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Wayne Hacker’s Problem Bank Solutions College Physics Version 1.0 (Draft) Wayne Hacker Copyright ©Wayne Hacker 2011. All rights reserved.
July 11, 2011
Physics Problem Bank Solutions
Copyright ©Wayne Hacker 2009. All rights reserved. 1
Contents 0 About the mathematical prerequisites material
I
11
0.1
What is a College Physics Course? . . . . . . . . . . . . . . . . . . . . .
11
0.2
What are the math-prerequise topics for college physics? . . . . . . . . .
11
0.3
What is a University Physics course? . . . . . . . . . . . . . . . . . . . .
11
0.4
What is a college physics course in the state of Arizona? . . . . . . . . .
12
Mathematical Prerequisites
1 Geometry 1.1
1.2
1.3
1.4
13 14
Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.1.1
Letter-based problems . . . . . . . . . . . . . . . . . . . . . . . .
14
1.1.2
Non-calculator-based problems . . . . . . . . . . . . . . . . . . . .
16
1.1.3
Calculator-based problems . . . . . . . . . . . . . . . . . . . . . .
19
Rectangles and rectangular solids . . . . . . . . . . . . . . . . . . . . . .
21
1.2.1
Letter-based problems . . . . . . . . . . . . . . . . . . . . . . . .
21
1.2.2
Non-calculator-based problems . . . . . . . . . . . . . . . . . . . .
22
1.2.3
Calculator-based problems . . . . . . . . . . . . . . . . . . . . . .
24
Right triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
1.3.1
Letter-based problems . . . . . . . . . . . . . . . . . . . . . . . .
26
1.3.2
Non-calculator-based problems . . . . . . . . . . . . . . . . . . . .
28
1.3.3
Calculator-based problems . . . . . . . . . . . . . . . . . . . . . .
30
Right triangle word problems . . . . . . . . . . . . . . . . . . . . . . . .
33
1.4.1
Non-calculator-based problems . . . . . . . . . . . . . . . . . . . .
33
1.4.2
Calculator-based problems . . . . . . . . . . . . . . . . . . . . . .
37
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2 Algebra 2.1
43
Evaluating functions with numerical arguments . . . . . . . . . . . . . .
43
2.1.1
Non-calculator based problems . . . . . . . . . . . . . . . . . . . .
43
2.1.2
Calculator-based problems . . . . . . . . . . . . . . . . . . . . . .
46
2.2
Evaluating functions with variable-expression arguments . . . . . . . . .
49
2.3
Evaluating functions of multiple variables . . . . . . . . . . . . . . . . . .
52
2.3.1
Functions of two variables . . . . . . . . . . . . . . . . . . . . . .
52
2.3.2
Functions of three variables . . . . . . . . . . . . . . . . . . . . .
57
Solving linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
2.4.1
Linear equations with numerical solutions . . . . . . . . . . . . .
62
2.4.2
Linear equations with variable-expression solutions . . . . . . . .
64
2.5
Solving systems of linear equations . . . . . . . . . . . . . . . . . . . . .
67
2.6
Solving quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . .
78
2.6.1
Factoring quadratic equations . . . . . . . . . . . . . . . . . . . .
78
2.6.2
Quadratic formula . . . . . . . . . . . . . . . . . . . . . . . . . .
82
Algebra word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
2.4
2.7
3 Graphs
II
88
3.1
Single points on graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
3.2
Matching graphs and equations . . . . . . . . . . . . . . . . . . . . . . .
96
Mathematical Preliminaries
107
Physics Problem Bank Solutions 4 Basic Trigonometry 4.1
4.2
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108
Arc-Length Problems using s = θ r . . . . . . . . . . . . . . . . . . . . . 108 4.1.1
Degrees to radians: formula . . . . . . . . . . . . . . . . . . . . . 108
4.1.2
Radians to degrees: formula . . . . . . . . . . . . . . . . . . . . . 110
4.1.3
Degrees to radians: calculator . . . . . . . . . . . . . . . . . . . . 112
4.1.4
Radians to degrees: calculator . . . . . . . . . . . . . . . . . . . . 114
4.1.5
Arc length to radians: pictures . . . . . . . . . . . . . . . . . . . 116
4.1.6
Arc length to radians: descriptions . . . . . . . . . . . . . . . . . 117
4.1.7
Radians to arc length: pictures . . . . . . . . . . . . . . . . . . . 119
4.1.8
Radians to arc length: descriptions . . . . . . . . . . . . . . . . . 120
4.1.9
Word problems: arc length and radians . . . . . . . . . . . . . . . 122
Basic Right-Triangle Trigonometry . . . . . . . . . . . . . . . . . . . . . 124 4.2.1
Basic trig functions: finding . . . . . . . . . . . . . . . . . . . . . 124
4.2.2
Basic trig functions: identifying . . . . . . . . . . . . . . . . . . . 127
4.2.3
Basic trig functions: calculator; radians . . . . . . . . . . . . . . . 133
4.2.4
Basic trig functions: calculator; degrees . . . . . . . . . . . . . . . 135
4.2.5
Using basic trig functions: formulas . . . . . . . . . . . . . . . . . 137
4.2.6
Using basic trig functions: calculator . . . . . . . . . . . . . . . . 139
4.2.7
Reference angles
4.2.8
Arc functions: definition . . . . . . . . . . . . . . . . . . . . . . . 148
4.2.9
Arc functions: definition; diagrams . . . . . . . . . . . . . . . . . 150
. . . . . . . . . . . . . . . . . . . . . . . . . . . 144
4.2.10 Arc functions: calculator; radians . . . . . . . . . . . . . . . . . . 151 4.2.11 Arc function: calculator; degrees . . . . . . . . . . . . . . . . . . . 153 4.2.12 Arc functions: calculator; diagram; radians . . . . . . . . . . . . . 154 4.2.13 Arc functions: calculator; diagram; degrees . . . . . . . . . . . . . 157 4.2.14 Word problems: basic trig functions . . . . . . . . . . . . . . . . . 160 4.2.15 Word problems: arc functions . . . . . . . . . . . . . . . . . . . . 163
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5 Introduction to measurement: dimensions, units, scientific notation, and significant figures 166 5.1
5.2
Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 5.1.1
Dimension or unit? . . . . . . . . . . . . . . . . . . . . . . . . . . 166
5.1.2
Dimensional consistency . . . . . . . . . . . . . . . . . . . . . . . 167
5.1.3
Practical questions involving dimensions . . . . . . . . . . . . . . 173
Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 5.2.1
Converting between different sets of units . . . . . . . . . . . . . . 176
5.3
Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
5.4
Significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
5.5
Determining derived units from equations . . . . . . . . . . . . . . . . . . 193
6 Introduction to Vectors 6.1
196
Identifying Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 6.1.1
Given vector, identify on figure . . . . . . . . . . . . . . . . . . . 196
6.1.2
Given vector on figure, identify formula . . . . . . . . . . . . . . . 201
6.1.3
Given direction, identify on figure . . . . . . . . . . . . . . . . . . 207
6.1.4
Given vectors on figure, name quadrant of sum/difference . . . . . 212
6.1.5
Mixing it up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
6.2
Geometric Vector Addition and Subtraction . . . . . . . . . . . . . . . . 220
6.3
Position vs. Displacement Vector Problems . . . . . . . . . . . . . . . . . 225
6.4
Finding Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . 229
6.5
Algebraic Vector Addition and Subtraction . . . . . . . . . . . . . . . . . 231
6.6
Concept vector and scalar questions . . . . . . . . . . . . . . . . . . . . . 235
6.7
Applications of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 6.7.1
Breaking vectors into components . . . . . . . . . . . . . . . . . . 240
6.7.2
Introduction to force vectors . . . . . . . . . . . . . . . . . . . . . 248
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III
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Kinematics
254
7 One-dimensional linear kinematics
255
7.1
Concept questions: kinematics . . . . . . . . . . . . . . . . . . . . . . . . 255
7.2
Qualitative kinematics: descriptions . . . . . . . . . . . . . . . . . . . . . 258
7.3
Qualitative kinematics: from graph . . . . . . . . . . . . . . . . . . . . . 266
7.4
Introducing the fundamental one-dimensional kinematic equations . . . . 271 7.4.1
Dimensional consistency of the fundamental equations
. . . . . . 272
7.5
Quantitative kinematics: horizontal . . . . . . . . . . . . . . . . . . . . . 273
7.6
Quantitative kinematics: vertical . . . . . . . . . . . . . . . . . . . . . . 287 7.6.1
The difference between average speed and average velocity in onedimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
7.7
Similarity problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
7.8
Lab-application problems: One-Dimensional Kinematics . . . . . . . . . . 300
7.9
7.8.1
Ball-drop apparatus
. . . . . . . . . . . . . . . . . . . . . . . . . 300
7.8.2
Spring-cannon apparatus . . . . . . . . . . . . . . . . . . . . . . . 300
Mixing It Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
8 One-dimensional rotational kinematics
310
8.1
Angular speed, period, and frequency . . . . . . . . . . . . . . . . . . . . 310
8.2
Fundamental equation 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
8.3
Fundamental equation 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
8.4
Fundamental equation 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
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9 Two-dimensional kinematics 9.1
9.2
The basics of velocity and acceleration in two dimensions . . . . . . . . . 328 9.1.1
The difference between average speed and average velocity in a plane328
9.1.2
Average acceleration in a plane . . . . . . . . . . . . . . . . . . . 329
9.1.3
Relative velocity in a plane
9.1.4
Graphical interpretation of velocity and acceleration . . . . . . . . 330
9.4
IV
Similarity problems . . . . . . . . . . . . . . . . . . . . . . . . . . 336
Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 9.3.1
General equations for projectile motion . . . . . . . . . . . . . . . 338
9.3.2
Velocity and acceleration of a projectile . . . . . . . . . . . . . . . 339
9.3.3
Special Case I: The Half-Parabola . . . . . . . . . . . . . . . . . . 341
9.3.4
Special Case II: The Full Parabola . . . . . . . . . . . . . . . . . 348
Lab application problems: Two-Dimensional Kinematics . . . . . . . . . 359 9.4.1
9.5
. . . . . . . . . . . . . . . . . . . . . 329
Uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 9.2.1
9.3
328
Spring-cannon apparatus . . . . . . . . . . . . . . . . . . . . . . . 359
Advanced-Level Problems (Kinematics) . . . . . . . . . . . . . . . . . . . 366
Newton’s Laws
10 Introduction to Newton’s Laws
372 373
10.1 Concept questions involving Newton’s three laws . . . . . . . . . . . . . . 373 10.2 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 10.3 Pushing and pulling objects on horizontal frictionless surfaces . . . . . . 380
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11 Applications of Newton’s Laws to Linear Motion
381
11.1 Newton’s laws in one-dimension . . . . . . . . . . . . . . . . . . . . . . . 381 11.1.1 One-dimensional kinematics and Newton’s second law . . . . . . . 381 11.1.2 Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 11.1.3 Apparent weight problems (The Elevator Equation) . . . . . . . . 395 11.1.4 Spring Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 402 11.2 Newton’s laws in two-dimensions . . . . . . . . . . . . . . . . . . . . . . 408 11.2.1 Static equilibrium problems . . . . . . . . . . . . . . . . . . . . . 408 11.2.2 Block and pulley frictionless systems . . . . . . . . . . . . . . . . 420 11.2.3 Accelerometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 11.2.4 Static and kinetic friction problems . . . . . . . . . . . . . . . . . 437 11.2.5 Stopping distance problems . . . . . . . . . . . . . . . . . . . . . 446 12 Applications of Newton’s Laws to Uniform Circular Motion
449
12.1 Centripetal force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449 12.1.1 Conceptual questions: centripetal force . . . . . . . . . . . . . . . 449 12.1.2 Horizontal Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 453 12.1.3 Vertical Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 12.2 Gravitation and circular orbits . . . . . . . . . . . . . . . . . . . . . . . . 473 12.2.1 Gravation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473 12.2.2 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 13 Applications of Newton’s Laws to Rotating Bodies
484
13.1 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 13.2 Angular acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 13.3 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
V
Work, Energy, and Momentum
509
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14 Work and mechanical energy
510
14.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 14.2 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 14.3 Kinetic energy (K.E.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514 14.4 Potential energy (P.E.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 14.5 Work-Mechanical-Energy Theorem . . . . . . . . . . . . . . . . . . . . . 519 14.5.1 Work-Kinetic-Energy Theorem 14.5.2 Work-Potential-Energy Theorem
. . . . . . . . . . . . . . . . . . . 519 . . . . . . . . . . . . . . . . . . 522
14.6 Conservation of mechanical energy (M.E.) . . . . . . . . . . . . . . . . . 529 15 Conservation of Linear Momentum
543
15.1 Computing Linear Momentum p = mv . . . . . . . . . . . . . . . . . . . 543 15.2 Applying conservation of momentum to isolated systems . . . . . . . . . 544 15.3 Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548 15.4 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553 16 Rotational kinetic energy and angular momentum
VI
Applications of Mechanics
17 Applications of Mechanics
568
578 579
17.1 Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 17.2 Mechanical waves and sound . . . . . . . . . . . . . . . . . . . . . . . . . 581 17.3 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584 17.4 Static Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 17.4.1 Pascal’s Law and Archimedes’s principle . . . . . . . . . . . . . . 587 17.4.2 Density, specific volume, specific weight . . . . . . . . . . . . . . . 587 17.4.3 Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589
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18 Introduction to Thermodynamics
593
18.1 Introductory Concepts and Definitions . . . . . . . . . . . . . . . . . . . 593 18.1.1 What is the study of thermodynamics? . . . . . . . . . . . . . . . 593 18.1.2 Defining Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 18.1.3 Closed systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594 18.1.4 Control volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594 18.1.5 Property, state, and process . . . . . . . . . . . . . . . . . . . . . 595 18.1.6 Extensive and intensive properties . . . . . . . . . . . . . . . . . . 595 18.1.7 Equilibrium, quasi-equilibrium, and processes . . . . . . . . . . . 595 18.1.8 Base SI units for mass, length, time, Force, energy, and pressure . 596 18.2 Temperature Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598
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List of Figures 1
A car is being driven erratically along a straight stretch of highway. The graph shows its position x as a function of the time t. . . . . . . . . . . 266
2
A car is being driven erratically along a straight stretch of highway. The graph shows its position x as a function of the time t. . . . . . . . . . . 268
3
Spring cannon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
4
Spring cannon and bucket . . . . . . . . . . . . . . . . . . . . . . . . . . 368
5
Spring cannon and bucket . . . . . . . . . . . . . . . . . . . . . . . . . . 369
6
Side view of the spring cannon firing ball into a bucket mounted on top of a glider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
7
Earth-mass system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528
8
Ball in the bucket. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540
9
The loop-the-loop. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541
10
Block sliding off a hemispherical piece of ice. . . . . . . . . . . . . . . . . 542
11
Bullet-wooden-block collision. . . . . . . . . . . . . . . . . . . . . . . . . 559
12
Ballistic Pendulum found in lab . . . . . . . . . . . . . . . . . . . . . . . 566
13
Ballistic Pendulum: Stage 1 . . . . . . . . . . . . . . . . . . . . . . . . . 566
14
Ballistic Pendulum: Stage 2 . . . . . . . . . . . . . . . . . . . . . . . . . 566
15
Ballistic Pendulum: Stage 3 . . . . . . . . . . . . . . . . . . . . . . . . . 567
16
Vertical glass tube attached to horizontal water pipe. . . . . . . . . . . . 592
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About the mathematical prerequisites material
0.1
What is a College Physics Course?
Typically, in the United States a college physics course is a physics course that is taught without the use of calculus. The prerequisites are high school algebra (a.k.a. Algebra II from high school, or equivalently intermediate algebra from a community college) and trigonometry. It is also assumed that the student has had an introduction to dimensions, units, and significant figures.
0.2
What are the math-prerequise topics for college physics?
College Physics Prerequisite Topics: • Geometry section 1 and all subsections therein • Algebra section 2 and all subsections therein • Graphing section 3 and all subsections therein • Basic Trigonometry section 4 and all subsections therein (except identities) • An introduction to vectors section 6 and all subsections therein (except the dot and cross-product) • Introduction to measurement and all subsections therein
0.3
What is a University Physics course?
Typically, in the United States a university physics course is a calculus-based physics course taken by engineering, physics, and math majors. The prerequisites are a highschool physics course and a minimum mathematical background of Calculus I from high school, or equivalently from a college, or university. Remember, this is an absolute minimum! A background of calculus II and a strong grasp of trigonometry will make the course more bearable. University Physics Prerequisite Topics: • Geometry section 1 and all subsections therein • Algebra section 2 and all subsections therein • Graphing section 3 and all subsections therein
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• Basic Trigonometry section 4 and all subsections therein • Calculus section ?? and all subsections therein • An introduction to vectors section 6 and all subsections therein • Introduction to measurement and all subsections therein
0.4
What is a college physics course in the state of Arizona?
Unfortunately, the University of Arizona and Pima community college only require highschool algebra as a prerequisite for college physics. As such, the physics 121 course taught at Pima presupposes no trigonometry background what-so-ever. However, a background in basic trigonometry will make the first few weeks of the course more bearable. College algebra is also a required co-requisite for this course.
13
Part I
Mathematical Prerequisites
14
1 1.1 1.1.1
Geometry Circles Letter-based problems
Problem 1. If the radius of a circle is R, what is its diameter? (a) R/2 (c) πR/2 (e) None of these Solution:
*(b) 2R (d) 2πR
You should know this by memory.
Problem 2. If the radius of a circle is R, what is its circumference? (a) R/2 (c) πR/2 (e) None of these Solution:
(b) 2R *(d) 2πR
You should know this by memory.
Problem 3. If the radius of a circle is R, what is its area? (a) πR *(c) πR2 (e) None of these Solution:
(b) (d)
2πR 2πR2
You should know this by memory.
Problem 4. If the diameter of a circle is D, what is its radius? (a) πD2 /2 *(c) D/2 (e) None of these Solution:
D = 2R
(b) 2πD2 (d) 2D ⇒
R = D/2
Problem 5. If the diameter of a circle is D, what is its circumference? *(a) πD (c) πD/2 (e) None of these Solution:
(b) 2πD (d) πD/4
C = 2πR and D = 2R; so C = 2π(D/2) = πD
Problem 6. If the diameter of a circle is D, what is its area? (a) πD (c) πD2 /2 (e) None of these Solution:
(b) 2πD *(d) πD2 /4
A = πR2 and D = 2R; so A = π(D/2)2 = πD2 /4.
15 Problem 7. If the area of a circle is A, what is its radius? r √ A 2A (a) (b) 2π π r √ A A *(d) (c) π π (e)
None of these r
Solution:
A = πR2
⇒
R=
A π
Problem 8. If the area of a circle is A, what is its diameter? r r A 2A (b) *(a) 2 π π √ √ 2 A 2A (d) (c) π π (e)
None of these r
Solution:
A = πR2 and D = 2R; so R =
A π
r ⇒
D=2
A π
Problem 9. If the area of a circle is A, what is its circumference? √ √ (a) 2π A *(b) 2 πA r r A 2A (d) (c) π π (e)
None of these r
Solution:
A = πR2 and C = 2πR; so R =
A π
r ⇒
C = 2π
√ A = 2 πA π
Problem 10. If the circumference of a circle is C, what is its radius? √ 2C 2C (a) (b) π π √ C C (c) *(d) 2π 2π (e)
None of these
Solution:
C = 2πR
⇒
R=
C 2π
16 Problem 11. If the circumference of a circle is C, what is its diameter? r √ C 2C (a) (b) 2π π C C (c) *(d) 2π π (e)
None of these
C C ⇒ D= 2π π Problem 12. If the circumference of a circle is C, what is its area? √ C2 2C *(b) (a) π 4π r 2C πC 2 (c) (d) π 4 Solution:
(e)
None of these
Solution:
1.1.2
C = 2πR and D = 2R; so R =
C C = 2πR and A = πR ; so R = 2π 2
⇒
A=π
C 2π
2 =
C2 4π
Non-calculator-based problems
Problem 13. What is the circumference of a circle with radius 6 cm? (a) 6π cm (b) 9π cm *(c) 12π cm (d) 36π cm (e) None of these Solution:
C = 2πR = 2π(6 cm) = 12π cm
Problem 14. What is the circumference of a circle with diameter 3 m? *(a) 3π m (b) 6π m (c) 9π m (d) 36π m (e) None of these Solution:
C = 2πR and D = 2R; so C = πD = 3π m
Problem 15. What is the area of a circle with diameter 5 m? 25π 2 5π 2 m *(b) m (a) 2 4 (c)
5π 2 m2
(e)
None of these
Solution:
(d)
2
10π 2 m2
A = πR and D = 2R; so A = π
D 2
2
=π
5m 2
2 =
25π 2 m 4
17 Problem 16. What is the area of a circle with radius 3 cm? 3π 9π (a) cm2 (b) cm2 4 4 9π cm2 *(d) 9π cm2 (c) 2 (e)
None of these
Solution:
A = πR2 = π(3 cm)2 = 9 cm2
Problem 17. What is the circumference of a circle with diameter 10 cm? (a) 5π cm *(b) 10π cm (c) 25π cm (d) 100π cm (e) None of these Solution:
C = 2πR and D = 2R; so C = πD = 10π m
Problem 18. What is the area of a circle with diameter 10 m? (a) 5π m2 (b) 10π m2 (c) (e)
25π cm2 4 None of these
Solution:
*(d)
25π cm2
A = πR2 and D = 2R; so A = π(D/2)2 = π(5 cm)2 = 25π cm2
Problem 19. What is the circumference of a circle with radius 4 cm? (a) 2π cm (b) 4π cm *(c) 8π cm (d) 16π cm (e) None of these Solution:
C = 2πR = 2π(4 cm) = 8π cm
Problem 20. What is the area of a circle with radius 6 cm? (a) 3π cm2 (b) 6π cm2 (c) 9π cm2 *(d) 36π cm2 (e) None of these Solution:
A = πR2 = π(6 cm)2 = 36π cm2
18 Problem 21. If the circumference of a circle is 5 cm, what is its area? 25 *(a) cm2 (b) 50π cm2 4π 25 25π (c) cm2 (d) cm2 2π 2 (e)
None of these
Solution:
C = 2πR and A = πR2 ; so
C R= 2π
⇒
A=π
C 2π
2 =
25 C2 = 4π 4π
Problem 22. If the area of a circle is 9 cm2 , what is its radius? √ (a) 3π cm2 (b) 3 π cm (c)
3 cm π
(e)
None of these
*(d)
3 √ cm π
r Solution:
A = πR2
⇒
R=
A = π
r
9 3 =√ π π
Problem 23. If the circumference of a circle is 4 cm, what is its diameter? 4 *(a) cm (b) 8 cm π 2 8 cm (d) √ cm2 (c) π π (e)
None of these
Solution:
C = 2πR and D = 2R; so R =
C 2π
⇒
D=
C 4 = π π
Problem 24. If the area of a circle is 4 m2 , what is its diameter? 2 4 (a) m *(b) √ m π π 16 (c) 16π m (d) m π (e)
None of these
Solution:
A = πR2 and D = 2R; so r r r A A 4 4 R= ⇒ D=2 =2 =√ π π π π
19 Problem 25. If the circumference of a circle is 10 m, what is its radius? r 10 10 (a) m (b) m π π r 5 5 *(c) m (d) m π π (e)
None of these
Solution:
C = 2πR
⇒
R=
C 10 5 = = 2π 2π π
Problem 26. If the area of a circle is 3 m2 , what is its circumference? √ √ 2 3 (a) √ m *(b) 2 3π m π r r 3 6 (c) m (d) m 2π π (e)
None of these
Solution:
1.1.3
A = πR2 and C = 2πR; so r r √ √ A A R= ⇒ C = 2π = 2 πA = 2 3π π π
Calculator-based problems
Problem 27. What is the circumference of a circle with radius 11.9 cm? Round your answer to the nearest 0.1 cm. *(a) 74.8 cm (b) 82.2 cm (c) 90.5 cm (d) 99.5 cm (e) None of these Solution:
C = 2πr = 2π(11.9 cm) = 74.8 cm
Problem 28. What is the circumference of a circle with radius 8.3 cm? Round your answer to the nearest centimeter. (a) 38 cm (b) 42 cm (c) 47 cm *(d) 52 cm (e) None of these Solution:
C = 2πr = 2π(8.3 cm) = 52 cm
20 Problem 29. What is the circumference of a circle with diameter 5.6 cm? Round your answer to the nearest centimeter. *(a) 18 cm (b) 19 cm (c) 21 cm (d) 23 cm (e) None of these Solution:
C = 2πr and d = 2r; so C = πd = π(5.6 cm) = 18 cm
Problem 30. What is the circumference of a circle with diameter 6.9 cm? Round your answer to the nearest centimeter. (a) 16 cm (b) 18 cm (c) 20 cm *(d) 22 cm (e) None of these Solution:
C = 2πr and d = 2r; so C = πd = π(6.9 cm) = 22 cm
Problem 31. What is the area of a circle with radius 1.1 cm? Round your answer to the nearest 0.1 cm2 . *(a) 3.8 cm2 (b) 4.2 cm2 (c) 4.6 cm2 (d) 5.1 cm2 (e) None of these Solution:
A = πr2 = π(1.1 cm)2 = 3.8 cm2
Problem 32. What is the area of a circle with radius 6.1 cm? Round your answer to the nearest 10 cm2 . (a) 90 cm2 (b) 100 cm2 (c) 110 cm2 *(d) 120 cm2 (e) None of these Solution:
A = πr2 = π(6.1 cm)2 = 120 cm2
Problem 33. What is the area of a circle with diameter 4.4 cm? Round your answer to the nearest cm2 . (a) 12 cm2 (b) 14 cm2 2 *(c) 15 cm (d) 17 cm2 (e) None of these 2 πd2 π(4.4 cm)2 d 2 = = = 15 cm2 Solution: A = πr and d = 2r; so A = π 2 4 4
21 Problem 34. What is the area of a circle with diameter 7.8 cm? Round your answer to the nearest cm2 . (a) 39 cm2 (b) 43 cm2 *(c) 48 cm2 (d) 53 cm2 (e) None of these 2 π(7.8 cm)2 d πd2 2 = = 48 cm2 Solution: A = πr and d = 2r; so A = π = 2 4 4 Problem 35. If the circumference of a circle is 4 cm, what is its area? (a)
4π cm2
(b)
16π cm2
*(c)
4 cm2 π
(d)
16 cm2 π
(e)
None of these
Solution:
C = 2πR and A = πR2 ; so C R= 2π
1.2 1.2.1
⇒
A=π
C 2π
2
42 4 C2 = = = 4π 4π π
Rectangles and rectangular solids Letter-based problems
Problem 36. What is the area of a square with sides of length s? (a) 4s *(c) s2 (e) None of these Solution:
(b) (d)
2πs 4πs2
You should know this by memory.
Problem 37. What is the area of a rectangle with length l and width w? (a) √ 2l + 2w l2 + w2 (c) (e) None of these Solution:
*(b) lw (d) l2 + w2
You should know this by memory.
Problem 38. What is the volume of a cube with sides of length s? (a) 8s *(c) s3 (e) None of these Solution:
(b) √ 6s2 (d) 3s
You should know this by memory.
22 Problem 39. What is the volume of a rectangular solid with length l, width w, and height h? (a) √ 2l + 2w + 2h (c) l2 + w2 + h2 (e) None of these Solution: 1.2.2
*(b) lwh (d) l3 + w3 + h3
You should know this by memory.
Non-calculator-based problems
Problem 40. What is the area of a square tabletop measuring 3 feet on a side? *(a) (c) (e)
9 ft2 12 ft2 None of these
Solution:
(b) 3π ft2 (d) 6π ft2
A = s2 = (3 ft)2 = 9 ft2
Problem 41. What is the area of a rectangular sheet of paper measuring 10 inches long by 8 inches wide? (a) 40 in2 (c) 40π in2 (e) None of these Solution:
*(b) 80 in2 (d) 80π in2
A = lw = (10 in)(8 in) = 80 in2
Problem 42. What is the volume of a cube measuring 2 inches on a side? (a) 4 in3 *(c) 8 in3 (e) None of these Solution:
(b) (d)
4π in3 8π in3
V = s3 = (2 in)3 = 8 in3
Problem 43. What is the volume of a rectangular box measuring 2 cm long by 3 cm wide by 5 cm high? (a) 10 cm3 *(c) 30 cm3 (e) None of these Solution:
(b) 20 cm3 (d) 60 cm3
V = lwh = (2 cm)(3 cm)(5 cm) = 30 cm3
23 Problem 44. What is the area of a square room measuring 5 m on a side? (a) (c) (e)
5π m2 25π 2 m 2 None of these
Solution:
(b)
25π m2
*(d)
25 m2
A = s2 = (5 m)2 = 25 m2
Problem 45. What is the volume of a rectangular room measuring 4 m long by 5 m wide by 3 m high? (a) 12 m3 (c) 30 m3 (e) None of these Solution:
(b) 24 m3 *(d) 60 m3
V = lwh = (4 m)(5 m)(3 m) = 60 m3
Problem 46. What is the volume of a cube measuring 10 cm on a side? 1000 (a) cm3 (b) 500 cm3 3 *(c)
1000 cm3
(e)
None of these
Solution:
(d)
2000 cm3
V = s3 = (10 cm)3 = 1000 cm3
Problem 47. What is the area of a rectangle measuring 6 m long by 2 m wide? *(a) (c) (e)
12 m2 48 m2 None of these
Solution:
(b) (d)
24 m2 6π m2
A = lw = (6 m)(2 m) = 12 m2
Problem 48. What is the area of a square tabletop measuring 3 feet on a side? *(a) (c) (e)
9 ft2 12 ft2 None of these
Solution:
(b) (d)
3π ft2 6π ft2
A = s2 = (3 ft)2 = 9 ft2
24 1.2.3
Calculator-based problems
Problem 49. What is the area of a square piece of ground whose sides are each 62 ft long? Round your answer to the nearest 100 ft2 . (a) 3100 ft2 *(c) 3800 ft2 (e) None of these Solution:
(b) 3500 ft2 (d) 4200 ft2
A = s2 = (62 ft)2 = 3800 ft2
Problem 50. What is the area of a square tabletop whose sides are each 2.7 ft long? Round your answer to the nearest 0.1 ft2 . *(a) 7.3 ft2 (c) 8.8 ft2 (e) None of these Solution:
(b) (d)
8.0 ft2 9.7 ft2
A = s2 = (2.7 ft)2 = 7.3 ft2
Problem 51. What is the area of a rectangular carpet measuring 2.4 m long by 1.7 m wide? Round your answer to the nearest 0.1 m2 . (a) 3.7 m2 (c) 4.5 m2 (e) None of these Solution:
*(b) (d)
4.1 m2 4.9 m2
A = lw = (2.4 m)(1.7 m) = 4.1 m2
Problem 52. What is the area of a rectangular windowpane measuring 21 cm wide by 27 cm high? Round your answer to the nearest 10 cm2 . (a) 510 cm2 (c) 620 cm2 (e) None of these Solution:
*(b) (d)
570 cm2 690 cm2
A = lw = (21 cm)(27 cm) = 570 m2
Problem 53. What is the volume of a cube measuring 13.4 cm on a side? Round your answer to the nearest 10 cm3 . (a) 2170 cm3 *(b) 2410 cm3 3 (c) 2650 cm (d) 2910 cm3 (e) None of these Solution:
V = s3 = (13.4 cm)3 = 2410 cm3
25 Problem 54. What is the volume of a cube measuring 0.88 ft on a side? Round your answer to the nearest 0.01 ft3 . (a) 0.50 ft3 (b) 0.55 ft3 (c) 0.61 ft3 *(d) 0.68 ft3 (e) None of these Solution:
V = s3 = (0.88 ft)3 = 0.68 ft3
Problem 55. A rectangular fish-tank is 53 cm long, 22 cm wide, and 24 cm high. What is its volume? Round your answer to the nearest 1000 cm3 (a) 20, 000 cm3 (c) 25, 000 cm3 (e) None of these Solution:
(b) 23, 000 cm3 *(d) 28, 000 cm3
V = lwh = (53 cm)(22 cm)(24 cm) = 28, 000 cm3
Problem 56. A room is 5.7 m long, 4.8 m wide, and 2.4 m high. What is its volume? Round your answer to the nearest cubic meter. *(a) (c) (e)
66 m3 79 m3 None of these
Solution:
(b) (d)
72 m3 87 m3
V = lwh = (5.7 m)(4.8 m)(2.4 m) = 66 m3
26
1.3 1.3.1
Right triangles Letter-based problems
Problem 57. In the right triangle at right, which of the following equations is true? (b) r = 21 (x + y) (d) r = 12 (x − y)
(a) r = x + y *(c) r2 = x2 + y 2 (e) None of these Solution:
r
y
x
You should know this by memory.
Problem 58. In the right triangle at right, which of the following equations is true? (a) a2 = b2 + c2 *(c) c2 = a2 + b2 (e) None of these Solution:
(b) b2 = a2 + c2 (d) a2 + b2 + c2 = 1
c
b
a
You should know this by memory.
Problem 59. In the right triangle at right, which of the following equations is true? 2
2
2
2
2
*(b) x = r − y (d) x2 = 12 (r2 − y 2 )
(a) x = r + y (c) x2 = 21 (r2 + y 2 ) (e) None of these Solution:
2
r
⇒
y
r 2 = x2 + y 2
x
x2 = r 2 − y 2
Problem 60. In the right triangle at right, which of the following equations is true? (a) a2 = 21 (c − b)2 (c) a2 = (c − b)2 (e) None of these Solution:
2
2
c =a +b
(b) a2 = 12 (c2 − b2 ) *(d) a2 = c2 − b2
c
⇒
2
2
a =c −b
b
2
a
2
Problem 61. In the right triangle at right, which of the following equations is true? 2
2
2
2
*(a) y = r − x (c) y 2 = 21 (x2 + r2 ) (e) None of these Solution:
2
2
r =x +y
2
2
(b) y = x − r (d) y 2 = 21 (x2 − r2 )
r
2
⇒
2
2
y =r −x
2
x
y
27 Problem 62. In the right triangle at right, which of the following equations is true? (a) b2 = 21 (c − a)2 *(c) b2 = c2 − a2 (e) None of these Solution:
2
2
c =a +b
b2 = 12 (c + a)2 b2 = c2 + a2
(b) (d)
c
⇒
2
2
b =c −a
Problem 63. In the right triangle following equations is true? p (b) *(a) r = x2 + y 2 p (c) r = (x + y)2 (d) (e) None of these Solution: Lengths like x, y, and
b
2
a
2
at right, which of the
r
r = x2 + y 2
y
r = 12 (x + y)
x
r are not negative. Hence √
r 2 = x2 + y 2
−−→
r=
p
x2 + y 2
Problem 64. In the right triangle at right, which of the following equations is true? √ (b) c = 12 a2 + b2 (a) c = 12 (a2 + b2 ) √ (c) c = a2 + b2 *(d) c = a2 + b2 (e) None of these Solution:
c
b
a
Lengths like a, b, and c are not negative. Hence √
c 2 = a2 + b 2
−−→
c=
√
a2 + b 2
Problem 65. In the right triangle at right, which of the following equations is true? p p (a) x = (r − y)2 (b) x = (r + y)2 p p *(c) x = r2 − y 2 (d) x = r2 + y 2 (e) None of these Solution:
r
x
Lengths like x, y, and r are not negative. Hence r 2 = x2 + y 2
−y 2
−−−→
√
x2 = r 2 − y 2
−−→
x=
p r2 − y 2
y
28 Problem 66. In the right triangle at right, which of the following equations is true? bc 2c (a) a = (b) a = 2 b √ c 2 − b2 (c) a = *(d) a = c2 − b2 2 (e)
c
b
a
None of these
Solution:
Lengths like a, b, and c are not negative. Hence c 2 = a2 + b 2
1.3.2
−b2
√
−−−→
a2 = c 2 − b 2
−−→
a=
√
c 2 − b2
Non-calculator-based problems
Problem 67. Find r in the right triangle at right. √ 7 *(b) 5 (a) (c) 7 (d) 25 (e) None of these Solution:
r
3
4
r is a length, so it must be positive. Hence √
r 2 = x2 + y 2
−−→ r =
p p √ √ x2 + y 2 = (4)2 + (3)2 = 16 + 9 = 25 = 5
Problem 68. Find r in the right triangle at right. √ (a) 3 (b) √9 (c) 9 *(d) 41 (e) None of these Solution:
r
4
5
r is a length, so it must be positive. Hence √
r 2 = x2 + y 2
−−→ r =
p p √ √ x2 + y 2 = (5)2 + (4)2 = 25 + 16 = 41
Problem 69. Find r in the right triangle at right. √ √ (a) 8 (b) √15 (c) 4 *(d) 34 (e) None of these Solution:
r
5
r is a length, so it must be positive. Hence √ 2
2
r =x +y
2
−−→ r =
p p √ √ x2 + y 2 = (5)2 + (3)2 = 25 + 9 = 34
3
29 Problem 70. Find x in the right triangle at right. √ (a) 1√ (b) √2 *(c) 3 (d) 5 (e) None of these Solution:
2
1
x
x is a length, so it must be positive. Hence r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p √ √ −−→ x = r2 − y 2 = (2)2 − (1)2 = 4 − 1 = 3 Problem 71. Find x in the right triangle at right. √ (a) 2 (b) √8 *(c) 4 (d) 34 (e) None of these Solution:
5
3
x
x is a length, so it must be positive. Hence r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p √ √ −−→ x = r2 − y 2 = (5)2 − (3)2 = 25 − 9 = 16 = 4 Problem 72. Find y in the right triangle at right. *(a) √ 3 (c) 21 (e) None of these Solution:
5
(b) √ 9 (d) 34
4
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ √ √ −−→ y = r2 − x2 = (5)2 − (4)2 = 25 − 16 = 9 = 3
y
30
Problem 73. Find y in the right triangle at right. √ (a) 1 *(b) 7 (c) 5 (d) 7 (e) None of these Solution:
4
y
3
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ √ √ −−→ y = r2 − x2 = (4)2 − (3)2 = 16 − 9 = 7 Problem 74. Find y in the right triangle at right. √ (a) √ 9 (b) √17 (c) 30 *(d) 51 (e) None of these Solution:
10
y
7
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ √ √ −−→ y = r2 − x2 = (10)2 − (7)2 = 100 − 49 = 51 1.3.3
Calculator-based problems
Problem 75. Find r in the right triangle at right. Round your answer to one decimal place. (a) 6.0 (c) 7.4 (e) None of these Solution:
r
(b) 6.6 *(d) 8.2
6.3
r is a length, so it must be positive. Hence √ 2
2
r =x +y
2
−−→ r =
p p x2 + y 2 = (6.3)2 + (5.2)2 = 8.2
5.2
31 Problem 76. Find r in the right triangle at right. Round your answer to the nearest integer. (a) 29 (c) 35 (e) None of these Solution:
r
*(b) 32 (d) 39
19
26
r is a length, so it must be positive. Hence √
r 2 = x2 + y 2
−−→ r =
p p x2 + y 2 = (26)2 + (19)2 = 32
Problem 77. Find r in the right triangle at right. Round your answer to two decimal places. (a) 0.46 (c) 0.56 (e) None of these Solution:
r
*(b) 0.51 (d) 0.62
0.28
0.43
r is a length, so it must be positive. Hence √
r 2 = x2 + y 2
−−→ r =
p p x2 + y 2 = (0.43)2 + (0.28)2 = 0.51
Problem 78. Find x in the right triangle at right. Round your answer to one decimal place. (a) 2.2 (c) 2.7 (e) None of these Solution:
3.5
(b) 2.4 *(d) 3.0
1.8
x
x is a length, so it must be positive. Hence r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (3.5)2 − (1.8)2 = 3.0 Problem 79. Find x in the right triangle at right. Round your answer to the nearest integer. (a) 33 *(c) 40 (e) None of these Solution:
74
(b) 36 (d) 44
x is a length, so it must be positive. Hence r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (74)2 − (62)2 = 40
x
62
32 Problem 80. Find x in the right triangle at right. Round your answer to two decimal places. (a) 0.69 (c) 0.85 (e) None of these Solution:
0.97
(b) 0.77 *(d) 0.95
0.20
x
x is a length, so it must be positive. Hence r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (0.97)2 − (0.20)2 = 0.95 Problem 81. Find y in the right triangle at right. Round your answer to two decimal places. (a) 0.10 (c) 0.13 (e) None of these Solution:
0.34
(b) 0.11 *(d) 0.14
y
0.31
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (0.34)2 − (0.41)2 = 0.14 Problem 82. Find y in the right triangle at right. Round your answer to one decimal place. (a) 5.2 *(c) 6.4 (e) None of these Solution:
7.6
(b) 5.8 (d) 7.0
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (7.6)2 − (4.1)2 = 6.4
4.1
y
33 Problem 83. Find y in the right triangle at right. Round your answer to the nearest integer. (a) 25 (c) 31 (e) None of these Solution:
52
(b) 28 *(d) 34
y
39
y is a length, so it must be positive. Hence r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (52)2 − (39)2 = 34
1.4
Right triangle word problems
1.4.1
Non-calculator-based problems
Problem 84. Pirates have buried their treasure in the desert, and you have obtained a copy of the directions to it. They tell you to begin at the entrance of the physics lab and go 2 km east, then to go north for another 5 km to the treasure. What is the straight-line distance from the physics lab to the treasure site? (a) √ 4 km (c) 21 km (e) None of these
(b) *(d)
7 km √ 29 km
Solution: Sketch a diagram, as at right. The route forms a right triangle, with the legs from the physics lab east to the turning point, and from the turning point north to the treasure. The straight line from the physics lab to the treasure is the hypotenuse. Use the theorem of Pythagoras: 2
2
r
r 5 km
2
r =x +y p p √ −−→ r = x2 + y 2 = (2 km)2 + (5 km)2 = 29 km √
treasure
r
lab
2 km
r
34 Problem 85. Pirates have buried their treasure in the desert, and you have obtained a copy of the directions to it. They tell you to begin at the entrance of the physics lab and go 3 km south, then to go west for another 4 km to the treasure. What is the straight-line distance from the physics lab to the treasure site? √ *(a) √ 5 km (b) 7 km 12 km (d) 7 km (c) (e) None of these Solution: Sketch a diagram, as at right. The route forms a right triangle, with the legs from the physics lab east to the turning point, and from the turning point north to the treasure. The straight line from the physics lab to the treasure is the hypotenuse. Use the theorem of Pythagoras: 2
2
lab
r
r 3 km
2
r =x +y p p −−→ r = x2 + y 2 = (4 km)2 + (3 km)2 = 5 km √
4 km treasure r
r
Problem 86. Your house is 5 km from the Pima Community College Missile Test Site. A missile is fired from the test site; it goes 3 km straight up, then explodes. How far is the missile from your house when it explodes? √ (a) √ 4 km (b) √8 km 15 km *(d) 34 km (c) (e) None of these Solution: Sketch a diagram, as at right. Your house, the test site, and the missile form a right triangle, with a horizontal leg from the house to the test site,and a vertical leg from the test site to the missile. The straight line from the house to the missile is the hypotenuse. Use the theorem of Pythagoras: 2
2
r 3 km
2
r =x +y p p √ −−→ r = x2 + y 2 = (5 km)2 + (3 km)2 = 34 km √
missiler
r
house
5 km
r
site
35 Problem 87. Your house is 7 miles from the Pima Community College Missile Test Site. A missile is fired from the test site; it goes 1 mile straight up, then explodes. How far is the missile from your house when it explodes? (a) 7.5 √ mi *(c) 50 mi (e) None of these
(b) 8 mi (d) 10 mi
Solution: Sketch a diagram, as at right. Your house, the test site, and the missile form a right triangle, with a horizontal leg from the house to the test site,and a vertical leg from the test site to the missile. The straight line from the house to the missile is the hypotenuse. Use the theorem of Pythagoras: 2
2
missile
r
r 1 mi
2
r =x +y p p √ −−→ r = x2 + y 2 = (7 mi)2 + (1 mi)2 = 50 mi √
Problem 88. A vertical telephone pole is stabilized by a diagonal guy wire anchored in the ground, as shown at right. The wire is 10 m long; it is attached to the pole 7 m above ground level. How far from the base of the pole does the wire meet the ground? √ (a) √ 9m *(b) √51 m (c) 93 m (d) 149 m (e) None of these
7 mi
r
r
site
house
@
10 m
@ @7 m @ x @@
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. This is a right triangle where the hypotenuse is 34 ft and the vertical side is 21 ft; we want to know the horizontal side x. Use Pythagoras: r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p √ √ −−→ x = r2 − y 2 = (10 m)2 − (7 m)2 = 100 − 49 m = 51 m
36 Problem 89. A physics instructor has locked himself out of his office, and tries to climb in through an upper window. He leans a 5-meter ladder against the side of the building so that the top of the ladder is 3 m above the ground. How far from the building is the bottom of the ladder? *(a) √ 4m (b) √ 8m 34 m (d) 44 m (c) (e) None of these Solution: Sketch a diagram, as at right. The ground, the building, and the ladder form a right triangle whose vertical side is 3 m and whose hypotenuse is 5 m. We want to know the horizontal side x. Use the theorem of Pythagoras: 5m
r 2 = x2 + y 2
3m
−y 2
−−−→ x2 = r2 − y 2 √ p p √ −−→ x = r2 − y 2 = (5 m)2 − (3 m)2 = 16 m = 4 m Problem 90. A blimp is attached to a cable 3 km long. The other end of the cable is fastened to the ground. The wind is strong enough to pull the cable into a straight line. When you are standing directly below the blimp, you are 2 km from the place where the cable is anchored in the ground. How high above ground level is the blimp? √ (a) 1 km *(b) √5 km (c) 5 km (d) 13 km (e) None of these
x
3 km
y
2 km
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. The system is a right triangle whose hypotenuse is 3 km and whose horizontal side is 2 km; we want to know the vertical side y. Use Pythagoras: r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ √ −−→ y = r2 − x2 = (3 km)2 − (2 km)2 = 5 km
37 1.4.2
Calculator-based problems
Problem 91. An airplane flies a route involving three cities. From Rio Cordaro, it flies 210 miles straight east to Hackerville. From Hackerville, it flies 140 miles straight north to San Pitucco. How far does it fly from San Pitucco to Rio Cordaro? Round your answer to the nearest 10 miles. *(a) 250 miles (b) 280 miles (c) 310 miles (d) 340 miles (e) None of these Solution: Sketch a diagram, as at right. The route forms a right triangle, with the legs from Rio Cordaro to Hackerville and from Hackerville to San Pitucco on either side of the right angle. The leg from San Pitucco to Rio Cordaro is the hypotenuse. Use the theorem of Pythagoras: 2
2
SP r
r 140 mi
2
r =x +y p p −−→ r = x2 + y 2 = (210 mi)2 + (140 mi)2 = 250 mi √
r
210 mi
RM
r
HV
Problem 92. An airplane flies a route involving three cities. From Rio Cordaro, it flies 140 miles straight east to Hackerville. From Hackerville, it flies 320 miles straight north to San Pitucco. How far does it fly from San Pitucco to Rio Cordaro? Round your answer to the nearest 10 miles. (a) 330 miles *(b) 350 miles (c) 370 miles (d) 390 miles (e) None of these Solution: Sketch a diagram, as at right. The route forms a right triangle, with the legs from Rio Cordaro to Hackerville and from Hackerville to San Pitucco on either side of the right angle. The leg from San Pitucco to Rio Cordaro is the hypotenuse. Use the theorem of Pythagoras: 2
2
r
r 320 mi
2
r =x +y p p −−→ r = x2 + y 2 = (140 mi)2 + (320 mi)2 = 350 mi √
SP
r
RM
140 mi
r
HV
38 Problem 93. A ship sails 58 miles straight south. It then sails another 44 miles straight west. At this point, what is its straight-line distance from its starting point? Round your answer to the nearest mile. *(a) (c) (e)
73 miles 88 miles None of these
(b) (d)
80 miles 97 miles
start
Solution: Sketch a diagram, as at right. The route forms a right triangle, with the southward and the westward legs on either side of the right angle. The final distance from the starting point is the hypotenuse. Use the theorem of Pythagoras: r 2 = x2 + y 2 √ p p −−→ r = x2 + y 2 = (58 mi)2 + (44 mi)2 = 73 mi
r
r 58 mi r
44 mi
finish
Problem 94. A ship sails 79 miles straight south. It then sails another 34 miles straight west. At this point, what is its straight-line distance from its starting point? Round your answer to the nearest mile. (a) 80 miles *(c) 86 miles (e) None of these
(b) 83 miles (d) 89 miles
start
Solution: Sketch a diagram, as at right. The route forms a right triangle, with the southward and the westward legs on either side of the right angle. The final distance from the starting point is the hypotenuse. Use the theorem of Pythagoras: r 2 = x2 + y 2 √ p p −−→ r = x2 + y 2 = (79 mi)2 + (34 mi)2 = 86 mi
r
r 79 mi r
finish
34 mi
39 Problem 95. A vertical telephone pole is stabilized by a diagonal guy wire anchored in the ground, as shown at right. The wire is 34 feet long; it is attached to the pole 21 feet above ground level. How far from the base of the pole does the wire meet the ground? Round your answer to the nearest foot. (a) 24 ft (c) 29 ft (e) None of these
*(b) (d)
27 ft 32 ft
@ @ @ 34 ft @ x @@
21 ft
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. This is a right triangle where the hypotenuse is 34 ft and the vertical side is 21 ft; we want to know the horizontal side x. Use Pythagoras: r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (34 ft)2 − (21 ft)2 = 27 ft Problem 96. A vertical telephone pole is stabilized by a diagonal guy wire anchored in the ground, as shown at right. The wire is 47 feet long; it is attached to the pole 28 feet above ground level. How far from the base of the pole does the wire meet the ground? Round your answer to the nearest foot. (a) 31 ft *(c) 38 ft (e) None of these
(b) 34 ft (d) 42 ft
@
28 ft
@ @ 47 ft @ x @@
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. This is a right triangle where the hypotenuse is 47 ft and the vertical side is 28 ft; we want to know the horizontal side x. Use Pythagoras: r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (47 ft)2 − (28 ft)2 = 38 ft
40 Problem 97. A physics instructor has locked himself out of his office, and tries to climb in through an upper window. He leans a 21-foot ladder against the side of the building so that the top of the ladder is 18 feet above the ground. How far from the building is the bottom of the ladder? Round your answer to the nearest foot. *(a) (c) (e)
11 ft 13 ft None of these
(b) (d)
12 ft 14 ft
Solution: Sketch a diagram, as at right. The ground, the building, and the ladder form a right triangle whose vertical side is 18 ft and whose hypotenuse is 21 ft. We want to know the horizontal side x. Use the theorem of Pythagoras: 2
2
r =x +y
21 ft
2
18 ft
−y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (21 ft)2 − (18 ft)2 = 11 ft
x
Problem 98. A physics instructor has locked himself out of his office, and tries to climb in through an upper window. He leans a 13-foot ladder against the side of the building so that the top of the ladder is 9 feet above the ground. How far from the building is the bottom of the ladder? Round your answer to the nearest foot. (a) 6 ft (c) 8 ft (e) None of these
(b) 7 ft *(d) 9 ft
Solution: Sketch a diagram, as at right. The ground, the building, and the ladder form a right triangle whose vertical side is 9 ft and whose hypotenuse is 13 ft. We want to know the horizontal side x. Use the theorem of Pythagoras: r 2 = x2 + y 2 −y 2
−−−→ x2 = r2 − y 2 √ p p −−→ x = r2 − y 2 = (13 ft)2 − (9 ft)2 = 9 ft
13 ft 9 ft x
41 Problem 99. A blimp is attached to a cable 890 meters long. The other end of the cable is fastened to the ground. The wind is strong enough to pull the cable into a straight line. When you are standing directly below the blimp, you are 660 meters from the place where the cable is anchored in the ground. How high above ground level is the blimp? Round your answer to the nearest 10 m. (a) 540 m (c) 660 m (e) None of these
*(b) (d)
600 m 720 m
890 m
y
660 m
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. The system is a right triangle whose hypotenuse is 890 m and whose horizontal side is 660 m; we want to know the vertical side y. Use Pythagoras: r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (890 m)2 − (660 m)2 = 600 m Problem 100. A blimp is attached to a cable 820 meters long. The other end of the cable is fastened to the ground. The wind is strong enough to pull the cable into a straight line. When you are standing directly below the blimp, you are 640 meters from the place where the cable is anchored in the ground. How high above ground level is the blimp? Round your answer to the nearest 10 m. (a) 420 m *(c) 510 m (e) None of these
(b) 460 m (d) 560 m
820 m
y
640 m
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. The system is a right triangle whose hypotenuse is 820 m and whose horizontal side is 640 m; we want to know the vertical side y. Use Pythagoras: r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (820 m)2 − (640 m)2 = 510 m
42 Problem 101. A kite is attached to a string 310 ft long. The wind is strong enough to stretch the string into a straight line. When you are standing directly under the kite, you are 270 ft from the person holding the string. How high above the ground is the kite? Round your answer to the nearest 10 ft. (a) 110 ft (c) 140 ft (e) None of these
(b) 120 ft *(d) 150 ft
310 m
y
270 m
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. The system is a right triangle whose hypotenuse is 310 m and whose horizontal side is 270 m; we want to know the vertical side y. Use Pythagoras: r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (310 m)2 − (270 m)2 = 150 m Problem 102. A kite is attached to a string 560 ft long. The wind is strong enough to stretch the string into a straight line. When you are standing directly under the kite, you are 330 ft from the person holding the string. How high above the ground is the kite? Round your answer to the nearest 10 ft. (a) 370 ft *(c) 450 ft (e) None of these
(b) 410 ft (d) 500 ft
560 m
y
330 m
Solution: We have a sketch already; we can label the sides that we know and the side that we want to find. The system is a right triangle whose hypotenuse is 560 m and whose horizontal side is 330 m; we want to know the vertical side y. Use Pythagoras: r 2 = x2 + y 2 −x2
−−−→ y 2 = r2 − x2 √ p √ −−→ y = r2 − x2 = (560 m)2 − (330 m)2 = 450 m
43
2 2.1
Algebra Evaluating functions with numerical arguments
2.1.1
Non-calculator based problems
Problem 103. If f (x) = x2 − 5, find f (5). (a) −5 (c) 9 (e) None of these Solution:
(b) *(d)
f (5) = 52 − 5 = 25 − 5 = 20
Problem 104. If f (x) = *(a) (c) (e)
5 7 17 7
f (2) =
2+3 5 = 7 7
Problem 105. If f (x) = (a)
5 7
*(c)
2
(e)
None of these
Solution:
f (4) =
(e)
3 4 5 4
x+2 , find f (4). 3 2 (b) 3 10 (d) 3
4+2 6 = =2 3 3
Problem 106. If f (x) =
(c)
x+3 , find f (2). 7 10 (b) 7 23 (d) 7
None of these
Solution:
(a)
0 20
x+1 , find f (2). 2 (b)
1
*(d)
3 2
None of these
Solution:
f (2) =
2+1 3 = 2 2
44 x2 + 1 , find f (2). Problem 107. If f (x) = 3 5 (a) (b) 1 9 5 *(c) (d) 3 3 (e)
None of these
Solution:
f (2) =
4+1 5 22 + 1 = = 3 3 3
Problem 108. If f (x) = (a) (c) (e)
1 9 5 3
x2 − 1 , find f (2). 3 4 (b) 9 *(d)
1
None of these
Solution:
f (2) =
4−1 3 22 − 1 = = =1 3 3 3
Problem 109. If f (x) = (a)
0
(c)
1 3
(e)
None of these
x2 − 3 , find f (3). 3 1 (b) 9 *(d)
2
32 − 3 9−3 6 = = =2 3 3 3 √ Problem 110. If f (x) = x2 − 5, find f (1). Solution:
f (3) =
(a) −2√ (c) − 6 (e) None of these
(b) *(d)
2 No real value
√ √ √ Solution: f (1) = 12 − 5 = 1 − 5 = −4. Since the quantity under the square root sign is negative, there is no real value.
