,SOLUTION MANUAL FOR Matter and Interactions
5e Chabay
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
,Chapter 1
Interactions and Motion
1.1-Q-01
150
Neutron Number
100
50
0
0 20 40 60 80 100
Atomic Number
By inspection, you can see that the number of neutrons increases faster as the atomic number increases.
1.2-Q-01
B, C, D, E, and F show evidence of an interaction. In the case of B, speed changes (and therefore velocity changes).
In the case of C through F, direction of motion changes (and therefore so does velocity). In the cases of A, velocity is
constant and therefore no net interaction is indicated.
1.2-Q-02
Here is a qualitative description of the diagram. During the first 4 minutes, the dots are evenly spaced since the
car’s speed is constant. During the next 4 minutes, the dots are successively farther apart since the car’s speed increases
during each minute. During the next 4 minutes, the dots are evenly spaced (approximately twice as far apart as during
the first 4 minutes) since the car’s speed is now constant once again (but a different constant than before). During the
1
, CHAPTER 1. INTERACTIONS AND MOTION
last 4 minutes, the dots are successively closer together since the car’s speed is decreasing. The dots must get closer
together faster than they got farther apart when the car first accelerated because the speed is decreasing at a greater rate
than it increased before.
1.3-Q-01
Reasons 1, 3, and 4 are true. Reason 2 is irrelevant. Reason 5 is correct only if one assumes that the spaceship
is indeed effectively infinitely far away from all other sources of gravitational attraction and is thus really only an
approximation, but a very good approximation.
1.3-Q-02
Observers 2, 4 may see something that appears to violate Newton’s first law because they are in reference frames
that are accelerating relative to Earth. These are not inertial reference frames, and Newton’s first law doesn’t hold for
such noninertial frames. Observers 1, 3, and 5 have constant velocity (magnitude and direction, relative to Earth) and
are thus in inertial reference frames so they will see Newton’s first law as not being violated.
1.3-Q-03
While you are walking and holding the book, the ball moves with a constant velocity (relative to an observer who is
standing at rest). When you stop, the ball continues moving with a constant velocity as it rolls across the book because
there is no net force on the ball to change its velocity, until it rolls off the book and then the net force on the ball is the
gravitational force by Earth which changes its velocity as it falls.
1.3-Q-04
4
Because nothing interacts with the spaceship, it will continue in a straight line and at a constant speed of 1 × 10 m∕s.
1.4-Q-01
a, c, and d are vectors. b is a scalar.
1.9-Q-01
Statements 1 and 5 are correct. Statements 2, 3, and 4 are incorrect.
1.10-Q-01
(a) 𝛾 is a scalar quantity.
(b) The minimum possible value of 𝛾 is 1.
(c) The minimum value is reached when the object’s speed is low, specifically when it is zero.
1-2
,CHAPTER 1. INTERACTIONS AND MOTION
(d) There is no maximum value for 𝛾.
(e) 𝛾 becomes large when an object’s speed is high.
(f) The approximation 𝛾 ≈ 1 applies when an object’s speed is low.
1.10-Q-02
The approximate formula for momentum may be used for (1), (2), (3) and (5) because in all of these cases, the
8
object or particle is moving with a speed much less than 3 × 10 m∕s. In case (5), the electron’s speed is one-hundredth
the speed of light. If a highly precise calculation is not needed, then even in this case, the approximate formula for
momentum may be used. As a rule of thumb, if an object’s speed is less than about 10% of the speed of light, then the
approximate formula may be used, except in cases where high precision (i.e. many significant figures) is needed.
1.4-P-01
Add the vector components.
𝑣⃗1 + 𝑣⃗2 = ⟨8, 12, −7⟩ m + ⟨−4, 0, 6⟩ m
= ⟨8 + −4, 12 + 0, −7 + 6⟩ m
= ⟨4, 12, −1⟩m
1.4-P-02
(a) The magnitude of a vector is indicated by the length of the arrow representing the vector. The arrows that have
the same magnitude as #
a have the same length as # a . Counting gridlines shows that | #
a | = 10 units (Note that we
# # #
don’t know what the unit is, and it doesn’t matter for answering this question.). So b , # c , d , #
e , and f have the
# #
same magnitude as #
a . You’ll need to use the Pythagorean theorem to prove this for b and d .
