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Assessment
Chapter Test B
Two-Dimensional Motion and Vectors
MULTIPLE CHOICE
In the space provided, write the letter of the term or phrase that best completes
each statement or best answers each question.
______ 1. Identify the following quantities as scalar or vector: the mass of an
object, the number of leaves on a tree, wind velocity.
a. vector, scalar, scalar
c. scalar, vector, scalar
b. scalar, scalar, vector
d. vector, scalar, vector
______ 2. A student walks from the door of the house to the end of the driveway
and realizes that he missed the bus. The student runs back to the
house, traveling three times as fast. Which of the following is the correct expression for the return velocity if the initial velocity is vstudent?
a. 3vstudent
b.
1
3
c.
1
3
vstudent
d. -3vstudent
vstudent
______ 3. An ant on a picnic table travels 3.0 101 cm eastward, then 25 cm
northward, and finally 15 cm westward. What is the magnitude of the
ant’s displacement relative to its original position?
a. 70 cm
c. 52 cm
b. 57 cm
d. 29 cm
______ 4. In a coordinate system, the magnitude of the x component of a vector
and q, the angle between the vector and x-axis, are known. The magnitude of the vector equals the x component
a. divided by the cosine of q.
b. divided by the sine of q.
c. multiplied by the cosine of q.
d. multiplied by the sine of q.
______ 5. Find the resultant of these two vectors: 2.00 102 units due east and
4.00 102 units 30.0° north of west.
a. 300 units 29.8° north of west
b. 581 units 20.1° north of east
c. 546 units 59.3° north of west
d. 248 units 53.9° north of west
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Chapter Test B continued
______ 6. In the figure at right, the magnitude of the ball’s velocity is least
at location
a. A.
b. B.
c. C.
d. D.
B
A
C
D
______ 7. In the figure at right, the horizontal component of the ball’s
velocity at A is
a. zero.
b. equal to the vertical component of the ball’s velocity at C.
c. equal in magnitude but opposite in direction to the horizontal
component of the ball’s velocity at D.
d. equal to the horizontal component of its initial velocity.
______ 8. A track star in the long jump goes into the jump at 12 m/s and
launches herself at 20.0° above the horizontal. What is the magnitude
of her horizontal displacement? (Assume no air resistance and that
ay g 9.81 m/s2.)
a. 4.6 m
c. 13 m
b. 9.2 m
d. 15 m
______ 9. A boat travels directly across a river that has a downstream current, v.
What is true about the perpendicular components of the boat’s velocity?
a. One component equals v; the other component equals zero.
b. One component is perpendicular to v; the other component
equals v.
c. One component is perpendicular to v; the other component
equals v.
d. One component is perpendicular to v; the other component
equals zero.
______10. A jet moving at 500.0 km/h due east is in a region where the wind is
moving at 120.0 km/h in a direction 30.00° north of east. What is the
speed of the aircraft relative to the ground?
a. 620.2 km/h
c. 588.7 km/h
b. 606.9 km/h
d. 511.3 km/h
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Chapter Test B continued
SHORT ANSWER
11. Briefly explain the triangle (or polygon) method of addition.
12. If the magnitude of a component vector equals the magnitude of the vector,
then what is the magnitude of the other component vector?
13. How can you use the Pythagorean theorem to add two vectors that are not
perpendicular?
14. Briefly explain why a basketball being thrown toward a hoop is considered
projectile motion.
PROBLEM
15. A cave explorer travels 3.0 m eastward, then 2.5 m northward, and finally
15.0 m westward. Use the graphical method to find the magnitude of the net
displacement.
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Chapter Test B continued
16. A dog walks 28 steps north and then walks 55 steps west to bury a bone. If
the dog walks back to the starting point in a straight line, how many steps will
the dog take? Use the graphical method to find the magnitude of the net
displacement.
17. A quarterback takes the ball from the line of scrimmage and runs backward
for 1.0 101 m then sideways parallel to the line of scrimmage for 15 m. The
ball is thrown forward 5.0 101 m perpendicular to the line of scrimmage.
The receiver is tackled immediately. How far is the football displaced from its
original position?