45 √ Problem 111. If f (x) = x2 + 2, find f (3). √ *(a) 11 (b) 5 (c) 25 (d) No real value (e) None of these √ √ √ Solution: f (3) = 32 + 2 = 9 + 2 = 11 √ Problem 112. If f (x) = x2 − 3, find f (2). (a) −1 √ (c) 5 (e) None of these Solution:
f (2) =
*(b) 1 (d) No real value √
22 − 3 =
√
4−3=
√
Problem 113. If f (x) = 3x + 2, find f (4). (a) 12 (c) 16 (e) None of these Solution:
*(b) (d)
14 18
f (4) = 3(4) + 2 = 12 + 2 = 14
Problem 114. If f (x) = 3x + 2, find f (7). (a) 15 *(c) 23 (e) None of these Solution:
(b) 21 (d) 27
f (7) = 3(7) + 2 = 21 + 2 = 23
Problem 115. If f (x) = x2 − 5, find f (5). (a) −5 (c) 9 (e) None of these Solution:
(b) *(d)
0 20
f (5) = 52 − 5 = 25 − 5 = 20
Problem 116. If f (x) = x2 − 5, find f (2). *(a) −1 (c) 9 (e) None of these Solution:
(b) 1 (d) 49
f (2) = 22 − 5 = 4 − 5 = −1
1=1
46 2.1.2
Calculator-based problems
Problem 117. If f (x) = 6.3x − 4.4, find f (3.9). (a) 14.70 (c) 18.15 (e) None of these Solution:
(b) *(d)
16.34 20.17
f (3.9) = (6.3)(3.9) − 4.4 = 20.17
Problem 118. If f (x) = 6.3x − 4.4, find f (2.1). (a) 7.15 *(c) 8.83 (e) None of these Solution:
(b) 7.95 (d) 9.71
f (2.1) = (6.3)(2.1) − 4.4 = 8.83
Problem 119. If f (x) = −2.3x + 7.5, find f (1.1). *(a) (c) (e)
4.97 6.01 None of these
Solution:
(b) (d)
5.47 6.62
f (1.1) = (−2.3)(1.1) + 7.5 = 4.97
Problem 120. If f (x) = −2.3x + 7.5, find f (4.6). (a) −2.77 (c) −3.39 (e) None of these Solution:
*(b) −3.08 (d) −3.73
f (4.6) = (−2.3)(4.6) + 7.5 = −3.08
Problem 121. If f (x) = *(a) (c) (e)
1.1 4.0 None of these
Solution:
f (5) =
5+3 8 = = 1.1 7 7
Problem 122. If f (x) = (a) 0.9 (c) 4.3 (e) None of these Solution:
f (9) =
x+3 , find f (5). Round your answer to one decimal place. 7 (b) 3.7 (d) 5.4
x+3 , find f (9). Round your answer to one decimal place. 7 *(b) 1.7 (d) 9.4
9+3 12 = = 1.7 7 7
47 x2 − 5 , find f (5). Problem 123. If f (x) = 4 (a) 0 (b) 1.25 *(c) 5 (d) 23.75 (e) None of these Solution:
f (5) =
52 − 5 =5 4
Problem 124. If f (x) = (a) 4 *(c) 19 (e) None of these Solution:
f (9) =
92 − 5 = 19 4
Problem 125. If f (x) = (a) −5.75 (c) −0.25 (e) None of these Solution:
f (2) =
x2 − 5 , find f (9). 4 (b) 15.25 (d) 79.75
(x − 5)2 , find f (2). 4 (b) −4.25 *(d) 2.25
(2 − 5)2 = 2.25 4
Problem 126. If f (x) = (a) −4.5 *(c) 1 (e) None of these
(x − 5)2 , find f (7). 4 (b) 0.75 (d) 11
(7 − 5)2 =1 4 √ Problem 127. If f (x) = x2 − 5, find f (7). Round your answer to one decimal place. Solution:
f (7) =
(a) 4.0 *(c) 6.6 (e) None of these Solution:
f (7) =
(b) 4.8 (d) No real value √
72 − 5 = 6.6
48 Problem 128. If f (x) = (a) 4.0 *(c) 8.7 (e) None of these
√ x2 − 5, find f (9). Round your answer to one decimal place. (b) 6.8 (d) No real value
√
92 − 5 = 8.7 √ Problem 129. If f (x) = x2 − 5, find f (−4). Round your answer to one decimal place.
Solution:
f (9) =
(a) −6.2 (c) 9.0 (e) None of these
*(b) 3.3 (d) No real value
p (−4)2 − 5 = 3.3 √ Problem 130. If f (x) = x2 − 5, find f (2). Round your answer to one decimal place. Solution:
f (−4) =
(a) −0.2 (c) 9.0 (e) None of these
(b) 3.0 *(d) No real value
√ Solution: f (2) = 22 − 5. Since the quantity under the square root sign is negative, there is no real value.
49
2.2
Evaluating functions with variable-expression arguments
Problem 131. If f (x) = 3x + 2, find f (a + 1). (a) 3a + 1 (c) 3a + 3 (e) None of these Solution:
(b) 3a + 2 *(d) 3a + 5
f (a + 1) = 3(a + 1) + 2 = 3a + 3 + 2 = 3a + 5
Problem 132. If f (x) = 3x + 2, find f (a − 5). (a) 3a − 3 *(c) 3a − 13 (e) None of these Solution:
(b) 3a − 7 (d) 3a − 15
f (a − 5) = 3(a − 5) + 2 = 3a − 15 + 2 = 3a − 13
Problem 133. If f (x) = x2 , find f (a + 4). (a) a2 + 4 (c) a2 + 16 (e) None of these Solution:
(b) a2 + 4a + 4 *(d) a2 + 8a + 16
f (a + 4) = (a + 4)2 = a2 + 8a + 16
Problem 134. If f (x) = x2 , find f (a − 3). (a) a2 + 3 (c) a2 + 9 (e) None of these Solution:
(b) *(d)
a2 − 3 a2 − 6a + 9
f (a − 3) = (a − 3)2 = a2 − 6a + 9
Problem 135. If f (x) = x2 , find f (a − 2). (a) a2 − 2 *(c) a2 − 4a + 4 (e) None of these Solution:
(b) a2 + 4 (d) a2 − 2a − 4
f (a − 2) = (a − 2)2 = a2 − 4a + 4
Problem 136. If f (x) = x2 − 2, find f (a + 3). *(a) a2 + 6a + 7 (c) a2 + 2a + 1 (e) None of these Solution:
(b) a2 + 7 (d) a2 − 10a + 25
f (a + 3) = (a + 3)2 − 2 = a2 + 6a + 9 − 2 = a2 + 6a + 7
50 Problem 137. If f (x) = x2 − 2, find f (a − 1). *(a) a2 − 2a − 1 (c) a2 − 6a − 9 (e) None of these Solution:
(b) (d)
a2 − 2a + 1 a2 − 6a + 9
f (a − 1) = (a − 1)2 − 2 = a2 − 2a + 1 − 2 = a2 − 2a − 1
Problem 138. If f (x) = x2 − 2, find f (a + 2). (a) a2 *(c) a2 + 4a + 2 (e) None of these Solution:
(b) a2 + 4 (d) a2 − 2a + 4
f (a + 2) = (a + 2)2 − 2 = a2 + 4a + 4 − 2 = a2 + 4a + 2
Problem 139. If f (x) = *(a) (c) (e)
a 4 a+9 4 None of these
Solution:
f (a + 3) =
(a + 3) − 3 a = 4 4
Problem 140. If f (x) = (a) (c) (e)
a−1 4 a+1 4
f (a − 1) =
Problem 141. If f (x) =
(c) (e)
x−3 , find f (a − 1). 4 a−4 *(b) 4 a+4 (d) 4
None of these
Solution:
(a)
x−3 , find f (a + 3). 4 4a + 9 (b) 4 4a − 9 (d) 4
a − 23 4 a + 23 4
(a − 1) − 3 a−4 = 4 4 x−3 , find f (a − 5). 4 a−8 *(b) 4 a+8 (d) 4
None of these
Solution:
f (a − 5) =
(a − 5) − 3 a−8 = 4 4
51 Problem 142. If f (x) = (x − 2)2 , find f (a + 2). *(a) a2 (c) a2 + 4a + 2 (e) None of these Solution:
(b) a2 + 4 (d) a2 − 2a + 4
f (a + 2) = ((a + 2) − 2)2 = (a)2 = a2
Problem 143. If f (x) = (x − 2)2 , find f (a − 1). (a) a2 − 2a − 1 (c) a2 − 4a + 3 (e) None of these Solution:
*(b) (d)
a2 − 6a + 9 a2 + 4a + 4
f (a − 1) = ((a − 1) − 2)2 = (a − 3)2 = a2 − 6a + 9
Problem 144. If f (x) = (x − 2)2 , find f (a − 2). (a) a2 (c) a2 − 4a − 6 (e) None of these Solution:
(b) *(d)
a2 − 4a + 2 a2 − 8a + 16
f (a − 2) = ((a − 2) − 2)2 = (a − 4)2 = a2 − 8a + 16
52
2.3 2.3.1
Evaluating functions of multiple variables Functions of two variables
Non-calculator based problems Problem 145. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 3 cm and length 2 cm, what is its volume? (a) 9π cm3 (b) 12π cm3 3 *(c) 18π cm (d) 36π cm3 (e) None of these Solution:
V = πr2 l = π(3 cm)2 (2 cm) = 18π cm3
Problem 146. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 2 cm and length 3 cm, what is its volume? (a) 9π cm3 *(b) 12π cm3 (c) 18π cm3 (d) 36π cm3 (e) None of these Solution:
V = πr2 l = π(2 cm)2 (3 cm) = 12π cm3
Problem 147. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 3 cm and length 4 cm, what is its volume? (a) 9π cm3 (b) 12π cm3 (c) 18π cm3 *(d) 36π cm3 (e) None of these Solution:
V = πr2 l = π(3 cm)2 (4 cm) = 36π cm3
Problem 148. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 6 at a speed of 9, what is its centripetal acceleration? 9 (a) (b) 4 4 27 *(c) (d) 54 2 (e)
None of these
Solution:
a=
v2 92 81 27 = = = . r 6 6 2
53 Problem 149. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 4 at a speed of 6, what is its centripetal acceleration? 9 4 (b) (a) 9 4 8 *(d) 9 (c) 3 (e)
None of these
Solution:
a=
v2 62 36 = = = 9. r 4 4
Problem 150. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 6 at a speed of 4, what is its centripetal acceleration? 4 9 (a) (b) 9 4 8 *(c) (d) 9 3 (e)
None of these
Solution:
a=
42 16 8 v2 = = = . r 6 6 3
Problem 151.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 5, and the gravitational acceleration is 10, what is the speed with which the object hits the ground? Round your answer to the nearest integer. √ (a) 5 (b) √50 *(c) 10 (d) 200 (e) None of these √ √ √ Solution: v = 2gh = 2 · 10 · 5 = 100 = 10
54 Problem 152.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 6, and the gravitational acceleration is 2, what is the speed with which the object hits the ground? Round your answer to the nearest integer. √ (b) 3 (a) √12 *(c) 24 (d) 9 (e) None of these √ √ √ Solution: v = 2gh = 2 · 2 · 6 = 24 Problem 153.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 3, and the gravitational acceleration is 5, what is the speed with which the object hits the ground? Round your answer to the nearest integer. √ *(a) √30 (b) 30 (c) 60 (d) 60 (e) None of these √ √ √ Solution: v = 2gh = 2 · 5 · 3 = 30 Calculator-based problems Problem 154. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 5.22 and length 7.80, what is its volume? Round your answer to the nearest integer. (a) 487 (c) 601 (e) None of these Solution:
(b) 541 *(d) 668
V = πr2 l = π(5.22)2 (7.80) = 668
Problem 155. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 3.03 and length 5.51, what is its volume? Round your answer to the nearest integer. (a) 116 (c) 143 (e) None of these Solution:
(b) 129 *(d) 159
V = πr2 l = π(3.03)2 (5.51) = 159
55 Problem 156. The formula for the volume of a cylinder is: V = πr2 l, where r is the radius of the cylinder and l is its length. If a cylinder has radius 6.99 and length 2.43, what is its volume? Round your answer to the nearest integer. *(a) (c) (e)
373 451 None of these
Solution:
(b) (d)
410 496
V = πr2 l = π(6.99)2 (2.43) = 373
Problem 157. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 4.8 at a speed of 13, what is its centripetal acceleration? Round your answer to the nearest integer. (a) 32 (c) 39 (e) None of these Solution:
a=
*(b) (d)
35 43
(13)2 v2 = = 35 r 4.8
Problem 158. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 4.4 at a speed of 19, what is its centripetal acceleration? Round your answer to the nearest integer. (a) 60 (c) 74 (e) None of these Solution:
a=
(b) 66 *(d) 82
(19)2 v2 = = 82 r 4.4
Problem 159. The formula for the centripetal acceleration of an object moving in a circle is: a = v 2 /r, where v is the object’s speed and r is the radius of the circle. (If you don’t know what centripetal acceleration is, don’t worry; you’ll learn about it during this course.) If an object is moving in a circle with radius 8.2 at a speed of 27, what is its centripetal acceleration? Round your answer to the nearest integer. (a) 70 *(c) 89 (e) None of these Solution:
a=
(b) 82 (d) 98
v2 (27)2 = = 89 r 8.2
56 Problem 160.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 23, and the gravitational acceleration is 32, what is the speed with which the object hits the ground? Round your answer to the nearest integer. (a) 31 (b) 35 *(c) 38 (d) 42 (e) None of these p p Solution: v = 2gh = 2(32)(23) = 38 Problem 161.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 39, and the gravitational acceleration is 9.8, what is the speed with which the object hits the ground? Round your answer to the nearest integer. (a) 20 (b) 22 (c) 25 *(d) 28 (e) None of these p p Solution: v = 2gh = 2(9.8)(39) = 28 Problem 162.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 120, and the gravitational acceleration is 32, what is the speed with which the object hits the ground? Round your answer to the nearest integer. (a) 71 (b) 79 *(c) 88 (d) 96 (e) None of these p p Solution: v = 2gh = 2(32)(120) = 88
57 Problem 163.√If an object is dropped from a height h, the speed with which it hits the ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If an object is dropped from a height of 56, and the gravitational acceleration is 9.8, what is the speed with which the object hits the ground? Round your answer to the nearest integer. *(a) (c) (e)
33 (b) 36 40 (d) 44 None of these p p Solution: v = 2gh = 2(9.8)(56) = 33 2.3.2
Functions of three variables
Non-calculator based problems Problem 164. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 3, a = 2, and t = 5? (a) 10 (c) 25 (e) None of these Solution:
*(b) 13 (d) 30
v = v0 + at = 3 + (2)(5) = 3 + 10 = 13
Problem 165. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 2, a = 3, and t = 5? *(a) (c) (e)
17 25 None of these
Solution:
(b) (d)
19 27
v = v0 + at = 2 + (3)(5) = 2 + 15 = 17
Problem 166. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 5, a = 2, and t = 3? *(a) (c) (e)
11 25 None of these
Solution:
(b) (d)
21 30
v = v0 + at = 5 + (2)(3) = 5 + 6 = 11
58 Problem 167. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 10, x = 5, and x0 = 4? *(a) (c) (e)
10 46 None of these
Solution:
(b) (d)
45 90
F = k(x − x0 ) = 10(5 − 4) = 10 · 1 = 10
Problem 168. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 5, x = 10, and x0 = 4? (a) 10 (c) 46 (e) None of these Solution:
*(b) (d)
30 60
F = k(x − x0 ) = 5(10 − 4) = 5 · 6 = 30
Problem 169. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 4, x = 10, and x0 = 5? (a) 10 (c) 35 (e) None of these Solution:
*(b) (d)
20 40
F = k(x − x0 ) = 4(10 − 5) = 4 · 5 = 20
Problem 170. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 2, its length is 3, and its density is 5? (a) 10π *(c) 60π (e) None of these Solution:
(b) 30π (d) 90π
m = πr2 lρ = π(22 )(3)(5) = π(4)(3)(5) = 60π
59 Problem 171. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 3, its length is 2, and its density is 5? (a) 10π (c) 60π (e) None of these Solution:
(b) 30π *(d) 90π
m = πr2 lρ = π(32 )(2)(5) = π(9)(2)(5) = 90π
Problem 172. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 3, its length is 5, and its density is 2? (a) 10π (c) 60π (e) None of these Solution:
(b) 30π *(d) 90π
m = πr2 lρ = π(32 )(5)(2) = π(9)(5)(2) = 90π
Calculator-based problems Problem 173. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 12, a = 1.8, and t = 14? Round your answer to the nearest integer. (a) 30 *(c) 37 (e) None of these Solution:
(b) 33 (d) 41
v = v0 + at = 12 + (1.8)(14) = 37
Problem 174. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 45, a = 2.9, and t = 8.3? Round your answer to the nearest integer. (a) 62 (c) 76 (e) None of these Solution:
*(b) (d)
69 84
v = v0 + at = 45 + (2.9)(8.3) = 69
60 Problem 175. The speed of an object is given by the formula: v = v0 + at, where v0 is the initial speed, a is the acceleration, and t is the time. What is the speed of an object if v0 = 28, a = −2.1, and t = 5.6? Round your answer to the nearest integer. (a) 15 (c) 18 (e) None of these Solution:
*(b) 16 (d) 20
v = v0 + at = 28 + (−2.1)(5.6) = 16
Problem 176. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 28, x = 19, and x0 = 12? Round your answer to the nearest integer. (a) 143 (c) 176 (e) None of these Solution:
(b) 159 *(d) 196
F = k(x − x0 ) = (28)(19 − 12) = 196
Problem 177. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 1700, x = 0.23, and x0 = 0.19? Round your answer to the nearest integer. *(a) (c) (e)
68 82 None of these
Solution:
(b) (d)
75 91
F = k(x − x0 ) = (1700)(0.23 − 0.19) = 68
Problem 178. The force exerted by a stretched spring is given by the formula: F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring, and x0 is its unstretched length. (You will learn about forces and spring constants during this course.) What is the force F if k = 2700, x = 0.41, and x0 = 0.34? Round your answer to the nearest integer. (a) 153 *(c) 189 (e) None of these Solution:
(b) 170 (d) 208
F = k(x − x0 ) = (2700)(0.41 − 0.34) = 189
61 Problem 179. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 1.7, its length is 5.1, and its density is 7.9? Round your answer to the nearest integer. (a) 296 *(c) 366 (e) None of these Solution:
(b) 329 (d) 402
m = πr2 lρ = π(1.7)2 (5.1)(7.9) = 366
Problem 180. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 0.32, its length is 1.1, and its density is 8.8? Round your answer to one decimal place. *(a) (c) (e)
3.1 3.8 None of these
Solution:
(b) (d)
3.5 4.2
m = πr2 lρ = π(0.32)2 (1.1)(8.8) = 3.1
Problem 181. The mass of a cylindrical weight is given by the formula: m = πr2 lρ, where r is the weight’s radius, l is its length, and ρ is its density. (You will learn about mass and density in this course. You will also learn a number of Greek letters, including ρ, which is “rho”.) What is the mass of such a weight if its radius is 0.55, its length is 1.8, and its density is 19.3? Round your answer to the nearest integer. *(a) (c) (e)
33 40 None of these
Solution:
(b) (d)
36 44
m = πr2 lρ = π(0.55)2 (1.8)(19.3) = 33
62
2.4 2.4.1
Solving linear equations Linear equations with numerical solutions
Problem 182. If 7x + 11 = 8, find x. Which of the following statements is true? (a) x < −1 (c) 0 ≤ x < 1 (e) None of these Solution: 7x + 11 = 8 Hence −1 ≤ x < 0
*(b) −1 ≤ x < 0 (d) x ≥ 1
−11
−−−→
÷7
7x = −3
−−→
x=−
3 7
Problem 183. If 9x − 11 = 7, find x. Which of the following statements is true? (a) x < −1 (c) 0 ≤ x < 1 (e) None of these Solution: 9x − 11 = 7 Hence x ≥ 1
(b) −1 ≤ x < 0 *(d) x ≥ 1
+11
−−−→
9x = 18
÷9
−−→
x=
18 =2 9
Problem 184. If 5x + 3 = 10, find x. Which of the following statements is true? (a) x < −1 (c) 0 ≤ x < 1 (e) None of these Solution: 5x + 3 = 10 Hence x ≥ 1
(b) −1 ≤ x < 0 *(d) x ≥ 1
−3
−−→
5x = 7
÷5
−−→
x=
7 5
Problem 185. If 3x − 2 = −9, find x. Which of the following statements is true? *(a) x < −1 (c) 0≤x 0, the larger solution is p √ 2.9 + (−2.9)2 − 4(0.82)(0.32) −b + b2 − 4ac x1 = = ≈ 3.4 ⇒ 0 ≤ x1 2a 2(0.82)
Problem 222. Use a calculator or equivalent to solve the equation: 1.7x2 + 8.9x − 0.77 = 0 There are two solutions, x1 and x2 , with x1 ≥ x2 . (It is possible that x1 = x2 .) Which of the following statements is true of the larger solution x1 ? (a) x1 < −2 *(c) 0 ≤ x1 < 2 (e) No real solution Solution:
(b) −2 ≤ x1 < 0 (d) 2 ≤ x1
We will need the quadratic formula to solve this equation: √ −b ± b2 − 4ac x= 2a
√ b2 − 4ac ≥ 0, if it’s not imaginary, the larger solution is p √ −8.9 + (8.9)2 − 4(1.7)(−0.77) −b + b2 − 4ac x1 = = ≈ 0.085 2a 2(1.7)
Since
⇒
0 ≤ x1 < 2
84
2.7
Algebra word problems
Problem 223. The length of a carpet is 3 feet greater than its width. The area of the carpet is 80 square feet. Which of the following equations describes the carpet’s width? *(a) w2 + 3w − 80 = 0 (c)
w2 − 20 = 0
(e)
None of these
Solution:
(b)
w2 − 3w + 80 = 0
(d) w2 + 20 = 0
If w is the carpet’s width, then the length is l = w + 3. Then A = lw = w(w + 3) = w2 + 3w = 80
⇒
w2 + 3w − 80 = 0
Problem 224. A window is twice as long as it is wide. The area of the window is 40 square feet. Which of the following equations describes the window’s length? (a)
3l2 − 40 = 0
(b)
l2 − 20 = 0
(c)
l2 − 120 = 0
*(d) l2 − 80 = 0
(e)
None of these
Solution: Let l be the window’s length and w its width. Since l = 2w, we know that w = l/2. Hence l2 l A = lw = l · = = 40 2 2
⇒
l2 = 80
⇒
l2 − 80 = 0
Problem 225. In the physics lab, you find two circular pieces of sheet metal. The radius of one of the circles is 3 centimeters greater than the radius of the other. The area of the larger circle is twice the area of the smaller one. Which of the following equations describes the radius of the smaller circle? *(a) r2 − 6r − 9 = 0 (b) r2 − 3r − 9 = 0 (c)
r2 + 6r + 9 = 0
(e)
None of these
(d) r2 + 3r + 9 = 0
Solution: Let r be the radius of the smaller circle. Then r + 3 is the radius of the larger one. The area of the small circle is πr2 ; the area of the large circle is π(r + 3)2 . Since the area of the large circle is twice the area of the small one, π(r + 3)2 = 2πr2 ⇒ (r + 3)2 = 2r2 ⇒ r2 + 6r + 9 = 2r2 ⇒ r2 − 6r − 9 = 0
85 Problem 226. A rectangular window is twice as long as it is wide. If its length were increased by 3 feet and its width were decreased by 1 foot, it would have the same area. Which of the following equations describes the window’s width? (a) w2 − w + 3 = 0 (c)
w2 + w − 3 = 0
(e)
None of these
*(b) w − 3 = 0 w2 − w − 3 = 0
(d)
Solution: Let w be the window’s width. Then its length is l = 2w. Its area is lw = (2w)w = 2w2 . Decreasing the width by 1 and increasing the length by 3 gives an area of (w − 1)(2w + 3). Hence 2w2 = (w − 1)(2w + 3) = 2w2 + w − 3
⇒
w−3=0
Problem 227. The distance from Tucson to Phoenix is 120 miles. You want to drive there and back at an average speed of 60 miles per hour. Because of traffic congestion, your average speed from Tucson to Phoenix is 40 mph. How fast do you have to drive on the return trip? (a)
60 miles per hour
(b)
80 miles per hour
(c)
90 miles per hour
*(d)
120 miles per hour
(e)
None of these
Solution: The round-trip distance is 240 miles. If you drive that distance at an average speed of 60 miles per hour, then the time that it will take is 240/60 = 4 hours. If your average speed from Tucson to Phoenix is 40 mph, then the time for that leg of the trip is 120/40 = 3 hours. Hence you have 4 − 3 = 1 hour for the return trip. This means you will have to cover 120 miles in 1 hour, so your speed must be 120 miles per hour. Problem 228. You drive 40 miles at 60 miles per hour, then bicycle an additional 10 miles at 12 miles per hour. What is your average speed for the entire trip? *(a)
33 31 miles per hour
(b)
36 miles per hour
(c)
50 miles per hour
(d)
50.4 miles per hour
(e)
None of these
Solution: Your average speed for the trip is the total distance divided by the total time. The total distance is 40 + 10 = 50 miles. The total time is 40 miles 10 miles 2 5 9 3 + = + = = hours 60 miles/hr 12 miles/hr 3 6 6 2 Hence the average speed is 50 miles 2 100 1 = 50 · = = 33 miles per hour 3 3 3 3 hours 2
86 Problem 229. A boat moves at 5 miles per hour in still water. It is launched in a river that flows at 3 miles per hour. From its launch point, it goes downstream for 4 miles, then turns around and comes back upstream to the launch point. How long does the round trip take? 8 4 hours (b) hours (a) 5 5 5 (c) 2 hours *(d) hours 2 (e)
None of these
Solution: The time for the round trip is the time that it takes to go downstream at 5 + 3 = 8 miles per hour, plus the time that it takes to come back upstream at 5 − 3 = 2 miles per hour. Thus t=
4 miles 4 miles 1 5 + = + 2 = hours 8 miles/hour 2 miles/hour 2 2
Problem 230. An airplane flies at a speed of 80 miles per hour in still air. On a day when the wind is blowing from the north at 20 miles per hour, the airplane flies 200 miles straight north, then turns around and returns to its starting point. What is its average speed on the round trip? (a)
64 miles per hour
(b)
66 23 miles per hour
*(c)
75 miles per hour
(d)
80 miles per hour
(e)
None of these
Solution: The average speed for the round trip is the total distance divided by the total time. The total distance is 2 × 200 = 400 miles. The total time is 200 200 200 200 10 16 + = + = +2= hours 80 − 20 80 + 20 60 100 3 3 Hence the average speed is 400 miles 3 = 400 · = 75 miles/hour 16 16 hours 3
87 Problem 231. A runner and a bicyclist start from the same point at the same time, with the runner going straight north and the bicyclist going straight south. The bicyclist is 7 miles per hour faster than the runner. At the end of two hours, the two are 60 miles apart. What is the bicyclist’s speed? (a)
11 21 miles per hour
(b)
14 miles per hour
*(c)
18 12 miles per hour
(d)
23 miles per hour
(e)
None of these
Solution: Let b be the bicyclist’s speed. Then the runner’s speed is b − 7. Since the two are going in opposite directions, after two hours the distance between them is 2b + 2(b − 7) = 4b − 14 = 60 miles
⇒
b=
60 + 14 37 1 = = 18 miles/hour 4 2 2
Problem 232. You drive from Smithtown to Jonesville at a speed of v, making the trip in time t. On the return trip, you are able to drive 10 miles per hour faster, which shortens your travel time by one hour. Which of the following equations is true? (a)
(v − 10)t = v(t + 1)
(b)
(v + 10)t = v(t − 1)
(c)
(v − 10)(t + 1) = vt
*(d)
(v + 10)(t − 1) = vt
(e)
None of these
Solution: On the first leg of the trip, your speed is v and your time is t. On the return leg, your speed is v + 10 and your time is t − 1. The distance is the same in each direction; so vt = (v + 10)(t − 1)
88
3 3.1
Graphs Single points on graphs y
Problem 233. The graph at right shows four points labelled with letters. The points are
6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2). -
Which of the four points is (3, 3)? (a) A (c) C
C
*(b) B (d) D
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5). Problem 234. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is (2, −5)? (a) A (c) C
(b) B *(d) D
C
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5).
89
Problem 235. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is (−6, 5)? *(a) A (c) C
C
(b) B (d) D
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5). Problem 236. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is (−4, −2)? (a) A *(c) C
(b) (d)
B D
C
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5).
90
Problem 237. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is A? (a) (3, 3) *(c) (−6, 5)
C
(b) (2, −5) (d) (−4, −2)
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5). Problem 238. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is B? *(a) (3, 3) (c) (−6, 5)
(b) (2, −5) (d) (−4, −2)
C
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5).
91
Problem 239. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is C? (a) (3, 3) (c) (−6, 5)
C
(b) (2, −5) *(d) (−4, −2)
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5). Problem 240. The graph at right shows four points labelled with letters. The points are
y 6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is D? (a) (3, 3) (c) (−6, 5)
*(b) (d)
(2, −5) (−4, −2)
C
x
r r
D
Solution: Each of the four points is in a different quadrant, so we can use that to identify them. Point A is in the second quadrant, with a negative x-coordinate and a positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5).
92 y6
Ar
Problem 241. The graph at right shows four points labelled with letters. The points are
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is A? (a) (2, 2) (c) (7, 2)
*(b) (3, 9) (d) (8, 8)
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B. Problem 242. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is B? (a) (2, 2) (c) (7, 2)
(b) (3, 9) *(d) (8, 8)
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B.
93
Problem 243. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is C? *(a) (c)
(2, 2) (7, 2)
(b) (d)
(3, 9) (8, 8)
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B. Problem 244. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is D? (a) (2, 2) *(c) (7, 2)
(b) (d)
(3, 9) (8, 8)
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B.
94
Problem 245. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is (2, 2)? (a) A *(c) C
(b) (d)
B D
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B. Problem 246. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is (3, 9)? *(a) A (c) C
(b) B (d) D
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B.
95
Problem 247. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is (7, 2)? (a) A (c) C
(b) B *(d) D
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B. Problem 248. The graph at right shows four points labelled with letters. The points are
y6
Ar
rB
(2, 2), (3, 9), (7, 2), and (8, 8). Which of the four points is (8, 8)? (a) A (c) C
*(b) B (d) D
C
r
r
D -
x
Solution: The point (3, 9) is the only one where the x-coordinate is small and the ycoordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinate and a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- and y-coordinates are both small; that point must be C. For the point (8, 8), the x- and y-coordinates are both large; that point must be B.
96
3.2
Matching graphs and equations
Problem 249. Which equation is shown on the graph at right? (a) *(b) (c) (d)
y y y y
y6
= 2x = 2x + 1 = −2x = −2x − 1
Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept.
-
x
The line in the figure has a positive slope (as x increases, y increases) and a positive y-intercept (when x = 0, y > 0). Of the answers given, only y = 2x + 1 has m > 0 and b > 0. Problem 250. Which equation is shown on the graph at right? *(a) (b) (c) (d)
y y y y
y6
= 2x = 2x + 1 = −2x = −2x − 1
Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept.
-
x
The line in the figure has a positive slope (as x increases, y increases) and a y-intercept of 0 (when x = 0, y = 0). Of the answers given, only y = 2x has m > 0 and b = 0. Problem 251. Which equation is shown on the graph at right? (a) (b) *(c) (d)
y y y y
= 2x = 2x + 1 = −2x = −2x − 1
Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept.
y6
A A A A
A A A
-
x
A A A A A A AA
The line in the figure has a negative slope (as x increases, y decreases) and a y-intercept of 0 (when x = 0, y = 0). Of the answers given, only y = −2x has m < 0 and b = 0.
97 Problem 252. Which equation is shown on the graph at right? (a) (b) (c) *(d)
y y y y
= 2x = 2x + 1 = −2x = −2x − 1
Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept.
y6
A A A A A
A A A A
-
x
A A A A AA
The line in the figure has a negative slope (as x increases, y increases) and a negative y-intercept (when x = 0, y < 0). Of the answers given, only y = −2x − 1 has m < 0 and b < 0. Problem 253. Which equation is shown on the graph at right? (The scale is the same for the x- and y-axes.) (a) (b)
y6
y = 3x y = −3x
*(c) y =
x 3
-
x
x 3 Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept. In this case, all of the possible answers have b = 0, so we must focus on the slope m. The line in the figure has a positive slope (as x increases, y increases), so it must be m = 3 or m = 1/3. Since the value of y increases more slowly than the value of x, m < 1. Hence m = 1/3; so y = x/3. (d)
y=−
98 Problem 254. Which equation is shown on the graph at right? (The scale is the same for the x- and y-axes.)
y 6
*(a) y = 3x (b)
y = −3x
-
x
(c)
y=
x 3
x 3 Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept. In this case, all of the possible answers have b = 0, so we must focus on the slope m. The line in the figure has a positive slope (as x increases, y increases), so it must be m = 3 or m = 1/3. Since the value of y increases more rapidly than the value of x, m > 1. Hence m = 3; so y = 3x. (d)
y=−
Problem 255. Which equation is shown on the graph at right? (The scale is the same for the x- and y-axes.) (a)
y = 3x
(b)
y = −3x
(c)
y=
x 3
y6
PP PP
PP PP P
PP P
x
PP P
x 3 Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept. In this case, all of the possible answers have b = 0, so we must focus on the slope m. The line in the figure has a negative slope (as x increases, y decreases), so it must be m = −3 or m = −1/3. Since the value of y decreases more slowly than the value of x increases, m > −1. Hence m = −1/3; so y = −x/3. *(d) y = −
99 Problem 256. Which equation is shown on the graph at right? (The scale is the same for the x- and y-axes.)
y6
B B B B
(a)
y = 3x
B B B
*(b) y = −3x (c)
y=
-
x
B B B
x 3
B
B x B 3 BB Solution: If the equation of a line is written in the form: y = mx + b, then m is the slope and b is the y-intercept. In this case, all of the possible answers have b = 0, so we must focus on the slope m. The line in the figure has a negative slope (as x increases, y decreases), so it must be m = −3 or m = −1/3. Since the value of y decreases more rapidly than the value of x increases, m < −1. Hence m = −3; so y = −3x.
y=−
(d)
Problem 257. The four graphs (a), (b), (c), and (d) below represent four different equations: y =x−1
y =x+1
y = −x + 1
y = −x − 1
Which of the four graphs represents y = x + 1? (a) @ y 6
(b)
y6
(c)
y6
*(d)
y6
@ @ @ @
@x @ @
@ @
-
x
@ @
-
x
-
x
@ @ @ @
Solution: The equation y = x + 1 is written in the form: y = mx + b. Here m = 1 > 0 and b = 1 > 0. Since m > 0, the graph should have a positive slope (as x increases, y increases). Since b > 0, the graph should have a positive y-intercept (when x = 0, y > 0). Only graph (d) slopes upward and crosses the y-axis above the x-axis.
100 Problem 258. The four graphs (a), (b), (c), and (d) below represent four different equations: y =x−1
y =x+1
y = −x + 1
y = −x − 1
Which of the four graphs represents y = x − 1? (a) @ y 6
(b)
y6
*(c)
y6
(d)
y6
@ @ @ @
-
@ @
@x @ @
-
-
x
@ @
-
x
x
@ @ @ @
Solution: The equation y = x − 1 is written in the form: y = mx + b. Here m = 1 > 0 and b = −1 < 0. Since m > 0, the graph should have a positive slope (as x increases, y increases). Since b < 0, the graph should have a negative y-intercept (when x = 0, y < 0). Only graph (c) slopes upward and crosses the y-axis below the x-axis. Problem 259. The four graphs (a), (b), (c), and (d) below represent four different equations: y =x−1
y =x+1
y = −x + 1
y = −x − 1
Which of the four graphs represents y = −x + 1? *(a) @ y 6
(b)
y6
(c)
y6
(d)
y6
@ @ @ @
@x @ @
@ @
-
x
@ @
-
x
-
x
@ @ @ @
Solution: The equation y = −x + 1 is written in the form: y = mx + b. Here m = −1 < 0 and b = 1 > 0. Since m < 0, the graph should have a negative slope (as x increases, y decreases). Since b > 0, the graph should have a positive y-intercept (when x = 0, y > 0). Only graph (a) slopes downward and crosses the y-axis above the x-axis.
101 Problem 260. The four graphs (a), (b), (c), and (d) below represent four different equations: y =x−1
y =x+1
y = −x + 1
y = −x − 1
Which of the four graphs represents y = −x − 1? (a) @ y 6
*(b)
y6
(c)
y6
(d)
y6
@ @ @ @
-
@x @ @
@ @
-
x
@ @
-
-
x
x
@ @ @ @
Solution: The equation y = −x − 1 is written in the form: y = mx + b. Here m = −1 < 0 and b = −1 < 0. Since m < 0, the graph should have a negative slope (as x increases, y decreases). Since b < 0, the graph should have a negative y-intercept (when x = 0, y < 0). Only graph (b) slopes downward and crosses the y-axis below the x-axis. Problem 261. Which equation is shown on the graph at right? (a)
y6
y = x2 + 1
*(b) y = x2 − 1
-
x
2
(c)
y = −x + 1
(d)
y = −x2 − 1
Solution: The graph has a negative y-intercept: it crosses the y-axis below the x-axis. This means that when x = 0, y < 0. That allows us to rule out equations (a) and (c). In the graph, when x has large absolute values, y > 0. This is consistent with (b), where the coefficient of x2 is positive; but not with (d), where the coefficient of x2 is negative.
102
Problem 262. Which equation is shown on the graph at right? (a)
y = x2 + 1
(b)
y = x2 − 1
y6
-
x
*(c) y = −x2 + 1 (d)
y = −x2 − 1
Solution: The graph has a positive y-intercept: it crosses the y-axis above the x-axis. This means that when x = 0, y > 0. That allows us to rule out equations (b) and (d). In the graph, when x has large absolute values, y < 0. This is consistent with (c), where the coefficient of x2 is negative, but not with (a), where the coefficient of x2 is positive. Problem 263. Which equation is shown on the graph at right?
y6
*(a) y = x2 + 1 (b)
y = x2 − 1
(c)
y = −x2 + 1
(d)
y = −x2 − 1
-
x
Solution: The graph has a positive y-intercept: it crosses the y-axis above the x-axis. This means that when x = 0, y > 0. That allows us to rule out equations (b) and (d). In the graph, when x has large absolute values, y > 0. This is consistent with (a), where the coefficient of x2 is positive, but not with (c), where the coefficient of x2 is negative. Problem 264. Which equation is shown on the graph at right? (a)
y = x2 + 1
(b)
y = x2 − 1
(c)
y = −x2 + 1
y6
-
x
*(d) y = −x2 − 1 Solution: The graph has a negative y-intercept: it crosses the y-axis below the x-axis. This means that when x = 0, y < 0. That allows us to rule out equations (a) and (c). In the graph, when x has large absolute values, y < 0. This is consistent with (d), where the coefficient of x2 is negative, but not with (b), where the coefficient of x2 is positive.
103 Problem 265. The four graphs (a), (b), (c), and (d) below are all drawn on the same scale. They represent four different equations: y = 2x2
y=
x2 2
y = −2x2
y=−
x2 2
Which of the four graphs represents y = 2x2 ? (a)
y6
(b)
-
x
y6
(c)
-
x
y6
*(d)
-
x
y6
-
x
Solution: All four equations have y-intercepts of zero, so that won’t help us. In graphs (a) and (b), y ≤ 0 for all values of x. They must represent the two equations where the coefficient of x2 is negative. That leaves us with graphs (c) and (d), which must represent the two equations with positive coefficients of x2 . Of these two equations, the curve of y = 2x2 will rise faster than the curve of y = x2 /2: for example, the first of these includes the point (1, 2), whereas the second includes the point (1, 21 ). Hence the graph of y = 2x2 is (d).
104 Problem 266. The four graphs (a), (b), (c), and (d) below are all drawn on the same scale. They represent four different equations: y = 2x2
y=
x2 2
y = −2x2
Which of the four graphs represents y = (a)
y6
(b)
-
x
y=−
x2 ? 2
y6
*(c) y 6
-
x
x2 2
(d)
-
x
y6
-
x
Solution: All four equations have y-intercepts of zero, so that won’t help us. In graphs (a) and (b), y ≤ 0 for all values of x. They must represent the two equations where the coefficient of x2 is negative. That leaves us with graphs (c) and (d), which must represent the two equations with positive coefficients of x2 . Of these two equations, the curve of y = 2x2 will rise faster than the curve of y = x2 /2: for example, the first of these includes the point (1, 2), whereas the second includes the point (1, 21 ). Hence the graph of y = x2 /2 is (c).
105 Problem 267. The four graphs (a), (b), (c), and (d) below are all drawn on the same scale. They represent four different equations: y = 2x2
y=
x2 2
y = −2x2
y=−
x2 2
Which of the four graphs represents y = −2x2 ? (a)
y6
*(b)
-
x
y6
(c)
-
x
y6
(d)
-
x
y6
-
x
Solution: All four equations have y-intercepts of zero, so that won’t help us. In graphs (c) and (d), y ≥ 0 for all values of x. They must represent the two equations where the coefficient of x2 is positive. That leaves us with graphs (a) and (b), which must represent the two equations with negative coefficients of x2 . Of these two equations, the curve of y = −2x2 will fall faster than the curve of y = −x2 /2: for example, the first of these includes the point (1, −2), whereas the second includes the point (1, − 21 ). Hence the graph of y = −2x2 is (b).
106 Problem 268. The four graphs (a), (b), (c), and (d) below are all drawn on the same scale. They represent four different equations: y = 2x2
y=
x2 2
y = −2x2
Which of the four graphs represents y = − *(a)
y6
(b)
-
x
y6
x
x2 2
x2 ? 2 (c)
-
y=−
y6
(d)
-
x
y6
-
x
Solution: All four equations have y-intercepts of zero, so that won’t help us. In graphs (c) and (d), y ≥ 0 for all values of x. They must represent the two equations where the coefficient of x2 is positive. That leaves us with graphs (a) and (b), which must represent the two equations with negative coefficients of x2 . Of these two equations, the curve of y = −2x2 will fall faster than the curve of y = −x2 /2: for example, the first of these includes the point (1, −2), whereas the second includes the point (1, − 21 ). Hence the graph of y = −x2 /2 is (a).
107
Part II
Mathematical Preliminaries
108
4 4.1 4.1.1
Basic Trigonometry Arc-Length Problems using s = θ r Degrees to radians: formula
Problem 269. An angle measures 29◦ . What is its measurement in radians? 29π 29π rad *(b) rad (a) 360 180 29 · 360 29 · 180 (c) rad (d) rad π π (e)
None of these
Problem 270. An angle measures 47◦ . What is its measurement in radians? 47 47π rad (b) rad *(a) 180 360π 47 47 (c) rad (d) rad 180π 2π (e)
None of these
Problem 271. An angle measures 61◦ . What is its measurement in radians? 61π 61π *(a) rad (b) rad 180 90 61 · 360 61 rad (d) rad (c) π π (e)
None of these
Problem 272. An angle measures 83◦ . What is its measurement in radians? 83 83π (a) rad *(b) rad π 180 83 · 360 83 (c) rad (d) rad π 2π (e)
None of these
109 Problem 273. An angle measures 19◦ . What is its measurement in radians? 19 19π (a) rad *(b) rad π 180 19 · 360 19π (c) rad (d) rad π 90 (e)
None of these
Problem 274. An angle measures 53◦ . What is its measurement in radians? 53 · 360 53 (a) rad (b) rad π 2π 53π 53π *(c) rad (d) rad 180 360 (e)
None of these
Problem 275. An angle measures 7◦ . What is its measurement in radians? 7 · 90 7 · 180 (a) rad (b) rad π π 7π 7 *(c) rad (d) rad 180 2π (e)
None of these
Problem 276. An angle measures 73◦ . What is its measurement in radians? 73 · 180 73π (a) rad *(b) rad π 180 73 · 360 73π (c) rad (d) rad π 360 (e)
None of these
Problem 277. An angle measures 43◦ . What is its measurement in radians? 43π 43 · 90 rad (b) rad *(a) 180 π 43π 43 (c) rad (d) rad 360 π (e)
None of these
110 Problem 278. An angle measures 59◦ . What is its measurement in radians? 59 · 180 59 (a) rad (b) rad π π 59π 59π *(c) rad (d) rad 180 360 (e)
4.1.2
None of these
Radians to degrees: formula
Problem 279. An angle measures 0.40 radians. What is its measurement in degrees? 0.40π 0.40π degrees (b) degrees (a) 360 180 (0.40)(360) (0.40)(180) (c) degrees *(d) degrees π π (e) None of these Problem 280. An angle measures 1.06 radians. What is its measurement in degrees? (a) (c) (e)
1.06π degrees 180 1.06 degrees 2π None of these
(b) *(d)
(1.06)(90) degrees π (1.06)(180) degrees π
Problem 281. An angle measures 0.88 radians. What is its measurement in degrees? *(a) (c) (e)
(0.88)(180) degrees π (0.88)(90) degrees π None of these
(b) (d)
(0.88)(360) degrees π (0.88) degrees 2π
Problem 282. An angle measures 0.04 radians. What is its measurement in degrees? (a) (c) (e)
(0.04)(360) degrees π 0.04π degrees 90 None of these
(b) *(d)
0.04π degrees 360 (0.04)(180) degrees π
111 Problem 283. An angle measures 0.15 radians. What is its measurement in degrees? 0.15 0.15π (a) degrees (b) degrees π 90 0.15 (0.15)(180) (c) degrees *(d) degrees 2π π (e) None of these Problem 284. An angle measures 1.34 radians. What is its measurement in degrees? *(a) (c) (e)
(1.34)(180) degrees π 1.34π degrees 180 None of these
(b) (d)
1.34 degrees π (1.34)(90) degrees π
Problem 285. An angle measures 0.77 radians. What is its measurement in degrees? 0.77 0.77π degrees (b) degrees (a) 2π 360 0.77 (0.77)(180) (c) degrees *(d) degrees π π (e) None of these Problem 286. An angle measures 0.12 radians. What is its measurement in degrees? *(a) (c) (e)
(0.12)(180) degrees π (0.12)(360) degrees π None of these
(b) (d)
0.12 degrees 2π 0.12π degrees 360
Problem 287. An angle measures 0.85 radians. What is its measurement in degrees? (0.85)(360) degrees π (0.85)(180) *(c) degrees π (e) None of these
(a)
(b) (d)
0.85 degrees π 0.85π degrees 90
Problem 288. An angle measures 0.34 radians. What is its measurement in degrees? (a) (c) (e)
0.34π degrees 360 (0.34)(360) degrees π None of these
*(b) (d)
(0.34)(180) degrees π 0.34π degrees 180
112 4.1.3
Degrees to radians: calculator
Problem 289. An angle measures 38◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 1.51 rad (c) 2.65 rad (e) None of these Solution:
*(b) 0.66 rad (d) 0.33 rad
38π = 0.66 rad 180
Problem 290. An angle measures 87◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 3.04 rad *(c) 1.52 rad (e) None of these Solution:
(b) 0.38 rad (d) 6.07 rad
87π = 1.52 rad 180
Problem 291. An angle measures 7◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 0.03 rad (c) 0.24 rad (e) None of these Solution:
*(b) (d)
0.12 rad 0.49 rad
7π = 0.12 rad 180
Problem 292. An angle measures 72◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 0.38 rad *(c) 1.26 rad (e) None of these Solution:
(b) 5.03 rad (d) 0.80 rad
72π = 1.26 rad 180
Problem 293. An angle measures 12◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 4.77 rad *(c) 0.21 rad (e) None of these Solution:
12π = 0.21 rad 180
(b) 0.05 rad (d) 0.42 rad
113 Problem 294. An angle measures 84◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 2.93 rad (c) 0.73 rad (e) None of these Solution:
*(b) 1.47 rad (d) 0.37 rad
84π = 1.47 rad 180
Problem 295. An angle measures 21◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 2.73 rad (c) 1.47 rad (e) None of these Solution:
(b) 0.18 rad *(d) 0.37 rad
21π = 0.37 rad 180
Problem 296. An angle measures 75◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. *(a) (c) (e)
1.31 rad 0.65 rad None of these
Solution:
(b) (d)
2.63 rad 0.76 rad
75π = 1.31 rad 180
Problem 297. An angle measures 63◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. (a) 0.55 rad (c) 0.91 rad (e) None of these Solution:
*(b) 1.10 rad (d) 3.09 rad
63π = 1.10 rad 180
Problem 298. An angle measures 42◦ . Use a calculator or equivalent to find its measurement in radians. Round your answer to the nearest 0.01 rad. *(a) (c) (e)
0.73 rad 1.47 rad None of these
Solution:
42π = 0.73 rad 180
(b) (d)
1.36 rad 3.10 rad
114 4.1.4
Radians to degrees: calculator
Problem 299. An angle measures 0.52 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. *(a) (c) (e)
30◦ 119◦ None of these
Solution:
(b) (d)
7◦ 15◦
(0.52)(180) = 30◦ π
Problem 300. An angle measures 1.14 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 131◦ (c) 7◦ (e) None of these Solution:
(b) 16◦ *(d) 65◦
(1.14)(180) = 65◦ π
Problem 301. An angle measures 0.11 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 57◦ (c) 1◦ (e) None of these Solution:
*(b) 6◦ (d) 13◦
(0.11)(180) = 6◦ π
Problem 302. An angle measures 1.52 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 348◦ *(c) 87◦ (e) None of these Solution:
(b) (d)
4◦ 10◦
(1.52)(180) = 87◦ π
Problem 303. An angle measures 0.37 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 5◦ (c) 42◦ (e) None of these Solution:
*(b) 21◦ (d) 11◦
(0.37)(180) = 21◦ π
115 Problem 304. An angle measures 0.89 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. *(a) (c) (e)
51◦ 102◦ None of these
Solution:
(b) (d)
7◦ 13◦
(0.89)(180) = 51◦ π
Problem 305. An angle measures 1.25 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 286◦ *(c) 72◦ (e) None of these Solution:
(b) 143◦ (d) 5◦
(1.25)(180) = 72◦ π
Problem 306. An angle measures 0.66 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 151◦ *(c) 38◦ (e) None of these Solution:
(b) (d)
4◦ 19◦
(0.66)(180) = 38◦ π
Problem 307. An angle measures 1.73 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 4◦ (c) 50◦ (e) None of these Solution:
*(b) 99◦ (d) 36◦
(1.73)(180) = 99◦ π
Problem 308. An angle measures 0.44 rad. Use a calculator or equivalent to find its measurement in degrees. Round your answer to the nearest degree. (a) 13◦ (c) 14◦ (e) None of these Solution:
(b) 6◦ *(d) 25◦
(0.44)(180) = 25◦ π
116 4.1.5
Arc length to radians: pictures
Problem 309. In the figure at right, what is the measure of the angle θ? √ (a) 41 rad (b) 3π rad *(c) 5/4 rad (d) 4π/5 rad (e) None of these
θ
Problem 310. In the figure at right, what is the measure of the angle θ? √ 19 π rad (a) 10π/9 rad (b) (c) 9π/10 rad *(d) 10/9 rad (e) None of these
θ
Problem 311. In the figure at right, what is the measure of the angle θ? (a) 5/3 rad (c) 4 rad (e) None of these
*(b) 3/5 rad (d) 4π rad
Problem 312. In the figure at right, what is the measure of the angle θ? √ 2 π rad *(a) 1 rad (b) (c) 3 rad (d) π rad (e) None of these
Problem 313. In the figure at right, what is the measure of the angle θ? √ (a) 14 rad (b) 5π/7 rad (c) 5/7 rad *(d) 7/5 rad (e) None of these
5
4
10
9
θ
3
5
9
θ
9
7
θ 5
117
Problem 314. In the figure at right, what is the measure of the angle θ? √ 10 π rad *(a) 1/3 rad (b) (c) 3 rad (d) π/3 rad (e) None of these
θ R @
(a) 4/7 rad (c) 4π/7 rad (e) None of these
4.1.6
θ
(b) 9/4 √ rad (d) 97 rad
Problem 316. In the figure at right, what is the measure of the angle θ? *(b) (d)
7/4 √ rad 33 π rad
1
3
Problem 315. In the figure at right, what is the measure of the angle θ? *(a) √ 4/9 rad (c) 65 π rad (e) None of these
4
9
7
B B B B B
θ 4
Arc length to radians: descriptions
Problem 317. A circle has a radius of 5 cm. A central angle θ intercepts an arc whose length is 4 cm. What is the measure of θ? (a) √ 3 rad (c) 41 π rad (e) None of these
(b) *(d)
4π/5 rad 4/5 rad
Problem 318. A circle has a radius of 13 cm. A central angle θ intercepts an arc whose length is 5 cm. What is the measure of θ? *(a) √ 5/13 rad (c) 194 π rad (e) None of these
(b) 5π/13 rad (d) 13π/5 rad
118 Problem 319. A circle has a radius of 3 cm. A central angle θ intercepts an arc whose length is 5 cm. What is the measure of θ? √ (a) 34 π rad *(b) 5/3 √ rad (c) 4π rad (d) 34 rad (e) None of these Problem 320. A circle has a radius of 6 cm. A central angle θ intercepts an arc whose length is 7 cm. What is the measure of θ? √ *(a) √ 7/6 rad (b) √85 rad (c) 85 π rad (d) 13 rad (e) None of these Problem 321. A circle has a radius of 6 cm. A central angle θ intercepts an arc whose length is 4 cm. What is the measure of θ? *(a) 2/3 rad (c) 2π/3 rad (e) None of these
(b) 3/2 √ rad (d) 2 13 π rad
Problem 322. A circle has a radius of 8 cm. A central angle θ intercepts an arc whose length is 4π cm. What is the measure of θ? (a) 1/2 rad (c) 2 rad (e) None of these
*(b) π/2 rad (d) 2π rad
Problem 323. A circle has a radius of 12 cm. A central angle θ intercepts an arc whose length is 4π cm. What is the measure of θ? (a) 3π rad (c) 3 rad (e) None of these
*(b) π/3 rad (d) 1/3 rad
Problem 324. A circle has a radius of 15π cm. A central angle θ intercepts an arc whose length is 3π cm. What is the measure of θ? (a) π/5 rad *(c) 1/5 rad (e) None of these
(b) 5π rad (d) 5 rad
Problem 325. A circle has a radius of 10 cm. A central angle θ intercepts an arc whose length is 15π cm. What is the measure of θ? (a) 2/3 rad (c) 3/2 rad (e) None of these
(b) *(d)
2π/3 rad 3π/2 rad
119 Problem 326. A circle has a radius of 6π cm. A central angle θ intercepts an arc whose length is 2 cm. What is the measure of θ? (a) 2π/3 rad *(c) 1/3π rad (e) None of these
4.1.7
(b) 3π rad (d) 3/2π rad
Radians to arc length: pictures
Problem 327. In the figure at right, if θ = 1.4 rad, what is the arc length s? *(a) 9.8 cm (c) π/5 cm (e) None of these Solution:
(b) 5π cm (d) 10π cm
θ 7 cm
(7 cm)(1.4) = 9.8 cm
Problem 328. In the figure at right, if θ = 1.2 rad, what is the arc length s? (a) 5 cm (c) π/5 cm (e) None of these Solution:
s
*(b) 7.2 cm (d) 14.4π cm
s
θ
6 cm
(6 cm)(1.2) = 7.2 cm
Problem 329. In the figure at right, if θ = 4/9 rad, what is the arc length s? (a) 27π/2 cm (c) 27π cm (e) None of these Solution:
*(b) 8/3 cm (d) 2/27 cm
θ
6 cm
(6 cm)(4/9) = 8/3 cm
Problem 330. In the figure at right, if θ = 2/3 rad, what is the arc length s? (a) 15π cm (c) 15π/2 cm (e) None of these Solution:
s
*(b) (d)
10/3 cm 2/15 cm
(5 cm)(2/3) = 10/3 cm
θ
5 cm
s
120 Problem 331. In the figure at right, if θ = 9/5 rad, what is the arc length s? (a) 9π/10 cm *(c) 18/5 cm (e) None of these Solution:
(b) 18π/5 cm (d) 20π/9 cm
s
C C C C
Cθ
(2 cm)(9/5) = 18/5 cm
Problem 332. In the figure at right, if θ = 5/4 rad, what is the arc length s? (a) 5π/8 cm (c) 12π/5 cm (e) None of these Solution: 4.1.8
2 cm
(b) 15π/4 cm *(d) 15/4 cm
s
θ
3 cm
(3 cm)(5/4) = 15/4 cm
Radians to arc length: descriptions
Problem 333. A circle has a radius of 10 cm. A central angle has measure θ = 3/5 rad. What is the length of the arc intercepted by the angle? (a) 6π/50 cm (c) 12π cm (e) None of these Solution:
*(b) 6 cm (d) 6π cm
(10 cm)(3/5) = 6 cm
Problem 334. A circle has a radius of 4 cm. A central angle has measure θ = 1/2 rad. What is the length of the arc intercepted by the angle? (a) π/4 cm (c) 2π cm (e) None of these Solution:
*(b) (d)
2 cm 4π cm
(4 cm)(1/2) = 2 cm
Problem 335. A circle has a radius of 5 cm. A central angle has measure θ = π/3 rad. What is the length of the arc intercepted by the angle? (a) 2/15 cm (c) 5/3 cm (e) None of these Solution:
*(b) 5π/3 cm (d) 10π/3 cm
(5 cm)(π/3) = 5π/3 cm
121 Problem 336. A circle has a radius of 6 cm. A central angle has measure θ = π/4 rad. What is the length of the arc intercepted by the angle? *(a) 3π/2 cm (c) π/24 cm (e) None of these Solution:
(b) 3/2 cm (d) π/12 cm
(6 cm)(π/4) = 3π/2 cm
Problem 337. A circle has a radius of 4 cm. A central angle has measure θ = 4/3 rad. What is the length of the arc intercepted by the angle? (a) 6π cm (c) 2π/3 cm (e) None of these Solution:
(b) *(d)
3 cm 16/3 cm
(4 cm)(4/3) = 16/3 cm
Problem 338. A circle has a radius of 10 cm. A central angle has measure θ = π/5 rad. What is the length of the arc intercepted by the angle? *(a) 2π cm (c) 50π cm (e) None of these Solution:
(b) π 2 /25 cm (d) 2 cm
(10 cm)(π/5) = 2π cm
Problem 339. A circle has a radius of 12 cm. A central angle has measure θ = 3/2 rad. What is the length of the arc intercepted by the angle? (a) π/8 cm (c) 8π cm (e) None of these Solution:
*(b) (d)
18 cm 1/8 cm
(12 cm)(3/2) = 18 cm
Problem 340. A circle has a radius of 10 cm. A central angle has measure θ = 2π/5 rad. What is the length of the arc intercepted by the angle? (a) 2π/50 cm (c) 25/π cm (e) None of these Solution:
(b) 2/25 cm *(d) 4π cm
(10 cm)(2π/5) = 4π cm
122 4.1.9
Word problems: arc length and radians
Problem 341. A wheel has a radius of 28 inches. It turns through an angle of 1.2 rad. How far does a point on the rim travel during the turn? Round your answer to the nearest inch. (a) 11 in *(b) 34 in (c) 53 in (d) 106 in (e) None of these Solution:
s = rθ = (28 in)(1.2) = 34 in
Problem 342. A wheel has a radius of 71 cm. It turns through an angle of 0.6 rad. How far does a point on the rim travel during the turn? Round your answer to the nearest cm. (a) 7 cm (b) 14 cm (c) 27 cm *(d) 43 cm (e) None of these Solution:
s = rθ = (71 cm)(0.6) = 43 cm
Problem 343. A pizza has a radius of 8.2 inches. A slice cut from the pizza has a central angle of 0.39 rad. What is the length of the crust on the slice? Round your answer to the nearest 0.1 in. (a) 0.5 in (b) 1.0 in (c) 2.0 in *(d) 3.2 in (e) None of these Solution:
s = rθ = (8.2 in)(0.39) = 3.2 in
Problem 344. A pie has a radius of 13 cm. A slice cut from the pie has a central angle of 0.29 rad. What is the length of the crust on the slice? Round your answer to the nearest 0.1 cm. (a) 1.2 cm (b) 2.7 cm *(c) 3.8 cm (d) 4.1 cm (e) None of these Solution:
s = rθ = (13 cm)(0.29) = 3.8 cm
Problem 345. A door measures 34 in from the hinge to the outer edge. It swings through an angle of 1.4 rad. How far does the outer edge travel? Round your answer to the nearest inch. (a) 45 in *(b) 48 in (c) 51 in (d) 54 in (e) None of these Solution:
s = rθ = (34 in)(1.4) = 48 in
123 Problem 346. A wheel has a radius of 63 cm. Two adjacent spokes meet at the center at an angle of 0.51 rad. How long is the rim between the two spokes? Round your answer to the nearest cm. *(a) 32 cm (b) 34 cm (c) 36 cm (d) 38 cm (e) None of these Solution:
s = rθ = (63 cm)(0.51) = 32 cm
Problem 347. A bicycle wheel has a radius of 27 inches. It turns so that a point on the rim travels 11 inches. What angle has the wheel turned through? Round your answer to the nearest 0.01 rad. (a) 0.31 rad (b) 0.35 rad (c) 0.38 rad *(d) 0.41 rad (e) None of these Solution:
θ = s/r = (11 in)/(27 in) = 0.41 rad
Problem 348. A pizza has a radius of 13 cm. A slice cut from the pizza has 10 cm of crust on the outside. What is the angle of the slice? Round your answer to the nearest 0.01 rad. (a) 0.70 rad (b) 0.74 rad *(c) 0.77 rad (d) 0.81 rad (e) None of these Solution:
θ = s/r = (10 cm)/(13 cm) = 0.77 rad
124
4.2 4.2.1
Basic Right-Triangle Trigonometry Basic trig functions: finding
Problem 349. In the figure at right, sin θ = ? (a) x/y *(c) y/r (e) None of these
r
(b) y/x (d) x/r
θ
x
Problem 350. In the figure at right, cos θ = ? r
*(b) x/r (d) y/x
x
r
(b) y/r (d) r/y
x
r
(b) r/x (d) y/x
φ
y
θ
x
Problem 353. In the figure at right, cos φ = ? r
r/x x/y
φ
y
θ
x
Problem 354. In the figure at right, tan φ = ? (a) y/r *(c) x/y (e) None of these
y
θ
Problem 352. In the figure at right, sin φ = ?