#
(b) Equal vectors must have both the same magnitude and the same direction. So #
a , #
c , and f are the only ones
meeting these criteria.
1.4-P-03
√
| #
v| = v2 + v2 + v2
𝑥 𝑦 𝑧
√
( )2 ( )2
= 8 × 106 + (0)2 + −2 × 107 m∕s
7
= 2.15 × 10 m∕s
1-3
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-04
#
a = ⟨5, 3, 0⟩ m
#
b = ⟨6, −9, 0⟩ m
#
c = ⟨−10, 3, 0⟩ m
√
#
| a | = a2 + a 2 + a 2
𝑥 𝑦 𝑧
√
= (5)2 + (3)2 + (0)2 m = 5.83 m
√
| #| 2 2 2
|b| = b + b + b
| | 𝑥 𝑦 𝑧
√
= (6)2 + (−9)2 + (0)2 m = 10.8 m
√
#
| c | = c2 + c 2 + c 2
𝑥 𝑦 𝑧
√
= (−10)2 + (3)2 + (0)2 m = 10.4 m
1.4-P-05
Extract components by counting gridlines.
(a) #
a = ⟨−4, −3, 0⟩
#
(b) b = ⟨−4, −3, 0⟩
#
(c) The statement is true. #
a and b have the same components, so the two vectors must be equivalent.
(d) #
c = ⟨4, 3, 0⟩
(e) The statement is true. Each component of #
c is the opposite of the corresponding component of #
a so the actual
vectors are opposites.
#
(f) d = ⟨−3, 4, 0⟩
#
(g) The statement is false because corresponding components of #
c and d are not opposites.
1.4-P-06
(a) See drawing.
1-4
,CHAPTER 1. INTERACTIONS AND MOTION
!p
(b) See drawing.
−!p
!p
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
1.4-P-07
𝑓 #
a = (2.0) ⟨0.02, −1.7, 30.0⟩
= ⟨0.04, −3.4, 60.0⟩
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
1.4-P-08
#
(a) d = ⟨−6, 3, 2⟩ m
#
(b) #
e = − d = −⟨−6, 3, 2⟩ m = ⟨+6, −3, −2⟩ m
#
(c) Take the position of the vector’s tail and add the vector d . ⟨−5, −2, 4⟩ m + ⟨−6, 3, 2⟩ m = ⟨−11, 1, 6⟩ m
#
(d) Take the position of the vector’s tail and add the vector − d . ⟨−1, −1, −1⟩ m + (−⟨−6, 3, 2⟩ m) = ⟨5, −4, −3⟩ m
1-5
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-09
Call ̂
n the direction of an arbitrary vector, then for the first vector we have
⟨2, 2, 2⟩
n= √
̂
(2)2 + (2)2 + (2)2
⟨2, 2, 2⟩
= √
12
⟨ ⟩
2 2 2
= √ ,√ ,√
12 12 12
≈ ⟨0.58, 0.58, 0.58⟩
and for the second vector we have the following.
⟨3, 3, 3⟩
n= √
̂
(3)2 + (3)2 + (3)2
⟨3, 3, 3⟩
= √
27
⟨ ⟩
3 3 3
= √ ,√ ,√
27 27 27
≈ ⟨0.58, 0.58, 0.58⟩
These directions are the same! How can that be? They’re the same because one vector is a multiple of the other.
⟨3, 3, 3⟩ = 23 ⟨2, 2, 2⟩. Of course you could also write ⟨2, 2, 2⟩ = 32 ⟨3, 3, 3⟩. When two vectors are multiples of each
other, their directions must be either parallel (if related by a positive multiple) or opposite (if related by a negative
multiple).
1.4-P-10
(a) See figure.
(b) See figure.
