N N 2
K 1
m v 2 1 m v 2 1 (2.014 u)(1.6605 10 27 kg/u)(1.531 107 m/s)2
final 2 D D 2 O O 2
1
(15.003 u)(1.6605 10 27 kg/u)(7.79 105 m/s)2 3.995 10 13 J
2
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic energy.
The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds
v1 and v2 . Conservation of momentum along the x direction ( px,initial px,final ) gives
0 m1v1 m2v2 or v1 v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 K2 92.2 keV
Because v1 v2 , it follows that K1 K 2 46.1 keV , so
v 2K1 2(46.1 103 eV)(1.602 10 19 J/eV) 1.49 106 m/s
m1 (4.00 u)(1.66051027 kg/u)
6
v v 1.49 10 m/s
2 1
7
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