(b) (d)
φ
*(a) y/r (c) x/r (e) None of these
y
θ
Problem 351. In the figure at right, tan θ = ?
*(a) x/r (c) y/r (e) None of these
φ
*(a) y/x (c) r/x (e) None of these
y
(a) r/y (c) r/x (e) None of these
φ
r
(b) r/x (d) y/x
θ
x
φ
y
125
Problem 355. In the figure at right, tan θ = ? (a) 4/3 (c) 4/5 (e) None of these
5
(b) 3/5 *(d) 3/4
θ
4
Problem 356. In the figure at right, sin φ = ? 5
(b) 4/3 *(d) 4/5
4
5
5/3 4/3
4
5
(b) 4/5 *(d) 3/5
4
5
3/5 5/4
φ
3
θ
4
Problem 360. In the figure at right, cos φ = ? 5
(b) 5/4 *(d) 3/5
φ
3
θ
4
Problem 361. In the figure at right, tan B = ? (a) c/b (c) b/a (e) None of these
3
θ
Problem 359. In the figure at right, tan φ = ?
(a) 5/3 (c) 4/3 (e) None of these
φ
(b) (d)
3
θ
Problem 358. In the figure at right, sin θ = ?
(a) 5/3 *(c) 4/3 (e) None of these
φ
(a) 5/4 (c) 4/3 (e) None of these
3
θ
Problem 357. In the figure at right, cos θ = ? (b) (d)
φ
*(a) 4/5 (c) 3/4 (e) None of these
3
(a) 5/4 (c) 3/5 (e) None of these
φ
a
*(b) b/c (d) a/c
C
b
B
c
126
Problem 362. In the figure at right, sin B = ? (a) c/b (c) b/c (e) None of these
a
*(b) b/a (d) a/c
C
b
Problem 363. In the figure at right, cos B = ? a
*(b) c/a (d) b/a
b
a
(b) c/b (d) a/c
B
C
b
Problem 365. In the figure at right, cos C = ? a
(b) a/b (d) c/b
B
C
b
Problem 366. In the figure at right, tan C = ? a
(b) b/c *(d) c/b
B
c
C
b
Problem 367. In the figure at right, cos β = ?
v
(b) w/u (d) v/u
β
w
α
u
Problem 368. In the figure at right, tan β = ? (a) v/w (c) u/v (e) None of these
c
(a) u/v *(c) w/v (e) None of these
c
(a) a/b (c) c/a (e) None of these
c
C
Problem 364. In the figure at right, sin C = ?
(a) a/c *(c) b/a (e) None of these
B
(a) a/b *(c) c/a (e) None of these
c
(a) b/c (c) a/b (e) None of these
B
v
*(b) u/w (d) w/v
α
u
β
w
127
Problem 369. In the figure at right, sin α = ? (a) u/w *(c) w/v (e) None of these
v
(b) v/w (d) w/u
α
u
Problem 370. In the figure at right, sin β = ?
v
(b) w/v (d) v/w
u
v
*(b) u/v (d) w/v
β
w
α
u
Problem 372. In the figure at right, tan α = ?
v
(b) v/w (d) v/u
β
w
4.2.2
w
α
Problem 371. In the figure at right, cos α = ?
*(a) w/u (c) u/v (e) None of these
β
(a) v/w (c) v/u (e) None of these
w
(a) w/u *(c) u/v (e) None of these
β
α
u
Basic trig functions: identifying
Problem 373. In the figure at right, y/x = ? (a) cos θ (c) sin θ (e) None of these
r
(b) 1/ cos θ *(d) tan θ
θ
x
Problem 374. In the figure at right, y/r = ? (a) 1/ sin θ (c) 1/ tan θ (e) None of these
y
r
*(b) sin θ (d) cos θ
θ
x
y
128
Problem 375. In the figure at right, x/r = ? (a) 1/ tan θ (c) sin θ (e) None of these
r
(b) 1/ cos θ *(d) cos θ
θ
x
Problem 376. In the figure at right, r/x = ? *(a) (c) (e)
1/ cos θ sin θ None of these
(b) (d)
r
1/ sin θ 1/ tan θ
θ
x
r
*(b) 1/ sin θ (d) 1/ tan θ
θ
x
r
(b) 1/ sin θ (d) sin θ
θ
x
r
(b) 1/ cos φ *(d) cos φ
y
x
r
*(b) 1/ sin φ (d) sin φ
φ
y
x
Problem 381. In the figure at right, x/y = ? (a) 1/ tan φ *(c) tan φ (e) None of these
φ
Problem 380. In the figure at right, r/x = ? (a) 1/ cos φ (c) tan φ (e) None of these
y
Problem 379. In the figure at right, y/r = ? (a) 1/ sin φ (c) tan φ (e) None of these
y
Problem 378. In the figure at right, x/y = ? *(a) 1/ tan θ (c) 1/ cos θ (e) None of these
y
Problem 377. In the figure at right, r/y = ? (a) sin θ (c) tan θ (e) None of these
y
r
(b) 1/ sin φ (d) sin φ
x
φ
y
129
Problem 382. In the figure at right, x/r = ? (a) 1/ tan φ (c) cos φ (e) None of these
r
*(b) sin φ (d) 1/ cos φ
x
r
(b) sin φ *(d) 1/ cos φ
x
r
(b) cos φ *(d) 1/ tan φ
1/ cos θ 1/ tan θ None of these
x
5
*(b) tan θ (d) sin θ
3
θ
4
5
*(b) 1/ tan θ (d) cos θ
φ
3
(b) (d)
φ
θ
4
5
cos θ sin θ
φ
3
θ
4
Problem 388. In the figure at right, 3/5 = ? (a) tan θ (c) 1/ sin θ (e) None of these
y
Problem 387. In the figure at right, 5/4 = ? *(a) (c) (e)
φ
Problem 386. In the figure at right, 4/3 = ? (a) sin θ (c) 1/ cos θ (e) None of these
y
Problem 385. In the figure at right, 3/4 = ? (a) 1/ cos θ (c) 1/ sin θ (e) None of these
φ
Problem 384. In the figure at right, y/x = ? (a) 1/ cos φ (c) 1/ sin φ (e) None of these
y
Problem 383. In the figure at right, r/y = ? (a) tan φ (c) 1/ tan φ (e) None of these
φ
5
(b) cos θ *(d) sin θ
θ
4
φ
3
130
Problem 389. In the figure at right, 5/3 = ? (a) sin θ (c) 1/ cos θ (e) None of these
5
(b) 1/ tan θ *(d) 1/ sin θ
cos θ 1/ tan θ None of these
θ
4
5
(b) 1/ sin θ (d) 1/ cos θ
4
5
*(b) tan φ (d) sin φ
4
5
(b) 1/ cos φ (d) tan φ
φ
3
θ
4
5
(b) tan φ *(d) 1/ cos φ
φ
3
θ
4
5
(b) 1/ cos φ (d) tan φ
φ
3
θ
4
Problem 395. In the figure at right, 4/5 = ? (a) tan φ (c) cos φ (e) None of these
3
θ
Problem 394. In the figure at right, 5/4 = ? (a) 1/ tan φ *(c) 1/ sin φ (e) None of these
φ
Problem 393. In the figure at right, 5/3 = ? (a) cos φ (c) 1/ tan φ (e) None of these
3
θ
Problem 392. In the figure at right, 3/5 = ? (a) sin φ *(c) cos φ (e) None of these
φ
Problem 391. In the figure at right, 4/3 = ? (a) 1/ cos φ (c) cos φ (e) None of these
3
Problem 390. In the figure at right, 4/5 = ? *(a) (c) (e)
φ
5
(b) 1/ sin φ *(d) sin φ
θ
4
φ
3
131
Problem 396. In the figure at right, 3/4 = ? *(a) (c) (e)
1/ tan φ 1/ cos φ None of these
(b) (d)
5
tan φ 1/ sin φ
(b) (d)
1/ sin B cos B None of these
4
c
tan A 1/ cos A
tan A 1/ cos A None of these
b
b
c
cos A 1/ sin A
B
a
A
b
c
(b) sin B *(d) 1/ cos B
B
a
A
b
c
*(b) sin B (d) 1/ sin B
B
a
A
b
Problem 402. In the figure at right, b/a = ? (a) 1/ tan B (c) 1/ sin B (e) None of these
a
A
Problem 401. In the figure at right, b/c = ? (a) tan B (c) 1/ cos B (e) None of these
B
Problem 400. In the figure at right, c/a = ? (a) 1/ tan B (c) 1/ sin B (e) None of these
a
A
c
(b) 1/ tan B (d) sin B
(b) (d)
B
Problem 399. In the figure at right, a/b = ? *(a) (c) (e)
3
θ
Problem 398. In the figure at right, c/b = ? *(a) (c) (e)
φ
Problem 397. In the figure at right, a/c = ? *(a) sin A (c) cos A (e) None of these
c
*(b) tan B (d) cos B
A
b
B
a
132
Problem 403. In the figure at right, u/v = ? (a) 1/ tan α (c) 1/ cos α (e) None of these
u
(b) sin α *(d) 1/ sin α
α
w
u
*(b) 1/ tan α (d) 1/ sin α
w
u
*(b) 1/ tan β (d) cos β
1/ sin β cos β None of these
v
α
w
u
*(b) sin β (d) 1/ cos β
β
v
α
w
u
(b) 1/ tan β (d) tan β
β
v
α
w
Problem 408. In the figure at right, v/u = ? (a) 1/ sin α (c) 1/ tan α (e) None of these
β
Problem 407. In the figure at right, u/w = ? *(a) (c) (e)
v
α
Problem 406. In the figure at right, w/u = ? (a) cos β (c) tan β (e) None of these
β
Problem 405. In the figure at right, v/w = ? (a) sin β (c) 1/ sin β (e) None of these
v
Problem 404. In the figure at right, w/v = ? (a) cos α (c) 1/ cos α (e) None of these
β
u
(b) tan α *(d) sin α
α
w
β
v
133 4.2.3
Basic trig functions: calculator; radians
Problem 409. Use a calculator or equivalent to find sin(0.35 rad). Round your answer to three decimal places. *(a) (c) (e)
0.343 2.603 None of these
(b) (d)
1.071 0.359
Problem 410. Use a calculator or equivalent to find cos(1.08 rad). Round your answer to three decimal places. (a) 1.871 (c) 0.815 (e) None of these
(b) *(d)
0.534 0.471
Problem 411. Use a calculator or equivalent to find tan(0.89 rad). Round your answer to three decimal places. (a) 1.589 (c) 1.339 (e) None of these
(b) *(d)
0.629 1.235
Problem 412. Use a calculator or equivalent to find sin(1.21 rad). Round your answer to three decimal places. (a) 0.353 (c) 0.211 (e) None of these
(b) *(d)
2.833 0.936
Problem 413. Use a calculator or equivalent to find cos(0.57 rad). Round your answer to three decimal places. *(a) (c) (e)
0.842 0.322 None of these
(b) (d)
1.853 1.188
Problem 414. Use a calculator or equivalent to find tan(1.39 rad). Round your answer to three decimal places. (a) 0.892 (c) 0.507 (e) None of these
*(b) (d)
5.471 1.971
134 Problem 415. Use a calculator or equivalent to find sin(0.74 rad). Round your answer to three decimal places. *(a) (c) (e)
0.674 1.354 None of these
(b) (d)
0.013 0.913
Problem 416. Use a calculator or equivalent to find cos(0.88 rad). Round your answer to three decimal places. (a) 1.569 *(c) 0.637 (e) None of these
(b) 0.988 (d) 0.771
Problem 417. Use a calculator or equivalent to find tan(0.97 rad). Round your answer to three decimal places. (a) 0.565 *(c) 1.459 (e) None of these
(b) 0.685 (d) 1.211
Problem 418. Use a calculator or equivalent to find sin(0.18 rad). Round your answer to three decimal places. *(a) (c) (e)
0.179 1.288 None of these
(b) (d)
5.495 0.984
Problem 419. Use a calculator or equivalent to find cos(0.33 rad). Round your answer to three decimal places. (a) 0.058 (c) 0.998 (e) None of these
*(b) (d)
0.946 3.086
Problem 420. Use a calculator or equivalent to find tan(0.41 rad). Round your answer to three decimal places. (a) 0.399 *(c) 0.435 (e) None of these
(b) 0.071 (d) 0.998
135 4.2.4
Basic trig functions: calculator; degrees
Problem 421. Use a calculator or equivalent to find sin 37◦ . Round your answer to three decimal places. (a) 0.754 *(c) 0.602 (e) None of these
(b) 0.644 (d) 1.327
Problem 422. Use a calculator or equivalent to find cos 81◦ . Round your answer to three decimal places. (a) 1.233 *(c) 0.156 (e) None of these
(b) 0.630 (d) 0.158
Problem 423. Use a calculator or equivalent to find tan 52◦ . Round your answer to three decimal places. (a) 0.163 (c) 0.781 (e) None of these
*(b) (d)
1.280 0.616
Problem 424. Use a calculator or equivalent to find sin 77◦ . Round your answer to three decimal places. (a) 1.026 (c) 0.331 (e) None of these
(b) *(d)
0.031 0.974
Problem 425. Use a calculator or equivalent to find cos 29◦ . Round your answer to three decimal places. *(a) (c) (e)
0.875 0.263 None of these
(b) (d)
1.127 1.143
Problem 426. Use a calculator or equivalent to find tan 23◦ . Round your answer to three decimal places. (a) 0.846 *(c) 0.424 (e) None of these
(b) 0.391 (d) 1.877
136 Problem 427. Use a calculator or equivalent to find sin 41◦ . Round your answer to three decimal places. *(a) (c) (e)
0.656 1.524 None of these
(b) (d)
0.869 0.159
Problem 428. Use a calculator or equivalent to find cos 52◦ . Round your answer to three decimal places. (a) 0.788 (c) 0.624 (e) None of these
*(b) (d)
0.616 1.269
Problem 429. Use a calculator or equivalent to find tan 69◦ . Round your answer to three decimal places. (a) 2.790 (c) 1.071 (e) None of these
(b) *(d)
0.654 2.605
Problem 430. Use a calculator or equivalent to find sin 16◦ . Round your answer to three decimal places. (a) 0.958 *(c) 0.276 (e) None of these
(b) 0.288 (d) 0.301
Problem 431. Use a calculator or equivalent to find cos 43◦ . Round your answer to three decimal places. (a) 0.367 (c) 0.498 (e) None of these
(b) *(d)
0.555 0.731
Problem 432. Use a calculator or equivalent to find tan 41◦ . Round your answer to three decimal places. (a) 6.304 (c) 0.987 (e) None of these
(b) *(d)
0.656 0.869
137 4.2.5
Using basic trig functions: formulas
Problem 433. In the figure at right, y = ? (a) r cos θ (c) r/ sin θ (e) None of these
*(b) r sin θ (d) r/ tan θ
r
y
Solution:
θ
y/r = sin θ; so y = r sin θ.
Problem 434. In the figure at right, x = ? (a) r sin θ (c) r/ cos θ (e) None of these
*(b) r cos θ (d) r tan θ
r
Solution:
θ
x
x/r = cos θ; so x = r cos θ.
Problem 435. In the figure at right, x = ? (a) y cos θ (c) y sin θ (e) None of these
(b) y tan θ *(d) y/ tan θ
Solution:
y
θ
x
y/x = tan θ; so x = y/ tan θ.
Problem 436. In the figure at right, r = ? (a) x tan θ (c) x sin θ (e) None of these
(b) x/ sin θ *(d) x/ cos θ
r
Solution:
θ
x
x/r = cos θ; so r = x/ cos θ
Problem 437. In the figure at right, r = ? *(a) y/ sin θ (c) y tan θ (e) None of these
(b) y/ cos θ (d) y/ tan θ
r
y
Solution:
θ
y/r = sin θ; so r = y/ sin θ.
Problem 438. In the figure at right, y = ? (a) x cos θ (c) x sin θ (e) None of these
*(b) x tan θ (d) x/ sin θ
Solution:
y/x = tan θ; so y = x tan θ.
θ
x
y
138 Problem 439. In the figure at right, s = ? (a) 5/ cos θ *(c) 5 sin θ (e) None of these
(b) 5/ tan θ (d) 5/ sin θ
5
s
Solution:
θ
s/5 = sin θ; so s = 5 sin θ.
Problem 440. In the figure at right, s = ? *(a) (c) (e)
3/ sin θ 3 tan θ None of these
(b) 3/ tan θ (d) 3 cos θ
s
3
Solution:
θ
3/s = sin θ; so s = 3/ sin θ.
Problem 441. In the figure at right, t = ? (a) 13 tan α (c) (cos α)/13 (e) None of these
(b) 13/ tan α *(d) 13/ cos α
t
Solution:
α
13
13/t = cos α; so t = 13/ cos α.
Problem 442. In the figure at right, t = ? (a) 8 sin α (c) 8/ tan α (e) None of these
(b) 8/ cos α *(d) 8 cos α
8
Solution:
α
t
t/8 = cos α; so t = 8 cos α.
Problem 443. In the figure at right, u = ? (a) (cos φ)/9 (c) 9 tan φ (e) None of these
*(b) 9/ tan φ (d) (tan φ)/9
φ
Solution:
9
u
9/u = tan φ; so u = 9/ tan φ.
Problem 444. In the figure at right, u = ? (a) (cos φ)/5 *(c) 5 tan φ (e) None of these
(b) 5/ sin φ (d) 5/ cos φ
φ
Solution:
u
5
u/5 = tan φ; so u = 5 tan φ.
Problem 445. In the figure at right, u = ? (a) 5 tan φ (c) 5/ tan φ (e) None of these
(b) (sin φ)/5 *(d) 5/ cos φ
u
Solution:
5/u = cos φ; so u = 5/ cos φ.
φ
5
139 Problem 446. In the figure at right, s = ? *(a) (c) (e)
9 cos φ (sin φ)/9 None of these
(b) 9/ cos φ (d) 9/ sin φ
9
φ
s
Solution:
s/9 = cos φ; so s = 9 cos φ.
Problem 447. In the figure at right, t = ? (a) (sin θ)/3 (c) 3 cos θ (e) None of these
*(b) 3/ tan θ (d) (tan θ)/3
θ
t
Solution:
3
3/t = tan θ; so t = 3/ tan θ.
Problem 448. In the figure at right, u = ? *(a) (c) (e)
8 sin α 8/ sin α None of these
(b) 8/ cos α (d) (tan α)/8
8
α
Solution:
u
u/8 = sin α; so u = 8 sin α.
Problem 449. In the figure at right, s = ? (a) (cos θ)/3 (c) (sin θ)/3 (e) None of these
*(b) 3 tan θ (d) 3 cos θ
3
φ
Solution:
θ
s
s/3 = tan θ; so s = 3 tan θ.
Problem 450. In the figure at right, u = ? (a) (cos θ)/13 *(c) 13/ sin θ (e) None of these
(b) (tan θ)/13 (d) 13/ cos θ
u
θ
Solution: 4.2.6
13
13/u = sin θ; so u = 13/ sin θ.
Using basic trig functions: calculator
Problem 451. In the figure at right, use a calculator or equivalent to find s. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 11.02 *(c) 3.27 (e) None of these Solution:
6
(b) 2.62 (d) 5.40
s/6 = cos 57◦ ; so s = 6 cos 57◦ = 3.27
◦ 57
s
140 Problem 452. In the figure at right, use a calculator or equivalent to find t. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 4.23 (c) 4.60 (e) None of these Solution: 11.78.
*(b) 11.78 (d) 12.80 ◦
23◦
t
◦
5/t = tan 23 ; so t = 5/ tan 23 =
Problem 453. In the figure at right, use a calculator or equivalent to find w. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 6.79 (c) 4.23 (e) None of these Solution:
8
*(b) 6.63 (d) 5.40
34◦
w
w/8 = cos 34◦ ; so w = 8 cos 34◦ = 6.63.
(a) 4.47 *(c) 7.79 (e) None of these
u
(b) 14.31 (d) 4.23
◦ 64
7
7/u = sin 64◦ ; so u = 7/ sin 64◦ = 7.79
Problem 455. In the figure at right, use a calculator or equivalent to find w. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 4.23 (c) 4.60 (e) None of these Solution: 10.24
Problem 454. In the figure at right, use a calculator or equivalent to find u. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.)
Solution:
5
◦ 52
*(b) 10.24 (d) 12.80
w/8 = tan 52◦ ; so w = 8 tan 52◦ =
w
8
141 Problem 456. In the figure at right, use a calculator or equivalent to find s. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 1.87 (c) 2.49 (e) None of these Solution:
s
*(b) 3.62 (d) 4.02
34◦
3
3/s = cos 34◦ ; so s = 3/ cos 34◦ = 3.62
Problem 457. In the figure at right, use a calculator or equivalent to find s. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 8.83 (c) 17.16 (e) None of these Solution:
8
*(b) 7.25 (d) 3.38
s
s/8 = sin 65◦ ; so s = 8 sin 65◦ = 7.25.
(a) 20.02 (c) 8.44 (e) None of these
*(b) (d)
13
7.08 10.90
a
33◦
a/13 = sin 33◦ ; so a = 13 sin 33◦ = 7.08.
Problem 459. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 2.67 (c) 3.06 (e) None of these Solution:
◦ 65
Problem 458. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.)
Solution:
a
*(b) 6.22 (d) 2.98
4/a = sin 40◦ ; so a = 4/ sin 40◦ = 6.22.
40◦
4
142 Problem 460. In the figure at right, use a calculator or equivalent to find t. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 11.12 *(c) 6.54 (e) None of these Solution:
(b) 15.31 (d) 5.29
36◦
9
t/9 = tan 36◦ ; so t = 9 tan 36◦ = 6.54.
Problem 461. In the figure at right, use a calculator or equivalent to find w. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 5.75 (c) 3.77 (e) None of these Solution:
5
(b) 4.77 *(d) 3.28
w/5 = cos 49◦ ; so w = 5 cos 49◦ = 3.28
◦ 60
(b) 4.16 (d) 15.01
◦
a
13
◦
13/a = tan 60 ; so a = 13/ tan 60 =
Problem 463. In the figure at right, use a calculator or equivalent to find t. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) 14.95 17.23 None of these
Solution: 14.95
w
(a) 11.26 *(c) 7.51 (e) None of these
*(a) (c) (e)
◦ 49
Problem 462. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.)
Solution: 7.51
t
(b) (d)
◦ 49
11.25 9.81
t/13 = tan 49◦ ; so t = 13 tan 49◦ =
t
13
143 Problem 464. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) (a) 5.60 (c) 12.59 (e) None of these Solution:
a
*(b) 13.95 (d) 9.77
8/a = cos 55◦ ; so a = 8/ cos 55◦ = 13.95
(a) 7.64 *(c) 1.92 (e) None of these
21◦
5
t/5 = tan 21◦ ; so t = 5 tan 21◦ = 1.92
(a) 1.75 (c) 4.42 (e) None of these
6
(b) 1.83 *(d) 5.74
s
s/6 = sin 73◦ ; so s = 6 sin 73◦ = 5.74
3.76 7.04 None of these
Solution:
◦ 73
Problem 467. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
t
(b) 4.18 (d) 2.36
Problem 466. In the figure at right, use a calculator or equivalent to find s. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.)
Solution:
8
Problem 465. In the figure at right, use a calculator or equivalent to find t. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.)
Solution:
◦ 55
(b) (d)
8
4.25 5.05
a/8 = sin 28◦ ; so a = 8 sin 28◦ = 3.76
28◦
a
144 Problem 468. In the figure at right, use a calculator or equivalent to find a. Round your answer to two decimal places. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
4.60 4.00 None of these
Solution: 4.2.7
(b) (d)
Problem 469. If θ = 320◦ , which quadrant is it in? (b) Quadrant II *(d) Quadrant IV
270◦ < θ < 360◦ ; so θ is in quadrant IV.
Problem 470. If θ = 109◦ , which quadrant is it in? (a) Quadrant I (c) Quadrant III (e) None of these Solution:
*(b) Quadrant II (d) Quadrant IV
90◦ < θ < 180◦ ; so θ is in quadrant II.
Problem 471. If θ = 194◦ , which quadrant is it in? (a) Quadrant I *(c) Quadrant III (e) None of these Solution:
(b) Quadrant II (d) Quadrant IV
180◦ < θ < 270◦ ; so θ is in quadrant III.
Problem 472. If θ = 32◦ , which quadrant is it in? *(a) (c) (e)
Quadrant I Quadrant III None of these
Solution:
(b) Quadrant II (d) Quadrant IV
0◦ < θ < 90◦ ; so θ is in quadrant I.
Problem 473. If θ = 5.91 rad, which quadrant is it in? (a) Quadrant I (c) Quadrant III (e) None of these Solution:
Reference angles
Solution:
6
3.86 7.15
a/6 = cos 40◦ ; so a = 6 cos 40◦ = 4.60
(a) Quadrant I (c) Quadrant III (e) None of these
(b) Quadrant II *(d) Quadrant IV
3π/2 < θ < 2π; so θ is in quadrant IV.
40◦
a
145 Problem 474. If θ = 0.32 rad, which quadrant is it in? *(a) (c) (e)
Quadrant I Quadrant III None of these
Solution:
(b) Quadrant II (d) Quadrant IV
0 < θ < π/2; so θ is in quadrant I.
Problem 475. If θ = 2.15 rad, which quadrant is it in? (a) Quadrant I (c) Quadrant III (e) None of these Solution:
*(b) Quadrant II (d) Quadrant IV
π/2 < θ < π; so θ is in quadrant II.
Problem 476. If θ = 3.40 rad, which quadrant is it in? (a) Quadrant I *(c) Quadrant III (e) None of these Solution:
(b) Quadrant II (d) Quadrant IV
π < θ < 3π/2; so θ is in quadrant III.
Problem 477. An angle with measure θ degrees is in quadrant II. What is the reference angle θref ? (a) θ − 180◦ (c) θ − 360◦ (e) None of these
*(b) 180◦ − θ (d) 360◦ − θ
Problem 478. An angle with measure θ degrees is in quadrant IV. What is the reference angle θref ? (a) θ − 180◦ (c) θ − 360◦ (e) None of these
(b) 180◦ − θ *(d) 360◦ − θ
Problem 479. An angle with measure θ degrees is in quadrant III. What is the reference angle θref ? *(a) θ − 180◦ (c) θ − 360◦ (e) None of these
(b) 180◦ − θ (d) 360◦ − θ
Problem 480. An angle with measure θ radians is in quadrant III. What is the reference angle θref ? (a) 2π − θ (c) π − θ (e) None of these
(b) θ − 2π *(d) θ − π
146 Problem 481. An angle with measure θ radians is in quadrant II. What is the reference angle θref ? (a) 2π − θ *(c) π − θ (e) None of these
(b) θ − 2π (d) θ − π
Problem 482. An angle with measure θ radians is in quadrant IV. What is the reference angle θref ? *(a) 2π − θ (c) π−θ (e) None of these
(b) θ − 2π (d) θ − π
Problem 483. If θ = 311◦ , what is θref ? (a) 31◦ (c) 41◦ (e) None of these Solution:
(b) 39◦ *(d) 49◦
270◦ < θ < 360◦ ; so θ is in quadrant IV; so θref = 360◦ − θ = 49◦ .
Problem 484. If θ = 122◦ , what is θref ? (a) 32◦ *(c) 58◦ (e) None of these Solution:
(b) (d)
42◦ 68◦
90◦ < θ < 180◦ ; so θ is in quadrant II; so θref = 180◦ − θ = 58◦ .
Problem 485. If θ = 195◦ , what is θref ? *(a) (c) (e)
15◦ 65◦ None of these
Solution:
(b) (d)
25◦ 75◦
180◦ < θ < 270◦ ; so θ is in quadrant III; so θref = θ − 180◦ = 15◦ .
Problem 486. If θ = 272◦ , what is θref ? (a) 2◦ (c) 72◦ (e) None of these Solution:
(b) *(d)
28◦ 88◦
270◦ < θ < 360◦ ; so θ is in quadrant IV; so θref = 360◦ − θ = 88◦ .
Problem 487. If θ = 163◦ , what is θref ? *(a) (c) (e)
17◦ 63◦ None of these
Solution:
(b) 27◦ (d) 73◦
90◦ < θ < 180◦ ; so θ is in quadrant II; so θref = 180◦ − θ = 17◦ .
147 Problem 488. If θ = 210◦ , what is θref ? (a) 10◦ (c) 40◦ (e) None of these Solution:
*(b) (d)
30◦ 60◦
180◦ < θ < 270◦ ; so θ is in quadrant III; so θref = θ − 180◦ = 30◦ .
Problem 489. If θ is in the first quadrant, which of the following is true? (a) sin θ ≤ 0 and cos θ ≤ 0 (c) sin θ ≥ 0 and cos θ ≤ 0 (e) None of these
(b) *(d)
sin θ ≤ 0 and cos θ ≥ 0 sin θ ≥ 0 and cos θ ≥ 0
Problem 490. If θ is in the second quadrant, which of the following is true? (a) sin θ ≤ 0 and cos θ ≤ 0 *(c) sin θ ≥ 0 and cos θ ≤ 0 (e) None of these
(b) (d)
sin θ ≤ 0 and cos θ ≥ 0 sin θ ≥ 0 and cos θ ≥ 0
Problem 491. If θ is in the third quadrant, which of the following is true? *(a) (c) (e)
sin θ ≤ 0 and cos θ ≤ 0 sin θ ≥ 0 and cos θ ≤ 0 None of these
(b) sin θ ≤ 0 and cos θ ≥ 0 (d) sin θ ≥ 0 and cos θ ≥ 0
Problem 492. If θ is in the fourth quadrant, which of the following is true? (a) sin θ ≤ 0 and cos θ ≤ 0 (c) sin θ ≥ 0 and cos θ ≤ 0 (e) None of these
*(b) (d)
sin θ ≤ 0 and cos θ ≥ 0 sin θ ≥ 0 and cos θ ≥ 0
Problem 493. An angle with measure θ degrees is in the third quadrant. What is sin θ? (a) sin(θ − 180◦ ) *(c) − sin(θ − 180◦ ) (e) None of these
(b) sin(360◦ − θ) (d) − sin(θ − 360◦ )
Solution: Since θ is in the third quadrant, the reference angle is θref = θ − 180◦ ; and sin θ ≤ 0. Hence sin θ = − sin θref = − sin(θ − 180◦ ). Problem 494. An angle with measure θ degrees is in the fourth quadrant. What is sin θ? (a) sin(θ − 180◦ ) (b) sin(360◦ − θ) (c) sin(θ + 180◦ ) *(d) − sin(360◦ − θ) (e) None of these Solution: Since θ is in the fourth quadrant, the reference angle is θref = 360◦ − θ; and sin θ ≤ 0. Hence sin θ = − sin θref = − sin(360◦ − θ).
148 Problem 495. An angle with measure θ degrees is in the second quadrant. What is sin θ? *(a) sin(180◦ − θ) (b) sin(180◦ + θ) (c) sin(360◦ − θ) (d) − sin(360◦ + θ) (e) None of these Solution: Since θ is in the second quadrant, the reference angle is θref = 180◦ − θ; and sin θ ≥ 0. Hence sin θ = sin θref = sin(180◦ − θ). Problem 496. An angle with measure θ degrees is in the second quadrant. What is cos θ? (a) cos(180◦ − θ) *(b) − cos(180◦ − θ) (c) cos(θ − 180◦ ) (d) − cos(360◦ + θ) (e) None of these Solution: Since θ is in the second quadrant, the reference angle is θref = 180◦ − θ; and cos θ ≤ 0. Hence cos θ = − cos θref = − cos(180◦ − θ). Problem 497. An angle with measure θ degrees is in the third quadrant. What is cos θ? (a) cos(θ − 180◦ ) *(c) − cos(θ − 180◦ ) (e) None of these
(b) cos(180◦ − θ) (d) cos(180◦ + θ)
Solution: Since θ is in the third quadrant, the reference angle is θref = θ − 180◦ ; and cos θ ≤ 0. Hence cos θ = − cos θref = − cos(θ − 180◦ ). Problem 498. An angle with measure θ degrees is in the fourth quadrant. What is cos θ? (a) cos(θ − 180◦ ) (b) cos(180◦ − θ) ◦ (c) cos(θ + 180 ) *(d) cos(360◦ − θ) (e) None of these Solution: Since θ is in the fourth quadrant, the reference angle is θref = 360◦ − θ; and cos θ ≥ 0. Hence cos θ = cos θref = cos(360◦ − θ). 4.2.8
Arc functions: definition
Problem 499. Complete the statement: If θ is an acute angle and sin θ = 0.3, then (a) θ = sin 0.3 *(c) θ = arcsin 0.3 (e) None of these
(b) θ = cos 0.3 (d) θ = 0.3/ sin
Problem 500. Complete the statement: If θ is an acute angle and cos θ = 0.4, then (a) θ = 1/ cos 0.4 *(c) θ = cos−1 0.4 (e) None of these
(b) (d)
θ = cos 0.4 θ = sin 0.4
149 Problem 501. Complete the statement: If θ is an acute angle and tan θ = 3.1, then *(a) θ = tan−1 3.1 (c) θ = 1/ tan 3.1 (e) None of these
(b) (d)
θ = 3.1/ tan θ = tan(1/3.1)
Problem 502. Complete the statement: If θ is an acute angle and sin θ = 0.9, then *(a) θ = sin−1 0.9 (c) θ = cos 0.9 (e) None of these
(b) (d)
θ = sin(1/0.9) θ = 1/ sin 0.9
Problem 503. Complete the statement: If θ is an acute angle and cos θ = 0.2, then *(a) θ = arccos 0.2 (c) θ = sin 0.2 (e) None of these
(b) θ = cos 0.2 (d) θ = cos(5)
Problem 504. Complete the statement: If θ is an acute angle and tan θ = 0.5, then *(a) θ = arctan 0.5 (c) θ = 1/ tan 0.5 (e) None of these
(b) θ = tan 2 (d) θ = 0.5/ tan
Problem 505. Complete the statement: If θ = arcsin 0.5, then (a) sin θ = 2 (c) θ = sin 2 (e) None of these
*(b) sin θ = 0.5 (d) cos θ = 2
Problem 506. Complete the statement: If θ = arccos 0.2, then (a) cos θ = 5 *(c) cos θ = 0.2 (e) None of these
(b) (d)
θ = cos 5 θ = cos 0.2
Problem 507. Complete the statement: If θ = arctan 2, then (a) tan θ = 0.5 (c) θ = 2/ tan (e) None of these
(b) θ = tan 0.5 *(d) tan θ = 2
Problem 508. Complete the statement: If θ = sin−1 0.4, then (a) sin θ = 2.5 *(c) sin θ = 0.4 (e) None of these
(b) 1/ sin θ = 0.4 (d) θ = 2.5
150 Problem 509. Complete the statement: If θ = cos−1 (1/3), then (a) cos θ = 3 *(c) cos θ = 1/3 (e) None of these
(b) 3θ = cos (d) sin θ = 3
Problem 510. Complete the statement: If θ = tan−1 (2/3), then (a) tan θ = 3/2 *(c) tan θ = 2/3 (e) None of these
4.2.9
(b) (d)
θ = 1/ tan(2/3) θ = tan(3/2)
Arc functions: definition; diagrams
Problem 511. Which statement is true of the figure at right? (a) θ = sin(5/3) (c) θ = arcsin(5/3) (e) None of these
5
(b) θ = sin(3/5) *(d) θ = arcsin(3/5)
4
5
(b) θ = arccos(5/4) *(d) θ = arccos(4/5)
φ
3
θ
4
5
(b) θ = tan(4/3) (d) θ = arctan(4/3)
φ
3
θ
4
Problem 514. Which statement is true of the figure at right? *(a) φ = sin−1 (4/5) (c) φ = sin−1 (5/4) (e) None of these
3
θ
Problem 513. Which statement is true of the figure at right? (a) θ = tan(3/4) *(c) θ = arctan(3/4) (e) None of these
φ
Problem 512. Which statement is true of the figure at right? (a) θ = cos(5/4) (c) θ = cos(4/5) (e) None of these
5
(b) φ = sin(4/5) (d) φ = sin(5/4)
θ
4
φ
3
151 Problem 515. Which statement is true of the figure at right? (a) φ = cos(4/5) (c) φ = cos(3/5) (e) None of these
5
(b) φ = cos−1 (4/5) *(d) φ = cos−1 (3/5)
4.2.10
φ
3
θ
4
Problem 516. Which statement is true of the figure at right? (a) φ = tan−1 (3/4) (c) φ = tan(3/4) (e) None of these
5
*(b) φ = tan−1 (4/3) (d) φ = tan(4/3)
φ
3
θ
4
Arc functions: calculator; radians
Problem 517. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 0.41. Round your answer to the nearest 0.01 radian. (a) 0.39 rad *(c) 1.15 rad (e) None of these Solution:
(b) 36.83 rad (d) No such angle
θ = cos−1 0.41 = 1.15 rad
Problem 518. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 0.83. Round your answer to the nearest 0.01 radian. (a) 1.48 rad *(c) 0.98 rad (e) None of these Solution:
(b) 0.59 rad (d) No such angle
θ = sin−1 0.83 = 0.98 rad
Problem 519. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 1.03. Round your answer to the nearest 0.01 radian. (a) 0.86 rad (c) 0.51 rad (e) None of these Solution:
*(b) (d)
0.80 rad No such angle
θ = tan−1 1.03 = 0.80 rad
Problem 520. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 0.77. Round your answer to the nearest 0.01 radian. (a) 0.65 rad (c) 0.35 rad (e) None of these Solution:
*(b) 0.69 rad (d) No such angle
θ = cos−1 0.77 = 0.69 rad
152 Problem 521. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 1.21. Round your answer to the nearest 0.01 radian. (a) 0.39 rad (c) 1.15 rad (e) None of these Solution:
(b) 36.83 rad *(d) No such angle
No such angle; sin θ > 1 is impossible.
Problem 522. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 0.22. Round your answer to the nearest 0.01 radian. *(a) (c) (e)
0.22 rad 18.45 rad None of these
Solution:
(b) 0.29 rad (d) No such angle
θ = tan−1 0.22 = 0.22 rad
Problem 523. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 1.03. Round your answer to the nearest 0.01 radian. (a) 1.17 rad (c) 0.86 rad (e) None of these Solution:
(b) 1.67 rad *(d) No such angle
No such angle; cos θ > 1 is impossible.
Problem 524. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 0.55. Round your answer to the nearest 0.01 radian. (a) 1.91 rad (c) 0.52 rad (e) None of these Solution:
*(b) 0.58 rad (d) No such angle
θ = sin−1 0.55 = 0.58 rad
Problem 525. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 1.71. Round your answer to the nearest 0.01 radian. *(a) (c) (e)
1.04 rad 89.48 rad None of these
Solution:
(b) 0.14 rad (d) No such angle
θ = tan−1 1.71 = 1.04 rad
153 4.2.11
Arc function: calculator; degrees
Problem 526. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 0.63. Round your answer to the nearest degree. (a) 46◦ (c) 32◦ (e) None of these Solution:
*(b) 39◦ (d) No such angle
θ = sin−1 0.63 = 39◦
Problem 527. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 1.19. Round your answer to the nearest degree. (a) 21◦ (c) 53◦ (e) None of these Solution:
(b) 50◦ *(d) No such angle
No such angle; cos θ > 1 is impossible.
Problem 528. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 1.30. Round your answer to the nearest degree. *(a) 52◦ (c) 15◦ (e) None of these Solution:
(b) (d)
34◦ No such angle
θ = tan−1 1.30 = 52◦
Problem 529. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 0.81. Round your answer to the nearest degree. *(a) 36◦ (c) 77◦ (e) None of these Solution:
(b) 39◦ (d) No such angle
θ = cos−1 0.81 = 36◦
Problem 530. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 0.46. Round your answer to the nearest degree. (a) 39◦ (c) 1◦ (e) None of these Solution:
*(b) 25◦ (d) No such angle
θ = tan−1 0.46 = 25◦
154 Problem 531. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 0.32. Round your answer to the nearest degree. (a) 3◦ (c) 27◦ (e) None of these Solution:
*(b) 19◦ (d) No such angle
θ = sin−1 0.32 = 19◦
Problem 532. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and sin θ = 0.67. Round your answer to the nearest degree. *(a) (c) (e)
42◦ 45◦ None of these
Solution:
(b) (d)
48◦ No such angle
θ = sin−1 0.67 = 42◦
Problem 533. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and cos θ = 0.96. Round your answer to the nearest degree. (a) 1◦ *(c) 16◦ (e) None of these Solution:
(b) (d)
2◦ No such angle
θ = cos−1 0.96 = 16◦
Problem 534. Use a calculator or equivalent to determine the value of θ if θ is an acute angle and tan θ = 0.02. Round your answer to the nearest degree. (a) 57◦ (c) 66◦ (e) None of these Solution: 4.2.12
*(b) 1◦ (d) No such angle
θ = tan−1 0.02 = 1◦
Arc functions: calculator; diagram; radians
Problem 535. In the figure at right, use a calculator or equivalent to find θ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
0.78 rad 1.53 rad None of these
Solution: rad
(b) (d)
7
1.32 rad 1.81 rad
cos θ = 5/7; so θ = cos−1 (5/7) = 0.78
θ
5
155 Problem 536. In the figure at right, use a calculator or equivalent to find θ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
0.41 rad 1.57 rad None of these
Solution: rad
(b) (d)
10
1.16 rad 0.31 rad
θ
sin θ = 4/10; so θ = sin−1 (4/10) = 0.41
Problem 537. In the figure at right, use a calculator or equivalent to find θ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) (a) 0.32 rad *(c) 0.63 rad (e) None of these Solution: 0.63 rad
(b) 0.34 rad (d) 1.34 rad
θ
11
tan θ = 8/11; so θ = tan−1 (8/11) =
(a) 0.90 rad *(c) 0.56 rad (e) None of these
(b) 0.32 rad (d) 0.68 rad
θ
8
Problem 539. In the figure at right, use a calculator or equivalent to find θ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.)
Solution: 0.68 rad
19
(b) 0.24 rad (d) 0.49 rad
sin θ = 12/19; so θ = sin−1 (12/19) =
5
tan θ = 5/8; so θ = tan−1 (5/8) = 0.56
(a) 0.91 rad *(c) 0.68 rad (e) None of these
8
Problem 538. In the figure at right, use a calculator or equivalent to find θ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.)
Solution: rad
4
θ
12
156 Problem 540. In the figure at right, use a calculator or equivalent to find φ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) (a) 0.46 rad (c) 0.66 rad (e) None of these Solution: rad
*(b) (d)
11
0.69 rad 0.48 rad
7
sin φ = 7/11; so φ = sin−1 (7/11) = 0.69
Problem 541. In the figure at right, use a calculator or equivalent to find φ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
1.03 rad 1.09 rad None of these
Solution: 1.03 rad
(b) (d)
1.07 rad 1.18 rad
10
14
(b) 0.84 rad (d) 1.13 rad
φ
9
cos φ = 9/14; so φ = cos−1 (9/14) =
Problem 543. In the figure at right, use a calculator or equivalent to find φ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) (a) 0.91 rad (c) 1.40 rad (e) None of these Solution: rad
6
Problem 542. In the figure at right, use a calculator or equivalent to find φ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.)
Solution: 0.87 rad
φ
tan φ = 10/6; so φ = tan−1 (10/6) =
(a) 1.04 rad *(c) 0.87 rad (e) None of these
φ
*(b) (d)
1.06 rad 0.85 rad
tan φ = 9/5; so φ = tan−1 (9/5) = 1.06
9
φ
5
157 Problem 544. In the figure at right, use a calculator or equivalent to find φ. Round your answer to 0.01 rad. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
1.42 rad 1.63 rad None of these
Solution: 1.42 rad 4.2.13
(b) (d)
20
1.29 rad 1.15 rad
φ
3
cos φ = 3/20; so φ = cos−1 (3/20) =
Arc functions: calculator; diagram; degrees
Problem 545. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 59◦ (c) 40◦ (e) None of these Solution:
5
*(b) 53◦ (d) 47◦
4
sin φ = 4/5; so φ = sin−1 (4/5) = 53◦
(a) 24◦ *(c) 49◦ (e) None of these
(b) (d)
φ
7
36◦ 57◦
8
tan φ = 8/7; so φ = tan−1 (8/7) = 49◦
Problem 547. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 52◦ *(c) 54◦ (e) None of these Solution:
φ
Problem 546. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.)
Solution:
(b) (d)
12
65◦ 67◦
cos φ = 7/12; so φ = cos−1 (7/12) = 54◦
φ
7
158 Problem 548. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 50◦ (c) 54◦ (e) None of these Solution:
9
tan φ = 9/7; so φ = tan−1 (9/7) = 52◦
(a) 60◦ *(c) 66◦ (e) None of these
(b) (d)
12
63◦ 69◦
φ
11
−1
sin φ = 11/12; so φ = sin (11/12) =
(a) 76◦ (c) 80◦ (e) None of these
5
*(b) 78◦ (d) 82◦
φ
1
cos φ = 1/5; so φ = cos−1 (1/5) = 78◦
Problem 551. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 47◦ (c) 53◦ (e) None of these Solution: 50◦
Problem 550. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.)
Solution:
7
*(b) 52◦ (d) 56◦
Problem 549. In the figure at right, use a calculator or equivalent to find φ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.)
Solution: 66◦
φ
13
*(b) 50◦ (d) 56◦
sin θ = 10/13; so θ = sin−1 (10/13) =
θ
10
159 Problem 552. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 21◦ (c) 25◦ (e) None of these Solution: 23◦
13
*(b) 23◦ (d) 27◦
θ
12
−1
cos θ = 12/13; so θ = cos (12/13) =
Problem 553. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) *(a) (c) (e)
40◦ 44◦ None of these
Solution: 40◦
(b) (d)
42◦ 46◦
θ
20
−1
tan θ = 17/20; so θ = tan (17/20) =
(a) 23◦ *(c) 27◦ (e) None of these
(b) (d)
9
25◦ 29◦
θ
8
cos θ = 8/9; so θ = cos−1 (8/9) = 27◦
Problem 555. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 6◦ (c) 10◦ (e) None of these Solution:
17
Problem 554. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.)
Solution:
*(b) 8◦ (d) 12◦
tan θ = 1/7; so θ = tan−1 (1/7) = 8◦
θ
7
1
160 Problem 556. In the figure at right, use a calculator or equivalent to find θ. Round your answer to the nearest degree. (The figure is not necessarily drawn to scale.) (a) 15◦ (c) 19◦ (e) None of these Solution: 4.2.14
7
*(b) 17◦ (d) 21◦
sin θ = 2/7; so θ = sin−1 (2/7) = 17◦
Word problems: basic trig functions
Problem 557. A tree casts a shadow 53 ft long. The sun is θ = 49◦ above the horizon. How tall is the tree? Round your answer to the nearest foot. (a) 46 ft (b) 51 ft (c) 56 ft *(d) 61 ft (e) None of these Solution: Let h be the height. Then h/53 = tan 49◦ ; so h = 53 tan 49◦ = 61 ft. Problem 558. You want to know the width of a river. You begin by standing directly across from a tree on the opposite bank. You then walk 400 ft straight downstream. From this new point, the tree is at an angle of θ = 67◦ to the upstream direction. How wide is the river? Round your answer to the nearest foot. (a) 170 ft (c) 1070 ft (e) None of these
*(b) 942 ft (d) 1413 ft
Solution: Let w be the width of the river. Then w/400 = tan 67◦ ; so w = 400 tan 67◦ = 942 ft.
θ
2
161 Problem 559. A prisoner is trying to escape by digging a tunnel under a wall 147 ft away. Unfortunately, his compass is inaccurate, and his tunnel is off course by θ = 19◦ . How long does the tunnel have to be to reach the wall? Round your answer to the nearest foot. (a) 149 ft (c) 153 ft (e) None of these
(b) 151 ft *(d) 155 ft
Solution: Let t be the length of the tunnel. Then 147/t = cos 19◦ ; so t = 147/ cos 19◦ = 155 ft. Problem 560. A highway runs directly east and west. An airplane flies across the highway in a direction θ = 34◦ north of east. What is the total distance that the airplane will fly before it is 21 miles north of the highway? Round your answer to the nearest tenth of a mile. (a) 25.3 miles (b) 29.4 miles (c) 33.5 miles *(d) 37.6 miles (e) None of these Solution: Let t be the total distance flown by the airplane. Then 21/t = sin 34◦ ; so t = 21/ sin 34◦ = 37.6 miles. Problem 561. A car is driving up a mountain road with a slope of θ = 4.5◦ . It goes 9000 ft along the road by the odometer. At the end of this time, how high is it above its starting point? Round your answer to the nearest 10 ft. (a) 680 ft (c) 740 ft (e) None of these
*(b) (d)
710 ft 770 ft
Solution: Let h be the height above the starting point. Then h/9000 = sin 4.5◦ ; so h = 9000 sin 4.5◦ = 710 ft.