1-6
,CHAPTER 1. INTERACTIONS AND MOTION
# #
(c) The magnitude of 2 f will be twice the magnitude of f .
# #
(d) The direction of 2 f is the same as that of f .
(e) See figure.
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
# #
(f) The magnitude of f ∕2 is half that of f .
# #
(g) The direction of f ∕2 is the same as that of f .
(h) Yes, multiplying a vector by a scalar changes the magnitude, assuming the scalar is neither 0 nor ±1.
(i)
# #
𝑎 f = −3 f
∴𝑎 = −3
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
#
Note that you must not attempt to solve for 𝑎 by dividing both sides by f because dividing by a vector is not
defined. Instead, what you are really doing here is solving the equation by visual inspection. You may have never
thought of this as a legitimate way of solving an equation, but this is a vector equation and the rules of ordinary
algebra do not always apply to vector equations. Until you learn how to correctly solve vector equations using the
rules of vector algebra (hopefully your instructor will show you), visual inspection is a perfectly legitimate way of
solving them.
1-7
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-11
The concept of writing a vector as a magnitude multiplying a direction is important and will appear many times in
later chapters. It also forces you to think about each part, magnitude and direction, individually.
#
a = | #
a| ⋅̂a
√
( )2 ( )2 ( )2
| #
a| = 400 m∕s2 + 200 m∕s2 + −100 m∕s2
| #
a | = 458.3 m∕s2
#
a
a = #
̂
|a|
⟨400, 200, −100⟩ m∕s2
=
458.3 m∕s2
∴ #
a = 458.3 m∕s2 ⟨0.873, 0.436, −0.218⟩
1.4-P-12
(a) See figure.
!g
(b)
√
| #
g| = (4)2 + (7)2 + (0)2 m
= 8.06 m
(c)
⟨4, 7, 0⟩ m
g=
̂
| #
a|
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
⟨4, 7, 0⟩ m
=
8.06 m
= ⟨0.496, 0.868, 0⟩
(d) See figure.
1-8
5e Chabay
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
,Chapter 1
Interactions and Motion
1.1-Q-01
150
Neutron Number
100
50
0
0 20 40 60 80 100
Atomic Number
By inspection, you can see that the number of neutrons increases faster as the atomic number increases.
1.2-Q-01
B, C, D, E, and F show evidence of an interaction. In the case of B, speed changes (and therefore velocity changes).
In the case of C through F, direction of motion changes (and therefore so does velocity). In the cases of A, velocity is
constant and therefore no net interaction is indicated.
1.2-Q-02
Here is a qualitative description of the diagram. During the first 4 minutes, the dots are evenly spaced since the
car’s speed is constant. During the next 4 minutes, the dots are successively farther apart since the car’s speed increases
during each minute. During the next 4 minutes, the dots are evenly spaced (approximately twice as far apart as during
the first 4 minutes) since the car’s speed is now constant once again (but a different constant than before). During the
1
, CHAPTER 1. INTERACTIONS AND MOTION
last 4 minutes, the dots are successively closer together since the car’s speed is decreasing. The dots must get closer
together faster than they got farther apart when the car first accelerated because the speed is decreasing at a greater rate
than it increased before.
1.3-Q-01
Reasons 1, 3, and 4 are true. Reason 2 is irrelevant. Reason 5 is correct only if one assumes that the spaceship
is indeed effectively infinitely far away from all other sources of gravitational attraction and is thus really only an
approximation, but a very good approximation.
1.3-Q-02
Observers 2, 4 may see something that appears to violate Newton’s first law because they are in reference frames
that are accelerating relative to Earth. These are not inertial reference frames, and Newton’s first law doesn’t hold for
such noninertial frames. Observers 1, 3, and 5 have constant velocity (magnitude and direction, relative to Earth) and
are thus in inertial reference frames so they will see Newton’s first law as not being violated.
