Solution Manual - Matter and Interactions 3rd Edition with complete solution

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1.X.1

(a) This smooth sailing ship’s motion constitutes motion with a constant velocity.
(b) Orbital motion does not constitute motion with a constant velocity; Moon’s direction in space continually changes.
(c) Tennis balls, and other projectiles, do not move with constant velocity.
(d) A velocity can have a magnitude of zero, so this is indeed motion with a constant velocity. The important thing is that
the velocity isn’t changing.
(e) A person on a Ferris wheel experiences a continual change in direction regardless of the Ferris wheel’s speed, and
regardless of whether the wheel is turning at a constant rate or a variable rate.




1.X.2

(a) The ball’s direction changes, so that’s evidence of a significant interaction.
(b) The baseball’s direction changed, so that’s evidence of a significant interaction.
(c) The satellite’s direction in space continually changes, so that’s evidence of a significant interaction.
(d) There is no evidence of significant interactions in this case.
(e) The particle’s changing direction is evidence of a significant interaction.




1.X.3

(a) Changing velocity (in this case, the velocity’s magnitude) indicates a net interaction.
(b) Changing velocity (in this case, the velocity’s magnitude) indicates a net interaction.
(c) Changing velocity (in this case, the velocity’s direction) indicates a net interaction.
(d) There is no sign of a net interaction.
(e) There is no sign of a net interaction.




1.X.4

(a) This statement is correct.
(b) This statement is incorrect.
(c) This statement is incorrect.
(d) This statement is correct.
(e) This statement is incorrect.
(f) This statement is incorrect.

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1.X.5

(a) This statement is incorrect.

(b) This statement is incorrect.

(c) This statement is incorrect.

(d) This statement is correct.

(e) This statement is correct.




1.X.6 Three numbers (signed) are needed to specify a 3D position vector.



1.X.7 One number (signed) is needed to specify a scalar.



1.X.8

r 
2  2  2
|~r| = rx + ry + rz
q
2 2 2
|~r| = (−3) + (−4) + (1) m
√
|~r| ≈ 26 m ≈ 5.10 m




1.X.9 A vector’s magnitude corresponds to the length of the arrow used to represent the vector. It’s impossible for an arrow
to have a negative length. Therefore, a vector cannot have a negative magnitude.



1.X.10 Within the framework of vector algebra, adding a vector and a scalar is not defined, and therefore has no meaning.
Therefore, the correct answer choice is (d). (NOTE: There is a mathematical language, called geometric algebra, used by
some physicists in which adding vectors and scalars is not only defined, but also useful. However, this text does not use
geometric algebra.)



1.X.11 Within the framework of vector algebra, dividing a scalar (or anything else for that matter) by a vector is not defined,
and therefore has no meaning. (NOTE: There is a mathematical language, called geometric algebra, used by some physicists
in which dividing by vectors is not only defined, but also useful. However, this text does not use geometric algebra.)



1.X.12 Simply multiply each component of ~a by f .

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f~a = (−3) h0.03, −1.4, 26.0i
f~a = h−0.09, 4.2, −78.0i




1.X.13 Simply divide each component of ~r by 2. Note that this is equivalent to multiplying ~r by 21 .


h2, −3, 5im
~r/2 =
2
~r/2 = h1, −1.5, 2.5im




1.X.14 The magnitude of 3~v will simply be three times the magnitude of ~v.

r
2  2  2
~
3v = |3| |~v| = 3 vx + vy + vz
q
2 2 2
~
3v = (3) (2) + (−3) + (5) m/s
√
~
3v = (3) 38 m/s ≈ 18.5 m/s




1.X.15 Any vector ~a and its opposite −~a will always have the same magnitude.



1.X.16


h0, 6, 0i h0, 6, 0i h0, 6, 0i
= q = = h0, 1, 0i
|h0, 6, 0i| 2 2 2 6
(0) + (6) + (0)




1.X.17 ~a = h400, 200, −100im/s2 . First, we need the magnitude of ~a.

r 2  2  2
|~a| = ax + ay + az
q
2 2 2
|~a| = (400) + (200) + (−100) m/s2
|~a| = 458 m/s2

Now we need to get the direction â.

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~a
â =
|~a|
2
h400, 200, −100i 
m/s

â = 2
= h0.873, 0.437, −0.218i
m/s
458  





1.X.18 If ~b and ~a are equal, then they must have equal components. So by must be 7.



1.X.19 If ~r1 and ~r2 are equal, then they must have the same magnitude and the same direction. However, these two vectors
have different directions so they are not equal even if they have the same magnitude. Note that the familiar expression equal
and opposite often seen referring to vectors in introductory physics textbooks is inherently oxymoronic; it contradicts itself.
Two vectors cannot be equal if they have opposite directions. Two vectors can indeed have equal magnitudes and opposite
directions though, and this is the correct way of articulating the relationship.



1.X.20 ~F1 = h300, 0, −200i and ~F2 = h150, −300, 0i

r 2 2 q
2
  2 2 2
~F = F1,x + F1,y + F1,z = (300) + (0) + (−200) = 361 N
1
r 2  2  2 q 2 2 2
~F = F2,x + F2,y + F2,z = (150) + (−300) + (0) = 335 N
2
q
D E 2 2 2
~F + ~F = F1,x + F2,x , F1,y + F2,y , F1,z + F2,z = (450) + (−300) + (−200) = 577 N
1 2


~F + ~F = 361 N + 335 N = 696 N
1 2




In general, it it not the case that ~F1 + ~F2 and ~F1 + ~F2 are equal. The only time it is true is when ~F1 and ~F2 have the
same direction.


D E D E
~ = 3 × 103 , −4 × 103 , −5 × 103 and B
1.X.21 A ~ = −3 × 103 , 4 × 103 , 5 × 103 so this is again very straightfoward.


(a)
D 3 3 3
E D 3 3 3
E
~ +B
A ~ = 3 × 10 , −4 × 10 , −5 × 10 + −3 × 10 , 4 × 10 , 5 × 10
= h0, 0, 0i

Note that technically, writing A
~ +B
~ = 0 is incorrect since the right hand side is a scalar and the left hand side is a
vector.

(b)

~ +B
A ~ = 0