162 Problem 562. From where you stand, a mountain peak has an elevation of θ = 12.5◦ . On a map, the horizontal distance between you and the peak is 4.11 miles. How high above you is the peak? Round your answer to the nearest 0.01 miles. (a) 0.83 miles (b) 0.87 miles *(c) 0.91 miles (d) 0.95 miles (e) None of these Solution: Let h be the height of the peak. Then h/4.11 = tan 12.5◦ ; so h = 4.11 tan 12.5◦ = 0.91 miles. Problem 563. You and a friend are trying to find out how high airplanes fly over his house. He phones you when a plane is directly overhead. At that moment, the airplane is at an elevation of θ = 36.2◦ , as seen from your house. If your house is 4100 ft from your friend’s house, how high is the airplane? Round your answer to the nearest 100 ft. *(a) 3000 ft (b) 3300 ft (c) 3600 ft (d) 3900 ft (e) None of these Solution: Let h be the height of the airplane. Then h/4100 = tan 36.2◦ ; so h = 4100 tan 36.2◦ = 3000 ft. Problem 564. A pole is supported by a diagonal guy wire. The wire is anchored in the ground 13.8 m from the base of the pole, and meets the ground at an angle of θ = 59◦ . How long is the wire? Round your answer to the nearest 0.1 m. (a) 25.8 m (b) 26.3 m *(c) 26.8 m (d) 27.3 m (e) None of these Solution: Let w be the length of the wire. Then 13.8/w = cos 59◦ ; so w = 13.8/ cos 59◦ = 26.8 m.
163 Problem 565. You are flying a kite in a wind strong enough to stretch the string into a straight line. The string is 71 m long, and makes an angle of θ = 51◦ with the horizontal. How high above the ground is the kite? Round your answer to the nearest meter. *(a) (c) (e)
55 m 59 m None of these
(b) (d)
57 m 61 m
Solution: Let h be the height of the kite. Then h/71 = sin 51◦ ; so h = 71 sin 51◦ = 55 m. 4.2.15
Word problems: arc functions
Problem 566. A tree is 73 ft tall. It casts a shadow 93 ft long. What is the sun’s angle of elevation θ? Round your answer to the nearest degree. *(a) (c) (e)
38◦ 42◦ None of these
(b) (d)
40◦ 44◦
Solution: 73/93 = tan θ; so θ = tan−1 (73/93) = 38◦ . Problem 567. A prisoner is trying to escape by digging a tunnel under a wall 113 ft away. Unfortunately, his compass is inaccurate, and the tunnel is off course by an angle of θ. He does not reach the wall until the tunnel is 128 ft long. What is the value of θ? Round your answer to the nearest degree. (a) 26◦ (c) 30◦ (e) None of these
*(b) 28◦ (d) 32◦
Solution: 113/128 = cos θ; so θ = cos−1 (113/128) = 28◦ .
164 Problem 568. A highway runs directly east and west. An airplane flies across the highway in a direction θ north of east. When the airplane has flown a total distance of 19 miles, it is 5 miles north of the highway. What is θ? Round your answer to the nearest degree. (a) 13◦ (c) 17◦ (e) None of these
*(b) 15◦ (d) 19◦
Solution: 5/19 = sin θ; so θ = sin−1 (5/19) = 15◦ . Problem 569. A car is driving up a mountain road with a slope of θ. It goes 12,000 ft along the road by the odometer. At the end of this time, it is 1800 ft above its starting point. What is θ? Round your answer to the nearest 0.1◦ . (a) 8.0◦ (b) 8.3◦ ◦ *(c) 8.6 (d) 8.9◦ (e) None of these Solution: 1800/12, 000 = sin θ; so θ = sin−1 (1800/12, 000) = 8.6◦ . Problem 570. A mountain peak is 7200 ft higher than your house. On the map, there is a horizontal distance of 53,000 ft from the house to the peak. What is the peak’s elevation θ as seen from the house? Round your answer to the nearest 0.1◦ . (a) 7.1◦ (b) 7.4◦ *(c) 7.7◦ (d) 8.0◦ (e) None of these Solution: 7200/53, 000 = tan θ; so θ = tan−1 (7200/53, 000) = 7.7◦ .
165 Problem 571. Your house is 3800 ft from your friend’s house. An airplane flies over the friend’s house at a height of 960 ft. When the plane is directly above your friend’s house, what is its elevation θ as seen from your house? Round your answer to the nearest degree. *(a) 14◦ (c) 16◦ (e) None of these
(b) 15◦ (d) 17◦
Solution: 960/3800 = tan θ; so θ = tan−1 (960/3800) = 14◦ . Problem 572. A pole is supported by a diagonal guy wire. The wire is 19.9 m long, and is anchored in the ground 15.2 m from the base of the pole. At what angle θ does the wire meet the ground? Round your answer to the nearest degree. (a) 36◦ *(c) 40◦ (e) None of these
(b) (d)
38◦ 42◦
Solution: 15.2/19.9 = cos θ; so θ = cos−1 (15.2/19.9) = 40◦ . Problem 573. You are flying a kite in a wind strong enough to stretch the string into a straight line. The string is 94 m long, and the kite is 75 m above the ground. What angle θ does the string make with the ground? Round your answer to the nearest degree. (a) 49◦ *(c) 53◦ (e) None of these Solution: 75/94 = sin θ; so θ = sin−1 (75/94) = 53◦ .
(b) (d)
51◦ 55◦
166
5
5.1 5.1.1
Introduction to measurement: dimensions, units, scientific notation, and significant figures Dimensions Dimension or unit?
Problem 574. What are the three fundamental dimensions? Solution:
length L, mass M , and time T .
Problem 575. What’s the difference between dimensions of the fundamental physical quantities: length, mass, time and the base units: meter, kilogram, second? Solution: Any physical quantity can be characterized by dimensions. You can think of a dimension as a measurable extent of some kind, such as length. The arbitrary magnitudes that we assign to assign to dimensions are called units. For example, in order to give a precise quantitative measurement of the length (a dimension) of a particular object, we need a standard reference length from which to compare our object’s length. Problem 576. For the given physical quantities, identify the three fundamental physical quantities: (a) mass (c) time (e) force (b) volume (d) density (f) length
Solution: quantities.
(a), (c) and (f). The others are known as secondary dimensions or derived
167 5.1.2
Dimensional consistency
In the following problems, determine which of the formulas could not be correct because it violates the consistency-of-dimensions principle discussed in class. Do not worry about the origin or application of the formulas: just focus on the issue of consistency of units. That is, based solely on consistency of units in an equation, could the given formula be correct? Definition: The Dimensionator [ ] is the “take the fundamental dimensions of” operator. Warning: Do not let the subscripts confuse you. The subscripts are only used to label variables, they are dimensionless numbers and letters and do not affect the outcome of dimensionator one bit. The dimensions of the fundamental quantities: {Length L, Mass M, Time T} are • d = distance; [d] = L • x = distance; [x] = L • R = radius; [R] = L • h = height; [h] = L • m = mass; [m] = M • t = time; [t] = T The dimensions of the derived quantities (i.e., the quantities derived from the fundamental quantities) are • r = rate (speed); [r] = L/T • v = velocity; [v] = L/T • a = acceleration; [a] = L/T 2 • A = area; [A] = L2 • V = volume; [V ] = L3 Problem 577. d = rt
168 Solution: Look at the units. d has units of (length); r has units of (length/time); and t has units of (time). In units, the formula looks like: length · (time) (length) = time Since the units of time on the right-hand side cancel, the two sides of the equation match, and the formula is dimensionally correct. In fact, this is the ubiquitous formula: distance = rate × time. Problem 578. v = x + t Solution: Look at the units. v has units of length/time ([v] = TL ); x has units of length ([x] = L); and t has units of time ([t] = T ). In units, the formula looks like: L =L+T . T Since the units on the two sides of the equation don’t match, the formula can’t be correct. In fact, every term is different! Problem 579. V = 34 πR2 Solution: Here V has units of (length)3 ; R has units of (length); and pure numbers, which don’t have units. In units, the formula looks like:
4 3
and π are
(length)3 = (length)2 Since the units on the two sides of the equation don’t match, the formula can’t be correct. Problem 580. V = πR2 h Solution: Here V has units of (length)3 ; R has units of (length); h has units of (length); and π is a pure number, which doesn’t have units. In units, the formula looks like: (length)3 = (length)2 (length) Since the units of time on the right-hand side cancel, the two sides of the equation match, and the formula is dimensionally correct. In fact, this is the formula for the volume of a cylinder. Problem 581. V = Ad Solution: Here V has units of (length)3 ; A has units of (length)2 ; and d has units of (length). In units, the formula looks like: (length)3 = (length)2 · (length) We get units of (length)3 on both sides of the equation. From the standpoint of consistency of units, this formula could be correct.
169 Problem 582. V = Ah Solution: Here V has units of (length)3 ; A has units of (length)2 ; and h has units of (length). In units, the formula looks like: (length)3 = (length)2 · (length) We get units of (length)3 on both sides of the equation. From the standpoint of consistency of units, this formula could be correct. Problem 583. V = Ax2 Solution: Here V has units of (length)3 ; A has units of (length)2 ; and x has units of (length). In units, the formula looks like: (length)3 = (length)2 · (length)2 = (length)4 Since the units on the two sides of the equation don’t match, the formula can’t be correct. Problem 584. v =
x t
Solution: Here v has units of In units, the formula looks like:
length ; time
x has units of (length); and t has units of (time).
length length = time time From the standpoint of consistency of units, this formula could be correct. Problem 585. v22 = v12 + x2 Solution: Here v2 and v1 have units of length ; x has units of (length). In units, the time formula looks like: (length)2 (length)2 = + (length)2 2 2 (time) (time) We can’t even say what the units are on the right-hand side of the equation, since we’re trying to add two quantities measured in different kinds of units. This formula doesn’t make any sense. Problem 586. v22 = v12 +
x2 t2
170 Solution: Here v2 and v1 have units of of (time). In units, the formula looks like:
length ; time
x has units of (length); and t has units
(length)2 (length)2 (length)2 = + (time)2 (time)2 (time)2 We can deal with the right-hand side, since we’re adding two quantities measured in the 2 same units. Both sides of the formula have units of (length) . From the standpoint of (time)2 consistency of units, this formula could be correct. Problem 587. Determine whether equations (i) and (ii) below are dimensionally consistent. x − x0 (ii) x = a(t + v)2 , (i) v = t where x = distance, v = velocity, t = time, and a = acceleration. *(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent (c) (i) and (ii) are both dimensionally consistent (d) (i) and (ii) are both dimensionally inconsistent Solution:
Can’t add velocity and time since they are dimensionally inconsistent.
Problem 588. Consider the formulas (i) and (ii). Are the formulas dimensionally consistent? Here x = distance, v = velocity, V = volume, and A = area. (i) V = Ax (ii) v22 = v12 + x2 *(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent (c) (i) and (ii) are both dimensionally consistent (d) (i) and (ii) are both dimensionally inconsistent Solution: In (i), V has units of (length)3 ; A has units of (length)2 ; and x has units of (length). In units, the formula looks like: (length)3 = (length)2 · (length) We get units of (length)3 on both sides of the equation. Hence (i) is dimensionally consistent. In (ii), v2 and v1 have units of length ; x has units of (length). In units, the formula looks time like: (length)2 (length)2 = + (length)2 (time)2 (time)2 Since the two terms on the right-hand side have different units, (ii) is not dimensionally consistent.
171 Problem 589. Consider the formulas (i) and (ii). Are the formulas dimensionally consistent? Here V = volume, r = distance, x = distance, and t = time. x − t0 (i) V = πxr2 (ii) t = 2 *(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent (c) (i) and (ii) are both dimensionally consistent (d) (i) and (ii) are both dimensionally inconsistent Solution: In (i), V has units of (length)3 ; π is dimensionless; and x and r both have dimensions of (length). In units, the formula looks like (length)3 = (length)(length)2 = (length)3 Hence (i) is dimensionally consistent. In (ii), t and t0 both have units of (time); x has units of (length); and 2 is dimensionless. In units, the formula looks like (time) = (length) − (time) The two terms on the right side are not dimensionally consistent; so the formula as a whole is not dimensionally consistent. Problem 590. Determine whether equations (i) and (ii) below are dimensionally consistent. x − x0 (ii) x2 = at2 + (x + vt)2 , (i) a = t2 where x = distance, v = velocity, t = time, and a = acceleration. *(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent (c) (i) and (ii) are both dimensionally consistent (d) (i) and (ii) are both dimensionally inconsistent Solution: Equation (i) is dimensional consistent. Equation (ii) is dimensionally inconsistent. To see this apply the dimensionator. 2 L L 2 2 2 2 2 [x] = [a][t] + ([x] + [vt]) ⇒ L = 2T + L + T = L + L2 T T The first term on the right-hand side of the equation is different from the other two.
172 Problem 591. Determine whether equations (i) and (ii) below are dimensionally consistent. (i) x = at2 + (1 + (vt)/x)2 (ii) vf2 = vi2 + 2ax2 , where x = distance, v = velocity, t = time, a = acceleration, r = distance, g = magnitude of gravity and µs is the coefficient of static friction. (a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent (c) (i) and (ii) are both dimensionally consistent *(d) (i) and (ii) are both dimensionally inconsistent Solution: (i) Dimensionally inconsistent. [x] = [a][t2 ] + (1 + [(vt)/x])2 becomes )2 = L + 1, where we’ve used the fact that [vt] = distance = L = TL2 T 2 + (1 + ([vt]=L) L L. Since L and 1 are dimensional inconsistent, the RHS of the equation is dimensional inconsistant. (ii) Dimensionally inconsistent. [vf2 ] = [vi2 ] = L2 /T 2 , but [2ax2 ] = (L/T 2 )(L2 ) = L3 /T 2 Problem 592. (Using dimensional analysis to get statistical-partial credit) Consider the following test question and multiple-choice answers. Without knowing how to solve the problem, can you identify the answer that cannot be a solution based solely on dimensional analysis? You are towing a crate of physics books down the hallway, using a rope that makes an angle of θ to the horizontal. The crate has a mass of m; the coefficient of kinetic friction between the crate and the floor is µk . As you go down the hallway, you are giving the crate a forward acceleration of a. What is the tension T in the rope, expressed in terms of the mass m, the acceleration a, the angle θ (dimensionless), µk (dimensionless), and the gravitational constant of acceration g? Note: Tension has units of force. *(a)
(mgµk + a) cos θ
(b)
(c)
m(gµk + a) cos θ
m(gµk + a) cos θ + µk sin θ
(d)
(mgµk + a)(cos θ + µk sin θ)
Solution: The answer is (a). Gravity g has units of acceleration, so each term in the factor (gµk + a) has units of acceleration and is therefore dimensionally consistent with acceleration. The factor (cos θ + µk sin θ) is composed of non-dimensional terms, so it’s dimensionless. Thus, the equation in (a) has units of acceleration. The equation in (b) is mass times acceleration (a.k.a force), which is consistent. Similarly, the equations in (c) and (d) have units of force.
173 5.1.3
Practical questions involving dimensions
Problem 593. In the construction business, concrete is often sold by the yard. What is the precise meaning of a yard of concrete? Solution: It wouldn’t make sense to sell something like concrete, which has volume, by units of length. A “yard” of concrete is actually a cubic yard. Problem 594. In the carpet business, carpet is sold by the yard. What do they really mean by a yard of carpet? Solution: It wouldn’t make sense to sell something like carpet, which has area, by units of length. A “yard” of carpet is actually a square yard. Problem 595. Name some physical phenomena that could be used as crude timing devices. Solution: There are lots of examples. The movement of shadows; the oscillations of a pendulum (a weight suspended from a string would do); the filling of a glass left under a dripping tap; the burning down of a candle... You’ve probably thought of different examples, and they’re probably just as good as these. Problem 596. Suppose your lab partner tells you that the volume of a cylinder is V = πr3 h. Could this formula be correct? Solution: In a cylinder, r is the radius, which has units of length; h is the height, which also has units of length; and π is a pure number, which has no units. On the right side of your lab partner’s formula, the units are (length)3 · (length) = (length)4 However, volume has units of (length)3 . The two sides of the formula don’t match. Clearly, your lab partner is trying to use you to bring down the curve so that he’ll get a better grade in the course. This, he hopes, will allow him to get into medical school, become a wealthy plastic surgeon, retire early, and pass the time relaxing in a gold-plated hot tub full of Scotch while you are still spending eight hours a day asking people if they want fries with that. You should consider asking for a new lab partner.
174
5.2
Units
Problem 597. What are the units of the number π? Solution: Pure numbers are dimensionless. One way you can see this for π in particular is this: π is defined as the ratio of a circle’s circumference to its diameter, and both of these are lengths. Suppose the circumference and the diameter are both measured in cm. Then cm circumference = π= diameter cm The cm in the numerator and denominator cancel, leaving no units. Problem 598. What are the units of the number e? Solution:
e is a pure number, and has no units.
Problem 599. What are the units of the number −5? Solution:
−5 is a pure number, and has no units.
Problem 600. Give an example from everyday life of something that’s about 1 m long. Solution: Find a tape measure and go looking for examples. Try the height of a low bookshelf, or the width of a wide door, or the length of your great-great grandfather’s cavalry saber, or... Problem 601. Give an example from everyday life of something that has a mass of about 1 kg. Solution: Check stuff in the kitchen. Your instructor’s local grocery store yielded a quart jar of mayo that weighed 946 g, which is pretty close. Problem 602. Give an example from everyday life of something that weighs about 1 N. Solution: You won’t find weights expressed in newtons on packages. Look for something that weighs about 100 g, or about 3 12 oz. The easiest way to remember what 1 newton feels like is to imagine holding a small apple, and then to remember the story of Newton and the apple falling on his head. Problem 603. Give an example from everyday life of something with a volume of about 1 l. Solution: This is easier, since soft drinks are often sold in 2 l bottles. Get one of those and pour out half. Some convenience stores sell 1 liter bottles.
175 Problem 604. Give an example from everyday life of a distance that’s about 1 km. Solution: The mile markers on Interstate 19 are actually kilometer markers. That is, the distance between markers is kilometers. If your neighborhood highways are marked in miles, you’ll have to find something else. A kilometer is about 0.6 miles. Problem 605. How much do you weigh in pounds? Round your answer to the nearest pound. Guess if you don’t have a scale handy. If that’s your weight in pounds, what is your mass in kg? Round to the nearest kg. Solution: This will depend on your actual weight. Suppose you weigh 150 lbs. Use the conversion factor 0.454 kg/lb. Then your weight is kg = 68 kg (150 lb) 0.454 lb Notice how the units work. We’re multiplying (lb) by (kg/lb). The pounds in the numerator and denominator cancel, leaving only kg. Problem 606. What is your height in inches? Round your answer to the nearest inch. Guess if you don’t have a tape measure handy. If that’s your height in inches, what is your height in cm? Round to the nearest cm. Solution: This will depend on your actual height. Suppose your height is 65 inches (5 ft 5 in). Use the conversion factor 2.54 cm/in. Then your height is cm = 165 cm (65 in) 2.54 in If we ignore the numbers and write this in terms of units alone, it’s cm (in) = cm in The inches in the numerator and the denominator cancel, leaving only cm in the numerator. Problem 607. How far is it in miles from where you live to your physics classroom? Round your answer to the nearest 0.1 mile. Guess if you haven’t measured it. If that’s the distance in miles, what is the distance in km? Round to the nearest 0.1 km. Solution: Your answer will depend on the distance from you to your classroom. Suppose it’s 2.3 miles. Use the conversion factor 1.61 km/mi. Then the distance in kilometers is km (2.3 mi) 1.61 = 3.7 km mi Again, look at the units: km = km mi · mi The miles in the numerator and the denominator cancel, leaving only km.
176 5.2.1
Converting between different sets of units
Problem 608. A box is 13 inches tall. What is its height in centimeters? Round to the nearest cm. Solution:
Use the conversion factor 2.54 cm/in. The height of the box is cm (13 in) 2.54 = 33 cm in
Notice that inches in the numerator and denominator cancel, leaving only cm. Problem 609. A bookshelf is 2.3 m tall. How tall is it in feet? Round to the nearest 0.1 ft. Solution:
Use the conversion factor 3.28 ft/m. The height of the bookshelf is ft (2.3 m) 3.28 = 7.5 ft m
Notice that meters in the numerator and denominator cancel, leaving only ft. Problem 610. Two towns are 37 miles apart. How far apart are they in kilometers? Round your answer to the nearest km. Solution:
Use the conversion factor 1.61 km/mi. The distance is km = 60 km (37 mi) 1.61 mi
Notice that miles in the numerator and denominator cancel, leaving only km. Problem 611. Two other towns are 113 km apart. How far apart are they in miles? Round your answer to the nearest mile. Solution: Again, use the conversion factor 1.61 km/mi. To make the units come out, we have to divide by the conversion factor. Equivalently, we have to multiply by the reciprocal: −1 km km mi km ÷ = km · = km · = mi mi mi km The distance is
(113 km) ·
1 mi 1.61 km
= 70 mi
Notice that we manipulated the given conversion factor in such a way that the unwanted units (km) cancelled out from the problem.
177 Problem 612. A house sits on a lot of 7260 square feet. What is the size of the lot in square meters? Round your answer to the nearest m2 . Solution: We’ll use the conversion factor 3.28 ft/m. Let’s look at the units to see what we need to do with the conversion factor. We have a measurement in ft2 . We want to convert it to m2 . Our conversion factor is in ft/m. If we divide by the square of the conversion factor, we get what we want: 2 −2 ft ft m2 2 2 ft ÷ = ft · = ft2 · 2 = m2 m m ft Hence the size of the lot is 2
7260 ft ·
1 m 3.28 ft
2
= 675 m2
Notice that we are using the one-factor here: 1 m 3.28 ft 1 m = 3.28 ft ⇒ 1 = = . 3.28 ft 1 m Problem 613. A pond has a surface area of 3900 m2 . What is its surface area in square feet? Round to the nearest 100 ft2 . Solution: Use the conversion factor 3.28 ft/m. We need to multiply by the square of the conversion factor to make the units turn out right: m2 ·
ft2 = ft2 m2
The surface area of the pond is 2 ft 3900 m · 3.28 = 42, 000 ft2 m 2
Problem 614. You are given a ticket for driving at 38 miles per hour in a school zone. What is your speed in meters per second? Round your answer to the nearest m/s. (a) 15 m/s (c) 19 m/s (e) None of these
*(b) (d)
17 m/s 20 m/s
178 Solution: Use the conversion factors: 1610 m/mi and 3600 s/hr. To make the units work out, we need to multiply by the first and the reciprocal of the second: m mi m s −1 mi m hr · · · · = = hr mi hr hr mi s s The speed for which you got the ticket is m 1 hr m mi · 1610 = 17 38 hr mi 3600 s s Problem 615. You are given a ticket for driving at 37 miles per hour in a school zone. What was your speed in m/s? Round your answer to the nearest m/s. Solution: Use the conversion factors: 1610 m/mi and 3600 s/hr. To make the units work out, we need to multiply by the first and the reciprocal of the second: mi m s −1 mi m hr m · · = · · = hr mi hr hr mi s s The speed for which you got the ticket is m mi 1 hr m · 1610 37 = 16.5 hr mi 3600 s s Problem 616. A water-balloon dropped from a tenth-floor window hits the ground at 24.5 m/s. What is its speed in miles per hour? Round your answer to the nearest 0.1 mile/hour. Solution: We’ll need to use two conversion factors: 1.61 km/mi = 1610 m/mi, and 3600 s/hr. We’ll proceed in steps. In the first step we’ll convert m/s to m/hr. Then in the second step we’ll convert m/hr to miles/hr. If we look at the units, we’ll see that we need to multiply by the reciprocal of the first conversion factor and multiply by the second: m m s Step 1: · = s hr hr m s m −1 m mi mi = Step 2: · · = s hr mi hr m hr The speed of the balloon at impact is m s 1 mi mi 24.5 · 3600 = 54.8 s hr 1610 m hr Problem 617. A water-balloon dropped from a 38th-floor window hits the ground at 86 m/s. What is its speed in ft/s? Round your answer to the nearest ft/s. (a) 26 ft/s (c) 257 ft/s (e) None of these
(b) *(d)
29 ft/s 282 ft/s
179 Solution:
Convert from SI system to British system: (86 m/s)(3.28 ft/m) = 282 ft/s.
Problem 618. An evil scientist grows a giant carrot with a mass of 60 kg. What is the carrot’s weight in pounds? Round your answer to the nearest pound. (a) 17 lb *(c) 132 lb (e) None of these Solution:
(b) 27 lb (d) 207 lb
(60 kg)(2.2 lb/kg) = 132 lb
Problem 619. A can of beer has a volume of 355 cm3 . What is its volume in cubic inches? Round your answer to the nearest 0.1 in3 . Solution: Use the conversion factor 2.54 cm/in. To get the units to work out, we need to multiple by the cube of the reciprocal of the conversion factor: 3 in3 in 3 = in3 = cm3 · cm · cm cm3 The volume of the beer-can is 3
355 cm ·
1 in 2.54 cm
3
= 21.7 in3
Problem 620. A pond has a volume of 325,000 ft3 . What is its volume in m3 ? Round your answer to the nearest 100 cubic meters. Solution: Use the conversion factor 3.28 ft/m. To get the units to work, we have to multiple by the cube of the reciprocal of the conversion factor: −3 3 ft 3 m 3 = ft · 3 = m3 ft · m ft The volume of the pond is 3
325, 000 ft ·
1m 3.28 ft
3
= 9200 m3
Problem 621. A river is flowing at 46,500 cubic feet per second. What is the river’s flow rate in cubic meters per second? Round your answer to the nearest 10 m3 /s.
180 Solution: Use the conversion factor 3.28 ft/m. To make the units work, we need to divide by the cube of the conversion factor: 3 m3 ft3 ft ft3 m3 ÷ = · 3 = s m s ft s From the previous problem it is clear that dividing by the conversion factor yields the same result as multiplying by the reciprocal. The flow rate of the river is 3 1m m3 ft3 · = 1320 46, 500 s 3.28 ft s Mass and weight. In the English system, the pound is a unit of force. In the metric system, the newton is a unit of force; the kilogram is a unit of mass. However, we often use grams and kilograms as units of weight instead of mass. For example, a one-pound box of crackers will inform you that it weighs 454 g, not 4.45 N. You should be able to convert between pounds and kilograms for everyday use; and between pounds and newtons for physics applications. Problem 622. A piano is pushed with a force of 450 N. What is the equivalent force in pounds? Round your answer to the nearest 10 lb. Solution: Use the conversion factor 4.45 N/lb. To make the units work, we need to divide by the conversion factor: N÷
lb N =N· = lb lb N
The force on the piano is 450 N ·
1 lb 4.45 N
= 100 lb
Problem 623. A box weighs 46 lb. What is the equivalent weight in newtons? Round your answer to the nearest N. Solution: Use the conversion factor 4.45 N/lb. To make the units work, we multiply by the conversion factor: N lb · =N lb The weight of the box is N = 205 N 46 lb · 4.45 lb Problem 624. A sack of potatoes has a mass of 12 kg. What is the weight of the potatoes in pounds? Round your answer to the nearest lb.
181 Solution: Use the conversion factor 2.2 lb/kg. To make the units work, multiply by the conversion factor: lb kg · = lb kg The weight of the potatoes is
lb 12 kg · 2.2 kg
= 26 lb
Problem 625. A sack of potatoes has a mass of 20 kg. What is the weight of the potatoes in pounds? Round your answer to the nearest lb. Solution: Fw = (20 kg)(2.2 lb/kg) = 44 lb. Problem 626. A German shepherd weighs 72 lb. What is the dog’s mass in kg? Round your answer to the nearest kg. Solution: Use the conversion factor 2.2 lb/kg. To make the units work, divide by the conversion factor: lb kg lb ÷ = lb · = kg kg lb The mass of the dog is 72 lb ·
1N 2.2 lb
= 33 kg
Problem 627. A cable weighs 5.2 lb/ft. What is the cable’s mass in kg/m? Round your answer to the nearest 0.1 kg/m. Solution: We need to convert lb to kg, and ft to m. Use the conversion factors 2.2 lb/kg and 3.28 ft/m. To make the units work, we have to multiply by the second conversion factor and to multiple by the reciprocal of the first: −1 lb lb ft kg kg lb ft · · = · · = ft m kg ft m lb m The linear density of the cable is 1 kg kg lb ft 5.2 · 3.28 = 7.8 ft m 2.2 lb m Problem 628. The pressure at the bottom of a swimming pool is 18 lb/in2 . What is the pressure in N/cm2 ? Round your answer to the nearest N/cm2 .
182 Solution: We need to convert pounds to newtons, and inches to centimeters. Use the conversion factors: 4.45 N/lb and 2.54 cm/in. To make the units work, multiply by the first conversion factor, and divide by the square of the second: lb N cm 2 lb N in2 N · ÷ = · = 2 2 · 2 in cm2 in lb in lb cm The pressure is 2 N 1 in N lb = 12 18 2 · 4.45 lb 2.54 cm cm2 in Problem 629. An engine burns fuel at a rate of 11.2 g/s. What is the consumption rate in kg/hr? Round your answer to the nearest 0.1 kg/hr. Solution: Use the conversion factors: 1000 g/kg and 3600 s/hr. To make the units work, multiply by the second factor and the reciprocal of the first: −1 g kg g s kg g s · · · = = · s hr kg s hr g hr The engine consumes fuel at a rate of s 1 kg kg g = 40.3 11.2 · 3600 s hr 1000 g hr Problem 630. You have just discovered the new element explodium. It has a density of 9.70 g/cm3 . The USA Today reporter says that their readers don’t understand all that centi-whatsis stuff, and wants the density in pounds per cubic foot. Round your answer to the nearest lb/ft3 . Solution: We need to convert grams to pounds, and cubic centimeters to cubic feet. For the former, we’ll use the conversion factor 454 g/lb. For the latter, we’ll combine two conversion factors: in cm cm 2.54 · 12 = 30.48 in ft ft To make the units work, we need to multiply by the cube of the second conversion factor, and divide by the first one: g cm 3 g g cm3 lb lb · ÷ = · = 3 3 · 3 3 cm ft lb cm g ft ft The density of explodium is 9.70
g cm 3 g lb · 30.48 ÷ 454 = 605 3 . 3 cm ft lb ft
183 For pedagogical purposes, we write the conversion out in one long string out products: 9.70
g g = 9.70 ·1·1·1 cm3 cm3 3 3 g 12 in 1 lb 2.54 cm = 9.70 · cm3 1 in 1 ft 454 g 3 3 9.70 · (2.54) · 12 lb = 454 ft3 lb = 605 3 ft
Notice that the second equality follows from the one-factor. Problem 631. Your male friend is trying to impress a pretty German girl at your local pub. He says to her ”I can bench press 330lbs”. She then asks ”How much is that in kilograms?”. He is dumbfounded. Can you save him by converting 330 lbs to kgs, so that she can be properly impressed? Solution: We need to convert lb to kg. Use the conversion factor 2.2 lb/kg. The weight in kilograms is 1 kg = 150 kg . 330 lb · 2.2 lb Problem 632. You have a measuring stick that has centimeters and inches on it. You notice that 1.00 in = 2.54 cm. Using only this information together with the relationship 1 mile = 5,280 ft., find a relationship between miles and kilometers. That is, 1 mile = how many kilometers? Solution: We will use a total of 5 conversion factors. That is, we will apply the one-factor 5 times (via multiplication). 1 mile = 1 mile · 1 · 1 · 1 · 1 · 1 5280ft 12in 2.54cm 1m 1km = 1 mile · 1mile 1ft 1in 100cm 1000m 5280 · 12 · 2.54 = km 105 = 1.61 km where the second equality follows from the one-factor. Thus, 1 mile = 1.61 km. Crudely, 1 mile ≈ 23 km. Problem 633. Use the relationship 1.00 in = 2.54 cm to find a relationship between a yard and a meter. Your answer should show that a yard is close in length to a meter.
184 Solution: we have
Using the relationship 1.00 in = 2.54 cm and a number of conversion factors 1 yd = 1 yd · 1 · 1 · 1 · 1 · 1 3ft 12in 2.54cm 1m = 1 yd · 1yd 1ft 1in 100cm 3 · 12 · 2.54 = m 100 = 0.914 m
Thus, 1m ≈ 1.09yd. So, 100 m is longer than a 1 football field.
5.3
Scientific Notation
For the following problems express the following numbers in scientific notation without using a calculator. Assume 3 significant figures in each number with more than 3 trailing zeros. Problem 634. 1 = 1 × 100 Problem 635. 12 = 1.2 × 101 Problem 636. 123 = 1.23 × 102 Problem 637. 1234 = 1.234 × 103 Problem 638. 12345 = 1.2345 × 104 Problem 639. 0.1 = 1. × 10−1 Problem 640. 0.12 = 1.2 × 10−1 Problem 641. 0.012 = 1.2 × 10−2 Problem 642. 0.0012 = 1.2 × 10−3 Problem 643. 0.00012 = 1.2 × 10−4 Problem 644. 395 = 3.95 × 102 Problem 645. 97, 000 = 9.70 × 104 Problem 646. .0000345 = 3.45 × 10−5 Problem 647. 10, 000, 000 = 1.00 × 107 Problem 648. .00000000500 = 5.00 × 10−9
185 For the following problems convert the expressions that are given in scientific notation into standard form without using a calculator. Problem 649. 1 × 100 = 1 Problem 650. 1.2 × 101 = 12 Problem 651. 1.23 × 102 = 123 Problem 652. 1.234 × 103 = 1234 Problem 653. 1.2345 × 104 = 12345 Problem 654. 1 × 10−1 = 0.1 Problem 655. 1.2 × 10−1 = 0.12 Problem 656. 1.2 × 10−2 = 0.012 Problem 657. 1.2 × 10−3 = 0.0012 Problem 658. 1.2 × 10−4 = 0.00012 Problem 659. 3.95 × 102 = 395 Problem 660. 9.70 × 104 = 97, 000 Problem 661. 3.45 × 10−5 = .0000345 Problem 662. 1.00 × 107 = 10, 000, 000 Problem 663. 5.00 × 10−9 = .00000000500 Problem 664. Convert the number 6.2 × 10−3 from scientific notation to standard form. (a) 0.00062 *(c) 0.0062 (e) None of these Solution:
(b) (d)
−6200 −62, 000
Move the decimal 3 places to the left.
Problem 665. Convert the number 9.70 × 10−4 to standard form. (a) 0.000097 (c) 0.0000970 (e) None of these
(b) *(d)
0.00097 0.000970
Solution: Since 9.70 × 10−4 has three significant figures, the standard-form number must also have three. By rule 4, the final zero in (d) is significant.
186 Problem 666. Convert the number 97,000 to scientific notation. Your answer should have three significant figures. (a) 9.70 × 103 (c) 9.7 × 103 (e) None of these
*(b) 9.70 × 104 (d) 9.7 × 104
Solution: Answers (b) and (d) are both equal to 97,000, but (d) only has two significant figures. For the following problems perform the following arithmetic without using a calculator. Problem 667.
(1.0 × 103 ) · (9.0 × 108 ) = 3.0 × 104 7 3.0 × 10
Problem 668.
(4.67 × 107 ) · (1.01 × 102 ) = 7.56 × 105 6.24 × 103
(7.3 × 102 ) · (2.23 × 105 ) = 6.6 × 103 Problem 669. 4 2.45 × 10 Problem 670. (3.4 × 107 ) − (3.0 × 106 ) = (3.4 × 107 ) − (0.30 × 107 ) = 3.1 × 107 . Problem 671. (3.4 × 103 ) + (3.0 × 104 ) = (.34 × 104 ) + (3.0 × 104 ) = 3.3 × 104 .
187
5.4
Significant figures
Problem 672. A useful approximation for the number of seconds in a year is π × 107 s. Determine the number of significant figures in this approximation given that there are 365.25 days in a year. Solution: Let’s start by using conversion factors to calculate how many seconds there are in a year. 1 year = 1 year · 1 · 1 · 1 · 1 24 hours 60 minutes 60 seconds 365.25 days = 1 year · 1 year 1 day 1 hour 1 minute = 365.25 · 24 · 3600 s = 31, 557, 600 s Let’s compare this to π × 107 = 31415926.59 s. The approximation is only accurate to 2 significant figures. Problem 673. A common approximation for π = 3.141592659 . . . is To how many significant figures is this approximation accurate?
22 . 7
That is, π ≈
22 . 7
Solution: To 8 decimal places 22 ≈ 3.142857143. The two expressions agree to the 7 hundreds place, and if we round off they agree to three decimal places (the thousands place), hence they agree to 4 significant figures. Problem 674. A common approximation for π = 3.141592659 . . . is . To how many significant figures is this approximation accurate? π ≈ 355 113
355 . 113
That is,
355 Solution: To 8 decimal places 113 ≈ 3.14159292. The two expressions agree to the 6th decimal place. Hence they agree to 7 significant figures.
For the following problems preform the indicated operations. Assume that all of the numbers come from real data with the correct number of significant figures. Be sure to express your answer in the correct number of significant figures. Please refer to rules 1-11 in the lecture notes. Problem 675. 566.3 + 32.50 + 2.197 = ? round off to one decimal place
Solution: 566.3 + 32.50 + 2.197 = 600.997 −−−−−−−−−−−−−−−−−→ Answer = 601.0 . The last zero is significant. Notice that we used rule 10 for addition and/or subtraction. Problem 676. 261.72 − 90.715 = ? Solution:
261.72 − 90.715 = 171.005
round off: rule 9
−−−−−−−−−→
Answer = 171.00
188 Problem 677. 523.498 − 32.907527 + 0.52 = ? Solution: 491.11
523.498 − 32.907527 + 0.52 = 491.110473
round off: 2 dec.
−−−−−−−−−→
Answer =
Problem 678. 225 + 1.087 − 45.3975 = ? Solution: There are two possible answers (even though I said all numbers come from measurements) If the number 225 is exact, then round off: 3 dec.
225 + 1.087 − 45.3975 = 180.6895
−−−−−−−−−→
Answer = 180.690
since the number 225 is exact so it does not affect the accuracy of the calculation. If 225 is a measured number, then it has 3 significant figures and zero decimal digits. Thus the sum must be rounded to 3 significant figures: 225 + 1.087 − 45.3975 = 181 Problem 679. 596.9/3.986 = ? Solution:
596.9/3.986 = 149.7491...
round off: 4 sig figs.
−−−−−−−−−−−→
Answer = 149.7
Problem 680. 505 × 0.0065 round off: 2 sig figs.
Solution: 505 × 0.0065 = 3.2825 −−−−−−−−−−−→ Answer = 3.3 Since the second factor in the product only has 2 significant figures, namely 65, the answer can only have two significant figures. Problem 681. 2.00 × 107 + 3.5 × 332 = ? Hint: First compute the exact numerical result, then express it the correct number of significant figures. Solution: 2.00 × 107 + 3.5 × 332 = 1376. The weakest link is 3.5, it is only know to 2 significant figures. Thus the answer = 1400. Problem 682. (3.7)3 = ? Solution: (3.7)3 = 50.653 ⇒ Answer = 51. Since x3 = x · x · x, the expression (3.7)3 is really just a multiplication. From rule 11, we must treat this as a multiplication. Problem 683. What is the average of 2.5212, 2.5214, 2.5213, 2.5212, and 2.5220? 2.5212 + 2.5214 + 2.5213 + 2.5212 + 2.5220 = 2.52142 Can Solution: Average = 5 only keep 5 significant figures, so Answer = 2.5214.
189 Problem 684. Calculate 2.00 × 1.07 + 3.5 × 332. Your answer should have the correct number of significant figures. *(a) (c) (e)
1200 1164 None of these
(b) (d)
1160 1164.14
Solution: Begin by calculating the products. 2.00 × 1.07 = 2.14; since both factors have 3 significant figures, so does the product. 3.5 × 332 = 1162; since 3.5 has only two significant figures, so does the product. That means that the 1162 is only accurate to the hundreds place. We add 2.14 (accurate to the one-hundredth place) to 1162 (accurate only to the hundreds place) and get 2.14+1162 = 1164.14. Since the sum is only accurate to the decimal place of the least accurate term, it’s only accurate to the hundreds place; so we round it to 1200. Problem 685. Evaluate the following expression. Round your answer to the correct number of significant figures: 1.24 + 24.038 + 6.925 (a) 32 *(c) 32.20 (e) None of these Solution:
(b) 32.2 (d) 32.203
Keep 2 decimal places (the weakest link).
Problem 686.
(7.3 × 102 ) · (2.23 × 105 ) 2.45 × 104
Solution: 6.6 × 103 . The first factor in the numerator has 2 significant digits; the other factor in the numerator, and the denominator, have 3. By rule 11, our answer should be rounded to 2 significant digits. Problem 687. How many significant digits are there in: m = 0.0507 kg? Solution: There are three significant digits. Starting from the left, the first two zeros are place holders. The first significant digit is 5, after that all of the digits are significant. Problem 688. How many significant digits are there in: m = 28.350 g? Solution: The last zero is significant, since it is to the right of the decimal point, and also to the right of a nonzero digit. Thus there are five significant digits. Problem 689. How many significant digits are there in: v = 1.022 m/s?
190 Solution: The zero is significant, since it comes between two nonzero digits. Thus there are four significant digits. Problem 690. How many significant digits are there in: r = 0.028 cm? Solution: Neither of the zeros is significant, since there are no nonzero digits to the left of them. The zeros are there only to give place value to the 2 and the 8. Thus there are two significant digits. Problem 691. How many significant digits are there in: h = 0.08 cm? Solution:
one
Problem 692. How many significant digits are there in: r = 1.028 cm? Solution:
4
Problem 693. How many significant digits are there in: r = 10.028 cm? Solution:
5
Problem 694. How many significant digits are there in: V = 44, 000 ft3 ? Problem 695. How many significant digits are there in: r = 0.0307 m? (a) 2 (c) 4 (e) None of these
*(b) 3 (d) 5
Solution: The zeros to the left of the nonzero digits are placeholders, and are not significant digits. Problem 696. How many significant digits are there in: v = 1.022 m/s? (a) 2 *(c) 4 (e) None of these Solution:
(b) 3 (d) 5
The zero is significant because it’s between two nonzero digits.
Problem 697. How many significant digits are there in: r = 0.02070 m? (a) 2 *(c) 4 (e) None of these
(b) 3 (d) 5
Solution: The zeros to the left of the nonzero digits are not significant since they are placeholders. There are 4 significant figures.
191 Problem 698. How many significant digits are there in: r = 0.001000 m? (a) 1 (c) 3 (e) None of these
(b) 2 *(d) 4
Solution: The zeros to the left of the first nonzero digit are not significant since they are placeholders. All of the zeros after the nonzero digit are significant figures. There are 4 significant figures. Problem 699. Use a calculator to determine the decimal value of 1/7. Round your answer to 3 significant digits. (a) 0.14 (c) 0.142 (e) None of these
(b) *(d)
0.15 0.143
Solution: The calculator should give 1/7 = 0.142857 . . . The initial zero is not significant, so you need to round to three decimal places. Since the 2 is followed by an 8, round it up to 3: 1/7 = 0.143. Solution: The zeros are not significant, since they do not follow a decimal point or precede a nonzero digit. They are there only to give place value to the two 4’s. Thus there are two significant digits. Problem 700. Use a calculator to determine the decimal value of 22/7. Round your answer to 3 significant digits. Solution: A calculator gives 3.142857... To get three significant digits, we round this to 3.14. Recall that 22/7 is a rational number approximation to π. Problem 701. Use a calculator to determine the decimal value of π. Round your answer to 3 significant digits. How does this answer compare to the previous answer for 22/7? Solution: A calculator gives 3.1415926... To get three significant digits, we round this to 0.143. Notice that the zero before the decimal point is not significant. Problem 702. Use a calculator to determine the decimal value of 1/237. Round your answer to 3 significant digits. Solution: A calculator gives 0.0042194... To get three significant digits, we round this to 0.00422. Notice that the three leading zeros are not significant. Problem 703. You have calculated the volume of a pond as 83,512.338 ft3 . Round this answer to 2 significant digits.
192 Solution: The first two digits are nonzero, so rounding to the thousands place is what we want: 84,000 ft3 . Notice that we don’t have a decimal point with zeros after it, since this would indicate that all of the zeros were significant. Problem 704. You have taken several measurements of the size of a very small particle. The mean of your measurements is 0.00013722 cm. Round this answer to 2 significant digits. Solution: The three zeros after the decimal point are not significant, since there’s no nonzero digit before them. We round to two digits after the string of zeros: 0.0014 cm. Problem 705. (Significant Figures and Conversion) If you were to convert the measured length 7.923 ft into yards by multiplying by the conversion factor (1 yd/3 ft), how many significant figures should the answer contain. Justify your answer. Solution: The answer should contain 4 significant figures since the original measured number contains 4 significant figures and we are multiplying it by the conversion, which is an exact number. Problem 706. Evaluate the following expression. Round your answer to the correct (7.3 × 102 ) · (2.23 × 105 ) number of significant figures: 2.45 × 104 (a) 6.64 × 103 (c) 6.6 × 102 (e) None of these
*(b) 6.6 × 103 (d) 6.64 × 10
Solution: 6.64449 × 103 → 6.6 × 103 . The first factor in the numerator has 2 significant digits; the other factor in the numerator, and the denominator, have 3. By rule 11, our answer should be rounded to 2 significant digits. Problem 707. Evaluate the following expression. Round your answer to the correct number of significant figures. (3.021 × 10−2 )(8.446 × 1011 ) 1.9 × 103 Solution: 1.3 × 107 . Doing this on a calculator yields 1.34291 × 107 . Since the denominator only has two significant figures, we must round the result likewise.
193
5.5
Determining derived units from equations
In the following problems use the given physical governing equation to determine the dimensions of the given derived quantity in terms of the fundamental dimensions of length, time, and mass, or in terms of other derived quantities as specified in the problem. Use the SI units when expressing the dimensions. Problem 708. Determine the expression for the unit of work (defined as a Joule J) in terms of the fundamental units from the definition of work (in one spacial dimension by a constant force) as force times distance W = F d, where W is the symbol for work (a method of transferring energy), F is the force exerted on the object and d is the distance that the force acts on the object. Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for work gives [W ] = [F d] = [F ][d] = N · m
⇒
1 Joule = 1N · m = 1J
A Joule is pronounced a Newton-meter. Problem 709. Determine the expression for the unit of kinetic energy (K.E.) in terms of the fundamental units from the definition of K.E. as K = 12 mv 2 where K is the symbol for kinetic energy, m is the mass of the object, v is the speed of the object. Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for K.E. gives m m2 m2 1 2 ⇒ [K] = 1 kg 2 = 1 kg 2 · m = 1N · m = 1J [K] = mv = [m][v]2 = kg 2 2 s s s Problem 710. Determine the expression for the unit of gravitational potential energy (P.E.) in terms of the fundamental units from the definition of P.E. as U = mgh, where U is the symbol for potential energy, m is the mass of the object, g is the magnitude of acceleration due to gravity, and h is the height of an object above some reference level. Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for gravitational P.E. gives m [U ] = [mgh] = [m][g][h] = kg 2 m ⇒ [U ] = 1N · m = 1J s Problem 711. Determine the expression for the unit of elastic potential energy (P.E.) stored in a spring in terms of the fundamental units from the definition of P.E. as U = 1 kx2 , where U is the symbol for potential energy, k is the spring constant, and and x is 2 the displacement from the equilibrium position.
194 Problem 712. Compare your results from the four previous expressions for work, kinetic, and potential energy. What do you conclude about the relation between the various units? Solution: All four expressions have the same units, namely Joules. The Joule is the SI unit for energy. This result says that work, kinetic energy, and potential all have units of energy. As we’ll discover later in the course, by the principle of conservation of energy: energy can neither be created nor destroyed, but energy can transform from one form to another via the mechanism of work. Problem 713. Determine the fundamental dimensions for density = mass/volume, denoted by ρ. mass = Solution: Start by writing ρ = density = volume [m] M volume. Using the dimensionator: [ρ] = [V ] = L3 .
m , V
where m is mass and V is
Problem 714. Express pressure, denoted by P , in terms of the fundamental dimensions. For now, just take pressure to be force/area. Note: In a later chapter we will use the mathematical expression for Newton’s second law to show that the dimensions of force L are [F ] = M 2 . For now, just take this as a fact. T = FA , where F is force and A is area. Start by writing P = pressure = force area [F ] [F ] L 1 Using the dimensionator: [P ] = = 2 =M 2 · 2 [A] L T L Solution:
Problem 715. Determine the expression for the unit of pressure (the Pascal Pa) in terms of the SI units for pressure and area from the equation defining pressure P = FA , where F is force and A is area. Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for pressure gives [F ] N N F = = 2 ⇒ 1 Pascal = 2 [P ] = A [A] m m Problem 716. Determine the units of the universal gas constant R from the equation for the ideal gas law: P V = nRT , where P is pressure, V is volume, n is moles (denoted mol), and T is temperature (with SI units of absolute temperature the kelvin K). Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the equation for the ideal gas law gives ÷[n][T ]
[P V ] = [nRT ] = [n][R][T ] −−−−−−−→ [R] =
[P ][V ] J = [n][T ] (mol) (K)
195 Problem 717. In the study of thermodynamics one of the common expressions that you encounter is pressure P times volume V . It turns out that this term has the dimensions of work = force × distance. To see this we write volume = area × length, then using the definitions of pressure and work we have pressure × volume = (force/area) × (area × length) = force × length = work. Verify our word equation using the fundamental dimensions for the product of pressure and volume: pressure × volume = P V and work as work = F d, where d is the distance that the force acts over. ] · L3 = [F ] · L. Thus, P V has units of force times Solution: [P V ] = [P ][V ] = [F L2 distance. Later in the course we will learn that this is the units of work.
Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for PV gives [P V ] = [P ][V ] =
[F ] N [V ] = 2 m3 [A] m
⇒
[P V ] = N · m = J
196
6
Introduction to Vectors
Unless stated otherwise, the direction of a vector in xy-space is the counterclockwise angle θ that it makes with the x-axis; 0 ≤ θ < 360◦ or 0 ≤ θ < 2π rad., and angle measurements in your answers should be in the same units as in the statement of the problem.
6.1 6.1.1
Identifying Vectors Given vector, identify on figure
Problem 718. Which of the vectors in the graph corresponds to −10ˆı − 5ˆ? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these
~ B AK
~ and C ~ have a Solution: Only B negative x-component and a positive ycomponent. In −6ˆı + 2ˆ, the magnitude of the x-component (6) is greater than the magnitude of the y-component (2). This ~ but not of B. ~ is true of C
6
A A
~ C
~ has a negative xSolution: Only D component and a negative y-component.
Problem 719. Which of the vectors in the graph corresponds to −6ˆı + 2ˆ? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these
y
A A A iP P PP A PP A
~ A -
x
~ D
~ B AK
y 6
A A
~ C
A A A iP P PP A PP A
~ D
~ A -
x
197 Problem 720. Which of the vectors in the graph corresponds to 4ˆı + 4ˆ? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these
~ B AK
~ C
~ has a negative xSolution: Only C component and a positive y-component.
~ A
A A A iP P PP A PP A
-
x
~ D
Problem 721. Which of the vectors in the graph corresponds to −ˆı + 2ˆ? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these
Problem 722. Which of the vectors in the graph corresponds to −3ˆı + ˆ? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these
6
A A
~ has a positive xSolution: Only A component and a positive y-component.
~ and C ~ have a Solution: Only B negative x-component and a positive ycomponent. In −ˆı + 2ˆ, the magnitude of the x-component (1) is smaller than the magnitude of the y-component (2). This ~ but not of C. ~ is true of B
y
~ B AK
y 6
A A
~ C
~ A
A A A iP P PP A PP A
-
x
~ D
y
~ B
6
~ i PP C PP
~ A
PP S S
-
x
S S S S S w S
~ D
198 Problem 723. Which of the vectors in the graph corresponds to 2ˆı + ˆ? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these
y
~ C
~ and B ~ have a Solution: Only A positive x-component and a positive ycomponent. In 2ˆı + ˆ, the magnitude of the x-component (2) is greater than the magnitude of the y-component (1). This ~ but not of B. ~ is true of A Problem 724. Which of the vectors in the graph corresponds to ˆı + 2ˆ? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these
~ has a positive xSolution: Only D component and a negative y-component.
~ iP P * A PP PP S S S S S S S w S
y
@ @
~ B AK
~ * A @ @ S S S S S S S w S
~ D
-
x
~ D
y 6
A A A
~ C
x
~ C I @ @
-
~ B
6
~ and B ~ have a Solution: Only A positive x-component and a positive ycomponent. In ˆı + 2ˆ, the magnitude of the x-component (1) is smaller than the magnitude of the y-component (2). This ~ but not of A. ~ is true of B Problem 725. Which of the vectors in the graph corresponds to 3ˆı − 4ˆ? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these
~ B
6
~ A
A A iP P PP A PP A S
-
x
S S S S S S w S
~ D
199 Problem 726. Which of the vectors in the graph corresponds to h−3, 1i? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these
~ B AK
6
A A
~ C
~ and C ~ have a Solution: Only B negative x-component and a positive ycomponent. In h−3, 1i, the magnitude of the x-component (3) is greater than the magnitude of the y-component (1). This ~ but not of B. ~ is true of C Problem 727. Which of the vectors in the graph corresponds to h2, 2i? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these
y
A A A iP P PP A PP A
~ A -
x
~ D
~ B AK
y 6
A A
~ C
~ has a positive xSolution: Only A component and a positive y-component.
A A A iP P PP A PP A
~ A -
x
~ D
Problem 728. Which of the vectors in the graph corresponds to h−1, 2i? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these ~ and C ~ have a Solution: Only B negative x-component and a positive ycomponent. In h−1, 2i, the magnitude of the x-component (1) is smaller than the magnitude of the y-component (2). This ~ but not of C. ~ is true of B
~ B AK
y 6
A A
~ C
A A A iP P PP A PP A HH HH j
~ D
-
x
~ A
200 Problem 729. Which of the vectors in the graph corresponds to h−2, −1i? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these
y
~ B AK
6
A A
~ C
~ has a negative xSolution: Only D component and a negative y-component.
A A A iP P PP A PP A A A ~ D A A
-
x
A A A AU
Problem 730. Which of the vectors in the graph corresponds to h1, 3i? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these
6
~ A
A A
~ has a positive xSolution: Only A component and a positive y-component.
Problem 731. Which of the vectors in the graph corresponds to h−4, −2i? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these ~ has a negative xSolution: Only C component and a negative y-component.
y
~ B AK
~ C
~ A
A A A iP P PP A PP AH H HH j
-
x
~ D
y
~ B AK
6
A A A A A
~ C
A A H A H j A HH A A A A A AU
-
x
~ A
~ D
201 Problem 732. Which of the vectors in the graph corresponds to h10, −5i? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these ~ and B ~ have a Solution: Only A positive x-component and a negative ycomponent. In h10, −5i, the magnitude of the x-component (10) is greater than the magnitude of the y-component (5). ~ but not of A. ~ This is true of B Problem 733. Which of the vectors in the graph corresponds to h3, −3i? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these
y 6
~ i PP C PP ~ D
PP H A H A HHH j H A A A A A ~ AU A
-
x
~ B
y 6
~ C I @
~ B
@ @
~ has a positive xSolution: Only A component and a negative y-component.
@ @
-
x
@ @
~ D
6.1.2
R @
~ A
Given vector on figure, identify formula
Problem 734. Which of the vectors ~ on the graph at below corresponds to A right? (a) ˆı + ˆ (b) −ˆı + ˆ *(c) ˆı − ˆ (d) −ˆı − ˆ (e) None of these ~ has a positive xSolution: A component and a negative y-component; so (c).
y 6
~ C I @
~ B
@ @ -
@ @
x
@ @
~ D
R @
~ A
202 Problem 735. Which of the vectors ~ on the graph at below corresponds to B right? *(a) ˆı + ˆ (b) −ˆı + ˆ (c) ˆı − ˆ (d) −ˆı − ˆ (e) None of these ~ has a positive xSolution: B component and a positive y-component; so (a).