1.3-Q-03
While you are walking and holding the book, the ball moves with a constant velocity (relative to an observer who is
standing at rest). When you stop, the ball continues moving with a constant velocity as it rolls across the book because
there is no net force on the ball to change its velocity, until it rolls off the book and then the net force on the ball is the
gravitational force by Earth which changes its velocity as it falls.
1.3-Q-04
4
Because nothing interacts with the spaceship, it will continue in a straight line and at a constant speed of 1 × 10 m∕s.
1.4-Q-01
a, c, and d are vectors. b is a scalar.
1.9-Q-01
Statements 1 and 5 are correct. Statements 2, 3, and 4 are incorrect.
1.10-Q-01
(a) 𝛾 is a scalar quantity.
(b) The minimum possible value of 𝛾 is 1.
(c) The minimum value is reached when the object’s speed is low, specifically when it is zero.
1-2
,CHAPTER 1. INTERACTIONS AND MOTION
(d) There is no maximum value for 𝛾.
(e) 𝛾 becomes large when an object’s speed is high.
(f) The approximation 𝛾 ≈ 1 applies when an object’s speed is low.
1.10-Q-02
The approximate formula for momentum may be used for (1), (2), (3) and (5) because in all of these cases, the
8
object or particle is moving with a speed much less than 3 × 10 m∕s. In case (5), the electron’s speed is one-hundredth
the speed of light. If a highly precise calculation is not needed, then even in this case, the approximate formula for
momentum may be used. As a rule of thumb, if an object’s speed is less than about 10% of the speed of light, then the
approximate formula may be used, except in cases where high precision (i.e. many significant figures) is needed.
1.4-P-01
Add the vector components.
𝑣⃗1 + 𝑣⃗2 = ⟨8, 12, −7⟩ m + ⟨−4, 0, 6⟩ m
= ⟨8 + −4, 12 + 0, −7 + 6⟩ m
= ⟨4, 12, −1⟩m
1.4-P-02
(a) The magnitude of a vector is indicated by the length of the arrow representing the vector. The arrows that have
the same magnitude as #
a have the same length as # a . Counting gridlines shows that | #
a | = 10 units (Note that we
# # #
don’t know what the unit is, and it doesn’t matter for answering this question.). So b , # c , d , #
e , and f have the
# #
same magnitude as #
a . You’ll need to use the Pythagorean theorem to prove this for b and d .
#
(b) Equal vectors must have both the same magnitude and the same direction. So #
a , #
c , and f are the only ones
meeting these criteria.
1.4-P-03
√
| #
v| = v2 + v2 + v2
𝑥 𝑦 𝑧
√
( )2 ( )2
= 8 × 106 + (0)2 + −2 × 107 m∕s
7
= 2.15 × 10 m∕s
1-3
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-04
#
a = ⟨5, 3, 0⟩ m
#
b = ⟨6, −9, 0⟩ m
#
c = ⟨−10, 3, 0⟩ m
√
#
| a | = a2 + a 2 + a 2
𝑥 𝑦 𝑧
√
= (5)2 + (3)2 + (0)2 m = 5.83 m
√
| #| 2 2 2
|b| = b + b + b
| | 𝑥 𝑦 𝑧
√
= (6)2 + (−9)2 + (0)2 m = 10.8 m
√
#
| c | = c2 + c 2 + c 2
𝑥 𝑦 𝑧
√
= (−10)2 + (3)2 + (0)2 m = 10.4 m
1.4-P-05
Extract components by counting gridlines.
(a) #
a = ⟨−4, −3, 0⟩
#
(b) b = ⟨−4, −3, 0⟩
#
(c) The statement is true. #
a and b have the same components, so the two vectors must be equivalent.
(d) #
c = ⟨4, 3, 0⟩
(e) The statement is true. Each component of #
c is the opposite of the corresponding component of #
a so the actual
vectors are opposites.
#
(f) d = ⟨−3, 4, 0⟩
#
(g) The statement is false because corresponding components of #
c and d are not opposites.
1.4-P-06
(a) See drawing.
1-4
,CHAPTER 1. INTERACTIONS AND MOTION
!p
(b) See drawing.
