University Physics with Modern Physics, 15th Edition – Solutions Manual by Hugh D. Young & Roger A. Freedman (ISBN 9780135216736) – Complete Worked-Out Answers

SOLUTIONS MANUAL

UNIVERSITY PHYSICS WITH MODERN PHYSICS
15TH EDITION

CHAPTER NO. 01: UNITS, PHYSICAL QUANTITIES AND
VECTORS



VP1.7.1. IDENTIFY: We know that the sum of three known vectors and a fourth unknown vector is zero. We
want to find the magnitude and direction of the unknown vector.
SET UP: The sum of their x-components and the sum of their y-components must both be zero.
Ax  Bx  Cx  Dx  0
Ay  By  Cy  Dy  0

The magnitude of a vector is w A  Ax2  Ay2 and the angle  it makes with the +x-axis is
Ay
  arctan .
A
EXECUTE: We use the results of Ex. 1.7. See Fig. 1.23 in the text.
Ax = 38.37 m, Bx = –46.36 m, Cx = 0.00 m, Ay = 61.40 m, By = –33.68 m, Cy = –17.80 m
Adding the x-components gives
38.37 m + (–46.36 m) + 0.00 m + Dx = 0  Dx = 7.99 m
Adding the y-components gives
61.40 m + (–33.68 m) + (–17.80 m) + Dy = 0  Dy = –9.92 m
D  Dx2  Dy2 = (7.99 m) 2 + (–9.92 m) 2 = 12.7 m
Dy
glo   arctan = arctan[(–9.92 m)/(7.99 m)] = –51°
Dx
Since D has a positive x-component and a negative y-component, it points into the fourth quadrant
making an angle of 51° below the +x-axis and an angle of 360° – 51° = 309° counterclockwise with the
+x-axis.
EVALUATE: The vector D has the same magnitude as the resultant in Ex. 1.7 but points in the opposite
direction. This is reasonable because D must be opposite to the resultant of the three vectors in Ex. 1.7
to make the resultant of all four vectors equal to zero.

VP1.7.2. IDENTIFY: We know three vectors A , B , and C and we want to find the sum S where
S = A – B + C . The components of – B are the negatives of the components of B .
SET UP: The components of S are
Sx  Ax  Bx  Cx
S y  Ay  By  Cy

The magnitude A of a vector A is A  Ax2  Ay2 and the angle  it makes with the +x-axis is
Ay
  arctan .
A

, EXECUTE: Using the components from Ex. 1.7 we have
Sx = 38.37 m – (–46.36 m) + 0.00 m = 84.73 m
Sy = 61.40 m – (–33.68 m) + (–17.80 m) = 77.28 m
S  S x2  S y2 = (84.73 m) 2 + (77.28 m) 2 = 115 m
Sy
  arctan = arctan[(77.28 m)/(84.73 m)] = 42°
Sx
Since both components of S are positive, S points into the first quadrant. Therefore it makes an angle
of 42° with the +x-axis.
EVALUATE:




Figure VP1.7.2

The graphical solution shown in Fig. VP1.7.2 shows that our results are reasonable.

VP1.7.3. IDENTIFY: We know three vectors A , B , and C and we want to find the sum T where
T = A + B + 2C .
SET UP: Find the components of vectors A , B , and C and use them to find the magnitude and
direction of T . The components of 2 C are twice those of C .
EXECUTE: Sx  Ax  Bx  2Cx and S y  Ay  By  2Cy
(a) Using the components from Ex. 1.7 gives
Tx = 38.37 m + (–46.36 m) + 2(0.00 m) = –7.99 m
Ty = 61.40 m + (–33.68 m) + 2(–17.80 m) = –7.88 m
(b) T  Tx2  Ty2 = (–7.99 m) 2 + (–7.88 m) 2 = 11.2 m
Ty
  arctan = arctan[(–7.88 m)/(–7.99 m)] = 45°
Tx
Both components of T are negative, so it points into the third quadrant, making an angle of 45° below
the –x-axis or 45° + 180° = 225° counterclockwise with the +x-axis, in the third quadrant.

, EVALUATE:




Figure VP1.7.3

The graphical solution shown in Fig. VP1.7.3 shows that this result is reasonable.
VP1.7.4. IDENTIFY: The hiker makes two displacements. We know the first one and their resultant, and we want
to find the second displacement.
SET UP: Calling A the known displacement, R the known resultant, and D the unknown vector, we
know that A + D = R . We also know that R = 38.0 m and R makes an angle R = 37.0° + 90° =
127° with the +x-axis. Fig. VP1.7.4 shows a sketch of these vectors.




Figure VP1.7.4

EXECUTE: From Ex. 1.7 we have Ax = 38.37 m and Ay = 61.40 m. The components of R are
Rx = R cos 127.0° = (38.0 m) cos 127.0° = –22.87 m
Ry = R sin 38.0° = (38.0 m) sin 127.0° = 30.35 m
Rx  Ax  Dx and Ry  Ay  Dy
Using these components, we find the components of D .
38.37 m + Dx = –22.87 m  Dx = –22.87 m
61.40 m + Dy = –31.05 m  Dy = –31.05 m
D  Dx2  Dy2 = (–61.24 m) 2 + (–31.05 m) 2 = 68.7 m
Dy
  arctan = arctan[(–31.05 m)/(–61.24 m)] = 27°
D

, Both components of D are negative, so it points into the third quadrant, making an angle of 27° + 180°
= 207° with the +x-axis.
EVALUATE: A graphical solution will confirm these results.
VP1.10.1. IDENTIFY: We know the magnitude and direction of two vectors. We want to use these to find their
components and their scalar product.
SET UP: Ax = A cos  A , Ay = A sin  A , Bx = B cos B , By = B sin B . We can find the scalar product
using the vector components or using their magnitudes and the angle between them.
A  B = Ax Bx  Ay By and A  B = AB cos  . Which form you use depends on the information you
have.
EXECUTE: (a) Ax = A cos  A = (5.00) cos(360° – 36.9°) = 4.00
Ay = A sin  A = (5.00) sin(360° – 36.9°) = –3.00
Bx = (6.40) cos(90° + 20.0°) = –2.19
By = (6.40)sin(90° + 20.0°) = 6.01
(b) Using components gives
A  B = Ax Bx  Ay By = (4.00)(–2.19) + (–3.00)(6.01) = –26.8
EVALUATE: We check by using A  B = AB cos  .
A  B = AB cos  = (5.00)(6.40) cos(20.0° + 90° + 36.9°) = –26.8
This agrees with our result in part (b).
VP1.10.2. IDENTIFY: We know the magnitude and direction of one vector and the components of another vector.
We want to use these to find their scalar product and the angle between them.
SET UP: The scalar product can be expressed as A  B = Ax Bx  Ay By and A  B = AB cos  . Which
form you use depends on the information you have.
EXECUTE: (a) Cx = C cos C = (6.50) cos 55.0° = 3.728
Cy = C sin C = (6.50) sin 55.0° = 5.324
Dx = 4.80 and Dy = –8.40
Using components gives C  D = Cx Dx  Cy Dy = (3.728)(4.80) + (5.324)(–8.40) = –26.8

(b) D  Dx2  Dy2 = (4.80)2 +(–8.40) 2 = 9.675
C  D = CD cos  , so cos  = C  D /CD = (–26.8)/[(6.50)(9.67)] = –0.426.  = 115°.
EVALUATE: Find the angle that D makes with the +x-axis.




Figure VP1.10.2