Problem 736. Which of the vectors ~ on the graph at below corresponds to C right? (a) ˆı + ˆ *(b) −ˆı + ˆ (c) ˆı − ˆ (d) −ˆı − ˆ (e) None of these ~ has a negative xSolution: C component and a positive y-component; so (b).
Problem 737. Which of the vectors ~ on the graph at below corresponds to D right? (a) ˆı + ˆ (b) −ˆı + ˆ (c) ˆı − ˆ *(d) −ˆı − ˆ (e) None of these ~ has a negative xSolution: D component and a negative y-component; so (d).
y 6
~ B
~ C I @
@ @ @ @
-
x
@ @
~ D
R @
~ A
y 6
~ C I @
~ B
@ @ @ @
-
x
@ @
~ D
R @
~ A
y 6
~ C I @
~ B
@ @ @ @
-
x
@ @
~ D
R @
~ A
203 y
Problem 738. Which of the vectors ~ on the graph at below corresponds to A right? (a) 3ˆı + ˆ (b) −3ˆı + ˆ *(c) 2ˆı − ˆ (d) −ˆı − 3ˆ (e) None of these
6
HH HH
~ has a positive xSolution: A component and a negative y-component; so (c).
H j H
~ A
6
~ A
~ has a positive xSolution: A component and a positive y-component; ~ has a small x-component so (a) or (b). A and a large y-component; so (a).
~ has a negative xSolution: A component and a positive y-component; ~ has a small x-component so (c) or (d). A and a large y-component; so (d).
x
y
Problem 739. Which of the vectors ~ on the graph at below corresponds to A right? *(a) 2ˆı + 6ˆ (b) 3ˆı + ˆ (c) 2ˆı − ˆ (d) −ˆı − 3ˆ (e) None of these
Problem 740. Which of the vectors ~ on the graph at below corresponds to A right? (a) 2ˆı − ˆ (b) ˆı − 2ˆ (c) −2ˆı + ˆ *(d) −ˆı + 2ˆ (e) None of these
-
-
x
y
~ A AK
6
A A A A A A A
-
x
204 Problem 741. Which of the vectors ~ on the graph at below corresponds to A right? (a) 6ˆı − 3ˆ (b) −3ˆı + 6ˆ *(c) −6ˆı + 3ˆ (d) 3ˆı − 6ˆ (e) None of these
y 6
~ Y HH A HH
H
H
-
x
~ has a negative xSolution: A component and a positive y-component; ~ has a large x-component so (b) or (c). A and a small y-component; so (c).
y
Problem 742. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−1, 2i (b) h−2, 1i (c) h−1, 2i *(d) h−2, −1i (e) None of these ~ has a negative xSolution: A component and a negative y-component; so (d).
Problem 743. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−1, 2i *(b) h1, −2i (c) h−2, 1i (d) h2, −1i (e) None of these
6
~ A
-
x
y 6
-
~ has a positive xSolution: A component and a negative y-component; ~ has a small x-component so (b) or (d). A and a large y-component; so (b).
x
A A A A A AU
~ A
205 Problem 744. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−6, −3i *(b) h−3, −6i (c) h6, −3i (d) h3, −6i (e) None of these
y
~ has a negative xSolution: A component and a negative y-component; ~ has a small x-component so (a) or (b). A and a large y-component; so (b).
6
-
Problem 745. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−5, −10i (b) h5, −10i *(c) h−10, 5i (d) h10, −5i (e) None of these
x
~ A
y 6
~ Y HH A HH
H
H
-
x
~ has a negative xSolution: A component and a positive y-component; so (c).
Problem 746. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−2, 4i *(b) h2, 4i (c) h−4, 2i (d) h4, 2i (e) None of these
y 6
~ A
~ has a positive xSolution: A component and a positive y-component; ~ has a small x-component so (b) or (d). A and a large y-component; so (b).
-
x
206 Problem 747. Which of the vectors ~ on the graph at below corresponds to A right? *(a) h−1, 2i (b) h−2, 1i (c) h1, −2i (d) h2, −1i (e) None of these
y 6
~ A AK A A A A A
-
x
~ has a negative xSolution: A component and a positive y-component; ~ has a small x-component so (a) or (b). A and a large y-component; so (a).
Problem 748. Which of the vectors ~ on the graph at below corresponds to A right? *(a) h2, 1i (b) h1, 2i (c) h−2, 1i (d) h−1, 2i (e) None of these
y 6
*
~ A -
x
~ has a positive xSolution: A component and a positive y-component; ~ has a large x-component so (a) or (b). A and a small y-component; so (a).
Problem 749. Which of the vectors ~ on the graph at below corresponds to A right? (a) h−2, −1i (b) h−1, −2i *(c) h2, −1i (d) h1, −2i (e) None of these ~ has a positive xSolution: A component and a negative y-component; ~ has a large x-component so (c) or (d). A and a small y-component; so (c).
y 6
HH HH
-
x
H j H
~ A
207 6.1.3
Given direction, identify on figure
Problem 750. Which of the vectors in the graph corresponds to the direction 20◦ east of north? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these Solution: Since we want “east of north”, we take the north direction as the reference line. Going 20◦ east of north ~ puts us in the first quadrant; so D.
Problem 751. Which of the vectors in the graph corresponds to the direction 20◦ east of south? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these Solution: Since we want “east of south”, we take the south direction as the reference line. Going 20◦ east of south ~ puts us in the fourth quadrant; so C.
Problem 752. Which of the vectors in the graph corresponds to the direction 20◦ west of north? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these Solution: Since we want “west of north”, we take the north direction as the reference line. Going 20◦ west of north ~ puts us in the second quadrant; so A.
N 6
~ AK A A A A
~ B
~ D
A A A A A A A AU
-
E
~ C
N 6
~ AK A A A A
~ B
~ D
A A A A A A A AU
-
E
~ C
N 6
~ AK A A A A
~ B
~ D
A A A A A A A AU
-
E
~ C
208 Problem 753. Which of the vectors in the graph corresponds to the direction 20◦ west of south? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these
N 6
~ AK A A A A
Solution: Since we want “west of south”, we take the south direction as the reference line. Going 20◦ west of south ~ puts us in the third quadrant; so B.
Problem 754. Which of the vectors in the graph corresponds to the direction 30◦ south of west? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these Solution: Since we want “south of west”, we take the west direction as the reference line. Going 30◦ south of west ~ or puts us in the third quadrant; so C ~ Since 30◦ < 45◦ , our vector should be D. closer to the west direction than to the ~ (D ~ would be more south direction; so C. ◦ like 60 south of west.) Problem 755. Which of the vectors in the graph corresponds to the direction 30◦ north of west? ~ ~ (a) A *(b) B ~ ~ (c) C (d) D (e) None of these Solution: Since we want “north of west”, we take the west direction as the reference line. Going 30◦ north of west ~ or puts us in the second quadrant; so A ~ Since 30◦ < 45◦ , our vector should be B. closer to the west direction than to the ~ (A ~ would be more north direction; so B. ◦ like 60 north of west.)
~ B
~ D
A A A A A A A AU
-
E
~ C
N 6
~ A AK A A ~ Y HH B A HH A H
~ C
H A ~
-
E
D
N 6
~ A AK ~ B
~ C
A A YH H H A HHA H A
~ D
-
E
209 Problem 756. Which of the vectors in the graph corresponds to the direction 60◦ south of west? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these Solution: Since we want “south of west”, we take the west direction as the reference line. Going 60◦ south of west ~ or puts us in the third quadrant; so C ◦ ◦ ~ Since 60 > 45 , our vector should be D. closer to the south direction than to the ~ (C ~ would be more west direction; so D. ◦ like 30 south of west.) Problem 757. Which of the vectors in the graph corresponds to the direction 60◦ north of west? ~ ~ *(a) A (b) B ~ ~ (c) C (d) D (e) None of these Solution: Since we want “north of west”, we take the west direction as the reference line. Going 60◦ north of west ~ or puts us in the second quadrant; so A ◦ ◦ ~ B. Since 60 > 45 , our vector should be closer to the north direction than to the ~ (B ~ would be more west direction; so A. ◦ like 30 north of west.) Problem 758. What is the direction of ~ in the graph at right? vector A (a) 30◦ north of east *(b) 30◦ south of east (c) 30◦ north of west (d) 30◦ south of west (e) None of these Solution: Of the solutions, only 30◦ south of east is in the fourth quadrant.
N 6
~ A AK A A ~ Y HH B H A H A
HH A
~ C
-
E
~
D
N 6
~ A AK A A ~ Y HH B H A H A
~ C
HH A ~
-
E
D
N 6
HH H
-
E
HH j H
~ A
210 Problem 759. What is the direction of ~ in the graph at right? vector A *(a) 30◦ north of east (b) 30◦ south of east (c) 30◦ north of west (d) 30◦ south of west (e) None of these
N 6
*
~ A -
E
Solution: Of the solutions, only 30◦ north of east is in the first quadrant.
Problem 760. What is the direction of ~ in the graph at right? vector A (a) 30◦ north of east (b) 30◦ south of east (c) 30◦ north of west *(d) 30◦ south of west (e) None of these Solution: Of the solutions, only 30◦ south of west is in the third quadrant.
Problem 761. What is the direction of ~ in the graph at right? vector A (a) 30◦ north of east (b) 30◦ south of east *(c) 30◦ north of west (d) 30◦ south of west (e) None of these Solution: Of the solutions, only 30◦ north of west is in the second quadrant.
N 6
~ A
-
E
N 6
~ Y HH A HH
H
H
-
E
211 Problem 762. What is the direction of ~ in the graph at right? vector A (a) 30◦ east of north (b) 60◦ east of north (c) 30◦ west of north *(d) 60◦ west of north (e) None of these Solution: 30◦ west of north and 60◦ west of north are both in the second quadrant. However, 30◦ west of north would be closer to the north direction than to the west direction. Hence: 60◦ west of north. Problem 763. What is the direction of ~ in the graph at right? vector A (a) 30◦ east of north (b) 60◦ east of north *(c) 30◦ west of north (d) 60◦ west of north (e) None of these Solution: 30◦ west of north and 60◦ west of north are both in the second quadrant. However, 60◦ west of north would be closer to the west direction than to the north direction. Hence: 30◦ west of north. Problem 764. What is the direction of ~ in the graph at right? vector A (a) 30◦ east of north *(b) 60◦ east of north (c) 30◦ west of north (d) 60◦ west of north (e) None of these Solution: 30◦ east of north and 60◦ east of north are both in the first quadrant. However, 30◦ east of north would be closer to the north direction than to the east direction. Hence: 60◦ east of north.
N 6
~ Y HH A HH
HH
-
E
N 6
~ A AK A A A A A
-
E
N 6
*
~ A -
E
212 Problem 765. What is the direction of ~ in the graph at right? vector A *(a) 30◦ east of north (b) 60◦ east of north (c) 30◦ west of north (d) 60◦ west of north (e) None of these
N 6
~ A
-
E
Solution: 30◦ east of north and 60◦ east of north are both in the first quadrant. However, 60◦ east of north would be closer to the east direction than to the north direction. Hence: 30◦ east of north.
6.1.4
Given vectors on figure, name quadrant of sum/difference
~ and B ~ are Problem 766. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ What quadrant label the vector A ~ ~ does A − B lie in? (a) Quadrant I (b) Quadrant II *(c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis ~ laSolution: Sketch a copy of −B, ~ 0 , with its tail at the head of A. ~ belled −B ~−B ~ has its tail at the tail of A ~ Then A 0 ~ and its head at the head of −B .
y 6
~ B
~ A ~0 −B
9
~−B ~ A
-
x
213 ~ and B ~ are Problem 767. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ What quadrant label the vector A ~ ~ does A + B lie in? (a) Quadrant I *(b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis ~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Then B ~ ~ ~ A + B has its tail at the tail of A and its ~ 0. head at the head of B
y 6
~ B
~0 B ~ A
~ and B ~ are Problem 769. The vectors A labelled on the figure at right. Sketch and ~ − A. ~ What quadrant label the vector B ~ −A ~ lie in? does B (a) Quadrant I *(b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis ~ laSolution: Sketch a copy of −A, 0 ~ ~ belled −A , with its tail at the head of B. ~ −A ~ has its tail at the tail of B ~ Then B 0 ~. and its head at the head of −A
~+B ~ A
-
)
~ and B ~ are Problem 768. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ What quadrant label the vector A ~−B ~ lie in? does A (a) Quadrant I (b) Quadrant II *(c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis ~ laSolution: Sketch a copy of −B, 0 ~ , with its tail at the head of A. ~ belled −B ~ ~ ~ Then A − B has its tail at the tail of A 0 ~. and its head at the head of −B
x
y 6
~ B
~ A
)
~ 0 −B
-
x
~−B ~ A
y 6
~ ~ 0 B −A
)
~ −A ~ B
) ~ A
-
x
214 ~ and B ~ are Problem 770. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ What quadrant label the vector A ~ ~ does A + B lie in? (a) Quadrant I (b) Quadrant II (c) Quadrant III (d) Quadrant IV *(e) It lies on the x- or y-axis ~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Then B ~ ~ ~ A + B has its tail at the tail of A and its ~ 0 : on the negative head at the head of B x-axis ~ and B ~ are Problem 771. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ What quadrant label the vector A ~+B ~ lie in? does A *(a) Quadrant I (b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis
y 6
~ B ~+B ~ A iP P
PP P
~0 B
iP P
P )
PP PP
-
x
~ A
y
~+B ~ A
6
~ B
~0 B
~ 1 A
x
~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Then B ~ ~ ~ A + B has its tail at the tail of A and its ~ 0. head at the head of B ~ and B ~ are Problem 772. The vectors A labelled on the figure at right. Sketch and ~ − A. ~ What quadrant label the vector B ~ −A ~ lie in? does B (a) Quadrant I *(b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis ~ laSolution: Sketch a copy of −A, 0 ~ ~ belled −A , with its tail at the head of B. ~ −A ~ has its tail at the tail of B ~ Then B 0 ~. and its head at the head of −A
y 6
~ ~ 0 B −A
)
~ −A ~ B
1
~ A -
x
215 ~ and B ~ are Problem 773. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ What quadrant label the vector A ~ ~ does A − B lie in? (a) Quadrant I (b) Quadrant II (c) Quadrant III *(d) Quadrant IV (e) It lies on the x- or y-axis
y 6
~ B 1 ~ A x ~0 −B
~ laSolution: Sketch a copy of −B, 0 ~ , with its tail at the head of A. ~ belled −B ~ ~ ~ Then A − B has its tail at the tail of A ~ 0. and its head at the head of −B ~ and B ~ are Problem 774. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ What quadrant label the vector A ~−B ~ lie in? does A *(a) Quadrant I (b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis
~−B ~ A
~−B ~ y A 6
B B B B B NB ~
~ laSolution: Sketch a copy of −B, 0 ~ , with its tail at the head of A. ~ belled −B ~ ~ ~ Then A − B has its tail at the tail of A 0 ~. and its head at the head of −B ~ and B ~ are Problem 775. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ What quadrant label the vector A ~+B ~ lie in? does A (a) Quadrant I (b) Quadrant II (c) Quadrant III *(d) Quadrant IV (e) It lies on the x- or y-axis ~ labelled Solution: Sketch a copy of B, 0 ~ ~ Then B , with its tail at the head of A. ~ +B ~ has its tail at the tail of A ~ and its A 0 ~. head at the head of B
MB B
~ B−B B B ~ 1B A 0
-
x
B
y 6
~ 1 A B B B B x ~ 0B B B B B ~+B ~ B A NB B NB ~ B
216 ~ and B ~ are Problem 776. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ What quadrant label the vector A ~ ~ does A − B lie in? (a) Quadrant I (b) Quadrant II (c) Quadrant III *(d) Quadrant IV (e) It lies on the x- or y-axis
y 6
~ B MB B B B B BP
~ laSolution: Sketch a copy of −B, 0 ~ , with its tail at the head of A. ~ belled −B ~ ~ ~ Then A − B has its tail at the tail of A ~ 0. and its head at the head of −B ~ and B ~ are Problem 777. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ What quadrant label the vector A ~+B ~ lie in? does A *(a) Quadrant I (b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis
~ laSolution: Sketch a copy of −A, 0 ~ ~ belled −A , with its tail at the head of B. ~ −A ~ has its tail at the tail of B ~ Then B 0 ~. and its head at the head of −A
x
PP q
~−B ~ A
~
BA B ~0 B−B B B NB
y 6
~ B MB
~+B ~ A
B B B B
MB B
~0 BB B BP PP x PP B qB ~ P A
~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Then B ~ ~ ~ A + B has its tail at the tail of A and its ~ 0. head at the head of B ~ and B ~ are Problem 778. The vectors A labelled on the figure at right. Sketch and ~ − A. ~ What quadrant label the vector B ~ −A ~ lie in? does B (a) Quadrant I *(b) Quadrant II (c) Quadrant III (d) Quadrant IV (e) It lies on the x- or y-axis
PP P
y
iP P ~0 6 P− PA PP B ~ MB B ~ −A ~ B B B B BP PP P
-
x
PP q
~ A
217 6.1.5
Mixing it up
Problem 779. Five vectors are graphed on the diagram at right and described below. Write the letter of each vector beside its description. ~ 3ˆı − ˆ E ~ −4ˆı + 2ˆ B ~ 2ˆı − 3ˆ D ~ ˆı + 4ˆ A ~ −3ˆı − 3ˆ C
y
~ 6 A ~ B YH H H HH H HH P J PPP PP q J J J J J ^ ~ ~
y
~ B AK
6 A A A
~ C
~ h3, −4i E ~ h−2, −1i D ~ h−2, 4i B
x
~ E
D
C
Problem 780. Five vectors are graphed on the diagram at right and described below. Write the letter of each vector beside its description. ~ h2, 2i A ~ h−3, 1i C
-
~ A
A A iP P PP A PP A S S ~
-
x
D
S S S S S w S
Problem 781. Five velocity vectors are graphed on the diagram at right and described below. Write the letter of each vector beside its magnitude and direction. ~ 14 m/s; 0.79 rad south of east E ~ 45 m/s; 1.11 rad south of east D ~ 41 m/s; 1.33 rad south of west C ~ 45 m/s; 0.46 rad north of west A ~ 36 m/s; 0.59 rad south of west B
~ E
N 6
~ A YH H H
~ B
HH H
HH A@ A@ R ~ A E + A A A A AU ~
C
-
E
~ D
218 Problem 782. Five velocity vectors are graphed on the diagram at right and described below. Write the letter of each vector beside its magnitude and direction. ~ 22 mph; 64◦ north of east E ~ 50 mph; 37◦ south of east C ~ 36 mph; 34◦ south of west A ~ 32 mph; 18◦ north of east D ~ 28 mph; 45◦ south of east B
N 6
~ E
~ A
1 Z @Z @Z @Z RZZ @ +
~ B
~ D -
E
Z
~ C ~ Z
y
~ is labelled Problem 783. The vector A on the figure at right. Sketch and label ~ the vector −A. ~ should be the same Solution: −A ~ in exactly the opposite dilength as A, rection.
6
~ A
~ −A
~ is labelled Problem 784. The vector A on the figure at right. Sketch and label ~ the vector −A. ~ should be the same Solution: −A ~ length as A, in exactly the opposite direction.
-
x
y 6
~ A PP i PP
PP P
PP q P
~ −A
x
219 Problem 785. Which of the vectors in the graph corresponds to ˆı − 3ˆ? ~ ~ (a) A (b) B ~ ~ *(c) C (d) D (e) None of these Solution: Since the tail is at the origin, the tip must be at the point (1, −3) (the 3rd quadrant). The magnitude of the y-component is greater than the xcomponent, so the answer must be C.
Problem 786. Five vectors are graphed on the diagram at right and described below. Write the letter of each vector beside its description. ~ = h2, 2i A ~ = h−3, 1i C
y 6
~ i PP A PP
~ B
PP P B PPP PP B q B B B
NB ~
C
6 A A A
~ = h3, −4i E ~ = h−2, −1i D ~ = h−2, 4i B
x
~ D
y
~ B AK
~ C
-
~ A
A A iP P PP A PP A S S ~
-
x
D
S S S S S w S
Problem 787. Which of the vectors in the graph corresponds to −6ˆı − 3ˆ? ~ ~ (a) A (b) B ~ ~ (c) C *(d) D (e) None of these ~ has negative x- and Solution: Only D y-components.
~ B AK
~ E
y 6
A A
~ C
A A A iP P PP A PP A
~ D
~ A -
x
220
6.2
Geometric Vector Addition and Subtraction y
~ and B ~ are Problem 788. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ label the vector A ~ labelled Solution: Sketch a copy of B, ~ 0 , with its tail at the head of A. ~ Then B ~ +B ~ has its tail at the tail of A ~ and its A 0 ~ head at the head of B .
6
~+B ~ A
~ 6 B
-
~0 B -
x
~ A
y
~ and B ~ are Problem 789. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ label the vector A ~ laSolution: Sketch a copy of −B, 0 ~ ~ belled −B , with its tail at the head of A. ~−B ~ has its tail at the tail of A ~ Then A 0 ~. and its head at the head of −B
6
~ 6 B ~ A J J J J J^ J
-
x
~0 −B ~−B ~ A
y
~ and B ~ are Problem 790. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ label the vector A
6
~ B 6
~ A
~0
−B
~−B ~ A
-
x
221 ~ Y~ , and Problem 791. The vectors X, ~ are labelled on the figure at right. Z Which of the following equations is true? ~ =X ~ − Y~ ~ = Y~ − X ~ (a) Z *(b) Z ~ =X ~ + Y~ ~ = −X ~ − Y~ (c) Z (d) Z (e) None of these Solution:
y 6
~ Z
Y~
] JJ J J J J
-
x
~ = Z. ~ Notice that Y~ + (−X) ?
~ X
y
~ Y~ , and Z ~ Problem 792. The vectors X, are labelled on the figure at right. Con~ From the figure it is sider the vector −Z. clear that
~ Z
~ = ±X ~ ± Y~ . −Z
Y~
6 ] JJ J J J J
Which of the following equations is true? ~ = −X ~ + Y~ ~ =X ~ + Y~ (a) −Z (b) −Z ~ = −X ~ − Y~ ~ =X ~ − Y~ (c) −Z (d) −Z (e) None of these ~ + (−Y~ ) = −Z. ~ Solution: X
~ and B ~ are Problem 793. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ label the vector A ~ labelled Solution: Sketch a copy of B, 0 ~ ~ Then B , with its tail at the head of A. ~ +B ~ has its tail at the tail of A ~ and its A 0 ~. head at the head of B
-
x
?
~ X
y 6
~+B ~ A 6
~ B YH H
~0 B
* H HH
~ A -
x
222 y
~ and B ~ are Problem 794. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ label the vector A ~ laSolution: Sketch a copy of −B, 0 ~ , with its tail at the head of A. ~ belled −B ~ ~ ~ Then A − B has its tail at the tail of A 0 ~. and its head at the head of −B
~ and B ~ are Problem 795. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ label the vector A ~ labelled Solution: Sketch a copy of B, ~ 0 , with its tail at the head of A. ~ Then B ~ +B ~ has its tail at the tail of A ~ and its A 0 ~ head at the head of B .
6
~ A
~ B KA A A A A
~0 −B -
-
x
~−B ~ A
y 6
~ A
~0 B
~+B ~ A I @ @
@ )
@
-
x
~ B
~ and B ~ are Problem 796. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ label the vector A ~ laSolution: Sketch a copy of −B, 0 ~ ~ belled −B , with its tail at the head of A. ~−B ~ has its tail at the tail of A ~ Then A 0 ~. and its head at the head of −B
y 6
~−B ~ A 6 ~0 −B PP S PP PP S q S S S S S w S
-
x
~ A
~ B
223 y
~ and B ~ are Problem 797. The vectors A labelled on the figure at right. Sketch and ~ + B. ~ label the vector A ~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Then B ~ ~ ~ A + B has its tail at the tail of A and its ~ 0. head at the head of B
6
~ B x B@ B@ B @ B @ ~ ~ B RA+B @ B BBN ~0
B
~ A
~ and B ~ are Problem 798. The vectors A labelled on the figure at right. Sketch and ~ − B. ~ label the vector A ~ laSolution: Sketch a copy of −B, ~ 0 , with its tail at the head of A. ~ belled −B ~−B ~ has its tail at the tail of A ~ Then A 0 ~ and its head at the head of −B .
y 6
~ B I @ @ @ @
~ A
@ 9 ~0 −B
-
x
~−B ~ A
~ B, ~ and C ~ Problem 799. The vectors A, are labelled on the figure at right. Sketch ~+B ~ + C. ~ and label the vector A ~ labelled Solution: Sketch a copy of B, 0 ~ ~ Sketch B , with its tail at the head of A. ~ labelled C ~ 0 , with its tail at a copy of C, 0 ~ . Then A ~ +B ~ +C ~ will have the head of B ~ its tail at the tail of A and its head at the ~ 0. head of C
y
~ BMB C
6 B B
~+B ~ +C ~ A
~ BB B 6 B B B S S S
~0 C -
x
6 S
~0 B
S SS w
~ A
224 ~ B, ~ and C ~ Problem 800. The vectors A, are labelled on the figure at right. Sketch ~+B ~ − C. ~ and label the vector A ~ labelled Solution: Sketch a copy of B, 0 ~ , with its tail at the head of A. ~ Sketch B 0 ~ ~ a copy of −C, labelled −C , with its tail ~ 0 . Then A ~+B ~ −C ~ will at the head of B ~ have its tail at the tail of A and its head ~ 0. at the head of −C
y I 6 @
~ C H Y HH H
~ B
~0 −C
~0 B
HH @ @ H I HH@ H
~ A @ H
1
~+B ~ −C ~ A x
225
6.3
Position vs. Displacement Vector Problems
Problem 801. On the graph below, draw and label the position vectors ~r1 and ~r2 corresponding to the points P1 and P2 on the graph, and the displacement vector ∆~r = ~r2 −~r1 . y 6
P2 r y X XXX∆~ XrXXrP1 BM B
~r2 B B
~r1 B B
Problem 802. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. On the axes at right, sketch and label the spider’s displacement vectors: ~ at 12:15 A ~ at 12:30 B ~ at 12:45 C ~ at 1:00 D Solution: Don’t forget that you’re measuring displacement from the 12:00 position, not from the center of the clock.
-
x
y 6
~
~ C
~
rD = 0 @ @ @ R @
~ ?B
-
x
~ A
226 Problem 803. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. Which of the vectors on the figure at right is in the direction of the spider’s displacement vector at 12:15? ~ ~ (a) C *(b) D ~ ~ (c) G (d) H (e) None of these
y 6
~ A
~ H
~ B
6
@ I @ @
~ G
~0
@
-
@ @
~ C x
@ @
F~
@ R @
?
~ D
~ E
Problem 804. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. Which of the vectors on the figure at right is in the direction of the spider’s displacement vector at 12:30? ~ (a) The zero vector ~0 (b) A ~ ~ (c) B *(d) E (e) None of these
y 6
~ A
~ H
~ B
6 @ I @ @
~ G
~0
@
@ @
-
~ C x
@ @
F~
@ R @
?
~ D
~ E
Problem 805. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. Which of the vectors on the figure at right is in the direction of the spider’s displacement vector at 12:45? ~ (a) B *(b) F~ ~ ~ (c) G (d) H (e) None of these
y 6
~ A
~ H
~ B
6 @ I @ @
~ G
@
~0 @ @
@ @
F~
@ R @
?
~ E
~ D
~ C x
227 Problem 806. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. Which of the vectors on the figure at right is in the direction of the spider’s displacement vector at 1:00? ~ *(a) The zero vector ~0 (b) A ~ ~ (c) B (d) E (e) None of these
y 6
~ A
~ H
~ B
6
@ I @ @
~ G
~0
@
-
@ @
~ C x
@ @
F~
@ R @
?
~ D
~ E
Problem 807. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. If the origin is at the center of the clock, which of the vectors on the figure at right is the spider’s position vector at 12:15? ~ ~ (a) B *(b) C ~ ~ (c) G (d) H (e) None of these
y 6
~ A
~ H
~ B
6 @ I @ @
~ G
~0
@
@ @
-
~ C x
@ @
F~
@ R @
?
~ D
~ E
Problem 808. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. If the origin is at the center of the clock, which of the vectors on the figure at right is the spider’s position vector at 12:30? ~ (a) The zero vector ~0 (b) A ~ *(c) E (d) F~ (e) None of these
y 6
~ A
~ H
~ B
6 @ I @ @
~ G
@
~0 @ @
@ @
F~
@ R @
?
~ E
~ D
~ C x
228 Problem 809. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. If the origin is at the center of the clock, which of the vectors on the figure at right is the spider’s position vector at 12:45? ~ (a) B (b) F~ ~ ~ *(c) G (d) H (e) None of these
y 6
~ A
~ H
~ B
6
@ I @ @
~ G
~0
@
-
@ @
~ C x
@ @
F~
@ R @
?
~ D
~ E
Problem 810. At exactly 12:00, a spider climbs onto the tip of a clock’s minute hand, where it remains for the next hour. If the origin is at the center of the clock, which of the vectors on the figure at right is the spider’s position vector at 1:00? ~ (a) The zero vector ~0 *(b) A ~ ~ (c) B (d) E (e) None of these
y 6
~ A
~ H
~ B
6 @ I @ @
~ G
@
~0 @ @
@ @
F~
@ R @
?
~ E
~ D
~ C x
229
6.4
Finding Components of Vectors
~ has magnitude 5.9 and direction θ = 0.34 rad. Find the xProblem 811. The vector A ~ and y-components of A. ~ cos θ = (5.9)(cos 0.34 rad) = 5.6 Solution: Ax = kAk ~ sin θ = (5.9)(sin 0.34 rad) = 2.0 Ay = kAk ~ has magnitude 83 and direction θ = 1.98 rad. Find the xProblem 812. The vector A ~ and y-components of A. ~ cos θ = (83)(cos 1.98 rad) = −33 Solution: Ax = kAk ~ Ay = kAk sin θ = (83)(sin 1.98 rad) = 76 ~ has magnitude 41 and direction θ = 3.51 rad. Find the xProblem 813. The vector A ~ and y-components of A. ~ cos θ = (41)(cos 3.51 rad) = −38 Solution: Ax = kAk ~ sin θ = (41)(sin 3.51 rad) = −15 Ay = kAk ~ has magnitude 7.7 and direction θ = 5.66 rad. Find the xProblem 814. The vector A ~ and y-components of A. ~ cos θ = (7.7)(cos 5.66 rad) = 6.3 Solution: Ax = kAk ~ sin θ = (7.7)(sin 5.66 rad) = −4.5 Ay = kAk ~ has magnitude 0.88 and direction 37◦ . Find the x- and Problem 815. The vector A ~ y-components of A. ~ cos θ = 0.88 cos 37◦ = 0.70 Solution: Ax = kAk ~ sin θ = 0.88 sin 37◦ = 0.53 Ay = kAk ~ has magnitude 1.28 and direction 98◦ . Find the x- and Problem 816. The vector A ~ y-components of A. ~ cos θ = 1.28 cos 98◦ = −0.18 Solution: Ax = kAk ~ sin θ = 1.28 sin 98◦ = 1.27 Ay = kAk ~ has magnitude 110 and direction 219◦ . Find the x- and Problem 817. The vector A ~ y-components of A. ~ cos θ = 110 cos 219◦ = −85 Solution: Ax = kAk ~ sin θ = 110 sin 219◦ = −69 Ay = kAk ~ has magnitude 37 and direction 304◦ . Find the x- and Problem 818. The vector A ~ y-components of A. ~ cos θ = 37 cos 304◦ = 21 Solution: Ax = kAk ~ sin θ = 37 sin 304◦ = −31 Ay = kAk
230 ~ = h5.1, 8.8i. Find the magnitude and direction of A. ~ Give the direction Problem 819. A in radians. q √ ~ Solution: kAk = Ax 2 + Ay 2 = 5.12 + 8.82 = 10.2 Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence: 8.8 −1 Ay θ = tan = tan−1 = 1.05 rad Ax 5.1 ~ = h16, −41i. Find the magnitude and direction of A. ~ Give the Problem 820. A direction in radians. q p ~ = Ax 2 + Ay 2 = 162 + (−41)2 = 44 Solution: kAk Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence: Ay 41 = 5.08 rad θ = 2π − tan−1 = 2π − tan−1 Ax 16 ~ = h−0.81, −1.13i. Find the magnitude and direction of A. ~ Give the Problem 821. A direction in degrees. q p ~ = Ax 2 + Ay 2 = (−0.81)2 + (−1.13)2 = 1.39 Solution: kAk Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence: 1.13 ◦ −1 Ay = 234◦ θ = 180 + tan = 180◦ + tan−1 Ax 0.81 ~ = h−216, 82i. Find the magnitude and direction of A. ~ Give the Problem 822. A direction in degrees. q p ~ = Ax 2 + Ay 2 = (−216)2 + 822 = 231 Solution: kAk Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence: 82 ◦ −1 Ay θ = 180 − tan = 180◦ − tan−1 = 159◦ Ax 216 ~ = 4.8ˆı + 7.0ˆ. Find the magnitude and direction of A. ~ Give the Problem 823. A direction in radians. q √ ~ = Ax 2 + Ay 2 = 4.82 + 7.02 = 8.5 Solution: kAk Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence: 7.0 −1 Ay θ = tan = tan−1 = 0.97 rad Ax 4.8 ~ = −120ˆı − 220ˆ. Find the magnitude and direction of A. ~ Give the Problem 824. A direction in radians. q p ~ = Ax 2 + Ay 2 = (−120)2 + (−220)2 = 251 Solution: kAk Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence: 220 −1 Ay θ = π + tan = π + tan−1 = 4.21 rad Ax 120
231 ~ = −29ˆı + 12ˆ. Find the magnitude and direction of A. ~ Give the Problem 825. A direction in degrees. q p ~ = Ax 2 + Ay 2 = (−29)2 + 122 = 31 Solution: kAk Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence: 12 ◦ −1 Ay θ = 180 − tan = 180◦ − tan−1 = 158◦ Ax 29 ~ = 0.16ˆı − 0.23ˆ. Find the magnitude and direction of A. ~ Give the Problem 826. A direction in degrees. q p ~ = Ax 2 + Ay 2 = 0.162 + (−0.23)2 = 0.28 Solution: kAk Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence: 0.23 ◦ −1 Ay θ = 360 − tan = 360◦ − tan−1 = 305◦ Ax 0.16
6.5
Algebraic Vector Addition and Subtraction
~ = 3ˆı − 2ˆ and B ~ = 4ˆı + 5ˆ. Find A ~ + B. ~ Problem 827. A ~+B ~ = (3 + 4)ˆı + (−2 + 5)ˆ = 7ˆı + 3ˆ Solution: A ~ = −ˆı + 7ˆ and B ~ = 2ˆı + 3ˆ. Find A ~ − B. ~ Problem 828. A ~−B ~ = (−1 − 2)ˆı + (7 − 3)ˆ = −3ˆı + 4ˆ Solution: A ~ = h8, 3i and B ~ = h1, 2i. Find A ~ + B. ~ Problem 829. A ~+B ~ = h8 + 1, 3 + 2i = h9, 5i Solution: A ~ = h2, 6i and B ~ = h−4, 5i. Find A ~ − B. ~ Problem 830. A ~−B ~ = h2 − (−4), 6 − 5i = h6, 1i Solution: A ~ = h4, −1i and B ~ = h−5, −3i. Find A ~ + B. ~ Problem 831. A (a) h−9, −4i (c) h3, −8i (e) None of these Solution:
(b) h1, 2i *(d) h−1, −4i
h4, −1i + h−5, −3i = h4 + (−5), −1 + (−3)i = h−1, −4i
~ = h−2, 5i and B ~ = h3, −3i. Find A ~ + B. ~ Problem 832. A (a) h5, 8i (c) h−5, 8i (e) None of these Solution:
*(b) h1, 2i (d) h−6, −15i
~ +B ~ = h−2, 5i + h3, −3i = h−2 + 3, 5 − 3i = h1, 2i. Add like components: A
232 ~ = h−2, 5i and B ~ = h7, −3i. Find A ~ − B. ~ Problem 833. A (a) h5, 2i *(c) h−9, 8i (e) None of these Solution: h−9, 8i.
(b) h−9, 2i (d) h5, −2i
~−B ~ = h−2, 5i − h7, −3i = h−2 − 7, 5 + 3i = Add like components: A
~ has magnitude 7.2 and direction 1.1 rad. B ~ has magnitude 5.5 and Problem 834. A ~ + B. ~ Give your answer in component form: hCx , Cy i. direction 2.9 rad. Find A ~ and B ~ in component form. Solution: We begin by putting A D E ~ ~ ~ A = kAk cos θA , kAk sin θA , = h(7.2)(cos 1.1 rad), (7.2)(sin 1.1 rad)i D E ~ = kBk ~ cos θB , kBk ~ sin θB , = h(5.5)(cos 2.9 rad), (5.5)(sin 2.9 rad)i B ~+B ~ = h(7.2)(cos 1.1 rad) + (5.5)(cos 2.9 rad), (7.2)(sin 1.1 rad) + (5.5)(sin 2.9 rad)i A = h−2.1, 7.7i ~ has magnitude 0.83 and direction 22◦ . B ~ has magnitude 1.42 and Problem 835. A ◦ ~ − B. ~ Give your answer in terms of unit vectors: Cxˆı + Cyˆ. direction 157 . Find A ~ and B ~ in unit-vector form: Solution: We begin by putting A ~ = kAk ~ cos θA ˆı + kAk ~ sin θA ˆ = 0.83 cos 22◦ ˆı + 0.83 sin 22◦ ˆ A ~ = kBk ~ cos θB ˆı + kBk ~ sin θB ˆ = 1.42 cos 157◦ ˆı + 1.42 sin 157◦ ˆ B ~−B ~ = (0.83 cos 22◦ − 1.42 cos 157◦ )ˆı + (0.83 sin 22◦ − 1.42 sin 157◦ )ˆ A = 2.08ˆı − 0.24ˆ ~ has magnitude 115 and direction 3.8 rad. B ~ has magnitude 303 and Problem 836. A ~ + B. ~ direction 5.1 rad. Find the magnitude and direction of A ~ and B ~ into component form; adding Solution: This will require three steps: putting A the two vectors; and then finding the magnitude and direction of the resultant. D E ~ = kAk ~ cos θA , kAk ~ sin θA , = h(115)(cos 3.8 rad), (115)(sin 3.8 rad)i A D E ~ ~ ~ B = kBk cos θB , kBk sin θB , = h(303)(cos 5.1 rad), (303)(sin 5.1 rad)i ~+B ~ = h(115)(cos 3.8 rad) + (303)(cos 5.1 rad), (115)(sin 3.8 rad) + (303)(sin 5.1 rad)i A = hCx , Cy i = h24, −351i
233 You should not use these rounded component values in calculating magnitude and direction. However, you’ll need to know their signs in order to make sure your angle is in the right quadrant. Since the x-component is positive and the y-component is negative, θ is in the fourth quadrant. q ~ + Bk ~ = Cx 2 + Cy 2 = 352 kA −1 Cy θ = 2π − tan = 4.78 rad Cx ~ has magnitude 57 and direction 43◦ . B ~ has magnitude 22 and direction Problem 837. A ◦ ~ ~ 117 . Find the magnitude and direction of A − B. ~ and B ~ in component form; finding C ~ =A ~ − B; ~ This will require three steps: putting A ~ and calculating the magnitude and direction of C. ~ = kAk ~ cos θA ˆı + kAk ~ sin θA ˆ = 57 cos 43◦ ˆı + 57 sin 43◦ ˆ A ~ = kBk ~ cos θB ˆı + kBk ~ sin θB ˆ = 22 cos 117◦ ˆı + 22 sin 117◦ ˆ B ~−B ~ = (57 cos 43◦ − 22 cos 117◦ )ˆı + (57 sin 43◦ − 22 sin 117◦ )ˆ A = Cxˆı + Cyˆ = 52ˆı + 19ˆ You should not use these rounded component values in calculating magnitude and direc~ However, you’ll need their signs to make sure you have θ in the right quadrant. tion of C. Since both components are positive, θ is in the first quadrant. q ~ ~ kA − Bk = Cx 2 + Cy 2 = 55 θ = tan−1
Cy = 20◦ Cx
~ = h3.3, 5.7i, what is the magnitude of A? ~ Round your answer to Problem 838. If A the nearest 0.1. (a) 5.3 (b) 5.9 *(c) 6.6 (d) 7.2 (e) None of these q p Solution: kAk = A2x + A2y = (3.3)2 + (5.7)2 = 6.6 ~ = h3, 4i, what is the magnitude of A? ~ Problem 839. If A *(a) (c) (e)
5 (b) 6 25 (d) 7 None of these q p Solution: kAk = A2x + A2y = (3)2 + (4)2 = 6.6
234 ~ = 3ˆı − 4ˆ, what is the magnitude of A? ~ Problem 840. If A (a) -1 (c) 4 (e) None of these Solution:
~ = kAk
(b) *(d)
2 5
p (3)2 + (−4)2 = 5
235
6.6
Concept vector and scalar questions
The first three problems deal with trigonometric functions. Problem 841. cos(5 m/s) = ? Solution: make sense.
We can only take the cosine of a pure number, so this question doesn’t
Problem 842. Consider the expression: sin(ω ∗ t∗ ) = ?, where t∗ has units of time. What must the units of ω ∗ be in order for the argument of the sine function to make sense? Solution:
[ω ∗ ] = T1 .
Problem 843. Why is an angle a dimensionless number when measured in radians? Solution: The angle in radians is the ratio of the arc intercepted by the angle to the radius of the circle: arc length angle in radians = radius The arc length and the radius are both measured in units of length; so length in the numerator and denominator cancel, leaving no units. Problem 844. True or false: If we add the scalars 1 m + 1 m, we get 2 m. Solution:
True. We add scalars as we do regular numbers: 1 + 1 = 2.
Problem 845. True or false: If we add two vectors, each with length 1 m, we get a vector with length 2 m. Solution: False. Suppose the first vector is 1 m long in the positive x-direction, and the second is 1 m long in the negative x-direction. Then the sum of the two vectors is zero. Unless the two vectors are pointing in exactly the same direction, we can’t find the magnitude of their sum by adding the magnitudes of the two vectors. Problem 846. Can you find a vector ~v = hv1 , v2 i such that ~v 6= 0, but k~v k = 0? Solution:
No. The magnitude of ~v is q k~v k = v12 + v22
Since it has to be the case that v12 ≥ 0 and v22 ≥ 0, the only way k~v k can be zero is if v12 = v22 = 0. But that means that v1 = v2 = 0, so ~v = 0. Problem 847. Is it possible for the magnitude of a vector to be smaller than any of its components?
236 Solution: No. For simplicity’s sake, let’s look at a vector with two components: ~v = hv1 , v2 i; and let’s suppose that v1 < v2 . The magnitude of ~v is q q k~v k = v12 + v22 ≥ v12 + 0 = |v1 | ≥ v1 Hence the magnitude of ~v is no smaller than its smallest component. Technically, we should prove this for an arbitrary vector ~v = hv1 , . . . , vn i, where vk is the smallest component. But the notation would get cumbersome, and the essentials of the proof are the same. Problem 848. True or false: Let ~v = ha, bi. If k~v k > 0, then a > 0 and b > 0. Solution:
False. Suppose a = −3 and b = −4. Then p √ k~v k = (−3)2 + (−4)2 = 25 = 5
In this example, k~v k > 0, and a < 0 and b < 0. ~ = Ax ˆı + Ay ˆ and makes an angle θ with the positive x-axis, express Problem 849. If A sin θ in terms of Ax and Ay . Solution:
Ay sin θ = p 2 Ax + A2y
~ = Ax ˆı + Ay ˆ and makes an angle θ with the positive x-axis, express Problem 850. If A cos θ in terms of Ax and Ay . Solution:
Ax cos θ = p 2 Ax + A2y
~ = Ax ˆı + Ay ˆ, then the equation for the magnitude Problem 851. True or false: If A satisfies A = Ax + Ay ? Solution:
False. A =
q
Ax 2 + Ay 2 6= Ax + Ay . For example, let Ax = Ay = −1.
~ = Ax ˆı + Ay ˆ and makes an angle θ with the positive x-axis, express Problem 852. If A θ in terms of Ax and Ay . Assume that Ax > 0 and Ay > 0. Ay −1 Solution: θ = tan . [Since Ax > 0 and Ay > 0, θ is in the first quadrant; so Ax we don’t have to worry about reference angles.] ~ = Ax ˆı + Ay ˆ, then the equation for the magnitude Problem 853. True or false: If A ~ ≥ |Ax |. Justify your answer. satisfies kAk q p ~ Solution: True. kAk = Ax 2 + Ay 2 ≥ Ax 2 = |Ax |
237 Problem 854. True or false: Let ~r = hx, yi = x ˆı + y ˆ. If k~rk > 0, then x > 0 and y > 0. Justify your answer. √ Solution: False. Let ~r = h−1, −1i. Then k~rk = 2 > 0. [There are many other counterexamples you could use.] Problem 855. Can you find a situation where the average velocity vector is equal to the zero vector, but the average speed is not zero? In symbols: Can we have ~vave =
∆~r ~ total distance travelled = 0 and speed = 6= 0 ? ∆t ∆t
Give an example, or explain why one doesn’t exist. Solution: A car drives 100 km straight north at 100 km/h, then quickly turns around and drives back to its starting point at 100 km/h. The total time it takes is ∆t = 2 h. Since it has returned to its starting point, ∆~r = ~0. The total distance travelled is 200 km, so the average speed is 100 km/h. [There are many other examples you could use.] Problem 856. Consider the statements (i) and (ii). Are the statements true or false? (i) If k is a scalar, then k + k = 2k (ii) Let ~v = ha, bi. If kvk > 0, then a > 0 and b > 0. Choose the correct answer from below. *(a) (i) is true; (ii) is false (c)
Both statements are true
(b)
(i) is false; (ii) is true
(d)
Both statements are false
Solution: (i) is true. (ii) is false: every nonzero vector has a positive magnitude, including those with negative components. For example, if ~v = h−3, 4i, then k~v k = p 2 2 (−3) + 4 = 5. Problem 857. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~v is a vector, then k2~v k = 2k~v k. (ii) Let ~v = ha, bi. If kvk = 0, then a = 0 and b = 0. Choose the correct answer from below. (a) (i) is true; (ii) is false *(c)
Both statements are true
(b)
(i) is false; (ii) is true
(d) Both statements are false
238 Problem 858. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~v is a vector, then k − ~v k = −k~v k. (ii) Let ~v = ha, bi. Then 2~v = h2a, 2bi. Choose the correct answer from below. (a) (i) is true; (ii) is false *(b) (c)
Both statements are true
(d)
(i) is false; (ii) is true Both statements are false
Solution: (ii) is true. (i) is false: the magnitude of a vector is always positive, unless the vector is ~0, in which case the magnitude is zero. Hence k − ~v k = k~v k. Problem 859. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~v is a vector, then k − ~v k = k~v k. (ii) Let ~v = ha, bi. Then −2~v = −2a − 2b. Choose the correct answer from below. *(a) (i) is true; (ii) is false (c)
Both statements are true
Solution:
(b)
(i) is false; (ii) is true
(d)
Both statements are false
(i) is true. (ii) is false: if ~v = ha, bi, then −2~v = h−2a, −2bi.
Problem 860. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u and ~v are vectors, then k~u + ~v k = k~uk + k~v k. √ (ii) Let ~v = ha, bi. Then k~v k = a2 + b2 . Choose the correct answer from below. (a) (i) is true; (ii) is false *(b) (c)
Both statements are true
(d)
(i) is false; (ii) is true Both statements are false
Solution: (ii) is true. (i) is false; for example, let ~u = h1, 0i and ~v = h−1, 0i. Then ~ ~u + ~v = 0; so k~uk = k~v k = 1, but k~u + ~v k = 0 6= 1 + 1. Problem 861. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u is a vector and k is a scalar, then kk~uk = kk~uk. √ (ii) Let ~v = ha, bi. Then k~v k = a + b. Choose the correct answer from below. (a) (i) is true; (ii) is false (b) (c)
Both statements are true
*(d)
(i) is false; (ii) is true Both statements are false
Solution: (i) is false. Let ~u = h1, 0i, and let k = −2. Then k~uk = 1; k~u = h−2, 0i, so kk~uk = 2 6= kk~uk = (−2)(1). (A true version of (i) would be: kk~uk = |k| · k~uk.) √ (ii) is also false. k~v k = a2 + b2 .
239 Problem 862. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u is a vector, k is a scalar, and k > 0, then kk~uk = kk~uk. (ii) Let ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i. Then v1 + v2 = ha1 + a2 , b1 + b2 i. Choose the correct answer from below. (a) (i) is true; (ii) is false *(c)
Both statements are true
(b)
(i) is false; (ii) is true
(d)
Both statements are false
Solution: Both statements are true. Statement (i) is a specific case of the general statement: If ~u is a vector and k is a scalar, then kk~uk = |k| · k~uk. If k > 0, then |k| = k. Problem 863. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u is a vector, then k − 2~uk = 2k~uk. (ii) Let ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i. Then v1 + v2 = ha1 + b1 , a2 + b2 i. Choose the correct answer from below. *(a) (i) is true; (ii) is false (c)
Both statements are true
(b)
(i) is false; (ii) is true
(d)
Both statements are false
Statement (i) is a specific case of the general statement: If ~u is a vector and k is a scalar, then kk~uk = |k| · k~uk. If k = −2, then |k| = 2. Statement (ii) is a garbled version of the true statement: if ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i, then ~v1 + ~v2 = ha1 + a2 , b1 + b2 i. Problem 864. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u is a vector, then ~u + (−~u) = ~0. (ii) Let ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i. Then v1 − v2 = ha1 − b1 , a2 − b2 i. Choose the correct answer from below. *(a) (i) is true; (ii) is false (c)
Both statements are true
(b)
(i) is false; (ii) is true
(d)
Both statements are false
Solution: (i) is true. (ii) is a garbled version of the true statement: if ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i, then v1 − v2 = ha1 − a2 , b1 − b2 i. Problem 865. Consider the statements (i) and (ii). Are the statements true or false? (i) If ~u is a vector, then ~u − 2~u = ~u. (ii) Let ~v1 = ha1 , b1 i and ~v2 = ha2 , b2 i. Then v1 − v2 = ha1 − a2 , b1 − b2 i. Choose the correct answer from below. (a) (i) is true; (ii) is false *(b) (c)
Both statements are true
Solution:
(d)
(i) is false; (ii) is true Both statements are false
(i) is false; a true version would be: ~u − 2~u = −~u. (ii) is true.