18. Vector A is 3.2 m in length and points along the positive y-axis. Vector B is
4.6 m in length and points along a direction 195° counterclockwise from the
positive x-axis. What is the magnitude of the resultant when vectors A and B
are added?
19. A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0
m/s. If the canyon below is 100.0 m deep, how far from the edge of the cliff
does the model rocket land? (ay g 9.81 m/s2)
20. A boat moves at 10.0 m/s relative to the water. If the boat is in a river where
the current is 2.00 m/s, how long does it take the boat to make a complete
round trip of 1000.0 m upstream followed by 1000.0 m downstream?
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TEACHER RESOURCE
5. d
Solution
vi,y vi sin q (12 m/s)(sin 30.0°) 6.0 m/s
Solution
x1 2.00 102 units
y1 0
x2 d2 cos q (4.00 102 units)(cos 30.0°) 3.46 102 units
y2 d2 sin q (4.00 102 units)
(sin 30.0°) 2.00 102 units
xtot x1 x2 (2.00 102 units) (3.46 102 units) 1.46 102 units
ytot y1 y2 0 (2.00 102 units) 2.00 102 units
2
d ( xtot)2 ( ytot)2
1
y 2ay(t)2 vi,yt 1
(g)(t)2
2
1
(9.81
2
vi,yt m/s2)(5.6 s)2 (6.0 m/s)(5.6 s)
y 120 m
h 120 m
20. 27 km/h north
Given
vpg velocity of plane to ground 145 km/h south
vpa velocity of plane to air 170.0
km/h south
d ( xtot
)2 (
ytot)2 (1.46
102
units)2
(2
.00 102 un
its)2
d 2.48 102 units
y
q tan1 y
Solution
vpg vpa vag
vag vpg vpa
vag 145 km/h 172 km/h 27 km/h
2.00 102 units
53.9°
tan1
1.46 102 units
d 2.48 102 units 53.9° north of west
vag 27 km/h north
Two-Dimensional Motion
and Vectors
6. b
7. d
8. b
CHAPTER TEST B (ADVANCED)
Solution
vx vi,x vi cos q (12 m/s)(cos 20.0°) 11 m/s
vi,y vi sin q (12 m/s)(sin 20.0°) 4.1 m/s
1. b
2. d
3. d
Given
x1 3.0 101 cm east
y1 25 cm north
x2 15 cm west
1
Solution
xtot x1 x2 (3.0 101 cm) (15 cm) 15 cm
ytot y2 25 cm
d2 (xtot)2 (ytot)2
d (xtot
) (
ytot) 2
2
y 0 2ay(t)2 vi,yt
2vi,y
2vi,y
2(4.1 m/s)
t ay
g
9.81 m/s
0.84 s
x vxt (11 m/s)(0.84 s) 9.2 m
9. c
10. b
Given
vpa velocity of plane relative to the
air 500.0 km/h east
vag velocity of air relative to the
ground 120.0 km/h 30.00° north
of east
(15
cm
)2 (
25 cm
)2
d
29 cm
4. a
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TEACHER RESOURCE
18. 4.9 m
Solution
vag,x vag cos q (120.0 km/h)
(cos 30.00°) 103.9 km/h
vag,y vag sin q (120.0 km/h)
(sin 30.00°) 60.0 km/h
vpg,x vpa vag,x 500.0 km/h 103.9 km/h 603.9 km/h
vpg,y 60.0 km/h
Given
d1 3.2 m along y-axis
d2 4.6 m at 195° counterclockwise
from x-axis
q1 0.0°
d1 3.2 m
q2 195°
d2 4.6 m
Solution
x1 0.0 m
y1 3.2 m
x2 d2 cos q (4.6 m)(cos 195°) 4.4 m
y2 d2 sin q (4.6 m)(sin 195°) 1.2 m
xtot x1 x2 (0 m) (4.4 m) 4.4 m
ytot y1 y2 (3.2 m) (1.2 m) 2.0 m
2
d (xtot)2 (ytot)2
2
vpg (v
(vpg,y
)2 pg,x)
2
(603.9
km/h)
(60.0
km
/h)2 606.9 km/h
11. The triangle method of adding vectors
12.
13.
