HW 1 – Solution
1.
REASONING AND SOLUTION
a. The SI unit for x is m. The SI units for the quantity vt are
FG IJ
H K
m
m
  (s) = m
(s) = m
s
s
Therefore, the units on the left hand side of the equation are consistent with the units
on the right hand side.
b. As described in part a, the SI units for the quantities x and vt are both m. The SI
units for the quantity
1
2
at2 are
FG m IJ (s
Hs K
2
2
)=m
Therefore, the units on the left hand side of the equation are consistent with the units
on the right hand side.
c. The SI unit for v is m/s. The SI unit for the quantity at is
FG m IJ (s) = m
Hs K s
2
Therefore, the units on the left hand side of the equation are consistent with the units
on the right hand side.
d. As described in part c, the SI units of the quantities v and at are both m/s. The SI
unit of the quantity
1
2
at3 is
FG m IJ (s ) = m ⋅ s
Hs K
3
2
Thus, the units on the left hand side are not consistent with the units on the right
hand side. In fact, the right hand side is not a valid operation because it is not
possible to add physical quantities that have different units.
e. The SI unit for the quantity v3 is m3/s3. The SI unit for the quantity 2ax3 is
FG m IJ (m
Hs K
2
2
)=
m3
s2
Therefore, the units on the left hand side of the equation are not consistent with the
units on the right hand side.
f. The SI unit for the quantity t is s. The SI unit for the quantity
m
=
(m / s 2 )
Fs I
mG J =
H mK
2
2x
is
a
s2 = s
Therefore, the units on the left hand side of the equation are consistent with the units
on the right hand side.
17. REASONING AND SOLUTION The equation A + B = C tells us that the vector C
is the resultant of the vectors A and B. The magnitudes of the vectors A, B, and C
are related by A + B = C. In other words, the length of the vector C is equal to
the combined lengths of vectors A and B. Therefore, the vectors A and B must point
in the same direction.
7. REASONING AND SOLUTION
a. F = [M][L]/[T]2; ma = [M][L]/[T]2 = [M][L]/[T]2 so F = ma
is dimensionally correct .
b. x = [L]; at3 = ([L]/[T]2)[T]3 = [L][T] so x = (1/2)at3
is not dimensionally correct .
c. E = [M][L]2/[T]2; mv = [M][L]/[T] so E = (1/2)mv
is not dimensionally correct .
d. E = [M][L]2/[T]2; max = [M]([L]/[T]2)[L] = [M][L]2/[T]2 so E = max
is dimensionally correct .
e. v = [L]/[T]; (Fx/m)1/2 = {([M][L]/[T]2)([L]/[M])}1/2 = {[L]2/[T]2}1/2 = [L]/[T]
so
v = (Fx/m)1/2 is dimensionally correct .
27.
SSM WWW REASONING AND SOLUTION The single force needed to
produce the same effect is equal to the resultant of the forces provided by the two
ropes. The figure below shows the force vectors drawn to scale and arranged tail to
head. The magnitude and direction of the resultant can be found by direct
measurement using the scale factor shown in the figure.
2900 N
2900 N
Scale:
1000 N
Resultant
a. From the figure, the magnitude of the resultant is 5600 N .
b. The single rope should be directed along the dashed line in the text drawing.
28. REASONING The displacement vector R for a straight-line
dash between the starting and finishing points can be found by
drawing the vectors A, B, and C to scale in both magnitude and Side of court
orientation, in a tail-to-head fashion, and connecting the tail of A
to the head of C. The magnitude of the resultant can be found
by measuring its length and making use of the scale factor.
Similarly, the direction of the resultant is found by measuring
the angle θ it makes with the side of the court.
SOLUTION By construction and measurement, the magnitude
of the resultant R is 20 m . The angle θ made with the side of
C
R
B
θ
A
the court is 15° .
29. REASONING The displacement Q needed for the bear to return to its starting point
is the vector that is equal in magnitude and opposite in direction to the actual
displacement R of the bear.
SOLUTION The first two rows of the following table gives components of the
individual displacements, A and B, of the bear.
The third row gives the
components of R = A + B . The directions due east and due north have been chosen
to be positive.
Displacement
East/West
Component
North/South
Component
A
B
–1563 m
(– 3348 m) cos 32°= –2839 m
0
(– 3348 m) sin 32°= 1774 m
R=A+B
–4402 m
1774 m
Thus, the components of the displacement Q needed for the bear to return to its
starting point are an eastward component of +4402 m and a southward component of
–1774 m.
a. From the Pythagorean theorem, we have
4402 m
Q = (4402 m) 2 + (–1774 m) 2 = 4746 m
θ
b. The angle θ is given by
θ = tan −1
FG 1774 m IJ =
H 4402 m K
Q
1774 m
21.9 ° , south of east
36. REASONING AND SOLUTION The horizontal component of the plane's velocity is
vx = v cos θ = (180 m/s) cos 34° = 150 m/s
37. REASONING AND SOLUTION
a. Ax = A cos 30.0° = (750 units)(0.866) =
650 units
Ay = A sin 30.0° = (750 units)(0.500) =
380 units
b. Ax' = A cos 40.0° = (750 units)(0.643) =
570 units
Ay' = –A sin 40.0° = –(750 units)(0.766) = –480 units
41. SSM REASONING The individual displacements of the golf ball, A, B, and C
are shown in the figure. Their resultant, R, is the displacement that would have been
needed to "hole the ball" on the very first putt. We will use the component method
to find R.
N
R
C
B
θ
20.0
o
E
A
SOLUTION
below.
The components of each displacement vector are given in the table
Vector
x Components
y Components
A
B
C
(5.0 m) cos 0° = 5.0 m
(2.1 m) cos 20.0° = 2.0 m
(0.50 m) cos 90.0° = 0
(5.0 m) sin 0° = 0
(2.1 m) sin 20.0° = 0.72 m
(0.50 m) sin 90.0° = 0.50 m
R = A+B+C
7.0 m
1.22 m
The resultant vector R has magnitude
R = ( 7 .0 m) 2 + ( 1.22 m) 2 = 7 .1 m
and the angle θ is
 1.22 m 
= 9.9°
7.0 m 
θ = tan −1 
Thus, the required direction is 9.9° north of east .
46. REASONING AND SOLUTION Since the resultant of the three vectors is zero, each
component of the resultant must be equal to zero.
The x components of the first two forces are
F1x = (166 N) cos 60.0° = 83.0 N
F2x = (284 N) cos 30.0° = 246 N
The x component of the resultant of the three vectors is, therefore,
(83 N) + (246 N) + F3x = 0
or F3x = – 329 N. The minus sign indicates that F3x points in the negative x
direction.
The y components of the first two forces are
F1y = (166 N) sin 60.0° = 144 N
F2y = (284 N) sin 30.0° = 142 N
The y component of the resultant of the three vectors is, therefore,
(144 N) + (142 N) + F3y = 0
or F3y = – 286 N. The minus sign indicates that F3y points in the negative y
direction.
The magnitude of F3 is, from
the Pythagorean theorem,
y
( 329 N) 2 + ( 286 N) 2 = 436 N
The angle made by F3 with the
negative x axis is
θ = tan
−1
LM 286 N OP =
N 392 N Q
F
3x
F
3y
θ
x
F
3
41.0 °
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END