240
6.7 6.7.1
Applications of Vectors Breaking vectors into components
Problem 866. An airplane is flying at a speed of V in a direction 22◦ north of east. Which expression gives the northward component of the plane’s velocity? (a) V cos 22◦ *(c) V sin 22◦ (e) None of these
(b) V / cos 22◦ (d) V / sin 22◦
Problem 867. An airplane is flying at a speed of V in a direction 22◦ north of east. Which expression gives the eastward component of the plane’s velocity? *(a) V cos 22◦ (c) V sin 22◦ (e) None of these
(b) V / cos 22◦ (d) V / sin 22◦
Problem 868. A car is driving at a speed of V up a hill that makes an angle of 22◦ with the horizontal. What is the vertical component of the car’s velocity? (a) V cos 22◦ *(c) V sin 22◦ (e) None of these
(b) V / cos 22◦ (d) V / sin 22◦
Problem 869. An airplane is flying at a speed of 190 mph in a direction 24◦ north of east. What is the northward component of the plane’s velocity? Round your answer to the nearest mph. *(a) (c) (e)
77 mph 161 mph None of these
Solution:
(b) 101 mph (d) 175 mph
vy = v sin θ = (190 mph) sin 24◦ = 77 mph
Problem 870. An airplane is flying at a speed of 310 mph in a direction 77◦ north of east. What is the eastward component of the plane’s velocity? Round your answer to the nearest mph. *(a) (c) (e)
70 mph 180 mph None of these
Solution:
(b) 155 mph (d) 294 mph
vx = v cos θ = (310 mph) cos 77◦ = 70 mph
241 Problem 871. An airplane is flying at 214 m/s in a direction 37◦ north of east. What is the eastward component of its velocity? Round your answer to the nearest m/s. (a) 129 m/s (c) 161 m/s (e) None of these Solution:
(b) 134 m/s *(d) 171 m/s
vE = (214) cos 37◦ = 171
Problem 872. An airplane is flying at 220 mph in a direction 30◦ north of east. What is the northward component of its velocity? Solution: You don’t need a calculator, since every good physics student knows that ◦ sin 30 = 1/2. Then vN = v sin θ = (220 mph) sin 30◦ = 110 mph Problem 873. An airplane is flying at 220 mph in a direction 30◦ east of north. What is the northward component of its velocity? Round your answer to the nearest mph. Solution: vN = v cos θ = (220 mph) cos 30◦ = 191 mph Problem 874. A hill has a slope of 0.30 radians above the horizontal. If you are driving up that hill at 22 km/hr, how fast are you gaining elevation? Solution:
You want the vertical component of your velocity: vy = v sin θ = (22 km/hr)(sin 0.30 rad) = 6.5 km/hr
Problem 875. A car is driving at 35 mph up a hill with a slope of 8◦ above the horizontal. What is the horizontal component of the car’s velocity? Round your answer to the nearest 0.1 mph. (a) 4.9 mph (c) 7.5 mph (e) None of these Solution:
(b) 5.1 mph *(d) 34.7 mph
vx = v cos θ = (35 mph) cos 8◦ = 34.7 mph
Problem 876. A gun is fired at an elevation of 39◦ above the horizontal. The shell emerges from the muzzle at 320 m/s. What is the vertical component of the shell’s velocity? Round your answer to the nearest m/s. (a) 85 m/s (c) 249 m/s (e) None of these Solution:
*(b) (d)
201 m/s 308 m/s
vy = v sin θ = (320 m/s) sin 39◦ = 201 m/s
242 Problem 877. (Lab Problem) A spring cannon is fired at an elevation of 34◦ above the horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontal component of the ball’s velocity? Round your answer to the nearest 0.1 m/s. (a) 8.1 m/s *(c) 11.9 m/s (e) None of these Solution:
(b) 9.7 m/s (d) 12.2 m/s
vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s
Problem 878. You are on an airplane flying upward at an angle of θ above the horizontal, at 110 m/s. You have a cold, and if you gain altitude faster than 25 m/s, your eardrums will explode. What is the maximum safe value for θ? Round your answer to the nearest degree. (a) 12◦ (c) 14◦ (e) None of these Solution:
*(b) 13◦ (d) 16◦
You know v and vy ; you want to know θ.
vy = v sin θ
⇒
vy sin θ = v
⇒
−1
θ = sin
hv i y
v
−1
= sin
25 m/s = 13◦ 110 m/s
Problem 879. (Lab Problem) A spring cannon is fired at an elevation of 34◦ above the horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontal component of the ball’s velocity? Round your answer to the nearest 0.1 m/s. (a) 8.1 m/s *(c) 11.9 m/s (e) None of these Solution:
(b) 9.7 m/s (d) 12.2 m/s
vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s
Problem 880. An bird is flying at 20 m/s in a direction 30◦ south of west. If we take the x-axis pointing east and the y-axis pointing north, then what is the southward component of its velocity ~v = hvx , vy i? Round your answer to the nearest m/s. √ 3 m/s (b) 10 m/s (a) 10 √ *(c) −10 3 m/s (d) −10 m/s (e) None of these Solution: The tip of the vector lies in the 3rd quadrant, so the y-component (the southward component) is negative. vy = vS = v sin θ = −20 sin(30◦ ) m/s = −10 m/s Problem 881. Your Civil War cannon has a muzzle velocity of v0 = 1220 ft/s. If it is fired at an elevation of 6◦ above the horizontal, what is the vertical component v0y of the velocity? Round your answer to the nearest ft/s. Solution: v0y = v0 sin θ = (1220 ft/s) sin 6◦ = 128 ft/s
243 Problem 882. An airplane is flying at 330 km/hr, in a direction 0.19 rad above the horizontal. At what speed is the airplane gaining altitude? Solution: We want the vertical component of the airplane’s velocity vector: vy = v sin θ = (330 km/hr)(sin 0.19 rad) = 62 km/hr Problem 883. A train is moving at 68 mph in a direction 22◦ north of west. What is the train’s speed westward? Solution: We want the westward component of the train’s velocity vector: vW = v cos θ = (68 mph)(cos 22◦ ) = 63 mph Problem 884. (Position) Pirates have buried their treasure on campus, and you have acquired a copy of their map. (Like everyone else, pirates need to be careful about that “Reply All” button.) It tells you to start at the main entrance to the physics building and take 120 steps in a direction 0.38 rad north of east; from there, you should go another 190 steps in a direction 0.33 rad south of east. How far from the entrance is the treasure, and in what direction? ~ and the second B. ~ We need Solution: We will call the first displacement vector A ~ ~ ~ to find the magnitude and direction of C = A + B. We begin by writing the vectors in component form: D E ~ ~ ~ A = kAk cos θA , kAk sin θA = h120 · cos 0.38 rad, 120 · sin 0.38 radi D E ~ ~ ~ B = kBk cos θB , kBk sin θB = h190 · cos(−0.33 rad), 190 · sin(−0.33 rad)i Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of ~ =A ~+B ~ are: C CE = 120 · cos 0.38 rad + 190 · cos 0.33 rad = 291 steps CN = 120 · sin 0.38 rad − 190 · sin 0.33 rad = −17 steps Since CE is positive and CN is negative, the direction is in the fourth quadrant: that is, south of east. We calculate p ~ = CE 2 + CN 2 = 292 steps kCk −1 CN south of east = 0.06 rad south of east θ = tan CE Problem 885. (Position) You have discovered the Lost Belgian Mine, which runs directly northward into a mountainside. From the opening, you go 47 m in a direction 0.28 rad below the horizontal. From there, you go another 66 m in a direction 0.19 rad below the horizontal. At this point, what is your horizontal distance from the opening,
244 and how far are you above or below it? Solution: We begin by finding the components of the two vectors: D E ~ = kAk ~ cos θA , kAk ~ sin θA = h47 m · cos(−0.28 rad), 47 m · sin(−0.28 rad)i A D E ~ = kBk ~ cos θB , kBk ~ sin θB = h66 m · cos(−0.19 rad), 66 m · sin(−0.19 rad)i B
Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of ~ =A ~+B ~ are: C Cx = 47 m · cos 0.28 rad + 66 m · cos 0.19 rad = 110 m Cy = −47 m · sin 0.28 rad − 66 m · sin 0.19 rad = −25 m Your horizontal distance from the entrance is 110 m, and you are 25 m below it. Problem 886. A goose is directly east of an airplane. The airplane is flying at a speed of 160 mph, in a direction 15◦ north of east. The goose is flying at a speed of 45 mph, in a direction θ north of west. What must θ be in order for the goose to collide with the airplane? Give your answer in degrees. Solution: Since the goose is directly east of the airplane, its northbound component of velocity must match the airplane’s if they are to collide. That means (45 mph)(sin θ) = (160 mph)(sin 15◦ ) Solving for θ gives us −1
θ = sin
(160 mph)(sin 15◦ ) 45 mph
= 67◦
Problem 887. You are tracking a submarine by determining its position when it sends radio messages. When you first detect the sub, it is 57 km west and 24 km north of you. When it sends its next message 37 min later, it is 51 km west and 13 km north of you. What are the magnitude and direction of the sub’s average velocity during that time? (Give the magnitude in km/hr, and the direction in radians, relative to east or west: e.g. θ rad north of east, with 0 < θ < π/2.) Solution: Don’t forget to convert 37 min to hours. A look at the numbers shows that the submarine is moving to the east and south. The eastbound component of the velocity is: 57 km − 51 km 60 min vE = · ≈ 9.7 km/hr 37 min 1 hr The southbound component of the velocity is: vS =
24 km − 13 km 60 min · ≈ 17.8 km/hr 37 min 1 hr
245 (We present these intermediate results because it’s a good idea to look at them and make sure that they seem reasonable. You should not use these rounded numbers in your later calculations.) The magnitude of the velocity is p The direction is −1
tan
vE 2 + vS 2 = 20 km/hr
vS vE
= 1.1 rad south of east
Problem 888. A cockroach has a tiny embedded radio transponder, allowing you to track its position and velocity. Just before your lab partner attempts to stomp on it, it is moving at 8 cm/s to the east and 5 cm/s to the south. 1.3 seconds later, it is scurrying away at 31 cm/s to the west and 19 cm/s to the south. What are the components of the roach’s average acceleration during that time? Solution: A look at the numbers shows that the roach is accelerating westward and southward. The westward component of acceleration is aW =
31 cm/s − (−8 cm/s) = 30 cm/s2 1.3 s
The southward component is aS =
19 cm/s − 5 cm/s = 11 cm/s2 1.3 s
Problem 889. A car is travelling eastward along a highway at 60 mph. It passes under a hawk that is flying northward at 42 mph. From the point of view of an observer in the car, what are the hawk’s speed and direction? Give the direction in degrees relative to north, e.g. θ◦ east of north. Solution: For an observer in the car, objects fixed to the ground are moving westward at 60 mph. To that observer, the hawk is moving westward at vW = 60 mph and northward at vN = 42 mph. In that frame of reference, the hawk’s speed is p p v = vW 2 + vN 2 = (60 mph)2 + (42 mph)2 = 73 mph The direction, relative to north, is vW 60 mph −1 −1 tan = tan = 55◦ west of north vN 42 mph
246 Problem 890. You can paddle a kayak in still water at 2.6 m/s. The Santa Cruz River is 55 m wide, and flows northward at 1.9 m/s. If you launch your kayak on the east bank, and point it directly westward, what will your speed and direction be relative to an observer standing on the bank? Give your direction in radians relative to west, e.g. θ rad south of west. Solution: Relative to a fixed observer on the bank, you are moving westward at vW = 2.6 m/s and northward at vN = 1.9 m/s. In that frame of reference, your speed is p p v = vW 2 + vN 2 = (2.6 m/s)2 + (1.9 m/s)2 = 3.2 m/s Your direction relative to west is vN 1.9 m/s −1 −1 tan = tan = 0.63 rad north of west vW 2.6 m/s The width of the river is irrelevant to the problem. Problem 891. An airplane can fly at 410 km/hr in still air. The pilot wishes to fly it from Smithpur to Jonesgorod, 530 km to the north. The wind is blowing at 55 km/hr from west to east. In what direction should the pilot point the plane to fly directly to Jonesgorod? How long will the journey take? Give your direction in radians relative to north, e.g. θ rad east of north. Solution: The pilot wants to fly directly north. He must point the airplane west of north, so that his still-air velocity has a westward component that offsets the 55 km/hr eastward wind. If θ is the pilot’s heading west of north, then vW = (410 km/hr) sin θ = 55 km/hr Solving this for θ yields −1
θ = sin
55 km/hr 410 km/hr
= 0.13 rad west of north
The northward component of the plane’s velocity is (410 km/hr) cos θ. To fly 530 km to the north requires a time of t=
530 km = 1.3 hr (410 km/hr) cos θ
Problem 892. A river has a current with a speed of 5 ft/s. You can swim at 4 ft/s in still water. At what angle to the cross-river direction do you need to swim to reach a point directly across the river from you? Solution: No solution. Whatever the angle is, your upstream component of velocity can be no greater than the total magnitude of your velocity: 4 ft/s. Thus you will find yourself going downstream at 1 ft/s or more.
247 Problem 893. A train is going directly north at 70 mph. An action-movie star is atop the train, running southward at 12 mph. Relative to someone standing on the ground, how fast is the action-movie star moving, and in what direction? Solution: Relative to someone on the ground, the AMS is moving northward at 70 − 12 = 58 mph.
248 6.7.2
Introduction to force vectors
Working definition: You can think of a force as a push or a pull. Since a push or pull has both magnitude and direction, it follows that a force is a vector. 1 We’ll learn more about forces later on in the course when you learn about Newton’s Laws. In particular, Newton’s 2nd Law Fnet = ma requires computing the net force (sometimes called the resultant force) acting on an object of mass m. Since a force is a vector, the net force is just the resultant vector. In more than one spatial dimension we find the resultant vector by breaking the individual force vectors acting on the object into component vectors and then finding the net force in each of the x and y directions. The purposes of these exercises is to give you an early exposure to the concept of breaking forces into their respective components and summing them (keeping proper ± sign conventions) to find the net force in the the x and y directions, denoted Fnet,x and Fnet,y . Problem 894. A rope is used to pull a block up a ramp. The ramp makes an angle of 25◦ to the horizontal; the rope makes an angle of 33◦ with the ramp. The rope is pulled with a force F~ whose magnitude is 317 N. What is the horizontal component of F~ ? Round your answer to the nearest newton. *(a) 168 N (b) 185 N (c) 203 N (d) 224 N (e) None of these
y6
Solution: Draw a free-body diagram and add the angles to find the net angle θ between F~ and the horizontal: θ = 25◦ +33◦ = 58◦ . The horizontal component is
y6
F = 317 N 7
1
x
7
Fy = 317 sin 58◦ = 269 N
-
F~
Fx = 317 cos 58◦ = 168 N The vertical component is
◦ 33
! ! !! ! ! !!L !! L ! L ! L! ! ! ◦ !! 25
◦ 33! !! ◦ ! ! 25
Actually, this is not obvious and must be proven mathematically.
θ = 58◦
-
x
249 Problem 895. A rope is used to pull a block up a ramp. The ramp makes an angle of 25◦ to the horizontal; the rope makes an angle of 33◦ with the ramp. The rope is pulled with a force F~ whose magnitude is 317 N. What is the vertical component of F~ ? Round your answer to the nearest newton. (a) 218 N (b) 242 N *(c) 269 N (d) 296 N (e) None of these
y6
Solution: Draw a free-body diagram and add the angles to find the net angle θ between F~ and the horizontal: θ = 25◦ +33◦ = 58◦ . The horizontal component is
y6
F = 317 N 7 ◦ 33
! ! !! ! ! !!L !! L ! L ! L! ! ! ◦ !! 25
7
θ = 58◦
Fy = 317 sin 58◦ = 269 N
◦ 33! !! ◦ ! 25 !
Problem 896. A rope is used to pull a block up a ramp. The ramp makes an angle of 22◦ to the horizontal; the rope makes an angle of 31◦ with the ramp. The rope is pulled with a force F~ whose magnitude is 428 N. What is the horizontal component of F~ ? Round your answer to the nearest newton. (a) 209 N (b) 232 N *(c) 258 N (d) 283 N (e) None of these
y6
Solution: Draw a free-body diagram and add the angles to find the net angle θ between F~ and the horizontal: θ = 22◦ +31◦ = 53◦ . The horizontal component is
y6
x
F = 428 N ◦ 31
! ! !! ! ! !!L !! L ! L ! L! ! ! ◦ !! 22
-
x
F~ 7
Fy = 428 sin 53◦ = 342 N
-
7
Fx = 428 cos 53◦ = 258 N The vertical component is
x
F~
Fx = 317 cos 58◦ = 168 N The vertical component is
-
◦ 31! !! ◦ ! ! 22
θ = 53◦
-
x
250 Problem 897. Two forces F~1 and F~2 are acting on a particle, as shown at right. F~1 makes an angle of 35◦ with the x-axis and has magnitude 87 N. F~2 makes an angle of 18◦ with the y-axis and has magnitude 181 N. Use the component method to find the net force in the x-direction. Round your answer to the nearest newton. (The picture is not drawn to scale.) (a) 12 N (b) 13 N *(c) 15 N (d) 16 N Solution:
y
F2 = 181 N AK A A A F2y
18◦ A A A
F1 = 87 N
Au
F2x
3 F1y ◦
35 F1x
x
Let F~ = F~1 + F~2 . Then Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ = 15 N
Problem 898. Two forces F~1 and F~2 are acting on a particle, as shown at right. F~1 makes an angle of 35◦ with the x-axis and has magnitude 87 N. F~2 makes an angle of 18◦ with the y-axis and has magnitude 181 N. Use the component method to find the net force in the y-direction. Round your answer to the nearest newton. (The picture is not drawn to scale.) (a) 200 N *(b) 222 N (c) 244 N (d) 269 N Solution:
y
F2 = 181 N AK A A A F2y
18◦ A A A
F2x
Let F~ = F~1 + F~2 . Then Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ = 222 N
Au
F1 = 87 N 3 F1y ◦
35 F1x
x
251 Problem 899. Two forces F~1 and F~2 are acting on a particle, as shown at right. F~1 makes an angle of 35◦ with the x-axis and has magnitude 87 N. F~2 makes an angle of 18◦ with the y-axis and has magnitude 181 N. Find the magnitude of the net force on the particle. Round your answer to the nearest newton. (The picture is not drawn to scale.) (a) 200 N *(b) 223 N (c) 245 N (d) 269 N Solution:
y
F2 = 181 N AK A A A F2y
18◦ A A A
F1 = 87 N
Au
F2x
3 F1y ◦
35 F1x
x
Let F~ = F~1 + F~2 . Then
Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ q F = Fx2 + Fy2 1/2 = ((87 N) cos 35◦ − (181 N) sin 18◦ )2 + ((87 N) sin 35◦ + (181 N) cos 18◦ )2 = 223 N Problem 900. Two forces F~1 and F~2 are acting on a particle, as shown at right. F~1 makes an angle of 35◦ with the x-axis and has magnitude 87 N. F~2 makes an angle of 18◦ with the y-axis and has magnitude 181 N. If F~net is the net force on the particle, find the angle that F~net makes with the x-axis. Round your answer to the nearest degree. (The picture is not drawn to scale.) (a) 83◦ *(b) 86◦ (c) 89◦ (d) 92◦ Solution:
y
F2 = 181 N AK A A A F2y
18◦ A A A
Au
F2x
F1 = 87 N 3 F1y ◦
35 F1x
Let F~ = F~1 + F~2 . Then Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ ≈ 15 N Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ ≈ 222 N
Since Fx and Fy are both positive, F~ is in the first quadrant. Hence Fy (87 N) sin 35◦ + (181 N) cos 18◦ −1 −1 θ = tan = tan = 86◦ Fx (87 N) cos 35◦ − (181 N) sin 18◦
x
252 Problem 901. You are pulling a crate of physics books down the hallway, using a rope that makes an angle of 34◦ to the horizontal. If you are pulling on the rope with a force of 440 N, what is the horizontal component of the force? Round your answer to the nearest newton. (a) 373 N (b) 233 N *(c) 365 N (d) 246 N (e) None of these Solution:
The horizontal component of the force is Fh = F cos θ = (440 N) cos 34◦ = 365 N
Problem 902. You are pulling a crate of physics books down the hallway, using a rope that makes an angle of 26◦ to the horizontal. If you are pulling on the rope with a force of 290 N, what is the horizontal component of the force? Round your answer to the nearest newton. (a) 127 N (b) 188 N (c) 221 N *(d) 261 N (e) None of these Solution:
The horizontal component of the force is Fh = F cos θ = (290 N) cos 26◦ = 261 N
Problem 903. Your car is stuck in a muddy road, and you and a friend are trying to pull it out with ropes. You are pulling with a force of 820 N in a direction 0.45 rad east of north; your friend is pulling with a force of 1040 N in a direction 0.32 rad west of north. What is the northward force on the car? Solution: We will use a coordinate system in which north is the positive x-direction and west is the positive y-direction. In that system, the angle at which you are pulling is −0.45 rad; the angle at which your friend is pulling is 0.32 rad. We will need to write the two force vectors in component form, then find their sum, then find the magnitude and direction of that sum. The two force vectors are: D E ~ = kAk ~ cos θA , kAk ~ sin θA = h820 N · cos(−0.45) rad, 820 N · sin(−0.45) radi A D E ~ = kBk ~ cos θB , kBk ~ sin θB = h1040 N · cos 0.32 rad, 1040 N · sin 0.32 radi B ~ =A ~+B ~ is: Remember that cos(−θ) = cos θ. Then the northerly component of C CN = 820 N · cos 0.45 rad + 1040 N · cos 0.32 rad = 1730 N We have rounded this number to 10 N, since that appears to be the accuracy of the original data.
253 Problem 904. A boat is held in the middle of a river by two ropes, one going to either bank. Each rope makes an angle of 79◦ to the upstream direction of the river. The current exerts a downstream force of 550 lbs on the boat. How much force does each rope have to exert to hold the boat in place? (It is the same for both ropes.) Solution: There are two ropes, each exerting a force with a magnitude of F . The upstream component of each force vector is F cos 79◦ . The two ropes working together must offset the downstream force of the current: 2F cos 79◦ = 550 lbs Solving this for F give us F =
550 lbs = 1440 lbs 2 cos 79◦
Problem 905. You are riding your bicycle straight north. The wind is exerting a force of 48 lbs on you, in a direction 39◦ east of south. What is the southward component of the wind’s force? Solution: The southward component is FS = F cos θ = (48 lbs)(cos 39◦ ) = 37 lbs Problem 906. A dietitian is pulling a crate of romaine lettuce, using a rope that makes an angle of 0.41 rad to the horizontal. To move the crate requires a horizontal force of 240 N. What is the force with which the dietitian needs to pull on the rope? Solution: The number we want is F , the magnitude of the force vector. The horizontal force is Fx = F cos θ = F (cos 0.41 rad) = 240 N Solving this for F give us F =
240 N Fx = = 262 N cos θ cos 0.41 rad
254
Part III
Kinematics
255
7 7.1
One-dimensional linear kinematics Concept questions: kinematics
Problem 907. Is it possible for an object to have nonzero velocity and zero speed? Give an example, or explain why it’s not possible. Solution: No. Speed is the magnitude of the velocity vector; and only the zero vector has a magnitude of zero. Problem 908. Is it possible for an object to have nonzero speed and zero velocity? Give an example, or explain why it’s not possible. Solution: No. Speed is the magnitude of the velocity vector; and the zero vector has a magnitude of zero. Problem 909. Is it possible for an object to have zero average velocity and nonzero average speed? Give an example, or explain why it’s not possible. Solution: Yes. Suppose you walk 100 ft north at 4 mph, then turn around and walk 100 ft south at 4 mph, returning to your starting point. Your average speed is 4 mph, since that’s the speed at which you were walking the whole time. (We’ll neglect the very short time it took you to turn around.) However, your average velocity is zero, because you returned exactly to your starting point. Problem 910. Is it possible for an object to have zero average speed and nonzero average velocity? Give an example, or explain why it’s not possible. Solution: No. Speed is always greater than or equal to zero, so the only way to have an average speed of zero is for the speed to be zero the whole time. That means that the velocity must also be zero the whole time; so the average velocity must be zero. Problem 911. You are given a ticket for driving at 37 mph in a school zone with a speed limit of 15 mph. At your hearing, you inform the judge that since you were going in reverse, your speed was −37 mph; and since −37 < 15, you were going below the speed limit and should not be penalized. Unfortunately for you, the policeman has taken physics. What does he tell the judge? Solution: The policeman informs the judge that your velocity vector may have been negative; but that speed, the magnitude of the velocity vector, must always be greater than or equal to zero. Thus although your velocity was negative, your speed was +37 mph, and greater than the legal limit.
256 Problem 912. You drive in a straight line from your home to your physics classroom. Your driveway is 30 m long; you drive down it at 5 m/s. You then drive 300 m down Smith Street at 15 m/s. After that, you drive 4.8 km down Jones Road to your classroom at 30 m/s. Once you get there, you realize that it’s Saturday, and that you don’t have to come to class. You turn around and return home along the same route: at 40 m/s along Jones Road; at 20 m/s along Smith Street; and at 10 m/s up your driveway, back to where you started. What is your average velocity for the round trip? Solution: Remember the distinction between speed and velocity. You returned to your starting point; so your net displacement for the whole trip is zero; so your average velocity is zero. If we’d asked for your average speed, this problem would have been a lot more work. Problem 913. Assume that the direction in which a car faces is positive. Give an example of a situation in which the car has: (a) positive velocity and positive acceleration (b) positive velocity and negative acceleration (c) negative velocity and positive acceleration (d) negative velocity and negative acceleration (e) positive velocity and zero acceleration (f) zero velocity and negative acceleration Solution: (a) The car is moving forward and speeding up. (b) The car is moving forward, but the driver is braking to slow down. (c) The car is moving backward and slowing down. Remember that “positive acceleration” means accelerating in the positive direction; it doesn’t mean “speeding up, regardless of the direction you’re going”. (d) The car is moving backward and speeding up. Remember that “negative acceleration” means accelerating in the negative direction, not “slowing down, regardless of the direction you’re going”. (e) The car is moving forward at a constant speed. (f) This is a little more complicated. The car was driving up a hill when the engine died; the car coasted forward until it stopped for an instant, then started rolling backward. At the instant that it stopped, the velocity was zero; during the whole time after the engine died, the acceleration was negative. We could not have come up with a situation in which the velocity was zero and the acceleration was nonzero for any positive amount of time. Such a situation can only last for an instant. The reason for this is that acceleration is the change in velocity, and if velocity is not changing (because it’s zero!), then the acceleration must also be zero. Here’s perhaps a better example not involving a car. Throw a ball straight up into the air. The acceleration that the ball experiences is due to the gravitational force between the ball and the earth. If we take the positive y-direction as up, then the acceleration always points downward towards the earth in the negative y- direction. When the ball reaches
257 its maximum height it must stop, hence it has zero velocity, before it turns around and falls back to earth. At no point along the balls trajectory did the earth quit pulling on the ball, and so the ball’s acceleration never vanished. In fact, it remained negative and constant. Problem 914. You and your lab partner are designing a fountain for the front of the physics building. You have a pressurized tank with a nozzle that squirts water out at 5.2 m/s. You claim that if that nozzle is pointed upward, it will produce a beautiful spray of water. Your lab partner disagrees: he says that since the acceleration due to gravity is 9.8, and 5.2 < 9.8, the water will barely dribble out of the nozzle and run back down the sides of the pipe. Is he right? Why or why not? Solution: Your lab partner is wrong. He is trying to compare two quantities that have different dimensions: v0 = 5.2 m/s, and g = 9.8 m/s2 . You can only write equations and inequalities when all terms have the same dimensions. Problem 915. A car is being driven erratically along a straight stretch of highway. The graph below shows its position x as a function of time t. For each interval between consecutive points (A-B, B-C, etc.), describe whether the velocity of the car is positive, negative, or zero; and whether the acceleration is positive, negative, or zero. x 6
s D s
C s B s
A
s
s
E
F s
GA A A A A As
H -
t Solution:
We will do these out of order, starting with the easiest one:
E-F: The position of the car does not change during this time. Therefore the car has a velocity of zero. Since the velocity does not change during the time period, the acceleration is also zero. A-B: The position as a function of time is a straight line. That means that the car is moving at a constant velocity. Since the value of x increases with time, the velocity is positive. Since the velocity is constant, the acceleration is zero.
258 C-D: Again, the position as a function of time is a straight line, so the velocity is constant; the value of x increases with time, so the velocity is positive; and since the velocity is constant, the acceleration is zero. Notice that the slope of the line from C to D is greater than the slope from A to B. That means that the velocity between C and D is greater than the velocity between A and B. G-H: The position as a function of time is a straight line, so the velocity is constant. The value of x decreases with time, so the velocity is negative. Since the velocity is constant, the acceleration is zero. B-C: The position as a function of time is not a straight line, so the velocity is not constant. That means some kind of acceleration is occurring. The value of x increases with time over the interval, so the velocity is positive. The velocity increases: since the slope from C to D is greater than the slope from A to B, the CD velocity is greater than the AB velocity. Hence the acceleration is positive. D-E: The position as a function of time is not a straight line, so the velocity is not constant. That means that there is a nonzero acceleration. The value of x increases with time, so the velocity is positive. The velocity decreases from the positive value on interval CD to the zero velocity on interval EF; so the acceleration is negative. F-G: The position as a function of time is not a straight line, so the velocity is not constant. That means that there is a nonzero acceleration. The value of x decreases with time, so the velocity is negative. The velocity decreases from zero on interval EF to some negative value on interval GH, so there is negative acceleration. (Again, remember that “negative acceleration” doesn’t mean “slowing down”; it means accelerating in the negative direction, which can mean slowing down while moving forward or speeding up while moving backward.)
7.2
Qualitative kinematics: descriptions
Problem 916. A car is facing in the positive direction. It is moving forward at 65 mph; the driver has just seen a deer on the road and applied the brakes. Which of the following describes the car’s situation? (a) positive velocity; positive acceleration (b) positive velocity; zero acceleration *(c) positive velocity; negative acceleration (d) negative velocity; positive acceleration (e) none of these Solution: The car is moving forward, so the velocity is positive; it is slowing down (that is, accelerating in the backward direction), so the acceleration is negative.
259 Problem 917. A car is facing in the positive direction. It is rolling backward down a hill, and its speed is increasing. Which of the following describes the car’s situation? (a) positive velocity; positive acceleration (b) positive velocity; negative acceleration (c) negative velocity; positive acceleration *(d) negative velocity; negative acceleration (e) none of these Solution: The car is moving backward, so the velocity is negative. It is speeding up in the backward direction, so the acceleration is negative. Careful! “Negative acceleration” and “slowing down” are not the same thing. In this case, negative acceleration means speeding up in the negative direction. Problem 918. A car is facing in the positive direction. It is rolling backward down a hill; the drive has just applied the brakes to slow it down. Which of the following describes the car’s situation? (a) positive velocity; zero acceleration (b) positive velocity; negative acceleration *(c) negative velocity; positive acceleration (d) negative velocity; negative acceleration (e) none of these Solution: The car is moving backward, so the velocity is negative. It is slowing down (that is, its velocity is becoming more positive), so the acceleration is positive. “Positive acceleration” is not the same thing as “speeding up”. Problem 919. A car is facing in the positive direction. It is moving forward, and the driver is speeding up to merge onto the freeway. Which of the following describes the car’s situation? *(a) positive velocity; positive acceleration (b) positive velocity; zero acceleration (c) zero velocity; positive acceleration (d) negative velocity; zero acceleration (e) none of these Solution: The car is moving forward, so the velocity is positive. It is speeding up in the forward direction, so the acceleration is positive.
260 Problem 920. A car is facing in the positive direction. The driver is going at a steady speed of 55 mph on a highway. Which of the following describes the car’s situation? *(a) positive velocity; zero acceleration (b) positive velocity; negative acceleration (c) positive velocity; positive acceleration (d) zero velocity; positive acceleration (e) none of these Solution: The car is moving forward, so the velocity is positive. It is moving at a constant speed, so the acceleration is zero. Problem 921. A car is facing in the positive direction. It was going up a hill when the engine failed; the car slowed down as it rolled up the hill, then started rolling backward. Which of the following describes the car’s situation at the point where it started rolling backward? (a) positive velocity; positive acceleration (b) negative velocity; negative acceleration (c) zero velocity; positive acceleration *(d) zero velocity; negative acceleration (e) none of these Solution: At the moment that the car stopped rolling forward and started rolling backward, its velocity was zero. Since its speed was changing from forward to backward, the acceleration was negative. Problem 922. In this problem, use upward as the positive direction. A rocket is flying straight upward; its speed increases as it rises. Which of the following describes the rocket’s situation? *(a) positive velocity; positive acceleration (b) positive velocity; negative acceleration (c) zero velocity; positive acceleration (d) zero velocity; negative acceleration (e) none of these Solution: The rocket is moving upward, so its velocity is positive. It is speeding up in the upward direction, so its acceleration is positive.
261 Problem 923. In this problem, use downward as the positive direction. A ball has been dropped from the top of a building; it has not yet reached the ground. Which of the following describes the ball’s situation? *(a) positive velocity; positive acceleration (b) positive velocity; negative acceleration (c) zero velocity; negative acceleration (d) negative velocity; positive acceleration (e) none of these Solution: The ball is moving downward, so the velocity is positive. It is accelerating downward under the influence of gravity, so the acceleration is negative. Problem 924. In this problem, use downward as the positive direction. A ball has been dropped from the top of a building, has hit the ground and bounced, and is now moving upward. Which of the following describes the ball’s situation? (a) positive velocity; positive acceleration (b) positive velocity; negative acceleration (c) positive velocity; zero acceleration *(d) negative velocity; positive acceleration (e) none of these Solution: The ball is moving upward (the negative direction), so the velocity is negative. Gravity is slowing it down as it moves upward, so the acceleration is positive. Problem 925. In this problem, use downward as the positive direction. An elevator has just left the 39th floor, heading downward and speeding up as it goes. Which of the following describes the elevator’s situation? *(a) positive velocity; positive acceleration (b) positive velocity; negative acceleration (c) negative velocity; negative acceleration (d) negative velocity; positive acceleration (e) none of these Solution: The elevator is moving downward (the positive direction), so its velocity is positive. It is speeding up in the downward direction, so its acceleration is positive.
262 Problem 926. In this problem, use downward as the positive direction. An elevator is going down from the 39th floor to the 3rd. It is now at the level of the fourth floor, and is slowing down as it approaches the third. Which of the following describes the elevator’s situation? (a) positive velocity; positive acceleration *(b) positive velocity; negative acceleration (c) negative velocity; negative acceleration (d) negative velocity; positive acceleration (e) none of these Solution: The elevator is moving downward (the positive direction), so the velocity is positive. It is slowing down as it moves downward, so the acceleration is negative. Problem 927. In this problem, use downward as the positive direction. An elevator is on its way from the 89th floor to the 4th. It has reached its maximum downward speed, and has not yet started slowing down. Which of the following describes the elevator’s situation? (a) positive velocity; negative acceleration *(b) positive velocity; zero acceleration (c) negative velocity; negative acceleration (d) negative velocity; positive acceleration (e) none of these Solution: The elevator is moving downward (the positive direction), so the velocity is positive. It is moving at a constant speed, so the acceleration is zero. Problem 928. In this problem, use downward as the positive direction. An elevator has just left the 3rd floor, heading upward and speeding up as it goes. Which of the following describes the elevator’s situation? (a) positive velocity; positive acceleration (b) positive velocity; negative acceleration *(c) negative velocity; negative acceleration (d) negative velocity; zero acceleration (e) none of these Solution: The elevator is moving upward (the negative direction), so its velocity is negative. It is speeding up in the negative direction, so its acceleration is negative.
263 Problem 929. If the upward direction is positive, which situation involves positive velocity and positive acceleration? (a) An elevator moving upward at a constant speed (b) An elevator speeding up as it moves downward *(c) An elevator speeding up as it moves upward (d) An elevator slowing down as it moves downward (e) none of these Solution: For velocity to be positive, the elevator must be moving upward. For acceleration to be positive, it must be speeding up as it goes upward. Problem 930. If the upward direction is positive, which situation involves negative velocity and negative acceleration? (a) An elevator moving upward at a constant speed *(b) An elevator speeding up as it moves downward (c) An elevator speeding up as it moves upward (d) An elevator slowing down as it moves downward (e) none of these Solution: In order for velocity to be negative, the elevator must be moving downward. In order for acceleration to be negative, it must be speeding up in the downward direction. Problem 931. If the upward direction is positive, which situation involves positive velocity and zero acceleration? *(a) An elevator moving upward at a constant speed (b) An elevator speeding up as it moves downward (c) An elevator speeding up as it moves upward (d) An elevator slowing down as it moves downward (e) none of these Solution: In order for velocity to be positive, the elevator must be moving upward. In order for acceleration to be zero, it must be moving at a constant speed. Problem 932. If the upward direction is positive, which situation involves negative velocity and positive acceleration? (a) An elevator moving upward at a constant speed (b) An elevator speeding up as it moves downward (c) An elevator speeding up as it moves upward *(d) An elevator slowing down as it moves downward (e) none of these Solution: In order for velocity to be negative, the elevator must be moving downward. In order for acceleration to be positive, it must be slowing down as it moves downward.
264 Problem 933. If north is the positive direction, which situation involves positive velocity and negative acceleration? (a) A car speeding up as it moves southward (b) A car moving southward at a constant speed *(c) A car slowing down as it moves northward (d) A car slowing down as it moves southward (e) none of these Solution: In order for the velocity to be positive, the car must be moving northward. In order for the acceleration to be negative, it must be slowing down as it moves in the positive direction. Problem 934. If north is the positive direction, which situation involves negative velocity and negative acceleration? *(a) A car speeding up as it moves southward (b) A car moving southward at a constant speed (c) A car slowing down as it moves northward (d) A car slowing down as it moves southward (e) none of these Solution: In order for velocity to be negative, the car must be moving southward. In order for acceleration to be negative, the car must be speeding up as it moves in the negative direction. Problem 935. If north is the positive direction, which situation involves negative velocity and positive acceleration? (a) A car speeding up as it moves southward (b) A car moving southward at a constant speed (c) A car slowing down as it moves northward *(d) A car slowing down as it moves southward (e) none of these Solution: In order for the velocity to be negative, the car must be moving southward. In order for the acceleration to be positive, it must be slowing down as it moves in the negative direction. Problem 936. If north is the positive direction, which situation involves negative velocity and zero acceleration? (a) A car speeding up as it moves southward *(b) A car moving southward at a constant speed (c) A car slowing down as it moves northward (d) A car slowing down as it moves southward (e) none of these Solution: In order for the velocity to be negative, the car must be moving southward. In order for acceleration to be zero, it must be moving at a constant speed.
265 Problem 937. If the positive direction is downward, which situation involves negative velocity and negative acceleration? (a) A rocket slowing down as it moves upward (b) A rocket moving downward at a constant speed (c) A rocket moving upward at a constant speed *(d) A rocket speeding up as it moves upward (e) none of these Solution: In order for the velocity to be negative, the rocket must be moving upward. In order for the acceleration to be negative, it must be speeding up as it moves in the negative direction. Problem 938. If the positive direction is downward, which situation involves positive velocity and zero acceleration? (a) A rocket slowing down as it moves upward *(b) A rocket moving downward at a constant speed (c) A rocket moving upward at a constant speed (d) A rocket speeding up as it moves upward (e) none of these Solution: In order for the velocity to be positive, the rocket must be moving downward. In order for the acceleration to be zero, it must be moving at a constant speed. Problem 939. If the positive direction is downward, which situation involves negative velocity and zero acceleration? (a) A rocket slowing down as it moves upward (b) A rocket moving downward at a constant speed *(c) A rocket moving upward at a constant speed (d) A rocket speeding up as it moves upward (e) none of these Solution: In order for the velocity to be negative, the rocket must be moving upward. In order for the acceleration to be zero, it must be moving at a constant speed. Problem 940. If the positive direction is downward, which situation involves negative velocity and positive acceleration? *(a) A rocket slowing down as it moves upward (b) A rocket moving downward at a constant speed (c) A rocket moving upward at a constant speed (d) A rocket speeding up as it moves upward (e) none of these Solution: In order for the velocity to be negative, the rocket must be moving upward. In order for the acceleration to be positive, it must be slowing down as it moves in the negative direction.
266
7.3
Qualitative kinematics: from graph x 6
s D
s
s
E
F s
GA
s
A A
C
A A As
s B s
H
A
-
t Figure 1: A car is being driven erratically along a straight stretch of highway. The graph shows its position x as a function of the time t. Problem 941. For the interval between points A and B in figure 1, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between A and B, the velocity is positive.
Problem 942. For the interval between points B and C in figure 1, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between B and C, the velocity is positive.
Problem 943. For the interval between points C and D in figure 1, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between C and D, the velocity is positive.
Problem 944. For the interval between points D and E in figure 1, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between D and E, the velocity is positive.
Problem 945. For the interval between points E and F in figure 1, the car’s velocity is: (a) positive Solution:
(b) negative
*(c) zero
Since x does not change between E and F, the velocity is zero.
Problem 946. For the interval between points F and G in figure 1, the car’s velocity is: (a) positive Solution:
*(b) negative
(c) zero
Since x decreases between F and G, the velocity is negative.
267 Problem 947. For the interval between points G and H in figure 1, the car’s velocity is: (a) positive Solution:
*(b) negative
(c) zero
Since x decreases between G and H, the velocity is negative.
Problem 948. For the interval between points A and B in figure 1, the car’s acceleration is: (a) positive
(b) negative
*(c) zero
Solution: The graph of x versus t is a straight line between A and B; so the velocity is constant. That means that acceleration is zero. Problem 949. For the interval between points B and C in figure 1, the car’s acceleration is: *(a) positive
(b) negative
(c) zero
Solution: The slope of the graph is greater between C and D than it is between A and B. This means that the velocity between C and D is greater than the velocity between A and B; so positive acceleration occurs between B and C. Problem 950. For the interval between points C and D in figure 1, the car’s acceleration is: (a) positive
(b) negative
*(c) zero
Solution: The graph of x versus t is a straight line between C and D; so the velocity is constant. Thus the acceleration is zero. Problem 951. For the interval between points D and E in figure 1, the car’s acceleration is: (a) positive
*(b) negative
(c) zero
Solution: The velocity between C and D is positive; the velocity between E and F is zero. Thus during the interval between D and E, the velocity decreases; so the acceleration is negative. Problem 952. For the interval between points E and F in figure 1, the car’s acceleration is: (a) positive Solution:
(b) negative
*(c) zero
The velocity is constant between E and F, so the acceleration is zero.
Problem 953. For the interval between points F and G in figure 1, the car’s acceleration is: (a) positive
*(b) negative
(c) zero
Solution: The velocity between E and F is zero. The velocity between G and H is negative. Hence negative acceleration occurs between F and G.
268 Problem 954. For the interval between points G and H in figure 1, the car’s acceleration is: (a) positive
(b) negative
*(c) zero
Solution: The graph of x versus t is a straight line between G and H; so the velocity is constant. Hence the acceleration is zero. x 6
s
H
s
s
A
B
s G s
s C@@
F @s
D
s
E
-
t Figure 2: A car is being driven erratically along a straight stretch of highway. The graph shows its position x as a function of the time t. Problem 955. For the interval between points A and B in figure 2, the car’s velocity is: (a) positive Solution:
(b) negative
*(c) zero
The value of x does not change between A and B; so the velocity is zero.
Problem 956. For the interval between points B and C in figure 2, the car’s velocity is: (a) positive Solution:
*(b) negative
(c) zero
Since x decreases between B and C, the velocity is negative.
Problem 957. For the interval between points C and D in figure 2, the car’s velocity is: (a) positive Solution:
*(b) negative
(c) zero
Since x decreases between C and D, the velocity is negative.
Problem 958. For the interval between points D and E in figure 2, the car’s velocity is: (a) positive Solution:
*(b) negative
(c) zero
Since x decreases between D and E, the velocity is negative.
Problem 959. For the interval between points E and F in figure 2, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between E and F, the velocity is positive.
269 Problem 960. For the interval between points F and G in figure 2, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between F and G, the velocity is positive.
Problem 961. For the interval between points G and H in figure 2, the car’s velocity is: *(a) positive Solution:
(b) negative
(c) zero
Since x increases between G and H, the velocity is positive.
Problem 962. For the interval between points A and B in figure 2, the car’s acceleration is: (a) positive Solution:
(b) negative
*(c) zero
The velocity is constant between A and B; so the acceleration is zero.
Problem 963. For the interval between points B and C in figure 2, the car’s acceleration is: (a) positive
*(b) negative
(c) zero
Solution: The velocity is zero between A and B, and negative between C and D. Hence negative acceleration occurs between B and C. Problem 964. For the interval between points C and D in figure 2, the car’s acceleration is: (a) positive
(b) negative
*(c) zero
Solution: The graph of x versus t is a straight line between C and D, so the velocity is constant. Thus the acceleration is zero. Problem 965. For the interval between points D and E in figure 2, the car’s acceleration is: *(a) positive
(b) negative
(c) zero
Solution: The velocity is negative between C and D, and positive between E and F. Thus positive acceleration occurs between D and E. Problem 966. For the interval between points E and F in figure 2, the car’s acceleration is: *(a) positive
(b) negative
(c) zero
Solution: The velocity is negative between D and E, and positive between F and G. Thus positive acceleration occurs between E and F. Problem 967. For the interval between points F and G in figure 2, the car’s acceleration is: (a) positive
(b) negative
*(c) zero
Solution: The graph of x versus t is a straight line between F and G, so the velocity is constant. Hence the acceleration is zero.
270 Problem 968. For the interval between points G and H in figure 2, the car’s acceleration is: (a) positive
*(b) negative
(c) zero
Solution: The slope of the line of x versus t decreases between G and H; so the velocity decreases; so negative acceleration occurs.
271
7.4
Introducing the fundamental one-dimensional kinematic equations
Assumptions: Unless instructed otherwise, assume that air resistance does not occur; and that projectiles (cannonballs, golf balls, arrows, etc.) are launched from ground level. All of the following problems in this problem set can be solved using one of the following three fundamental equations: In practice there are three different sets of subscripts used on physical variables: 1. (subscripts: i for initial values, f for final values) Example: initial position xi ; final position xf 2. (subscripts: zero (pronounced nought) for initial values, no subscript for final values) Example: initial position x0 ; final position x 3. (subscripts: 1 for initial values, 2 for final values) Example: initial position x1 ; final position x2 The general fundamental kinematic equations in the first type of subscript notation are: Fundamental Equation 1: vf = vi + a(tf − ti ) Fundamental Equation 2: vf2 = vi2 + 2a(xf − xi ) 1 Fundamental Equation 3: xf = xi + vi (tf − ti ) + a(tf − ti )2 2 where a is acceleration, v is velocity, x is position, t is time, and the subscript i and f denotes initial and final values, respectively. Typically, we take the initial time ti = 0 since, in practice, when you time an experiment you always start with the stopwatch set to zero. Warning: These equations are only valid if the acceleration is constant. Assume acceleration is constant unless stated otherwise. The general fundamental kinematic equations in the second type of subscript notation are:
272 Fundamental formulas for motion in 1-D Equation 1: Equation 2: Equation 3:
v = v0 + at (use when: position not in problem statement) 2 2 v = v0 + 2a(x − x0 ) (use when: time not in problem statement) 1 x = x0 + v0 t + at2 (use when: want position as a fcn of time) 2
The general fundamental kinematic equations in the third type of subscript notation are: Fundamental formulas for motion in 1-D Equation 1: Equation 2: Equation 3:
v2 = v1 + at2 v22 = v12 + 2a(x2 − x1 ) 1 x2 = x1 + v1 t2 + at22 2
This final form has the potential of leading to much confusion. For this reason we will not use this form. But you should be aware that it can be found in textbooks everywhere. 7.4.1
Dimensional consistency of the fundamental equations
Problem 969. Verify that the 1st fundamental equation is dimensionally consistent. Solution: Start by rewriting the equation as ∆v = a∆t. Taking the units of the right-hand side: [a∆t] = (L/T2 )T]= L/T, the units for velocity. Problem 970. Verify that the 2nd fundamental equation is dimensionally consistent. Solution: Start by rewriting the equation as vf2 − vi2 = 2a∆x. Taking the units of the right and left-hand side yields [vf2 − vi2 ] = [2a∆x]], which leads to (L/T)2 = (L/T2 )L. Problem 971. Verify that the 3rd fundamental equation is dimensionally consistent. Solution:
Start by rewriting the equation as 1 ∆x = vi ∆t + a(∆t)2 2
÷∆t
−−−−−−→
∆x 1 = vi + a∆t ∆t 2
As we’ve already seen, a∆t has units of velocity. Thus the equation is dimensionally self consistent.
273
7.5
Quantitative kinematics: horizontal
Problem 972. Beginning at rest, a car accelerates down a straight road at 1.03 m/s2 for 6.7 s. What is the car’s final speed? Round your answer to the nearest 0.1 m/s. *(a) (c) (e)
6.9 m/s 8.4 m/s None of these
(b) (d)
7.6 m/s 9.2 m/s
Solution: We know acceleration and time (and initial velocity, which is zero); we want the final velocity. v = v0 + at = 0 m/s + (1.03 m/s2 )(6.7 s) = 6.9 m/s Problem 973. Beginning at rest, a car accelerates down a straight road at 2.18 m/s2 for 14.8 s. What is the car’s final speed? Round your answer to the nearest 0.1 m/s. (a) 6.8 m/s (c) 35.2 m/s (e) None of these
*(b) (d)
32.3 m/s 70.3 m/s
Solution: We know acceleration and time (and initial velocity, which is zero); we want the final velocity. v = v0 + at = 0 m/s + (2.18 m/s2 )(14.8 s) = 32.3 m/s Problem 974. A boat can accelerate at 2.9 m/s2 . If it begins at rest, how long must it accelerate before it reaches a speed of 14.4 m/s? Round your answer to the nearest 0.1 s. *(a) 5.0 s (b) 5.5 s (c) 6.0 s (d) 6.6 s (e) None of these Solution: We know the initial velocity (which is zero), the acceleration, and the final velocity. We want to know the time. v = v0 + at = at
⇒
t=
14.4 m/s v = = 5.0 s a 2.9 m/s2
274 Problem 975. A boat is initially going northward at 7.4 m/s. It accelerates at 0.36 m/s2 northward for 22 s. At the end of this time, how far north has it travelled? Round your answer to the nearest 10 m. (a) 200 m *(c) 250 m (e) None of these
(b) 220 m (d) 270 m
Solution: We know the initial position (zero), the initial velocity, the acceleration, and the time. We want to know the final position. 1 1 x = x0 + v0 t + at2 = (7.4 m/s)(22 s) + (0.36 m/s2 )(22 s)2 = 250 m 2 2 Problem 976. A car can accelerate at 1.31 m/s2 . If it begins at rest, how long must it accelerate before it reaches a speed of 20.4 m/s? Round your answer to the nearest 0.1 s. *(a) 15.6 s *(b) 17.1 s (c) 18.8 s (d) 20.7 s (e) None of these Solution: We know the initial velocity (which is zero), the acceleration, and the final velocity. We want to know the time. v = v0 + at = at
⇒
t=
20.4 m/s v = = 15.6 s a 1.31 m/s2
Problem 977. Beginning at rest, a car accelerates forward for 5.2 s, reaching a speed of 15.0 m/s. What is the car’s acceleration? Round your answer to the nearest 0.1 m/s2 . (a) 2.6 m/s2 (c) 3.2 m/s2 (e) None of these
*(b) 2.9 m/s2 (d) 3.5 m/s2
Solution: We know the initial speed (which is zero), the time, and the final speed. We want to know the acceleration. v = v0 + at = at
⇒
a=
v 15.0 m/s = = 2.9 m/s2 t 5.2 s
275 Problem 978. A ship is initially moving northward at 12.1 m/s. It accelerates northward at 1.91 m/s2 for 4.9 s. What is its final speed? Round your answer to the nearest 0.1 m/s. (a) 15.7 m/s (c) 45.9 m/s (e) None of these
*(b) (d)
21.5 m/s 56.6 m/s
Solution: We know the initial velocity, the acceleration, and the time. We want to know the final velocity. v = v0 + at = 12.1 m/s + (1.91 m/s2 )(4.9 s) = 21.5 m/s Problem 979. A car is initially moving northward at 5.9 m/s. It accelerates northward at 0.88 m/s2 for 12.1 s. What is its final speed? Round your answer to the nearest 0.1 m/s. (a) 13.4 m/s *(c) 16.5 m/s (e) None of these
(b) 14.9 m/s (d) 18.2 m/s
Solution: We know the initial velocity, the acceleration, and the time. We want to know the final velocity. v = v0 + at = 5.9 m/s + (0.88 m/s2 )(12.1 s) = 16.5 m/s Problem 980. A ship is initially moving eastward at 4.2 m/s. It accelerates over the course of 27 s, at the end of which it is moving eastward at 10.3 m/s. What is the ship’s acceleration? Round your answer to the nearest 0.01 m/s2 . (a) 0.20 m/s2 (c) 0.25 m/s2 (e) None of these
*(b) 0.23 m/s2 (d) 0.27 m/s2
Solution: We know the initial velocity, the final velocity, and the time. We want to know the acceleration. v = v0 + at
⇒
a=
10.3 m/s − 4.2 m/s v − v0 = = 0.23 m/s2 t 27 s
276 Problem 981. A train can accelerate at 1.04 m/s2 . It is initially travelling eastward at 10.0 m/s. How long must it accelerate before it reaches an eastward speed of 30.8 m/s? Round your answer to the nearest 0.1 s. (a) 18.0 s *(c) 20.0 s (e) None of these
(b) 19.1 s (d) 24.2 s
Solution: We know the initial velocity, the final velocity, and the acceleration. We want to know the time. v = v0 + at
⇒
t=
30.8 m/s − 10.0 m/s v − v0 = = 20.0 s a 1.04 m/s2
Problem 982. A train can accelerate from a complete stop to a speed of 31 m/s in 145 s. What is the train’s acceleration? Assume constant acceleration. Round your answer to two significant figures. (a) 0.15 m/s2 (c) 0.48 m/s2 (e) None of these
*(b) 0.21 m/s2 (d) 0.68 m/s2
Solution: Write down what information you are given, what you want, and which fundamental equation you need to use. vi = 0 m/s Want: a Which equation: Fundamental equation (1) Given: vf = 31 m/s ∆t = 145 s Solving for a in equation 1: a =
vf − vi 31 = = .21. Here we’ve used the fact that a = a. ∆t 145
Problem 983. A ship is initially moving northward at 16.7 m/s. It accelerates southward at 2.24 m/s2 for 3.8 s. What is its final speed? Round your answer to the nearest 0.1 m/s. *(a) (c) (e)
8.2 m/s 25.2 m/s None of these
(b) 21.7 m/s (d) 32.9 m/s
Solution: Careful! Notice that the initial velocity is northward, but that the acceleration is southward. We will take the northward direction to be positive, so the acceleration is negative. We know initial velocity, acceleration, and time; we want to know final velocity. v = v0 + at = 16.7 m/s + (−2.24 m/s2 )(3.8 s) = 8.2 m/s The final velocity is positive, which means that the ship is moving northward. However, the problem only asks for speed, which doesn’t include the direction.
277 Problem 984. A car is initially moving northward at 9.5 m/s. It accelerates southward at 3.34 m/s2 for 5.9 s. What is its final speed? Round your answer to the nearest 0.1 m/s. (a) 4.3 m/s *(c) 10.2 m/s (e) None of these
(b) 7.6 m/s (d) 29.2 m/s
Solution: Careful! Notice that the initial velocity is northward, but that the acceleration is southward. We will take the northward direction to be positive, so the acceleration is negative. We know initial velocity, acceleration, and time; we want to know final velocity. v = v0 + at = 9.5 m/s + (−3.34 m/s2 )(5.9 s) = −10.2 m/s The final velocity is negative, which means that the car is moving southward. The problem only asks for speed, which doesn’t include the direction; so 10.2 m/s is the correct answer. Problem 985. A train is initially rolling backward at 6.9 m/s. It accelerates forward at 0.58 m/s2 . How long must it accelerate before it is moving forward at 9.2 m/s? Round your answer to the nearest second. (a) 22 s *(c) 28 s (e) None of these
(b) 25 s (d) 31 s
Solution: We will use forward as the positive direction, so the train’s initial velocity will be negative. We know initial velocity, final velocity, and acceleration; we want to know time. v = v0 + at
⇒
t=
9.2 m/s − (−6.9 m/s) v − v0 = 28 s = a 0.58 m/s2
Problem 986. A ship is initially moving backward at 5.6 m/s. It accelerates over the course of 63 s, at the end of which it is moving forward at 11.0 m/s. What is the ship’s acceleration? Round your answer to the nearest 0.01 m/s2 . (a) 0.09 m/s2 (c) 0.24 m/s2 (e) None of these
(b) 0.11 m/s2 *(d) 0.26 m/s2
Solution: We will use forward as the positive direction, so the ship’s initial velocity will be negative. We know initial velocity, final velocity, and time; we want to know acceleration. v = v0 + at
⇒
a=
v − v0 11.0 m/s − (−5.6 m/s) = = 0.26 m/s2 t 63 s
278 Problem 987. Beginning at rest, a car accelerates down a straight road at 3.3 m/s2 for 9.8 s. At the end of this time, how far is the car from its starting point? Round your answer to the nearest 10 m. (a) 130 m (b) 140 m *(c) 160 m (d) 170 m (e) None of these Solution: We know acceleration, time, and initial velocity and position (both of which are zero). We want to know final position. 1 1 1 x = x0 + v0 t + at2 = at2 = (3.3 m/s2 )(9.8 s)2 = 160 m 2 2 2 Problem 988. Beginning at rest, a car accelerates down a straight road at 5.6 m/s2 for 2.9 s. At the end of this time, how far is the car from its starting point? Round your answer to the nearest meter. *(a) 24 m (b) 26 m (c) 28 m (d) 31 m (e) None of these Solution: We know acceleration, time, and initial velocity and position (both of which are zero). We want to know final position. 1 1 1 x = x0 + v0 t + at2 = at2 = (5.6 m/s2 )(2.9 s)2 = 24 m 2 2 2 Problem 989. A boat is initially going northward at 15.3 m/s. It accelerates at 0.93 m/s2 northward for 8.3 s. At the end of this time, how far north has it travelled? Round your answer to the nearest 10 m. (a) 150 m (c) 170 m (e) None of these
*(b) (d)
160 m 180 m
Solution: We know the initial position (zero), the initial velocity, the acceleration, and the time. We want to know the final position. 1 1 x = x0 + v0 t + at2 = (15.3 m/s)(8.3 s) + (0.93 m/s2 )(8.3 s)2 = 160 m 2 2
279 Problem 990. A boat is initially going northward at 8.8 m/s. It accelerates at 3.1 m/s2 northward for 9.9 s. At the end of this time, how far north has it travelled? Round your answer to the nearest 10 m. *(a) 240 m (b) 260 m (c) 280 m (d) 300 m (e) None of these Solution: We know the initial position (zero), the initial velocity, the acceleration, and the time. We want to know the final position. 1 1 x = x0 + v0 t + at2 = (8.8 m/s)(9.9 s) + (3.1 m/s2 )(9.9 s)2 = 240 m 2 2 Problem 991. A train is initially rolling southward at 7.9 m/s. It accelerates northward at 0.41 m/s2 for 15 s. At the end of this time, where is it relative to its starting point? Round your answer to the nearest meter. (a) 65 m south (c) 65 m north (e) None of these
*(b) 72 m south (d) 72 m north
Solution: We will take northward to be the positive direction, so the initial velocity will be negative and the acceleration will be positive. We know the initial position (zero), the initial velocity, the acceleration, and the time; we want to know the final position. 1 1 x = x0 + v0 t + at2 = (−7.9 m/s)(15 s) + (0.41 m/s2 )(15 s)2 = −72 m 2 2 Since the value of x is negative, it is south of the starting point. Problem 992. A boat is initially drifting southward at 4.2 m/s. It accelerates northward at 0.84 m/s2 for 12 s. At the end of this time, where is it relative to its starting point? Round your answer to the nearest meter. (a) 10 m south *(c) 10 m north (e) None of these
(b) 12 m south (d) 12 m north
Solution: We will take northward to be the positive direction, so the initial velocity will be negative and the acceleration will be positive. We know the initial position (zero), the initial velocity, the acceleration, and the time; we want to know the final position. 1 1 x = x0 + v0 t + at2 = (−4.2 m/s)(12 s) + (0.84 m/s2 )(12 s)2 = 10 m 2 2 Since the value of x is positive, it is north of the starting point.