14.
15.
requires that you align the vectors, one
after the other, tail to nose, by moving
them parallel and perpendicular to
their original orientations. The resultant vector is an arrow drawn from the
tail of the first vector to the tip of the
last vector.
The magnitude of the other component vector is zero.
Resolve each vector into perpendicular components and add the components that lie along the same axis. The
resultant vectors can be added by
using the Pythagorean theorem
because they are perpendicular.
Objects sent into the air and subject to
gravity exhibit projectile motion.
12.2 m
d (xtot
)2 (
ytot)2 (4.4
m)2 (2.0 m
)2 4.9 m
19. 226 m
Given
v 50.0 m/s horizontally
y 100.0 m
Solution
vi,x vx 50.0 m/s horizontally
vi,y 0
1
y 2ay(t)2
2y
(t)2 ay
Solution
d (12.0
m
) (
2.5 m) 12.2 m
2
2
2y
2y
a
(g)
2(100.0 m)
4.52 s
9.8
1 m/s 16. 62 steps
t Solution
y
2
d (28
steps)
(55 st
eps)2 2
62 steps
x vxt (50.0 m/s)(4.52 s) 17. 43 m
Given
x1 1.0 101 m
y1 15 m
x2 5.0 101 m
226 m
20. 208 s
Given
vrg velocity of river to ground 2.00 m/s downstream
vbr velocity of boat to river 10.00 m/s
x1 1000.0 m downstream
x2 1000.0 m downstream
vbg velocity of boat
Solution
xtot x1 x2 (1.0 101 m)
(5.0 101 m) 4.0 101 m
ytot y1 15 m
d2 (xtot)2 (ytot)2
d (xtot
)2 (
ytot)2 (4.0 101 m
)2 (1.5 101 m
)2
4.3 101 m
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TEACHER RESOURCE
Solution
downstream
vbg vbr vrg
vbg 10.00 m/s 2.00 m/s 12.00 m/s
t1 x1/vbg 1000.0 m/12.00 m/s 83.33 s
upstream
vbg vbr vrg
vbg 10.00 m/s 2.00 m/s 8.00 m/s
1000.0 m
t2 x2\vbg 125 s
8.00 m/s
t t1 t2 83.33 s 125 s Solution
Fnet, x Fx max (32 kg)(0.50 m/s2) 16 N
25. 10.4 N
Given
m 1.10 kg
a 15.0°
g 9.81 m/s2
Solution
Fnet, y Fy Fn Fy 0
q 180° 90° 15.0° 75.0°
Fn Fy Fg sin q mg sin q
Fn (1.10 kg)(9.81 m/s2)(sin 75.0°)
10.4 N
208 s
Forces and the Laws
of Motion
Forces and the Laws
of Motion
CHAPTER TEST A (GENERAL)
CHAPTER TEST B (ADVANCED)
1.
2.
3.
4.
5.
6.
7.
8.
9.
19.
20.
21.
22.
23.
c
10. d
d
11. c
d
12. a
c
13. d
c
14. d
c
15. b
c
16. d
b
17. c
d
18. d
Forces exerted by the object do not
change its motion.
An object at rest remains at rest and
an object in motion continues in
motion with constant velocity unless it
experiences a net external force.
F is the vector sum of the external
forces acting on the object.
In most cases, air resistance increases
with increasing speed.
27 N, to the right
1.
2.
3.
4.
Given
Fy 60.0 N
q 30.0°
Solution
5.
6.
7.
11.
12.
Fy
cos q F
Fy
60.6 N
F 70.0 N
cos q
cos 30.0°
c
8. a
d
9. c
d
10. a
b
a
Given
Fg 1.0 102 N
q 20.0°
Given
F1 102 N, to the right
F2 75 N, to the left
Solution
Fnet F1 F2
Fnet F1 F2 102 N 75 N 27 N
Fnet 27 N, to the right
d
a
c
b
Solution
Fy Fn Fg,y 0
Fn Fg, y Fg cos q (1.0 102 N)(cos 20.0°) 94 N
13. b
Given
Fg, book 5 N
m s 0.2
24. 16 N
Given
m 33 kg
a 0.50 m/s2
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