−!p
!p
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
1.4-P-07
𝑓 #
a = (2.0) ⟨0.02, −1.7, 30.0⟩
= ⟨0.04, −3.4, 60.0⟩
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
1.4-P-08
#
(a) d = ⟨−6, 3, 2⟩ m
#
(b) #
e = − d = −⟨−6, 3, 2⟩ m = ⟨+6, −3, −2⟩ m
#
(c) Take the position of the vector’s tail and add the vector d . ⟨−5, −2, 4⟩ m + ⟨−6, 3, 2⟩ m = ⟨−11, 1, 6⟩ m
#
(d) Take the position of the vector’s tail and add the vector − d . ⟨−1, −1, −1⟩ m + (−⟨−6, 3, 2⟩ m) = ⟨5, −4, −3⟩ m
1-5
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-09
Call ̂
n the direction of an arbitrary vector, then for the first vector we have
⟨2, 2, 2⟩
n= √
̂
(2)2 + (2)2 + (2)2
⟨2, 2, 2⟩
= √
12
⟨ ⟩
2 2 2
= √ ,√ ,√
12 12 12
≈ ⟨0.58, 0.58, 0.58⟩
and for the second vector we have the following.
⟨3, 3, 3⟩
n= √
̂
(3)2 + (3)2 + (3)2
⟨3, 3, 3⟩
= √
27
⟨ ⟩
3 3 3
= √ ,√ ,√
27 27 27
≈ ⟨0.58, 0.58, 0.58⟩
These directions are the same! How can that be? They’re the same because one vector is a multiple of the other.
⟨3, 3, 3⟩ = 23 ⟨2, 2, 2⟩. Of course you could also write ⟨2, 2, 2⟩ = 32 ⟨3, 3, 3⟩. When two vectors are multiples of each
other, their directions must be either parallel (if related by a positive multiple) or opposite (if related by a negative
multiple).
1.4-P-10
(a) See figure.
(b) See figure.
1-6
,CHAPTER 1. INTERACTIONS AND MOTION
# #
(c) The magnitude of 2 f will be twice the magnitude of f .
# #
(d) The direction of 2 f is the same as that of f .
(e) See figure.
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
# #
(f) The magnitude of f ∕2 is half that of f .
# #
(g) The direction of f ∕2 is the same as that of f .
(h) Yes, multiplying a vector by a scalar changes the magnitude, assuming the scalar is neither 0 nor ±1.
(i)
# #
𝑎 f = −3 f
∴𝑎 = −3
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
#
Note that you must not attempt to solve for 𝑎 by dividing both sides by f because dividing by a vector is not
defined. Instead, what you are really doing here is solving the equation by visual inspection. You may have never
thought of this as a legitimate way of solving an equation, but this is a vector equation and the rules of ordinary
algebra do not always apply to vector equations. Until you learn how to correctly solve vector equations using the
rules of vector algebra (hopefully your instructor will show you), visual inspection is a perfectly legitimate way of
solving them.
1-7
, CHAPTER 1. INTERACTIONS AND MOTION
1.4-P-11
The concept of writing a vector as a magnitude multiplying a direction is important and will appear many times in
later chapters. It also forces you to think about each part, magnitude and direction, individually.
#
a = | #
a| ⋅̂a
√
( )2 ( )2 ( )2
| #
a| = 400 m∕s2 + 200 m∕s2 + −100 m∕s2
| #
a | = 458.3 m∕s2
#
a
a = #
̂
|a|
⟨400, 200, −100⟩ m∕s2
=
458.3 m∕s2
∴ #
a = 458.3 m∕s2 ⟨0.873, 0.436, −0.218⟩
1.4-P-12
(a) See figure.
!g
(b)
√
| #
g| = (4)2 + (7)2 + (0)2 m
= 8.06 m
(c)
⟨4, 7, 0⟩ m
g=
̂
| #
a|
Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/
⟨4, 7, 0⟩ m
=
8.06 m
= ⟨0.496, 0.868, 0⟩
(d) See figure.
1-8