280 Problem 993. A car is initially 30 m south of an intersection and moving northward at 6.0 m/s. It accelerates northward at 2.1 m/s2 for 8.3 s. At the end of this time, where is it with respect to the intersection? Round your answer to the nearest meter. (a) 92 m south *(c) 92 m north (e) None of these
(b) 104 m south (d) 104 m north
Solution: We will take northward to be the positive direction, so the initial position is negative and the initial velocity and acceleration are positive. We know the initial position, initial velocity, acceleration, and time; we want to know the final position. 1 1 x = x0 + v0 t + at2 = −30 m + (6.0 m/s)(8.3 s) + (2.1 m/s2 )(8.3 s)2 = 92 m 2 2 Since x is positive, it is north of the intersection. Problem 994. Beginning at rest, a ship accelerates at 0.45 m/s2 until it has gone 150 m. At the end of this time, how fast is the ship moving? Round your answer to the nearest m/s. (a) 9 m/s *(c) 12 m/s (e) None of these
(b) 10 m/s (d) 13 m/s
Solution: We know the initial position and initial velocity (both zero), the acceleration, and the final position. We want to know the final velocity. We have a formula that we use when time isn’t in the problem statement. v 2 = v02 + 2a(x − x0 ) = 2ax √ 1/2 = 12 m/s ⇒ v = 2ax = 2(0.45 m/s2 )(150 m) Problem 995. Beginning at rest, a train accelerates at 0.13 m/s2 until it has gone 220 m. At the end of this time, how fast is the train moving? Round your answer to the nearest 0.1 m/s. (a) 6.1 m/s *(c) 7.6 m/s (e) None of these
(b) 6.8 m/s (d) 8.3 m/s
Solution: We know the initial position and initial velocity (both zero), the acceleration, and the final position. We want to know the final velocity. We have a formula that we use when time isn’t in the problem statement. v 2 = v02 + 2a(x − x0 ) = 2ax √ 1/2 ⇒ v = 2ax = 2(0.13 m/s2 )(220 m) = 7.6 m/s
281 Problem 996. Beginning at rest, a boat accelerates at 0.13 m/s2 until it has reached a speed of 5.6 m/s. How far has the boat travelled in this time? Round your answer to the nearest meter. (a) 88 m (b) 98 m (c) 109 m *(d) 121 m (e) None of these Solution: We know the initial position and initial velocity (both zero), the acceleration, and the final velocity. We want to know the final position. We have a formula that we use when time isn’t in the problem statement. v 2 = v02 + 2a(x − x0 ) = 2ax
⇒
x=
(5.6 m/s)2 v2 = = 121 m 2a 2(0.13 m/s2 )
Problem 997. A train is moving at 31 m/s when the engineer is informed that there is a puppy on the track in front of him. If the train is capable of decelerating at 0.80 m/s2 , how far does it go before it comes to a stop? Round your answer to the nearest 10 m. (a) 490 m *(c) 600 m (e) None of these
(b) 540 m (d) 660 m
Solution: We will take the forward direction to be positive, so the acceleration will be negative; and we will take the train’s initial position as x0 = 0. We know the initial position, the initial velocity, the final velocity (zero), and the acceleration; we want to know the final position. We have a formula to use when time is not in the statement of the problem. v 2 = v02 + 2a(x − x0 ) ⇒ v02 = −2ax (31 m/s)2 v2 = 600 m ⇒ x= 0 = −2a −2(−0.80 m/s2 ) Problem 998. A train can accelerate from a complete stop to a speed of 31 m/s in 145 s. What is the train’s acceleration? Assume constant acceleration. Round your answer to two significant figures. (a) 0.15 m/s2 (c) 0.48 m/s2 (e) None of these
*(b) 0.21 m/s2 (d) 0.68 m/s2
Solution: Write down what information you are given, what you want, and which fundamental equation you need to use. vi = 0 m/s Given: vf = 31 m/s Want: a Which equation: Fundamental equation (1) ∆t = 145 s
282
Solving for a in equation 1: a =
vf − vi 31 = = .21. Here we’ve used the fact that a = a. ∆t 145
Problem 999. A car is initially moving at 18.9 m/s. It accelerates forward at 2.09 m/s2 until it has reached a speed of 29.3 m/s. How far has it gone in this time? Round your answer to the nearest meter. (a) 108 m *(b) 120 m (c) 132 m (d) 145 m (e) None of these Solution: We know the initial position (zero), the initial velocity, the final velocity, and the acceleration. We want to know the final position. We have a formula that we can use when time isn’t in the problem statement. v 2 = v02 + 2a(x − x0 ) = v02 + 2ax (29.3 m/s)2 − (18.9 m/s)2 v 2 − v02 = = 120 m ⇒ x= 2a 2(2.09 m/s2 ) Problem 1000. During World War I, the Germans used the Paris Gun to shell that city from 75 miles away. The barrel of the gun was 28 m long, and the shells had a muzzle velocity of 1600 m/s. Assuming that the acceleration of the shell was constant over the whole length of the barrel, what was that acceleration? Round your answer to the nearest 1000 m/s2 . (a) 46,000 m/s2 (c) 55,000 m/s2 (e) None of these
(b) (d)
50,000 m/s2 61,000 m/s2
Solution: We know the initial position and velocity (both zero), and the final position and velocity. We want to know the acceleration. We have a formula to use when time is not included in the problem statement. v 2 = v02 + 2a(x − x0 ) = 2ax (1600 m/s)2 v2 = = 46, 000 m/s2 ⇒ a= 2x 2(28 m)
283 Problem 1001. A ship has a maximum speed in still water of 13.0 m/s. To be operated legally on the Rhine River, it must be able to come to a complete stop in 350 m. What acceleration would the ship need to do this? Round your answer to the nearest 0.01 m/s2 . (a) 0.20 m/s2 *(c) 0.24 m/s2 (e) None of these
(b) (d)
0.22 m/s2 0.27 m/s2
Solution: We know the initial velocity, the final velocity (zero), the initial position (zero), and the final position. We want to know the acceleration. We have a formula that we can use when time isn’t in the problem statement. v 2 = v02 + 2a(x − x0 ) ⇒ v02 = −2ax v2 (13.0 m/s)2 ⇒ a=− 0 =− = −0.24 m/s2 2x 2(350 m) The negative sign is because we’ve taken the direction of the ship’s movement to be positive; to stop it, the acceleration must be in the negative direction. Since the problem doesn’t ask for the direction of the acceleration, it’s enough to give the magnitude: 0.24 m/s2 . Problem 1002. A radio-controlled airplane needs to reach a speed of 20 m/s to take off. Assuming a constant acceleration and a runway length of 20 m, what is the minimum acceleration that the plane needs if it starts from rest? (a) 1 m/s2 (c) 10 m/s2 (e) None of these
(b) (d)
5 m/s2 20 m/s2
Solution: Given: vi = 0 (the plane starts from rest), vf = 20 m/s, and ∆x = 20 m. Want: amin . We don’t know time, nor do we want it, so we’ll use fundamental kinematic equation 2. vf2
=
vi2
+ 2a∆x = 0 + 2a∆x
⇒
amin
vf2 202 = = m/s2 = 10 m/s2 ≈ g . 2∆x 2(20)
Thus, the plane must be able to pull one g in order to take off. Problem 1003. A certain bi-plane needs to reach a speed of 50 m/s to take off. Assuming the plane starts from rest with a constant acceleration of 2.5 m/s2 , what is the minimum runway length the jet needs to takeoff? (a) 250 m (c) 750 m (e) None of these Solution:
*(b) (d)
500 m 1000 m
Given: vi = 0 (the plane starts from rest), vf = 50 m/s, and a = 5/2 m/s2 .
284 Take the initial position to be xi = 0. Want: x. We don’t know time, nor do we want it, so we’ll use fundamental kinematic equation 2. vf2 = vi2 + 2ax = 0 + 2ax
⇒
x=
vf2 502 = m = 500m . 2a 5
Problem 1004. A young physics student buys a pre-made remote-controlled air plane. The instructions say that the plane needs a minimum speed of 10 m/s to takeoff. The instructions also give the thrust-to-weight ratio of the plane when it’s at full throttle. From this information he computes the the maximum acceleration to be 3 m/s2 . Allowing for friction between the wheels and the runway the student estimates that the maximum acceleration on the ground is 2.5 m/s2 . Assuming the plane starts from rest and has a constant acceleration 2.5 m/s2 (he keeps the throttle wide open at takeoff), what is the minimum runway length the jet needs to takeoff? (a) 10 m (c) 50 m (e) None of these
*(b) (d)
20 m 100 m
Solution: Given: vi = 0 (the plane starts from rest), vf = 10 m/s, and a = 5/2 m/s2 . Take the initial position to be xi = 0. Want: x. We don’t know time, nor do we want it, so we’ll use fundamental kinematic equation 2. vf2 = vi2 + 2ax = 0 + 2ax
⇒
x=
vf2 100 = m = 20m . 2a 5
Problem 1005. A typical airline jet needs to reach a speed of 360 km/h to take off. Assuming a constant acceleration and a short runway length of 1.0 km, what is the minimum acceleration that the jet needs if it starts from rest? Write your answer in terms of g, where g ≈ 10 m/s2 . Solution: The jet begins at x0 = 0, with initial speed v0 = 0. Hence the relationship between the speed and the distance travelled is: v 2 = 2ax. Solving for a, we get: 2 (360 km/h)(1000 m/km) 1 g v2 = · = 5.0 m/s2 ≈ . a= 2x 3600 s/h 2(1 km)(1000 m/km) 2
285 Problem 1006. An overloaded transport jet needs to reach a speed of 500 km/h to take off. Assuming a constant acceleration and a runway length of 2.0 km, what is the minimum acceleration that the jet needs if it starts from rest? Write your answer in terms of g, where 1g = 9.8 m/s2 . Round your answer to the nearest 0.01g. Solution: The jet begins at x0 = 0, with initial speed v0 = 0. Hence the relationship between the speed and the distance travelled is: v 2 = 2ax. Solving for a, we get: 2 (500 km/h)(1000 m/km) v2 1 = = 4.8 m/s2 a= · 2x 3600 s/h 2(2 km)(1000 m/km) [Don’t use this rounded value of a in further calculations; we only present it because it’s a good idea to check intermediate results to make sure they seem reasonable.] The answer we want is g = 0.49g a=a 9.8 m/s2
286 Problem 1007. (This problem requires solving a system of equations) A policeman is lurking behind a billboard beside the highway when someone speeds by him at 38 m/s. The policeman immediately accelerates at a constant rate of 6.7 m/s2 until he has caught up with the speeder. How long does it take for the policeman to catch up? Round your answer to the nearest second. *(a) (c) (e)
11 s 13 s None of these
(b) (d)
12 s 15 s
Solution: We will use the formula for position as a function of time: x = x0 +v0 t+ 12 at2 . We will use the position of the billboard as x0 = 0. We need to use the formula twice: once for the policeman, once for the speeder. For the speeder, x0 = 0, v0,s = 38 m/s, and as = 0. For the policeman, x0 = 0, v0,p = 0, and ap = 6.7 m/s2 . Hence their positions are: 1 xs = x0 + v0,s t + as t2 = v0,s t 2 1 2 1 2 xp = x0 + v0,p t + ap t = ap t 2 2 When the policeman catches up with the speeder, their positions will be the same: so xs = xp . Equating the expressions: 1 v0,s t = ap t2 2
⇒
ap t2 − v0,s t = 0
⇒
t (ap t − 2v0,s ) = 0
There are two solutions: t = 0 and t =
2v0,s 2(38 m/s) = 11 s = ap 6.7 m/s2
The first solution refers to the fact that both were at the billboard at t = 0; the second solution is the time at which the policeman catches the speeder. Note: In general, the velocity of the police officer and the speeder at the position xfinal = xs = xp will not be the same vs,f 6= vp,f .
287
7.6
Quantitative kinematics: vertical
Note: By convention, we will use y in place of x to denote position. Problem 1008. An astronaut drops a watermelon from the top of a high cliff on the Moon, where the surface gravity is 1.63 m/s2 . After 3.3 s, how fast is the watermelon moving? Round your answer to the nearest 0.1 m/s. (a) 3.9 m/s (c) 4.8 m/s (e) None of these
(b) 4.4 m/s *(d) 5.4 m/s
Solution: We will take the positive direction to be downward, which will save us having to worry about negative signs. We know the initial velocity (zero), the acceleration, and the time; we want to know the final velocity. v = v0 + at = at = (1.63 m/s2 )(3.3 s) = 5.4 m/s Problem 1009. Using a stopwatch, you determine that a television dropped from a rooftop takes 2.7 s to reach the ground. How fast is it going when it hits the ground? Round your answer to the nearest m/s. (a) 21 m/s *(c) 26 m/s (e) None of these
(b) 24 m/s (d) 29 m/s
Solution: We will take the positive direction and the acceleration to be downward. We know the initial velocity (zero), the acceleration, and the time. We want to know the final velocity. v = v0 + at = at = (9.8 m/s2 )(2.7 s) = 26 m/s Problem 1010. An astronaut hurls a watermelon straight downward from the top of a high cliff on the Moon, where the surface gravity is 1.63 m/s2 . The initial speed of the watermelon is 11.8 m/s downward. After 2.4 s, how fast is the watermelon moving? Round your answer to the nearest 0.1 m/s. (a) 11.5 m/s (c) 14.1 m/s (e) None of these
(b) 12.7 m/s *(d) 15.7 m/s
Solution: We will take the positive direction to be downward, so the acceleration and the initial velocity will be positive. We know the initial velocity, the acceleration, and the time; we want to know the final velocity. v = v0 + at = 11.8 m/s + (1.63 m/s2 )(2.4 s) = 15.7 m/s
288 Problem 1011. While walking in the desert, you discover an old mine. To find out how deep it is, you drop a rock down the shaft. Using the stopwatch on your cell phone, you time how long it takes before you hear the rock hit the bottom. If it takes 3.2 s for the rock to reach the bottom, how deep is the shaft? Round your answer to the nearest meter. *(a) 50 m (b) 55 m (c) 61 m (d) 67 m (e) None of these Solution: We will take the positive direction to be downward, so that the acceleration and the final position will be positive. We will take the initial position to be zero. Then we know the initial position, the initial velocity (zero), the acceleration, and the time. We want to know the final position. 1 1 1 y = y0 + v0 t + at2 = at2 = (9.8 m/s2 )(3.2 s)2 = 50 m 2 2 2 Problem 1012. A banker drops a sack of pennies from the top of a tall building. After 1.5 s, how far has the sack of pennies fallen? Round your answer to the nearest 0.1 m. (a) 9.9 m (c) 12.1 m (e) None of these
*(b) (d)
11.0 m 13.3 m
Solution: We will take the positive direction to be downward, so that the acceleration and the final position will be positive. We will take the initial position to be zero. Then we know the initial position, the initial velocity (zero), the acceleration, and the time. We want to know the final position. 1 1 1 y = y0 + v0 t + at2 = at2 = (9.8 m/s2 )(1.5 s)2 = 11.0 m 2 2 2 Problem 1013. An astronaut drops a watermelon from the top of a tall cliff on Mercury, where the surface gravity is 3.7 m/s2 . After 2.4 s, how far has the watermelon fallen? Round your answer to the nearest 0.1 m. *(a) (c) (e)
10.7 m 12.9 m None of these
(b) (d)
11.7 m 14.2 m
Solution: We will take the positive direction to be downward, so that the acceleration and the final position will be positive. We will take the initial position to be zero. Then we know the initial position, the initial velocity (zero), the acceleration, and the time. We want to know the final position. 1 1 1 y = y0 + v0 t + at2 = at2 = (3.7 m/s2 )(2.4 s)2 = 10.7 m 2 2 2
289 Problem 1014. A dietitian drops a head of iceberg lettuce from a bridge. How long does it take for the lettuce to reach the river 31.2 m below? Round your answer to the nearest 0.1 s. (a) 2.0 s (b) 2.3 s *(c) 2.5 s (d) 2.8 s (e) None of these Solution: We will take the positive direction to be downward, which will make the acceleration and the final position positive. We will take the initial position to be zero. Then we know the initial position and velocity (both zero), the acceleration, and the final position; we want to know the time. 1 1 y = y0 + v0 t + at2 = at2 2 2 r 1/2 2(31.2 m) 2y ⇒ t= = = 2.5 s a 9.8 m/s2 Problem 1015. You are on the surface of Ganymede, where the acceleration due to gravity is 1.35 m/s2 . You throw an egg straight upward at 25 m/s. How long does it take for the egg to reach the ground? Round your answer to the nearest second. (a) 30 s *(c) 37 s (e) None of these
(b) 33 s (d) 41 s
Solution: We will use upward as the positive direction and ground level as y0 = 0. We know the initial and final position (both zero), the initial velocity, and the acceleration. We want to know the time. 1 y = y0 + v0 t + at2 2
⇒
t = 0 or t =
2v0 t + at2 = 0
⇒
t(2v0 + at) = 0
−2v0 −2(25 m/s) = 37 s = a −1.35 m/s2
The first solution refers to the fact that at t = 0, you threw the egg upward from ground level; the second, t = 37 s, is the time at which the egg strikes the ground on its way back down.
290 Problem 1016. An astronaut on Planet X drops a hammer from a cliff 6.8 m high. The hammer hits the ground after 1.3 s. What is the surface gravity of Planet X? Round your answer to the nearest 0.1 m/s2 . *(a) (c) (e)
8.0 m/s2 9.7 m/s2 None of these
(b) (d)
8.9 m/s2 10.7 m/s2
Solution: We will take the positive direction to be downward. We will take the top of the cliff to be y0 = 0. We know the initial position and velocity (both zero), the final position, and the time. We want to know the acceleration. 1 1 y = y0 + v0 t + at2 = at2 2 2 2y 2(6.8 m) ⇒ a= 2 = = 8.0 m/s2 t (1.3 s)2 Problem 1017. A physics student angrily hurls his cell phone downward from a bridge at 19.9 m/s. After 2.2 s, how far below the bridge will the phone be? Round your answer to the nearest meter. *(a) 67 m (b) 74 m (c) 82 m (d) 90 m (e) None of these Solution: We will take down to be the positive direction, and the bridge to be y0 = 0. Then we know the initial position, the initial velocity, and the time; we want to know the final position. 1 1 y = y0 + v0 t + at2 = (19.9 m/s)(2.2 s) + (9.8 m/s2 )(2.2 s)2 = 67 m 2 2 Problem 1018. A banker drops a sack of pennies from the top of a building 18.8 m tall. How fast is the sack moving when it strikes the ground? Round your answer to the nearest 0.1 m/s. (a) 15.5 m/s *(c) 19.2 m/s (e) None of these
(b) 17.3 m/s (d) 21.1 m/s
Solution: We will take the top of the building to be y0 = 0, and downward as the positive direction. We know the initial and final position, the initial velocity (zero), and the acceleration. We want the final velocity. We have a formula to use when time is not part of the problem statement. v 2 = v02 + 2a(y − y0 ) = 2ax q p ⇒ v = 2ay = 2(9.8 m/s2 )(18.8 m) = 19.2 m/s
291 Problem 1019. If we ignore air drag, what is the speed of a hail stone that falls from rest at the top of a cumulus cloud that is 10,000 m high? (a) 126 m/s (b) 13.2 m/s (c) 67.6 m/s (d) 443 m/s (e) None of these Solution: We will take the positive direction to be downward in the direction of motion and take the height of 10 km to be the initial position y0 . This means that the velocity is positive, and the acceleration is positive ay = g. Let’s write down what we’re given, what we want, and which equation we need to use. vf = 0 (The hail stone is initially at rest at the maximum height) a = g = 9.8 m/s2 Given: y0 = 0 yf = 104 m Notice that we must convert the height to meters. Want: vf = ? (final velocity) Which equation: equation (2) Notice that we don’t know time; nor do we want time, so we can’t use fundamental equations (1), or (3). Substituting v0 = 0, y0 = 0, and a = g into equation (2) and solving for v0 we arrive at vf2 = 0 + 2gyf √ p −−−−−−→ vf = 2gyf q evaluate −−−−→ v0 = 2(9.8m/s2 )(104 m) ≈ 443 m/s (that’s almost 1000 mph!) Problem 1020. A ball is thrown straight upward from ground level, y = 0. At what speed must the ball be thrown if it is to reach a height of ymax = 10 meters above the ground? Round your answer to the nearest m/s. p p Solution: v = 2gymax = 2(9.8 m/s)(10 m) = 14 m/s Problem 1021. You are standing on a bridge over a canyon. To find out how deep the canyon is, you drop a rock off the bridge and time its fall. The rock hits the river at the canyon bottom after 7.2 s. How deep is the canyon? Round your answer to the nearest meter. *(a) 254 m (b) 359 m (c) 508 m (d) 718 m (e) None of these Solution: Given: yi = 0 (take y-axis pointing downward in the direction of motion), vi = 0, ay = g, and tf = 7.2 s. Want: yf . Use fundamental equation 3: y = 12 gt2fall .
292 7.6.1
The difference between average speed and average velocity in onedimension
Problem 1022. (Speed vs. Velocity) An object is launched straight up into the air from the ground with an initial vertical velocity of 30 m/s. The object rises to a height of approximately 45 m above the ground in 3 seconds; it then falls back to the ground in 3 more seconds, impacting with a speed of 30 m/s. Approximately what is the average speed of the object during its time in the air? (a) 0 m/s *(c) 15 m/s (e) None of these
Solution:
(b) 5 m/s (d) 30 m/s
The definition of average speed is total distance travelled total time distance up + distance down = time to go up + time to go down 2distance up = (since yup = ydown = 45 m; 2time up 45 m = 15 m/s = 3s
average speed =
tup = tdown = 3 s)
Problem 1023. (Speed vs. Velocity) An object is launched straight upward from ground level with an initial vertical velocity of about 30 m/s. In roughly 3 seconds, the object reaches a maximum height of approximately 45 m above the ground; it takes another 3 seconds to fall back down, striking the ground at 30 m/s. If vave is the average velocity of the object during its 6-second flight, then the magnitude of the average velocity vave is: *(a) (c) (e)
0 m/s 15 m/s None of these
(b) 5 m/s (d) 30 m/s
Solution: Recall the definition of average velocity the change in location (the displacement) over the change in time. It is a vector quantity, which in one spatial dimension is a quantity that can be positive or negative denoting which direction the object is traveling with respect to the x-axis. Let (ti , yi ) be the initial time and position, respectively, and
293 (tf , yf ) be the final time and position. displacement change in time yf − yi = tf − ti yi − yi = (Since the ball starts and stops at the same point yi = yf ) tf − ti = 0 (since ∆y = yf − yi = 0 and ∆t = tf − ti > 0)
average velocity =
294
7.7
Similarity problems
The Idea: With all similarity problems there is always two similar problems (hence the name!) that are modeled by the same equation/equations. The two similar situations are sometimes called experiments, and the equation/equations that describes the situation (the model that governs the behavior of the particle) is/are referred to as the governing equation/equations. Some of the variables/parameters have the same value in both experiments. These parameters are known as common parameters. The variables/parameters that are situation dependent (i.e., different between the two experiments) are labelled by subscripts: a subscript 1 is given to the parameters from the first experiment and a subscript 2 is given to the parameters from the second experiment. The parameters that are in common to both experiments are the link between the two experiments. We typically don’t give these common parameters subscripts. Problem-Solving Algorithm: Write down what your given, what you want, and determine the governing equation that describes both situations. Identify all of the variables/parameters that are in common to both situations and put them on the right-hand side (RHS) of the equation, and put all of the variables/parameters that are situationdependent on the LHS of the equation. Then equate the two situations through the common RHS of the equation of the re-written governing equation. This is your common link between the two situations! Problem 1024. (Similarity Problem) A train has a constant forward acceleration of a. Starting at rest, it reaches a speed of v1 at time t1 . At time t2 = 2t1 , how fast is the train moving? √ *(b) 2v1 (a) √2 v1 (c) 2 2 v1 (d) 4v1 (e) None of these Solution: We have two similar situations that can both be modeled by one governing equation. In both situations we have an object (the train) starting from rest vi = 0 under constant acceleration a = const. In both situations we are given time and in the second situation we want the final velocity. The equation describing the motion (i.e., the governing equation), is fundamental kinematic equation 1 vf = vi + atf (we’re not given distance, nor do we want it). Taking ti = 0, vi = 0, a = const, and dropping the f subscripts yields v = at (The governing equation) , where we recognize vi = 0 and a as common parameters (i.e., parameters that are in common to both situations). We’ll now use a to relate the velocity and time from each of the similar situations. Rewriting the governing equation, by putting the common parameter a on the right and the situation-dependent parameters on the left yields: v2 t2 2t1 v1 · t2 =a= −−−−−→ v2 = v1 = v1 = 2v1 . t1 t2 t1 t1
295 Problem 1025. (Similarity Problem) A physics student drops a cell phone off a very high bridge. At time t1 after being released, the phone has fallen a distance y1 . How far has the phone fallen at time t2 = 3t1 ? √ (b) 3y1 (a) √3 y1 (c) 3 3 y1 *(d) 9y1 (e) None of these Solution: We will take the positive direction as downward, and the bridge level as y0 = 0. We know that the initial velocity is v0 = 0. Hence y = y0 + v0 t + 21 at2 = 12 gt2 is our governing equation for both cases. We are given t1 , y1 , and t2 . At time t = t1 , y1 = 12 gt21 , at time t = t2 = 3t1 , y2 = 12 gt22 , and g/2 is a constant that is in common to y g both situations. Re-writing the governing equation 2 = (notice that the right-hand t 2 side of the equation is constant). We want y2 , thus equating the two equations gives y1 g y2 = = 2 2 t1 2 t2
· t22
−−−−−→
2 2 t2 3t1 y2 = y1 = y1 = 9y1 . t1 t1
Problem 1026. (Similarity Problem) An object dropped from a building with height y1 takes time t1 to reach the ground. How long would it take for an object dropped from a building with height y2 = 2y1 ? √ (b) 2t1 *(a) √2 t1 (c) 2 2 t1 (d) 4t1 (e) None of these Solution: We will take downward as the positive direction, and the top of the building as y0 = 0. The initial velocity is zero v0 = 0. Then fundamental equation (3) reduces to y = 21 gt2 . Solving this equation for time gives the governing equation for the fall time as a function of height. 1 y = gt2 2 · g2 2 −−−−−→ t2 = y g r √ 2y −−−−−−→ tfall = . g We’re given: y1 , t1 , and y2 . We want: t2 . Re-writing the governing equation with common parameters on the RHS of the equation and the situation-dependent parameters on the LHS of the equation yields r √ ÷ y tfall 2 −−−−−−→ = . y g Equating the two situations via the common parameters yields r r √ √ √ · y2 tfall,1 g tfall,2 y2 y2 =2y1 = √ −−−−−−→ tfall,2 = tfall,1 −−−−−−−−→ tfall,2 = 2tfall,1 . √ = y1 2 y2 y1
296 Problem 1027. (Similarity Problem) An object dropped from a building with height y1 strikes the ground with speed v1 . If an object is dropped from a building with height y2 = 4y1 , how fast is it moving when it hits the ground? *(a) 2v1 (c) 8v1 (e) None of these
(b) 4v1 (d) 16v1
Solution: We will take the positive direction as downward, and the top of the building as y0 = 0. The initial velocity is zero v0 = 0. Since time isn’t in the statement of the problem, we use the fundamental formula (2): v 2 = v02 +2a(y −y0 ); in this case, v 2 = 2gy. Taking √ the square root of this equation gives v as a function of y. The governing equation is v = 2gy. Note that 2g is the same for both situations. This will be are common parameters. Putting all of the parameters that are situation dependent on the left-hand side of the equation, and the common parameters on the right-hand side of the governing equation yields: p v = 2gy √ p ÷ y v −−−−−−→ √ = 2g (similarity form of governing equation) y Substituting the variables into this equation and using the common link (the RHS of the governing equation) to equate the two situations yields r r √ p · y2 v1 v2 y2 4y1 −−−−−−→ v2 = v1 = v1 = 2v1 √ = 2g = √ y1 y2 y1 y1 Problem 1028. (Similarity Problem) An object dropped from a certain height on Planet X takes time tX to reach the ground. An object dropped from the same height on Planet Y takes time tY to fall. Which equation describes the relation between the surface gravities gX and gY of the two planets? (a)
tY gY = gX tX
(c)
gY tX = gX tY
(e)
None of these
(b)
gY t2 = 2Y gX tX
*(d)
gY t2X = 2 gX tY
Solution: An object falling in a gravity of g for a time of t falls for a distance of 1 2 x = 2 gt . The distance fallen, x, is the same on both planets. Hence gX t2X = 2x = gY t2Y = ⇒
gY t2 = X gX t2Y
297 Problem 1029. (Similarity Problem) Your friend’s hot rod can go from 0 to 60 mph in 6 seconds. Your father’s minivan can only do 0 to 60 mph in 12 seconds. Assuming constant acceleration for each vehicle, what is the ratio of the hot rod’s acceleration ahr to the minivan’s acceleration amv ? 1 ahr 1 ahr = (b) =√ (a) amv 2 amv 2 √ ahr ahr = 2 *(d) =2 (c) amv amv (e)
None of these
Solution: Since each vehicle starts from rest v0 = 0 and accelerates until it reaches the final velocity v = 60 mph, we have ∆v = v − v0 = v = 60 m/hr. Because of the constant acceleration assumption, we can use the equation v = at and the common parameter v to find a relationship between the ratio of accelerations and the times: ahr thr = v = amv tmv ahr tmv 2 thr ÷amv thr −−−−−−− −→ = = =2 amv thr thr v Notice that by taking a ratio, we avoided having to convert the acceleration a = = t 60 mph into ft/s2 or some other “standard” measure of acceleration. 6s Problem 1030. (Similarity Problem) A spherical balloon has a radius of r when filled with 0.50 gallons of water. If you fill the balloon with 1.0 gallons of water, what will its radius be? Your answer should be exact. Solution: The volume of a sphere is proportional to the cube of the radius. If the radius of the one-gallon balloon is R, then: 1.0 gal 0.50 gal = r3 R3 √ Hence R3 = 2.0 r3 ; so R = 3 2 r.
R3 1.0 = = 2.0 r3 0.50
⇒
Problem 1031. (Similarity Problem) An unloaded train has an acceleration of au . Starting from a standstill, it can accelerate to a speed of v in time tu . A fully loaded train has an acceleration of al = 0.4au . How long does it take the loaded train to accelerate from zero to v? Solution: Since the initial velocity is zero, we can write: v = au tu for the unloaded train. For the loaded train, the acceleration is al = 0.40au . Let tl be the time that it takes for the loaded train to accelerate to v. Then al tl = v = au tu
÷a
−−−−l−→
tl =
au tu tu au = = 2.5 tu al 0.40au
298 Problem 1032. (Similarity Problem) The Smith Building is twice as tall as the Jones Building. A sack of pennies dropped from the top of the Jones Building takes time tJ to reach the ground. How long does it take for a similar sack dropped from the top of the Smith Building? Your answer should be exact. (Hint: Use the top of the building as the origin, and let downward be the positive y-direction.) Solution: Since y0 = 0 and v0 = 0, distance fallen as a function of time is: y = 21 gt2 . Since the Smith Building is twice as tall as the Jones Building, yS = 2yJ . Hence: √ 1 1 yJ = gtJ 2 ⇒ yS = gtS 2 = 2yJ = gtJ 2 ⇒ tS 2 = 2tJ 2 ⇒ tS = 2 tJ 2 2 Problem 1033. (Similarity Problem) The Smith Building is three times as high as the Jones Building. A watermelon dropped from the top of the Smith Building takes ts seconds to hit the ground. How long does it take for a watermelon dropped from the top of the Jones Building? Your answer should be exact. Solution: We will use the downward direction as positive, and take the top of each building to be x = 0. Then x0 = 0 and v0 = 0, so we can write: r 1 2 2x ⇒ t= x = gt 2 g Thus the falling time is proportional to the square root of the height of the building. We use the subscripts s and j to designate Smith and Jones: √ √ 3tj 3 ts tj = √ ts ⇒ tj = √ =p xs xs 3 xs /3 Problem 1034. (Similarity Problem) Two identical balls are launched vertically into the air with the same initial velocity v0 . One of them is launched from the earth; the other from the surface of Planet X, which has 1/10 of the earth’s gravitational attraction (gearth = 10gX ). The ball thrown on the earth reaches a maximum height of HE ; the ball launched on Planet X reaches a maximum height of HX . What is the relationship between HE and HX ? (Assume that air drag has no effect.) Solution: We can use the formula: v 2 = v02 + 2a(x − x0 ). The balls are launched from ground level, so we can write x0 = 0; the maximum height x = H is reached when v = 0. Hence the formula becomes: 0 = v0 2 + 2aH. Since v0 is the same on both planets, we can write: 2gearth HE = 2gX HX . Solving for HE yields HE =
gX HX gX HX HX = = gearth 10gX 10
Problem 1035. (Similarity Problem) Two identical balls are launched vertically into the air with the same initial velocity v0 . One of them is launched from the earth; the other from the surface of Planet X, which has 1/10 of the earth’s gravitational attraction (gE = 10gX ). The ball thrown on the earth reaches its maximum height after time tE ; the ball launched on Planet X reaches its maximum height after time tX . What is the relationship between tE and tX ? (Assume that air drag has no effect.)
299 Solution: For the two similar experiments we are given: vE,i = vX,i = v0 (initial velocity), vE,f = vX,f = 0 (final velocity), and gE = 10gX ), where we have used the fact when the ball reaches its maximum height, then its velocity is zero. The common parameters are: the initial and final velocities. Using the fundamental kinematic equation 1 (vf = vi − gt) with the y axis pointing upward yields: gE tE = vE,i = vX,i = gX tX
⇒
gX tE = . tX gE
300
7.8
Lab-application problems: One-Dimensional Kinematics
This is a mixture of problem types based on actual equipment found in instructional physic labs everywhere. 7.8.1
Ball-drop apparatus
Problem 1036. (Lab Problem) In a ball-drop apparatus, a ball is held in place above a drop pad. When a knob is turned, the ball is released to fall onto the pad; an electrical circuit through the drop apparatus and the pad is used to measure the time it takes for the ball to fall. If the ball is initially 0.85 m above the pad, how long should it take for the ball to fall to the pad? Round your answer to the nearest 0.01 s. (a) 0.37 s m (c) 0.46 s (e) None of these
*(b) (d)
0.42 s 0.50 s
Solution: We will take the positive direction to be downward, which will make the acceleration and the final position positive. We will take the initial position to be zero. Then we know the initial position and velocity (both zero), the acceleration, and the final position; we want to know the time. 1 1 y = y0 + v0 t + at2 = at2 2 2 r 1/2 2y 2(0.85 m) ⇒ t= = = 0.42 s a 9.8 m/s2 7.8.2
Spring-cannon apparatus
Problem 1037. (Lab Problem) In this problem, use upward as the positive direction. A spring cannon has been pointed straight up, and a ball has been fired from it. The ball has not yet reached the highest point in its flight. Which of the following describes the ball’s situation? (a) positive velocity; positive acceleration *(b) positive velocity; negative acceleration (c) positive velocity; zero acceleration (d) zero velocity; negative acceleration (e) none of these Solution: The ball hasn’t reached its highest point, so it is still moving upward. Hence the velocity is positive. It is slowing down as it moves upward, so its acceleration is negative.
301 Problem 1038. (Lab Problem) In this problem, use upward as the positive direction. A spring cannon has been pointed straight up, and a ball has been fired from it. The ball has reached the highest point in its flight, and its speed is increasing as it falls back toward the ground. Which of the following describes the ball’s situation? (a) negative velocity; positive acceleration *(b) negative velocity; negative acceleration (c) negative velocity; zero acceleration (d) zero velocity; positive acceleration (e) none of these Solution: The ball is moving downward, so its velocity is negative. It is speeding up in the downward direction, so its acceleration is negative. Problem 1039. (Lab Problem) A physics student shoots a ball straight upward from a spring cannon, with an initial speed of 15.1 m/s. How long does it take for the ball to reach the highest point in its flight? Round your answer to the nearest 0.1 s. *(a) (c) (e)
1.5 s 1.9 s None of these
(b) (d)
1.7 s 2.1 s
Solution: We will take the positive direction to be upward. This means that the initial velocity is positive, and the acceleration is negative. We know the initial velocity, the final velocity (zero at the highest point in the flight), and the acceleration; we want to know the time. ⇒
v = v0 + at
t=
−15.1 m/s v − v0 = = 1.5 s a −9.8 m/s2
Problem 1040. (Lab Problem) A physics student on Pluto shoots a ball straight upward from a spring cannon, with an initial speed of 9.1 m/s. The ball reaches the highest point in its flight after 11.2 s. What is Pluto’s surface gravity? Round your answer to the nearest 0.01 m/s2 . (a) 0.59 m/s2 (c) 0.73 m/s2 (e) None of these
(b) 0.66 m/s2 *(d) 0.81 m/s2
Solution: We will take the positive direction to be downward, so the initial velocity will be negative and the acceleration will be positive. We know the initial velocity, the final velocity (zero), and the time; we want to know the acceleration. v = v0 + at
⇒
a=
−v0 9.1 m/s v − v0 = = = 0.81 m/s2 t t 11.2 s
302 Problem 1041. (Lab Problem) A student shoots a ball straight upward from a spring cannon. The ball reaches the highest point in its flight after 1.2 s. What was the ball’s initial speed? Round your answer to the nearest 0.1 m/s. (a) 10.6 m/s (c) 12.9 m/s (e) None of these
*(b) (d)
11.8 m/s 14.2 m/s
Solution: We will take the positive direction to be upward, so the acceleration will be negative and the initial velocity positive. We know the final velocity (zero), the acceleration, and the time; we want to know the initial velocity. v = v0 + at
⇒
v0 = v − at = −at = (9.8 m/s2 )(1.2 s) = 11.8 m/s
Problem 1042. (Lab Problem) A physics student shoots a ball from a spring cannon straight upward at 15.0 m/s. After 1.1 s, how high will the ball be? Round your answer to the nearest 0.1 m. (a) 9.5 m *(b) 10.6 m (c) 11.6 m (d) 12.8 m (e) None of these Solution: We will take upward to be the positive direction; so the acceleration will be negative. We will take ground level to be y0 = 0. We know initial position, initial velocity, acceleration, and time; we want to know final position. 1 1 y = y0 + v0 t + at2 = (15.0 m/s)(1.1 s) + (−9.8 m/s2 )(1.1 s)2 = 10.6 m 2 2 Problem 1043. (Lab Problem) A physics student shoots a ball from a spring cannon straight upward at 15.0 m/s. After 2.3 s, how high will the ball be? Round your answer to the nearest 0.1 m. *(a) 8.6 m (b) 9.4 m (c) 10.4 m (d) 11.4 m (e) None of these Solution: We will take upward to be the positive direction; so the acceleration will be negative. We will take ground level to be y0 = 0. We know initial position, initial velocity, acceleration, and time; we want to know final position. 1 1 y = y0 + v0 t + at2 = (15.0 m/s)(2.3 s) + (−9.8 m/s2 )(2.3 s)2 = 8.6 m 2 2
303 Problem 1044. (Lab Problem) A physics student shoots a ball from a spring cannon straight upward. After 1.2 s, the ball is 15.0 m high. What was the initial velocity of the ball? *(a) 18.4 m/s (b) 20.2 m/s (c) 22.2 m/s (d) 24.5 m/s (e) None of these Solution: We will take upward to be the positive direction; so the acceleration will be negative. We will take ground level to be y0 = 0. We know initial position, final position, acceleration, and time; we want to know initial velocity. 1 1 y = y0 + v0 t + at2 = v0 t + at2 2 2 2 ⇒ 2y = 2v0 t + at ⇒ v0 =
2(15.0 m) − (−9.8 m/s2 )(1.2 s)2 2y − at2 = = 18.4 m/s 2t 2(1.2 s)
304 Problem 1045. (Lab Problema ) A physics student shoots a ball straight upward from a spring cannon. Using a meter stick, the student measures the ball’s maximum height. If the height was found to be 53 cm, what was the ball’s initial velocity as it left the muzzle of the cannon? Round your answer to the nearest 0.1 m/s. (a) 2.6 m/s (c) 2.9 m/s (e) None of these
*(b) 3.2 m/s (d) 3.7 m/s
Solution: We will take the positive direction to be upward in the direction of motion. This means that the initial velocity is positive, and the acceleration is negative a = −g. Let’s write down what we’re given, what we want, and which equation we need to use. vf = 0 (The ball is instantaneous at rest at the maximum height) a = −g = −9.8 m/s2 Given: ymax = 53 cm = .53 m y0 = 0 (We’ll that the mouth of the cannon as our origin) Notice that we must convert the height to meters (or acceleration to cm/s2 ). Want: v0 = ? (initial velocity) Which equation: equation (2) Notice that we don’t know time; nor do we want time, so we can’t use fundamental equations (1), or (3). Substituting v = 0, a = −g into equation (2) and solving for v0 we arrive at 02 = v02 − 2gymax + 2gymax
−−−−−→ v02 = 2gymax √ p −−−−−−→ v0 = 2gymax q evaluate −−−−→ v0 = 2(9.8m/s2 )(.53m) = 3.2 m/s a This is a very accurate method for determining the initial velocity of the ball, especially when you shoot the spring cannon at a high angle of inclination (θ0 ≥ 60◦ ).
305 Problem 1046. (Lab Problem) A physics student is designing a spring cannon. He wants to fire a ball straight upward from ground level; and he wants the ball to have an upward speed of 10.2 m/s when it reaches a height of 6.8 m above the ground. What must the ball’s initial speed be? Round your answer to the nearest 0.1 m/s. (a) 12.5 m/s *(c) 15.4 m/s (e) None of these
(b) 13.9 m/s (d) 16.9 m/s
Solution: We will take upward as the positive direction and ground level as y0 = 0. We know the initial and final position, the acceleration (negative), and the final velocity. We want the initial velocity. We have a formula we can use when time isn’t in the statement of the problem. v 2 = v02 + 2a(y − y0 ) = v02 + 2ay q p 2 ⇒ v0 = v − 2ay = (10.2 m/s)2 − 2(−9.8 m/s2 )(6.8 m) = 15.4 m/s Problem 1047. (Lab Problem) A physics student shoots a ball straight upward from a spring cannon, with an initial velocity of 18.2 m/s. At what time will the ball reach a height of 10.4 m on its way up? Round your answer to the nearest 0.01 s. (a) 0.51 s (c) 0.63 s (e) None of these
(b) 0.57 s *(d) 0.71 s
Solution: We will take upward as the positive direction and ground level as y0 = 0. Then acceleration is negative. We know initial position and velocity, final position, and acceleration; we want to know time. 1 1 y = y0 + v0 t + at2 = v0 t + at2 2 2 2 ⇒ at + 2v0 t − 2y = 0 We will use the quadratic formula to solve this for t. As often happens with quadratic equations, we will get two solutions. This is because the ball will pass through most points on its trajectory twice: once on the way up, and again on the way back down. The problem asks when the ball reaches y on the way up; so we want the smaller of the two times. p p −2v0 ± 4v02 + 8ay −v0 ± v02 + 2ay t= = 2a q a −18.2 m/s ±
= = 0.71 s or 3.01 s
(18.2 m/s)2 + 2(−9.8 m/s2 )(10.4 m) −9.8 m/s2
306 Problem 1048. (Lab Problem) A physics student shoots a ball straight upward from a spring cannon, with an initial velocity of 14.3 m/s. At what time will the ball reach a height of 8.8 m on its way back down, after it has reached its maximum height? Round your answer to the nearest 0.01 s. (a) 1.83 s (c) 2.24 s (e) None of these
*(b) 2.04 s (d) 2.46 s
Solution: We will take upward as the positive direction and ground level as y0 = 0. Then acceleration is negative. We know initial position and velocity, final position, and acceleration; we want to know time. 1 1 y = y0 + v0 t + at2 = v0 t + at2 2 2 2 ⇒ at + 2v0 t − 2y = 0 We will use the quadratic formula to solve this. As often happens with quadratic equations, we will get two solutions. This is because the ball will pass through most points on its trajectory twice: once on the way up, and again on the way back down. The problem asks when the ball reaches y on the way down; so we want the larger of the two times. p p −v0 ± v02 + 2ay −2v0 ± 4v02 + 8ay = t= 2a q a −14.3 m/s ±
=
(14.3 m/s)2 + 2(−9.8 m/s2 )(8.8 m) −9.8 m/s2
= 0.88 s or 2.04 s Problem 1049. A student shoots a ball straight upward from a spring cannon. The ball reaches the highest point in its flight after 1.5 s. What was the ball’s initial speed? Round your answer to the nearest meter per second. (a) 11 m/s *(c) 15 m/s (e) None of these
(b) 13 m/s (d) 17 m/s
Solution: We will take the positive direction to be upward, so the acceleration will be negative and the initial velocity positive. We know the final velocity (zero), the acceleration, and the time; we want to know the initial velocity. v = v0 + at
⇒
v0 = v − at = −at = (9.8 m/s2 )(1.5 s) = 15 m/s
307
7.9
Mixing It Up
Problem 1050. Harry Potter’s magic flying broomstick is described as having an acceleration of 150 miles per hour in 10 seconds. What is this acceleration in ft/s2 ? What is it in terms of g? Solution:
We’ll begin by converting miles per hour to ft/s: 150 mi 5280 ft 1 hr · · = 220 ft/s hr 1 mi 3600 s
The broomstick attains this speed in 10 s; so we divide by 10 s to get acceleration: a=
220 ft/s = 22 ft/s2 10 s
Since g = 32 ft/s2 , a=
22 ft/s2 = 0.69 g 32 ft/s2
Problem 1051. Using a stopwatch, you determine that a television dropped from a rooftop takes 2.7 s to reach the ground. How fast is it going when it hits the ground? Give your answer in m/s. Solution:
For constant acceleration, v = v0 + at = 0 + (9.8 m/s2 )(2.7 s) = 26 m/s
Problem 1052. An experimental aircraft is capable of accelerating horizontally at 4.2 g. How long does it take for the aircraft to reach a horizontal speed of 5600 ft/s? Solution: For constant acceleration, v = at; so t = v/a. We need to convert the acceleration to units of ft/s2 : a = 4.2 g = (4.2)(32 ft/s2 ) = 134.4 ft/s2 Now that we have consistent units, we can calculate the time: t=
5600 ft/s v = = 42 s a 134.4 ft/s2
Notice that we’ve rounded this answer to two significant digits, since that’s all we had in the statement of the problem. Problem 1053. You are on the surface of the Moon, where the acceleration due to gravity is 1.62 m/s2 . You throw an egg straight upward at 25 m/s. How long does it take for the egg to reach the ground?
308 Solution: We can use the formula: x = x0 + v0 t + 12 at2 . We know x0 = 0, since we’re throwing the egg from ground level; v0 = 25 m/s; a = −1.62 m/s2 (taking the upward direction as positive); and x = 0, since we want to know the time at which the egg reaches ground level. Plugging these numbers into the formula gives us 0 = −0.81t2 + 25t = −t(0.81t − 25) Solving this yields 25 = 31 s 0.81 The first solution refers to the fact that at t = 0, you were throwing the egg up from the surface. The second solution is the time at which the egg reaches x = 0 on its way back down. Hence the answer is: 31 s. t=0s
and
t=
Problem 1054. A train is moving at 32 m/s when the engineer is informed that a school bus full of orphans and puppies is on the tracks ahead of him. If the train is capable of decelerating at 0.80 m/s2 , how long does it take for the train to come to a halt? How far does it go after the brakes are applied? Solution: We’ll use: v = v0 + at, with v = 0 m/s, v0 = 32 m/s and a = −0.80 m/s2 . (We’ve taken the forward direction of the train as positive.) Then 0 = 32 − 0.80t
so
t=
32 m/s = 40 s 0.8 m/s2
Now that we know the time, we can use the formula: x = x0 + v0 t + 21 at2 . We’ll use x0 = 0, v0 = 32 m/s, a = −0.80 m/s2 , and t = 40 s. Then 1 x = 0 + (32 m/s)(40 s) − (0.8 m/s2 )(40 s)2 = 640 m 2 Problem 1055. Your lab partner accidentally rides his canoe over Horseshoe Falls, which is 53 m high. How long does it take for him to reach the bottom? What is his downward speed when he gets there? Solution: We’ll take the edge of the falls as x = 0, and the downward direction as positive. Then we can use the formula: x = x0 + v0 t + 12 at2 , with x0 = 0, v0 = 0, a = g = 9.8 m/s2 , and x = 53 m. We get 1 53 = 0 + 0t + (9.8 m/s2 )t2 = 4.9t2 2 p Hence t2 = 53/4.9; so t = 53/4.9 = 3.3 ps. Since v0 = 0, we can write v = at = 9.8 53/4.9 = 32 m/s.
309 Problem 1056. A ship has a maximum speed in still water of 8.8 m/s. To be operated legally on the Rhine River, it must be able to come to a complete stop in 350 m. What acceleration would the ship need to do this? Solution:
Again, we’ll need two formulas to solve this: 1 x = x0 + v0 t + at2 2
v = v0 + at
We know that x = 350 m, x0 = 0, v = 0, and v0 = 8.8 m/s. We want to calculate a. We’ll begin by solving the second equation for t: ⇒
0 = v0 + at
t=−
v0 a
We substitute this into the first equation: x=0−
v0 2 v0 2 v0 2 + =− a 2a 2a
We solve this for a: a=−
(8.8 m/s)2 v0 2 =− = −0.11 m/s2 2x (2)(350 m)
Problem 1057. A ball is thrown straight up from ground level, which we take to be y = 0. (a) At what speed must it be thrown to reach a height ymax meters above the ground? (b) How long does it take the ball to reach the maximum height tup ? Solution: (a) Use fundamental equation 2 with vf = 0, since when the ball reaches its highest point the its velocity is zero. p 0 = vf2 = vi2 − 2gymax ⇒ vi = 2gymax (b) Use fundamental equation one vf = vi + gt with vf = 0. Then tup = − vgi =
q
2ymax g
310
8 8.1
One-dimensional rotational kinematics Angular speed, period, and frequency
Problem 1058. A Ferris wheel has a radius of 12 m. If a point on the outer rim is moving at 4 m/s, what is its angular speed? *(a) (c) (e)
1/3 rad/s 2π/3 rad/s None of these
Solution:
v = rω
(b) 3 rad/s (d) 6π rad/s
⇒
ω=
4 m/s 1 v = = rad/s. r 12 m 3
Problem 1059. A reflector is attached to the spoke of a bicycle wheel, 20 cm from the center. If the wheel is turning at 18 rad/s, how fast is the reflector moving? Round your answer to two significant figures. (a) 110 cm/s (c) 640 cm/s (e) None of these Solution:
*(b) (d)
360 cm/s 1100 cm/s
v = rω = (20 cm)(18 rad/s) = 360 cm/s
Problem 1060. A bug is clinging to the tip of a wind turbine blade, 45 meters from the center. The blade turns through a complete circle every 7.5 s. What is the bug’s angular speed? Round your answer to two significant figures. *(a) (c) (e)
0.84 rad/s 19 rad/s None of these
Solution:
(b) 6.0 rad/s (d) 38 rad/s
If T is the period (the time that it takes to make a complete circle), then ω=
2π rad 2π rad = = 0.84 rad/s T 7.5 s
The 45-m radius of the circle is irrelevant to this problem. Problem 1061. The orbit of Jupiter’s moon Callisto has a radius of 1.9 × 109 m. The moon makes a complete circle around Jupiter in 1.4 × 106 s. What is Callisto’s orbital speed? Round your answer to two significant figures. (a) 6700 m/s *(c) 8500 m/s (e) None of these Solution:
v=
(b) 7400 m/s (d) 9100 m/s
2πr 2π · 1.9 × 109 m = = 8500 m/s T 1.4 × 106 s
311 Problem 1062. Planet X orbits its star at a distance of 1.5 × 1011 m. Its orbital speed is 3.0 × 104 m/s. How long does it take for the planet to make a complete circle around the star? Round your answer to two significant figures. (a) 2.3 × 107 s (c) 2.8 × 107 s (e) None of these
(b) 2.5 × 107 s *(d) 3.1 × 107 s
Solution: In making a complete circle, the planet goes a distance of 2πr; if it does so at speed v, then the time that it takes is 2π · 1.5 × 1011 2πr = 3.1 × 107 s T = = 4 v 3.0 × 10 m Problem 1063. An airplane propeller is turning at a rate of 40 revolutions per second. What is the propeller’s angular speed? Round your answer to two significant figures. (a) 6.4 rad/s (c) 130 rad/s (e) None of these Solution:
(b) *(d)
13 rad/s 250 rad/s
1 revolution = 2π rad; so ω=
(40 rev)(2π rad/rev) = 250 rad/s 1s
Problem 1064. A Ferris wheel takes 40 s to make a complete circle. What is its angular speed? Round your answer to two significant figures. (a) 0.025 rad/s (c) 6.4 rad/s (e) None of these
*(b) 0.16 rad/s (d) 13 rad/s
If T is the time that it takes to make a complete circle, then ω=
2π rad 2π rad = = 0.16 rad/s T 40 s
Problem 1065. An electric fan turns at a rate of 24 revolutions per second. How long does it take for the fan to turn through 60 radians? Round your answer to two significant figures. *(a) (c) (e)
0.40 s 1.4 s None of these
(b) 0.73 s (d) 2.5 s
Solution: Careful! We need to convert revolutions to radians. Since there are 2π radians in a revolution, if f is the angular speed in revolutions/s, then 2πf is the angular speed in rad/s. θ = ωt
⇒
t=
θ θ 60 rad = = = 0.40 s ω 2πf (2π rad/rev)(24 rev/s)
312 Problem 1066. A wind turbine turns with an angular speed of 1.1 rad/s. How long does it take for the turbine to make 10 complete revolutions? Round your answer to two significant figures. (a) 9.1 s (c) 31 s (e) None of these Solution:
(b) *(d)
17 s 57 s
We need to convert revolutions to radians, at 2π rad/rev. Then θ = ωt
⇒
t=
(10 rev)(2π rad/rev) θ = = 57 s ω 1.1 rad/s
Problem 1067. (Similarity Problem) Two bugs are clinging to a spoke of a turning bicycle wheel. The first bug is 10 cm from the center of the wheel; the second bug is 20 cm from the center. If the first bug is experiencing an angular speed of ω1 , what is the second bug’s angular speed? (a) √ ω1 /2 2 ω1 (c) (e) None of these Solution: itself.
*(b) ω1 (d) 2ω1
Both bugs experience the same angular speed, which is that of the wheel
Problem 1068. (Similarity Problem) Two bugs are clinging to a spoke of a turning bicycle wheel. The first bug is 10 cm from the center of the wheel; the second bug is 20 cm from the center. If the first bug is moving at a speed of v1 , what is the speed of the second bug? (a) v√1 /2 (c) 2 v1 (e) None of these
(b) v1 *(d) 2v1
Solution: Both bugs experience the same angular velocity ω, which is that of the wheel. Then v = ωr ⇒ v1 = (10 cm)ω and v2 = (20 cm)ω = 2v1 Problem 1069. (Similarity problem) A physics professor is riding on a moving merrygo-round. At a distance of r1 > 0 from the center, his angular speed is ω1 . At what distance from the center will his angular speed be 2ω1 ? √ (a) 2 r1 (b) 2r1 (c) 4r1 *(d) No solution (e) None of these Solution: No matter where the professor goes on the merry-go-round, his angular velocity will be the same: that of the merry-go-round itself. Since the merry-go-round is moving and r1 > 0, ω1 6= 0. Hence there is no distance at which his angular velocity will be 2ω1 6= ω1 .
313 Problem 1070. (Similarity problem) A physics professor is riding on a moving merrygo-round. At a distance of r1 > 0 from the center, his speed is v1 . At what distance from the center will his speed be 2v1 ? √ 2 r1 *(b) 2r1 (a) (c) 4r1 (d) No solution (e) None of these Solution: The professor’s angular velocity is always ω, the angular velocity of the merry-go-round. In the equation v = ωr, v and r change between the two situations; ω doesn’t. Separating the changing from the unchanging parameters, we get ω=
8.2
v1 2v1 v = = r r1 r2
⇒
r2 =
2v1 r1 = 2r1 v1
Fundamental equation 1
Problem 1071. A ceiling fan is initially at rest. When it is switched on, it experiences an angular acceleration of 2.2 rad/s2 . How long does it take before the fan is turning at 7.5 rad/s? Round your answer to two significant figures. (a) 1.6 s (c) 7.5 s (e) None of these
*(b) (d)
3.4 s 17 s
Solution: We know ω0 = 0, ω, and α. We want to know t. Use the first fundamental kinematic equation: ω = ω0 + αt = αt
⇒
t=
7.5 rad/s ω = 3.4 s = α 2.2 rad/s2
Problem 1072. An airplane propeller is initially turning at 220 rad/s. When the engine is switched off, it takes the propeller 18 seconds to come to a complete stop. What is the propeller’s angular acceleration? Round your answer to two significant figures. (a) −8.9 rad/s2 (c) −11 rad/s2 (e) None of these
(b) −9.9 rad/s2 *(d) −12 rad/s2
Solution: We know ω0 , ω = 0 rad/s, and t. We want to know α. Use the first fundamental kinematic equation: ω = 0 = ω0 + αt
⇒
α=−
ω0 220 rad/s =− = −12 rad/s2 t 18 s
314 Problem 1073. A wheel is initially at rest. It experiences an angular acceleration of 13 rad/s2 for 22 s. At the end of this time, what is its angular speed? Round your answer to two significant figures. *(a) (c) (e)
290 rad/s 3100 rad/s None of these
(b) 950 rad/s (d) 6300 rad/s
Solution: We know ω0 = 0, α, and t. We want to know ω. Use the first fundamental kinematic equation. ω = ω0 + αt = αt = (13 rad/s2 )(22 s) = 290 rad/s Problem 1074. A wind turbine is turning clockwise at 0.85 rad/s. The wind strengthens, causing the turbine to experience a clockwise angular acceleration of 0.044 rad/s2 . After 15 seconds, how fast is the turbine turning? Round your answer to two significant figures. (a) 0.56 rad/s (c) 4.1 rad/s (e) None of these
*(b) (d)
1.5 rad/s 6.1 rad/s
Solution: We know ω0 , α, and t. We want to know ω. Use the first fundamental kinematic equation. Careful! Notice that the problem specifies “clockwise”. When you see a direction specified, you should check the problem with extra care to make sure that all of your signs are right. There are no negative quantities in this particular problem, but there might be on the exam... ω = ω0 + αt = 0.85 rad/s + (0.044 rad/s2 )(15 s) = 1.5 rad/s Problem 1075. A disc is spinning at an initial angular speed of 10 rad/s. It is subjected to an angular acceleration of 2 rad/s2 for 5 seconds. At the end of this time, how fast is the disc spinning? (a) 10 rad/s (c) 35 rad/s (e) None of these
*(b) (d)
20 rad/s 60 rad/s
Solution: We know ω0 , α, and t; we want to know ω. Use the first fundamental kinematic equation. ω = ω0 + αt = 10 rad/s + (2 rad/s2 )(5 s) = 20 rad/s
315 Problem 1076. A propeller is initially turning at 20 rad/s. When the power to it is increased, it speeds up, reaching a speed of 48 rad/s after 4 seconds. What was its angular acceleration? *(a) (c) (e)
7 rad/s2 80 rad/s2 None of these
(b) (d)
12 rad/s2 192 rad/s2
Solution: We know ω0 , ω, and t. We want to know α. Use the first fundamental kinematic equation. ω = ω0 + αt
⇒
α=
48 rad/s − 20 rad/s ω − ω0 = = 7 rad/s2 t 4s
Problem 1077. A fan is initially turning at 120 rad/s. When it is switched off, it experiences an angular acceleration of −5 rad/s2 . After 10 seconds, how fast is it turning? (a) 24 rad/s *(c) 70 rad/s (e) None of these
(b) 50 rad/s (d) 96 rad/s
Solution: We know ω0 , α, and t. We want to know ω. Use the first fundamental kinematic equation. ω = ω0 + αt = 120 rad/s + (−5 rad/s2 )(10 s) = 70 rad/s Problem 1078. A circular saw is initially turning at 450 rad/s. It is subjected to an angular acceleration of −30 rad/s2 until it is turning at 300 rad/s. How long does it take before it reaches this speed? (a) 5/3 s (c) 10/3 s (e) None of these
*(b) (d)
5s 10 s
Solution: We know ω0 , ω, and α. We want to know t. Use the first fundamental kinematic equation. Watch your signs! ω = ω0 + αt
⇒
t=
300 rad/s − 450 rad/s ω − ω0 = =5s α −30 rad/s2
316 Problem 1079. A windmill is turning at 10 rad/s in a clockwise direction. The wind suddenly shifts, producing an angular acceleration of 0.5 rad/s2 in a counterclockwise direction. After 12 sec, what is the windmill’s angular velocity? (a) 6 rad/s counterclockwise (c) 6 rad/s clockwise (e) None of these
(b) *(d)
16 rad/s counterclockwise 4 rad/s clockwise
Solution: For no particularly good reason, we’ll take clockwise as the positive direction. We know ω0 , α, and t. We want to know ω. Use the first fundamental kinematic equation. ω = ω0 + αt = 10 rad/s + (−0.5 rad/s2 )(12 s) = 4 rad/s Since ω is positive, it’s clockwise. Problem 1080. A flywheel is turning at 60 rad/s in a counterclockwise direction. It is then subjected to an angular acceleration of 5 rad/s2 in a clockwise direction. After 20 seconds, what is the flywheel’s angular velocity? (a) 30 rad/s counterclockwise (c) 100 rad/s clockwise (e) None of these
*(b) 40 rad/s clockwise (d) 40 rad/s counterclockwise
Solution: We’ll arbitrarily choose clockwise as the positive direction. We know ω0 , α, and t. We want to know ω. Use the first fundamental kinematic equation. ω = ω0 + αt = −60 rad/s + (5 rad/s2 )(20 s) = 40 rad/s Since ω is positive, it’s turning clockwise. Problem 1081. Starting from rest, a wheel is subjected to an angular acceleration of 3.5 rad/s2 for 5 seconds; after that, it turns at a constant angular speed. At that constant speed, how long does it take for the wheel to make one revolution? Round your answer to two significant figures. (a) 0.32 s (c) 0.39 s (e) None of these
*(b) (d)
0.36 s 0.43 s
Solution: We’ll need to do this in two parts: first, calculate the final angular velocity; then calculate the period (the time that it takes for the wheel to make one revolution). We know ω0 = 0, α and t. We can get ω from the first fundamental kinematic equation, then calculate the period using 2π = ωT . ω = ω0 + αt = αt 2π = ωT
⇒
T =
2π 2π 2π = = = 0.36 s ω αt (3.5 rad/s2 )(5 s)
317 Problem 1082. A shaft is initially turning at 50 revolutions per second. When a brake is applied to it, it slows down and comes to a halt after 5.0 seconds. What angular acceleration does the shaft experience while it is slowing down? Round your answer to two significant figures. (a) −3.2 rad/s2 (c) −31 rad/s2 (e) None of these
(b) −6.4 rad/s2 *(d) −63 rad/s2
Solution: We need to convert revolutions per second to radians per second, using 2π rad/revolution. Then we’ll know ω0 , ω = 0, and t; we want to know α. Use the first fundamental kinematic equation. We’ll use f (for “frequency”) to signify the revolutions per second. ω = 0 = ω0 + αt
⇒
α=−
ω0 2πf0 (2π rad/rev)(50 rev/s) =− =− = −63 rad/s2 t t 5.0 s
Problem 1083. A merry-go-round is turning at a constant rate, with a period (the time that it takes for it to make one complete revolution) of 8.0 seconds. The operator increases the power to it, applying an angular acceleration of 0.25 rad/s for 5.0 seconds. At the end of this time, what is the merry-go-round’s period? Round your answer to two significant figures. *(a) (c) (e)
3.1 s 6.8 s None of these
(b) (d)
4.6 s 9.3 s
Solution: We will need to convert back and forth between period and angular speed. Since a complete circle is 2π radians, the conversion is ω = 2π/T , where T is the period. When we’ve done that with the initial period T0 , we’ll know ω0 , α, and t; we’ll use those to calculate the final angular speed ωf , which we’ll then convert to the final period Tf . From the first fundamental kinematic equation, ωf = ω0 + αt =
2π + αt T0
We convert ωf to Tf : Tf =
2π 2π 2πT0 2π(8 s) = = = = 3.1 s 2 2π ωf 2π + αtT0 2π + (0.25 rad/s )(5.0 s)(8.0 s) + αt T0
318 Problem 1084. A Ferris wheel is turning at a rate of one revolution every 10 seconds. The operator applies a brake to it for 8 seconds, after which it is turning at a rate of one revolution every 15 seconds. What angular acceleration does the Ferris wheel experience while it is slowing down? Round your answer to two significant figures. (a) −0.011 rad/s2 (c) −0.064 rad/s2 (e) None of these
*(b) −0.026 rad/s2 (d) −0.16 rad/s2
Solution: We will need to convert back and forth between period and angular speed. Since a complete circle is 2π radians, the conversion is ω = 2π/T , where T is the period. When we’ve done that with the initial period T0 and the final period Tf , we’ll know ω0 , ωf , and t; we’ll use these to calculate α. From the first fundamental kinematic equation: ωf = ω0 + αt 2π 2π − 2π(T0 − Tf ) 2π(10 s − 15 s) ωf − ω0 Tf T0 = = = = −0.026 rad/s2 ⇒ α= t t T0 Tf t (10 s)(15 s)(8 s)
8.3
Fundamental equation 2
Problem 1085. Starting from rest, a flywheel is subjected to an angular acceleration of 2.5 rad/s2 . After it has turned through 11 radians, how fast is it turning? Round your answer to two significant figures. *(a) (c) (e)
7.4 rad/s 9.0 rad/s None of these
(b) (d)
8.2 rad/s 9.9 rad/s
Solution: We know ω0 = 0, θ0 = 0, θ, and α. We want to know ω. We neither know nor care about the time. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) 2
⇒ ω = 2αθ
⇒
√ ω=
q 2αθ = 2(2.5 rad/s2 )(11 rad) = 7.4 rad/s2
319 Problem 1086. A windmill is initially held at rest by a brake. When the brake is released, it is subjected to an angular acceleration of 3 rad/s2 . What angle will it turn through before it reaches an angular speed of 12 rad/s? (a) 4 rad (c) 16 rad (e) None of these
(b) *(d)
12 rad 24 rad
Solution: We know ω0 = 0, α, θ0 = 0, and ω. We want to know θ. We neither know nor care about time. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) ⇒ ω 2 = 2αθ ω2 (12 rad/s)2 ⇒ θ= = = 24 rad 2α 2(3 rad/s2 ) Problem 1087. A physics instructor is riding on a merry-go-round, which is initially turning at 3 rad/s. He ignites a rocket that he has attached to the merry-go-round, producing an angular acceleration of 8 rad/s2 . How fast will the merry-go-round be turning after it has turned through 15 rad? Round your answer to the nearest rad/s. (a) 12 rad/s (c) 15 rad/s (e) None of these
(b) *(d)
13 rad/s 16 rad/s
Solution: We know ω0 , θ0 = 0, α, and θ. We want to know ω. We neither know nor care about the time. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) = 2αθ q q ⇒ ω = ω02 + 2αθ = (3 rad/s)2 + 2(8 rad/s2 )(15 rad) = 16 rad/s Problem 1088. A wheel is turning at a rate of 12 rad/s. When a brake is applied to it, it experiences an angular acceleration of α. If the wheel turns through 9 rad before coming to a halt, what is α? (a) −1/3 rad/s2 (c) (e)
−2 rad/s2 None of these
(b)
−4/3 rad/s2
*(d) −8 rad/s2
Solution: We know ω0 , ω = 0, θ0 = 0, and θ. We want to know α. We neither know nor care about t. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) ⇒ −ω02 = 2αθ ω2 (12 rad/s)2 = −8 rad/s2 ⇒ α=− 0 =− 2θ 2(9 rad)
320 Problem 1089. A propeller is turning at 30 rad/s. The power to it is increased, so that after it has turned through 20 radians, it is turning at 45 rad/s. What angular acceleration did the propeller experience? Round your answer to 2 significant figures. (a) 21 rad/s2 (c) 25 rad/s2 (e) None of these
(b) 23 rad/s2 *(d) 28 rad/s2
Solution: We know ω0 , ω, θ0 = 0, and θ. We want to know α. We neither know nor care about t. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) = 2αθ (45 rad/s)2 − (30 rad/s)2 ω 2 − ω02 = = 28 rad/s2 ⇒ α= 2θ 2(20 rad) Problem 1090. A shaft is initially turning at 10 rad/s. A brake is applied, producing an angular acceleration of −0.5 rad/s2 . What angle does the shaft turn through before it has slowed down to 6 rad/s? (a) 36 rad (c) 56 rad (e) None of these
(b) 45 rad *(d) 64 rad
Solution: We know ω0 , ω, α, and θ0 = 0. We want to know θ. We neither know nor care about t. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) = 2αθ ⇒ θ=
ω 2 − ω02 (6 rad/s)2 − (10 rad/s)2 = = 64 rad 2α 2(−0.5 rad/s2 )
Problem 1091. A windmill is initially spinning at a constant rate. The wind abruptly dies, slowing the windmill down at a rate of 0.25 rad/s2 . After the wind dies, the windmill turns through 30 rad before stopping. What was the windmill’s initial angular speed? Round your answer to two significant figures. *(a) (c) (e)
3.9 rad/s 6.0 rad/s None of these
(b) (d)
4.8 rad/s 7.5 rad/s
Solution: Watch your signs! We want ω0 to be positive, so α will be negative. We know α, θ0 = 0, θ, and ω = 0. We want to know ω0 . We neither know nor care about t. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) ⇒ ω02 = −2αθ q √ ω = −2αθ = −2(−0.25 rad/s2 )(30 rad) = 3.9 rad/s
321 Problem 1092. A steam turbine is initially turning at a constant rate. The steam pressure to it is increased, producing a positive angular acceleration of 20 rad/s2 . After it has turned through 30 rad, it has an angular speed of 50 rad/s. What was its initial angular speed? Round your answer to two significant figures. (a) 18 rad/s *(c) 36 rad/s (e) None of these
(b) 28 rad/s (d) 42 rad/s
Solution: We know θ0 = 0, θ, α, and ω. We want to know ω0 . We neither know nor care about t. Use the second fundamental kinematic equation. ω 2 − ω02 = 2α(θ − θ0 ) = 2αθ ⇒ ω 2 − 2αθ = ω02 q √ ⇒ ω0 = ω 2 − 2αθ = (50 rad/s)2 − 2(20 rad/s2 )(30 rad) = 36 rad/s Problem 1093. A wind turbine is initially spinning at a constant rate, making one revolution every 4.5 seconds. The wind increases, subjecting the turbine to an angular acceleration of 0.2 rad/s. After the turbine has completed 3 revolutions, it is once again turning at a constant rate. What is its new period of revolution? Round your answer to the nearest 0.1 s. *(a) 2.0 s (b) 2.7 s (c) 3.5 s (d) 4.6 s (e) None of these Solution: We need to convert between the period T (the time that it takes for the turbine to make one complete circle) and the angular speed ω. Since one complete circle is 2π rad, ωT = 2π. We know the initial period T0 (from which we can get ω0 ), the angular acceleration, θ0 = 0, and θ = (3 rev)(2π rad/rev) = 6π rad. We want the final period Tf , which we can get from the final angular speed ωf . We neither know nor care about t. Use the second fundamental kinematic equation. ωf2 − ω02 = 2α(θ − θ0 ) = 2αθ ⇒ ωf2 = ω02 + 2αθ ⇒
4π 2 4π 2 + 2αθT02 4π 2 = + 2αθ = Tf2 T02 T02
4π 2 T02 4π 2 + 2αθT02 2πT0 2π(4.5 s) q ⇒ Tf = p = = 2.0 s 2 4π 2 + 2αθT02 2 2 4π + 2(0.2 rad/s )(6π rad)(4.5 s) ⇒ Tf2 =
322 Problem 1094. A Ferris wheel is turning at a rate of one revolution every 8.0 seconds. The operator applies a brake while the wheel makes two revolutions; at the end of this time, it is turning at a rate of one revolution every 15.0 sec. What is the magnitude of the angular acceleration produced by the brake? Round your answer to two significant figures. α? *(a)
0.018 rad/s2
(b)
0.11 rad/s2
(c) (e)
6.4 rad/s2 None of these
(d)
40 rad/s2
Solution: We need to convert between the period T (the time that it takes to make one complete revolution) and the angular speed ω. Since one revolution is 2π radians, ωT = 2π. We know the inital period T0 , from which we can get ω0 ; the final period, Tf , from which we can get ωf ; and the number of revolutions turned, from which we can get the angle ∆θ = (2 revolutions)(2π rad/rev) = 4π rad. We want to know α. We neither know nor care about t. Use the second fundamental kinematic equation. 2α∆θ = ωf2 − ω02 = ⇒ α=
4π 2 (T02 − Tf2 ) 4π 2 4π 2 = − Tf2 T02 Tf2 T02
2π 2 (T02 − Tf2 ) 2π 2 [(8.0 s)2 − (15.0 s)2 ] = = −0.018 rad/s2 Tf2 T02 ∆θ (15.0 s)2 (8.0 s)2 (4π rad)
Since the question asks for the magnitude of α, we use the absolute value: 0.018 rad/s2 .
8.4
Fundamental equation 3
Problem 1095. A fan is initially turning at 12 rad/s. The power to the fan is increased, applying a rotational acceleration of 3 rad/s2 . What angle does the fan turn through over the next 2 sec? *(a) 30 rad (b) 36 rad (c) 42 rad (d) 45 rad (e) None of these Solution: We know θ0 = 0, ω0 , α, and t. We want to know θ. Use the third fundamental kinematic equation. 1 1 θ = θ0 + ω0 t + αt2 = (12 rad/s)(2 s) + (3 rad/s2 )(2 s)2 = 30 rad 2 2
323 Problem 1096. Starting at rest, a wheel experiences an angular acceleration of 2 rad/s2 . How much time elapses before the wheel has turned through 16 rad? √ 8s (a) 2 s (b) *(c) 4 s (d) 8 s (e) None of these Solution: We know θ0 = 0, ω0 = 0, α, and θ. We want to know t. Use the third fundamental kinematic equation. r 1/2 2θ 1 2 1 2 2(16 rad) =4s θ = θ0 + ω0 t + αt = αt ⇒ t= = 2 2 α 2 rad/s2 Problem 1097. A shaft is turning at 15 rad/s. A brake is applied, producing an angular acceleration of −0.25 rad/s2 . What angle does the shaft turn through over the next 4 seconds? (a) 49 rad (b) 52 rad *(c) 58 rad (d) 62 rad (e) None of these Solution: We know θ0 = 0, ω0 , α, and t. We want to know θ. Use the third fundamental kinematic equation. 1 1 θ = θ0 + ω0 t + αt2 = (15 rad/s)(4 s) + (−0.25 rad/s2 )(4 s)2 = 58 rad 2 2 Problem 1098. A propeller is spinning at 40 rad/s. When the power is switched off, it experiences a constant angular acceleration α. In the next 3 seconds, the propeller turns through 60 rad. What is α? Round your answer to the nearest rad/s2 . (a) −10 rad/s2 (c) −12 rad/s2 (e) None of these
(b) −11 rad/s2 *(d) −13 rad/s2
Solution: We know θ0 = 0, θ, ω0 , and t. We want to know α. Use the third fundamental kinematic equation. 1 1 θ = θ0 + ω0 t + αt2 = ω0 t + αt2 2 2 2[60 rad − (40 rad/s)(3 s)] 2(θ − ω0 t) = = −13 rad/s2 ⇒ α= t2 (3 s)2
324 Problem 1099. A windmill is turning at 12 rad/s. When the wind abruptly dies, it slows down at a rate of 3 rad/s2 . How long does it take for the windmill to turn through 16 rad? Round your answer to two significant figures. (a) 1.5 s (c) 1.9 s (e) None of these
*(b) 1.7 s (d) 2.1 s
Solution: We know θ0 = 0, ω0 , α, and θ. We want to know t. Use the third fundamental kinematic equation. Watch your signs; since the windmill is slowing down, we’ll take α as negative. 1 1 θ = θ0 + ω0 t + αt2 = ω0 t + αt2 2 2
⇒
α 2 t + ω0 t − θ = 0 2
Since the solution choices don’t include any integers, it’s likely that we won’t be able to factor this. We’ll use the quadratic formula. p p −ω0 ± ω02 − 4(α/2)(−θ) −ω0 ± ω02 + 2αθ t= = 2(α/2) α q −12 rad/s ± (12 rad/s)2 + 2(−3 rad/s2 )(16 rad) = −3 rad/s2 = 1.7 s or 6.3 s There are two positive solutions; we use the smaller one. Why? The equation doesn’t “know” that the windmill will slow down and remain stopped; as far as it’s concerned, the angular acceleration of −3 m/s2 will continue forever. If that were the case, then the windmill would turn through 16 rad as it slowed down; would come to a halt and then reverse its direction; and would then speed up in that backward direction, passing through the 16-rad mark a second time as it turned backward. Problem 1100. A turbine is turning at some constant angular velocity. The power to it is increased, producing an angular acceleration of 2.0 rad/s2 . Over the next 10 seconds, the turbine turns through 300 radians. What was its original angular velocity? Round your answer to two significant figures. (a) 16 rad/s *(c) 20 rad/s (e) None of these
(b) 18 rad/s (d) 22 rad/s
Solution: We know α, θ0 = 0, θ, and t. We want to know ω0 . Use the third fundamental kinematic equation. 1 1 θ = θ0 + ω0 t + αt2 = ω0 t + αt2 2 2 2θ − αt2 2(300 rad) − (2 rad/s2 )(10 s)2 ⇒ ω0 = = = 20 rad/s 2t 2(10 s)
325 Problem 1101. A winch is turning clockwise at 3 rad/s. The operator reverses the power to it, producing a counterclockwise angular acceleration of 0.50 rad/s2 for 10 seconds. At the end of this time, what is the net angle through which the winch has turned? *(a) (c) (e)
5 rad clockwise 55 rad clockwise None of these
(b) 5 rad counterclockwise (d) 55 rad counterclockwise
Solution: We know θ0 = 0, ω0 , α, and t. We want to know θ. Use the third fundamental kinematic equation. We’ll use clockwise as the positive direction, so α will be negative. 1 1 θ = θ0 + ω0 t + αt2 = (3 rad/s)(10 s) + (−0.50 rad/s2 )(10 s)2 = 5 rad 2 2 Since this is positive, it’s 5 rad clockwise. Problem 1102. A fan is turning at some constant angular velocity. It is subjected to a clockwise angular acceleration of 0.4 rad/s2 for 6 seconds, during which it turns 8 radians clockwise. What was its initial angular velocity? Round your answer to two significant figures. *(a) (c) (e)
0.13 rad/s clockwise 0.13 rad/s counterclockwise None of these
(b) 0.50 rad/s clockwise (d) 0.50 rad/s counterclockwise
Solution: We know θ0 = 0, θ, α, and t. We want to know ω0 . Use the fourth fundamental kinematic equation. We’ll use clockwise as the positive direction. 1 1 θ = θ0 + ω0 t + αt2 = ω0 t + αt2 2 2 2 2θ − αt 2(8 rad) − (0.4 rad/s2 )(6 s)2 ⇒ ω0 = = = 0.13 rad/s 2t 2(6 s) Since this number is positive, it’s 0.13 rad/s clockwise.
326 Problem 1103. A windmill is initially turning at a rate of one revolution per 4.8 seconds. The wind strengthens, applying an angular acceleration of 0.2 rad/s2 . Over the next 8 seconds, what angle does the windmill turn through? Round your answer to two significant figures. (a) 6.4 rad *(c) 17 rad (e) None of these
(b) 10 rad (d) 27 rad
Solution: We need to convert from the period T (the time that it takes for the windmill to make one revolution) to the angular speed. Since one revolution equals 2π rad, ωT = 2π. Then we know T0 , from which we can get ω0 ; θ0 = 0; α; and t. We want to know θ. Use the third fundamental kinematic equation. 1 2 2πt 1 2 2π(8 s) (0.2 rad/s2 )(8 s)2 + αt = + = 17 rad θ = ω0 t + αt = 2 T0 2 4.8 s 2 Problem 1104. A merry-go-round is turning at a rate of one revolution per 5.5 sec. The operator applies a brake, producing an angular acceleration α. In the next 4.0 seconds, the merry-go-round turns half a revolution. What is the magnitude of α? Round your answer to two significant figures. (a)
0.098 rad/s2
*(b)
0.18 rad/s2
(c) (e)
0.59 rad/s2 None of these
(d)
0.93 rad/s2
Solution: We need to convert from revolutions to radians, and from the period T (the time that it takes to make one revolution) to the angular speed. Since one revolution is 2π rad, ωT = 2π. We know the initial period T0 , from which we can calculate ω0 ; θ0 = 0; θf = 1/2 revolution = π rad; and t. We want to know α. Use the third fundamental kinematic equation. 1 θf = ω0 t + αt2 2 2(θf − ω0 t) 2(θf − 2πt/T0 ) 2(θf T0 − 2πt) ⇒ α= = = 2 2 t t T0 t2 2[(π rad)(5.5 s) − 2π(4.0 s)] = = −0.18 rad/s2 (5.5 s)(4.0 s)2 Since we’re asked for the magnitude, we use the absolute value: 0.18 rad/s2 .
327 Problem 1105. A water wheel is turning at a rate of one revolution per 7.5 seconds. The current to the wheel increases, applying an angular acceleration of 0.5 rad/s. How long does it take for the wheel to turn through two revolutions? Round your answer to two significant figures. (a) 4.5 s *(c) 5.6 s (e) None of these
(b) 5.0 s (d) 6.2 s
Solution: We need to convert the period T (the time that it takes to make one revolution) to the angular speed ω. Since one revolution is 2π rad, ωT = 2π. We know the initial period T0 , from which we can calculate ω0 ; α; and θ = 4π rad. We want to know t. Use the third fundamental kinematic equation.
⇒
1 2 1 αt + ω0 t − θ = 0 θ = ω0 t + αt2 ⇒ 2p 2 p p −ω0 ± ω02 − 4(α/2)(−θ) −ω0 ± ω02 + 2αθ −2π ± 4π 2 + 2αθT02 = = t= 2(α/2) α αT0 q −2π ± 4π 2 + 2(0.5 rad/s2 )(4π)(7.5 s)2 = = 5.6 s (0.5 rad/s2 )(7.5 s)
There’s also a negative root, but we ignore it. Problem 1106. An evil physics instructor tries to fail as many students as possible by speeding up the clock during an exam. He secretly installs a device that applies a constant angular acceleration α to the minute hand. At the beginning of the test, the hand is moving at the correct speed; because of the angular acceleration, it makes a complete circle in only 40 minutes. What is the value of α? Round your answer to two significant figures. (a)
8.7 × 10−8 rad/s2
*(b)
7.3 × 10−7 rad/s2
(c) (e)
6.6 × 10−5 rad/s2 None of these
(d)
4.2 × 10−4 rad/s2
Solution: We need to convert revolutions and minutes and hours to radians and seconds. We can easily determine ω0 , θ, and t. We want α. Use the first fundamental kinematic equation. 2(θ − ω0 t) 1 θ = ω0 t + αt2 ⇒ α = 2 t2 We can calculate ω0 from the fact that at the correct speed, a minute hand circles the clock in T0 = 1 hr = 3600 s; so ω0 = 2π/T0 . We need to convert t to seconds: t = 40 min = 2400 s. The minute hand makes a complete circle, so θ = 2π rad. Hence α=
2(θ − 2πt/T0 ) 2(θT0 − 2πt) 2[2π(3600 s) − 2π(2400 s)] = = = 7.3 × 10−7 rad/s2 2 2 2 t t T0 (2400 s) (3600 s)
328
9 9.1 9.1.1
Two-dimensional kinematics The basics of velocity and acceleration in two dimensions The difference between average speed and average velocity in a plane
Problem 1107. (Speed vs. Velocity) A race car goes around a track at a constant speed of 200 mph (It has cruise control). What is the car’s average speed and velocity over one lap on the race track? Solution: The average speed is 200 mph, but the average velocity is zero! The initial and final positions are the same, so ∆x = xf − xi = xi − xi = 0
⇒
vave =
∆x . ∆t
The point of this exercise is to emphasize that one must be careful when using long-time averages with vector quantities. The average velocity may give you a result that you were not intending to get. Problem 1108. (Speed vs. Velocity) If we throw a ball straight up in the air and it comes back down to the same spot, what is the average velocity of the ball over one complete trip up and back down? Solution: so
The average velocity is zero. The initial and final positions are the same,
∆y . ∆t The point of this exercise is to emphasize that one must be careful when using average velocity, because you may get a result that you were not intending to get. ∆y = yf − yi = yi − yi = 0
⇒
vave =
329 9.1.2
Average acceleration in a plane
Problem 1109. A circular room has a radius of 4 m and a small window facing north. A fly moving at 6 m/s flies directly south through the window and into the room. Over the next 20 s, he flies three times counterclockwise around the perimeter of the room, then flies directly north out the window at 6 m/s. What are the magnitude and direction of the fly’s average acceleration between the time he enters the room and the time he leaves? Solution: Since we asked for the magnitude and direction of the average acceleration, we must’ve meant the acceleration vector. (Unlike “velocity” and “speed”, we don’t have a convenient way to distinguish between the acceleration vector and its magnitude.) The only data that matter are the fly’s initial and final velocity, and the time over which it changed: from 6 m/s southward to 6 m/s northward, over 20 s. The difference between the final and initial velocity is 12 m/s northward; the change occurred over 20 s, so the average acceleration is 12 m/s = 0.6 m/s2 northward 20 s 9.1.3
Relative velocity in a plane
Problem 1110. A river has a current with a speed of 5 ft/s. You can swim at 4 ft/s in still water. At what angle to the cross-river direction do you need to swim to reach a point directly across the river from you? Solution: No solution. Whatever the angle is, your upstream component of velocity can be no greater than the total magnitude of your velocity: 4 ft/s. Thus you will find yourself going downstream at 1 ft/s or more. Problem 1111. A train is going directly north at 70 mph. An action-movie star is atop the train, running southward at 12 mph. Relative to someone standing on the ground, how fast is the action-movie star moving, and in what direction? Solution: Relative to someone on the ground, the AMS is moving northward at 70 − 12 = 58 mph.
330 9.1.4
Graphical interpretation of velocity and acceleration
Problem 1112. On the graph below, draw the velocity vectors at the points P1 and P2 on the curve. Assume the particle is moving with uniform speed from left to right. y6 r
P2
P1 r
-
x
Solution: Since the particle is moving with uniform speed, your vectors should have the same length. Since it’s moving from left to right, they should have positive x-components—that is, they should point up and to the left, not down and to the right. Problem 1113. The graph at right shows the top view of the path of a particle moving along a horizontal surface. The particle is moving from the lower left to the upper right. Which of the labelled vectors is the velocity vector at point P ? ~ (a) A *(b) ~ (c) C (d) (e) None of these
y6 ~ A
~ B @ I @
P
@s @ @
~ D
R @
~ C
~ B ~ D -
x
Solution:
The velocity vector is tangent to the curve in the direction of motion.
331 Problem 1114. The graph at right shows the path of a particle moving from the lower left to the upper right. Which of the labelled vectors is the acceleration vector at point P ? ~ (a) A (b) ~ *(c) C (d) (e) None of these Solution: The acceleration vector always points into the turn (on the concave side of the curve).
y6 ~ B
~ A @ I @
P @s @
~ B ~ D
@
~ D
R @
~ C
-
x
332
9.2
Uniform circular motion
Acceleration in uniform circular motion: v2 , r where v is the tangential speed of the object and r is the radius. arad =
Speed in uniform circular motion: v=v=
circumference of circle 2πr = = 2π r f , total time Tperiod
where the frequency f is defined as f ≡
1 Tperiod
.
Problem 1115. A freeway on-ramp has a curve with a radius of 59 m. You attempt to drive around the curve at 16 m/s. What is the centripetal acceleration on your car? Round your answer to the nearest 0.1 m/s2 . *(a) 4.3 m/s2 (c) 5.3 m/s2 (e) None of these Solution:
a=
(b) 4.8 m/s2 (d) 5.8 m/s2
(16 m/s)2 v2 = = 4.3 m/s2 r 59 m
Problem 1116. A ceiling fan has blades 81 cm long. A spider is clinging to the tip of one of the blades, which are turning so that the spider moves at 10.2 m/s. What is the spider’s radial acceleration? Round your answer to the nearest m/s2 . (a) 105 m/s2 *(c) 128 m/s2 (e) None of these Solution:
(b) (d)
117 m/s2 143 m/s2
Don’t forget to convert 81 cm to 0.81 m. a=
v2 (10.2 m/s)2 = = 128 m/s2 r 0.81 m
333 Problem 1117. A toddler’s parents have placed him on a merry-go-round, 2.7 m from the center. If the toddler experiences a radial acceleration greater than 1.2 m/s2 , it will fall off, prompting the angry parents to sue the merry-go-round operator. How fast can the operator allow the toddler to go without fear of legal action? Round your answer to the nearest 0.1 m/s. (a) 1.5 m/s *(c) 1.8 m/s (e) None of these Solution:
(b) 1.6 m/s (d) 2.0 m/s
We know r and a; we want to know v. q √ v2 ⇒ v = ar = (1.2 m/s2 )(2.7 m) = 1.8 m/s a= r
Problem 1118. A curve in a highway has a radius of 93 m. A truck will roll over if it experiences a centripetal acceleration greater than 1.9 m/s2 . How fast can the truck drive around the curve? Round your answer to the nearest m/s. (a) 11 m/s *(c) 13 m/s (e) None of these Solution:
(b) 12 m/s (d) 15 m/s
We know a and r; we want to know v. q √ v2 ⇒ v = ar = (1.9 m/s2 )(93 m) = 13 m/s a= r
Problem 1119. You are designing a highway that will include a curve. Cars will skid off the curve if they experience a centripetal acceleration greater than 2.2 m/s2 . You want cars to be able to drive the curve safely at 20 m/s. What must the radius of the curve be? Round your answer to the nearest 10 m. *(a) (c) (e)
180 m 220 m None of these
Solution:
(b) (d)
200 m 240 m
We know a and v; we want to know r. a=
v2 r
⇒
r=
v2 (20 m/s)2 = 180 m = a 2.2 m/s2
334 Problem 1120. An airplane is moving at 330 m/s when the pilot realizes that he is going in the wrong direction and needs to make a U-turn. If the pilot experiences a radial acceleration greater than 49 m/s2 , he will lose consciousness and crash the plane. What is the minimum safe radius of the turn? Round your answer to the nearest 100 m. (a) 1600 m (c) 2000 m (e) None of these Solution:
(b) *(d)
1800 m 2200 m
We know a and v; we want to know r. a=
v2 r
⇒
r=
v2 (330 m/s)2 = = 2200 m a 49 m/s2
Problem 1121. A bug is standing atop the blade of a ceiling fan, which is turning at 1.7 revolutions per second. The bug will lose its grip on the fan and go sailing off into the room if it experiences a radial acceleration greater than 91 m/s2 . How far from the center of the fan can the bug go without losing its hold? Round your answer to the nearest centimeter. (a) 58 cm (b) 65 cm (c) 72 cm *(d) 80 cm (e) None of these Solution: We can’t use our formula directly, since we know a but we don’t know r or v. However, there is a relationship between r and v, and we can substitute that into our formula. If the bug is standing at a distance r from the center of the fan, then every time the fan makes a complete revolution, the bug travels a distance of 2πr. We know that the fan is turning at a frequency of f = 1.7 revolutions/second. Hence the bug’s speed is v = 2πrf . Now we can use the formula: 4π 2 r2 f 2 v2 = = 4π 2 rf 2 a= r r
⇒
a 91 m/s2 r= 2 2 = 2 = 0.80 m = 80 cm 4π f 4π (1.7 s−1 )2
Problem 1122. A windmill has three blades, each 37 m long; the blades make a complete circle every 5.2 s. A bug is clinging to the tip of one of the blades. How much radial acceleration does the bug experience? Ignore the effect of gravity. Solution: The radius of the circle traced by the bug is r = 37 m. The bug makes a complete circle in t = 5.2 s. Hence the bug’s speed is v = 2πr/t. Its centripetal acceleration is v2 4π 2 r2 4π 2 r 4π 2 (37 m) ac = = = = = 54 m/s2 r rt2 t2 (5.2 s)2
335 Problem 1123. A windmill has three blades, each 8.3 m long; the blades make a complete circle every 4.1 s. A bug is clinging to the tip of one of the blades. How much radial acceleration does the bug experience? Ignore the effect of gravity. Round your answer to the nearest m/s2 . (a) 10 m/s2 (b) 0.1 m/s2 (c) 6 m/s2 *(d) 19 m/s2 (e) None of these Solution: The bug is moving in a circle with radius r = 8.3 m, going around it once every T = 4.1 s. The circumference of the circle is 2πr; so the bug’s speed is v = 2πr/T . The centripetal acceleration is 4π 2 r2 4π 2 r 4π 2 (8.3 m) v2 = 2 = = = 19 m/s2 2 2 r T r T (4.1 s)
a=
Problem 1124. A ceiling fan is turning at a rate of 1.8 revolutions per second. A spider is clinging to the blade of the fan. If he experiences centripetal acceleration greater than 2.7 g, he will lose his grip on the blade and go sailing off. How far from the center of the fan can the spider safely go? Give your answer in centimeters. Solution:
Since we want the answer in centimeters, we will convert: g = 980 cm/s2 .
If the spider is at a distance r from the center of the fan, which is turning at f revolutions per second, his speed is v = 2πrf The centripetal acceleration is ac =
4π 2 r2 f 2 v2 = = 4π 2 rf 2 r r
We know ac and f , and we want to find r: ac (2.7)(980cm/s2 ) r= 2 2 = = 21 cm 4π f 4π 2 (1.8/s)2 Problem 1125. A highway has a curve with radius of curvature 72 m. A truck will skid off the highway if it experiences a radial acceleration of more than 2.0 m/s2 . How fast can the truck safely go on the curve? Solution:
The centripetal acceleration is ac =
v2 r
Solving this for v yields v=
√
q ac r = (2.0 m/s2 )(72 m) = 12 m/s
336 9.2.1
Similarity problems
For all vehicles-going-around-circular-curve problems: Although it is not obvious at this point, the vehicle’s ability to stay on the road depends on the centripetal force provided by the friction and the tires. It turns out that there is a maximum centripetal force that the tires can supply to the road to keep the car from slidding. There is a direct correlation between force and acceleration, so it follows that there is a maximum centripetal acceleration for which the car can remain on the road. If this centripetal acceleration is exceeded, then the car will begin to slid. Problem 1126. (Similarity Problem) Two bugs are clinging to the second hand of a very large clock: one 3 ft from the center of the clock, the other 6 ft from the center. If the first bug experiences a radial acceleration of a, what radial acceleration does the second bug experience? Solution:
The centripetal acceleration of a body in circular motion is a=
v2 r
Be careful! Don’t forget that the two bugs are moving at different speeds. Since the second bug is twice as far from the center of the clock, he is moving twice as fast as the first bug. The second bug’s centripetal acceleration is (2v)2 2v 2 = = 2a 2r r Problem 1127. (Similarity Problem) A highway has a curve with radius r1 . Large trucks must drive the curve at a speed of v1 or less to keep from skidding off the road. What would the radius of the curve have to be in order for trucks to drive it safely at αv1 , where α > 0 is a real number? √ (a) α r1 *(b) α2 r1 √ (c) α α r1 (d) αr1 (e) None of these Solution: In the first case we are given v1 is the maximum speed of the truck on a curve with radius r1 . In the second case let v2 be the maximum speed of the truck on a curve with radius r2 . We want to solve for r2 in the case v2 = αv1 . The equation governing uniform circular motion is ac = v 2 /r. Since the centripetal acceleration must be the same for both cases, we can relate the two cases to one another via the centripetal acceleration ac as follows: 2 2 v12 v22 v2 αv1 = ac = ⇒ r2 = r1 = r1 = α2 r1 r1 r2 v1 v1
337 Problem 1128. (Similarity Problem) A highway has a curve with a radius of r1 . Large trucks must drive the curve at a speed of v1 or less to keep from skidding off the road. If the highway were rebuilt so that the curve had twice radius, how fast could trucks safely drive around the curve? √ (b) 2v1 *(a) √2 v1 (c) 2 2 v1 (d) 4v1 (e) None of these Solution: We are given: r1 , v1 , and r2 = 2r1 , and we want v2 . Since the centripetal acceleration ac must be the same for both situations the governing equation is already in the standard similarity form: v 2 /r = ac . Equating the two situations through this equation, and using the centripetal acceleration as the common link gives: v2 v12 = ac = 2 r1 r2
·r2
−→
v22
=
r2 v12 r1
√
=
2v12
−−−−−−→
v2 =
√
2v1
Problem 1129. (Similarity Problem) A highway has a curve with radius r1 . Large trucks must drive the curve at a speed of 20 mph or less to keep from skidding off the road. What would the radius of the curve have to be in order for trucks to drive it safely at 60 mph? √ (b) 3r1 (a) √3 r1 *(d) 9r1 (c) 3 3 r1 (e) None of these Solution: Let v1 = 20 mph be the maximum speed of the truck on a curve with radius r1 . The governing equation is in standard similarity form v 2 /r = ac . The maximum centripetal acceleration that a truck can experience without sliding off the road is ac = v12 /r1 . Now we want to know how large r2 must be for a truck going at a speed of v2 = 60 mph = 3v1 to experience an acceleration of ac . Equating the two situations through the common parameter in the governing equation yields 2 ·v22 v22 r1 r2 v2 v12 flip eqn = ac = −−−−−−−→ = 2 −−−−−→ r2 = r1 = 9r1 2 r1 r2 v1 v2 v1 Problem 1130. (Similarity Problem) A highway has a quarter-circle curve with a radius of r. Large trucks must take the curve at a speed of 25 mph or less to avoid skidding off the road. What would the radius of the curve have to be in order for trucks to drive around it safely at 50 mph? Solution:
The centripetal acceleration of a body in circular motion is a=
v2 r
We want to double the value of v and change the value of the radius so that a remains the same. Thus we need to increase the radius to 4r. STOP
338
9.3
Projectile motion
~ = Axˆı + Ay ˆ, then from the definition of the trig functions about the right triangle: If A
p
~ A = A
= A2x + A2y Ay −1 θ0 = tan Ax 9.3.1
General equations for projectile motion
Let the initial time be t0 = 0, the initial position be (x0 , y0 ), and the initial velocity be ~v0 = v0xˆı + v0y ˆ. Typically we take the initial position to be (x0 , y0 ) = (0, h), where h is the height of the mechanism doing the launching. The general equations governing motion in 2-D are the set of fundamental equations in 1-D repeated in each of the x and y directions. Substituting (x0 , y0 ) = (0, h) into the fundamental equations (1) and (3) in each coordinate direction, and taking ax = 0 (since there is no force acting in the horizontal direction) yields
velocity x-component: position x-component:
vx = v0x x = v0x t
(9.1a) (9.1b)
If we take our y-axis pointing upward, then ay = −g and we have velocity y-component: position y-component:
vy = v0y − gt 1 y = h + v0y t − gt2 2
(9.2a) (9.2b)
If the initial velocity is given in terms of magnitude k~v0 k = v0 , and direction θ0 , as measured in the counterclockwise direction from the x-axis, then we must convert this information into information about the components of the vector in order to use the above formulas. ( v0x = v0 cos θ0 (x-component of initial velocity) v0y = v0 sin θ0 (y-component of initial velocity)
339 9.3.2
Velocity and acceleration of a projectile
Problem 1131. The following four diagrams show the trajectory of a cannonball. Which diagram correctly shows the acceleration vectors at three points along the trajectory? r
y6
~0
6
r
(a)
?
r
r -
y6 r
r Q
7
(b) s Q
-
-
x
r
y6
? r Z ~ Z
r
(c)
x
r
y6
r
*(d)
? r
?
? -
-
x
x
(e) None of these Solution: At every point, the acceleration is 9.8 m/s2 , directed straight down. Your vectors should all be of the same length. The answer is (d). Problem 1132. The following four diagrams show the trajectory of a cannonball. Which diagram correctly shows the velocity vectors at three points along the trajectory? r
y6
~0
6
r
(a)
?
r
r -
y6 r
r Q
7
-
-
x
r
y6
r ?
r Z ~ Z
(c)
*(b) s Q
x
r
y6
r
(d)
? r
?
? -
x
(e) None of these Solution:
Only figure (b) has vectors that are tangent to the curve.
-
x
340 Problem 1133. Draw the acceleration vector due to gravity at the points A, B, and C along the curve of the trajectory of the cannonball. y6
Br Ar ?
rC
? ? -
x
Solution: At every point, the acceleration is 9.8 m/s2 , directed straight down. Your vectors should all be of the same length.
341 9.3.3
Special Case I: The Half-Parabola
Note: For the case of the half-parabola the initial height is the maximum height and the ball is fired horizontally. Setting h = ymax > 0 and θ0 = 0, which leads to v0y = 0 and v0x = v0 , in the general equations reduces them to
velocity x-component: position x-component:
vx = v0 x = v0 t
(9.3a) (9.3b)
If we take our y-axis pointing upward, then ay = −g and we have velocity y-component: position y-component:
vy = −gt 1 y = ymax − gt2 2
(9.4a) (9.4b)
Note: For this case, it would also make sense to take the y-axis pointing downward. Then the time to fall is just a one-dimensional problem. The time to fall, denoted by tfall , is found from the fundamental equation 3, with the y-axis pointing downward so that ay = g. Once again v0y = 0, so the equation becomes r 1 2 2ymax ymax = gtfall ⇒ tfall = . 2 g Problem 1134. A skateboarder inadvertently rides off the edge of a flat roof at 8.2 m/s. The roof is 23 m high. How long does it take for the skateboarder to reach the ground? Round your answer to the nearest 0.1 s. (a) 1.8 s *(c) 2.2 s (e) None of these
(b) 1.9 s (d) 2.4 s
Solution: Since the skateboarder is riding off a flat roof, we assume that his angle of launch is θ0 = 0. Hence there will be no vertical component of his initial velocity. That means that the time it takes for him to reach the ground is the same as if he were dropped from roof level: r 1/2 2y 2(23 m) = = 2.2 s t= g 9.8 m/s2 The skateboarder’s initial velocity has nothing to do with this problem.
342 Problem 1135. A bicyclist rides off a flat roof at 13.4 m/s. The roof is 3.6 m above the ground. How far from the edge of the building does the bicyclist land? Round your answer to the nearest meter. *(a) 11 m (b) 12 m (c) 14 m (d) 15 m (e) None of these Solution: Since the bicyclist is riding off a flat roof, we assume that the elevation is θ0 = 0. Then the vertical component of the initial velocity is v0y = 0, and the horizontal component is v0x = v = 13.4 m/s. The horizontal velocity does not change, since there’s no horizontal acceleration. Hence the horizontal distance that the bicycle travels is the initial horizontal velocity times the time it takes for the bicycle to fall to the ground: r 1/2 2y 2(3.6 m) = (13.4 m/s) = 11 m x = v0x t = v0x g 9.8 m/s2 Problem 1136. You fire your potato gun horizontally from shoulder height, which is 140 cm above the ground. The potato strikes the ground 15 m away from you. What was the initial speed of the potato? Round your answer to the nearest m/s. (a) 25 m/s (c) 31 m/s (e) None of these
*(b) (d)
28 m/s 34 m/s
Solution: Since you fired the potato gun horizontally, the angle of elevation is θ0 = 0. This means that there is no vertical component to the initial velocity. The potato’s initial velocity was all horizontal; and since there’s no horizontal acceleration, the horizontal p velocity is constant. The time it takes for the potato to reach the ground is t = 2y/g, where y = 1.4 m is the initial height. (Don’t forget to convert 140 cm to 1.4 m.) During this time, the potato travels a horizontal distance of x = 15 m. Hence the initial velocity was s r x x g 9.8 m/s2 = v0 = v0x = = p x= (15 m) = 28 m/s t 2y 2(1.4 m) 2y/g
343 Problem 1137. You fire your potato gun horizontally from shoulder height, which is 150 cm above the ground. The potato strikes the ground 11 m away from you. What was the initial speed of the potato? Round your answer to the nearest m/s. *(a) (c) (e)
20 m/s 24 m/s None of these
(b) (d)
22 m/s 26 m/s
Solution: Since you fired the potato gun horizontally, the angle of elevation is θ0 = 0. This means that there is no vertical component to the initial velocity. The potato’s initial velocity was all horizontal; and since there’s no horizontal acceleration, the horizontal p velocity is constant. The time it takes for the potato to reach the ground is t = 2y/g, where y = 1.5 m is the initial height. (Don’t forget to convert 150 cm to 1.5 m.) During this time, the potato travels a horizontal distance of x = 11 m. Hence the initial velocity was s r xmax g 9.8 m/s2 xmax =p = xmax = (11 m) = 20 m/s v0 = v0x = tfall 2y 2(1.5 m) 2y/g Problem 1138. Your potato gun has a muzzle velocity of 27 m/s. You shoot it horizontally from a window 12 m above the ground. How far from the building does the potato strike the ground? Round your answer to the nearest meter. (a) 32 m (c) 52 m (e) None of these
*(b) (d)
42 m 62 m
Solution: Since you’re shooting the gun horizontally, the vertical component of the initial velocity is v0y = 0, and the horizontal component is v0x = v = 27 m/s. The range xmax is the product of the horizontal velocity v0x and the time tfall that it takes for the potato to reach ground level (i.e., xmax = v0x tfall , where tfall is the time it takes the object to fall from its initial position at y = y0 to its final position on the ground at y = 0). To find tfall , we use the formula y = y0 + v0y t − 12 gt2 . We use y0 = 12 m, y = 0, and t = tfall . The formula simplifies to: 0 = y0 − 21 gt2fall . Solving for tfall yields r tfall =
2y0 g
Then the range is r x = v0x tfall = v0x
√ 2y0 (27 m/s) 2 · 12 m q = = 42 m g 2 9.8 m/s
344 Problem 1139. Your potato gun has a muzzle velocity of 10 m/s. You shoot it horizontally from a window 5 m above the ground. How far from the building does the potato strike the ground? Approximate gravity as g = 10 m/s2 . Round your answer to the nearest meter. *(a) 10 m (b) 12 m (c) 14 m (d) 8 m (e) None of these Solution: Since you’re shooting the gun horizontally, the vertical component of the initial velocity is v0y = 0, and the horizontal component is v0x = v = 27 m/s. The range xmax is the product of the horizontal velocity v0x and the time tfall that it takes for the potato to reach ground level (i.e., xmax = v0x tfall , where tfall is the time it takes the object to fall from its initial position at y = y0 to its final position on the ground at y = 0). To find tfall , we use the formula y = y0 + v0y t − 12 gt2 . We use y0 = 12 m, y = 0, and t = tfall . The formula simplifies to: 0 = y0 − 21 gt2fall . Solving for tfall yields r tfall =
2y0 g
Then the range is r x = v0x tfall = v0x
s 2y0 2·5 m = 10 m = (10 m/s) g 10 m/s2
Problem 1140. (Similarity Problem) You shoot your potato gun horizontally from shoulder height, which is 150 cm. The potato lands 25 m away from you. You would like the potato to go 75 m, so you climb up a ladder and fire the gun horizontally from there. How high off the ground must the gun be? Solution: This is a similarity problem. Since the gun is fired horizontally, this is a half-parabola case. Step 1 is always write down what you’re given, what you want, and any assumptions. Step 1: h1 = 150 cm = 1.5 m (convert to meters right away) Given: xmax,1 = 25 m xmax,2 = 75 m = 3 xmax,1 Notice that we must convert the height to meters (or acceleration to cm/s2 ). Want: h2 = ? (initial velocity)
345 Assumptions: The potato gun always produces the same initial speed for every potato. This is a big assumption that is probably not true. Step 2: Determine the equation that governs the motion of the particle in both experiments. Then determine which of the variables in the governing equation are held fixed for both experiments (we call these the common parameters) and which variables are varied. Put all of the equations that vary on the left-hand side of the governing equation (LHS) and all of the parameters that are held fixed over the two experiments on the RHS of the governing equation. Lastly, solve for the desired variable. Governing equation: We need an equation that gives xmax in terms of height. We’ve already derived such an equation from our fundamental equations: s xmax = v0
2h g
Common parameters: v0 , g, 2 Next, we rearrange the governing equation and put all of the parameters that are varied over the experiments on the LHS of the governing equation and all of the parameters that are held fixed over the two experiments on the RHS of the governing equation. s r √ xmax 2h 2 · h √ = v0 −−−−−→ xmax = v0 g g h Now we equate the results from the two experiments through the common parameters:
flip equation
−−−−−−−−−−→ (
)2
−−−−−−−→ ·x2max,2
r xmax,1 2 xmax,2 √ = √ = v0 g h1 h2 √ √ h1 h2 = xmax,1 xmax,2 h1 2 xmax,1
−−−−−−−→ h2 =
h2 2 xmax,2 2 xmax,2 =
xmax,1
h1 =
3xmax,1 xmax,1
2
h1 = 32 h1
Discussion: The range of a projectile fired horizontally is proportional to the time that it takes it to fall from its initial height to the ground. This time is proportional to the square root of the initial height. Since you want your potato’s range to be three times
346 greater, you must increase the fall time by a factor of 3; so you must increase the initial height by a factor of 32 = 9. Therefore you must fire the gun from a height of 9 · 3/2m = 27/2 m = 13.5m ≈ 41 ft. You are going to need a very tall ladder. Try the local fire department... Problem 1141. A movie stuntman wants to ride a motorcycle off a flat roof, sail across a gap 24 m wide, and land on another roof 3.6 m below the first one. How fast does he need to be moving when he leaves the first roof? *(a) (c) (e)
28 m/s 34 m/s None of these
(b) (d)
31 m/s 37 m/s
Solution: The stuntman is going off a flat roof, with no mention of a ramp; so we assume that the elevation is θ0 = 0. There is no vertical component of initial velocity, and v0 = v0x . p The time it takes for the motorcycle to fall y = 3.6 m is t = 2y/g. During this time, the motorcycle needs to cover a horizontal distance of x = 24 m. Hence the initial (horizontal) velocity must be s r x g 9.8 m/s2 x = x= (24 m) = 28 m/s v= =p t 2y 2(3.6 m) 2y/g Problem 1142. While an unmanned rover is exploring the surface of Planet X, it goes off the edge of a cliff at 11.2 m/s. The cliff is 3.1 m high; the rover strikes the ground 18.8 m from the base of the cliff. What is the surface gravity of Planet X? Round your answer to the nearest 0.1 m/s2 . *(a) 2.2 m/s2 (c) 2.6 m/s2 (e) None of these
(b) 2.4 m/s2 (d) 2.9 m/s2
Solution: Since no elevation is mentioned, we assume that θ0 = 0 and thus that v0y = 0 and v0x = v0 . We can calculate the time of fall from the horizontal velocity and the horizontal distance that the rover travels as it falls: t = x/v0 . During this time t, the rover falls a distance of y. 1 y = gX t2 2
⇒
gX =
2y 2y 2yv02 2(3.1 m)(11.2 m/s2 ) = = = = 2.2 m/s2 t2 (x/v0 )2 x2 (18.8 m)2
347 Problem 1143. A ball rolls off the edge of a horizontal table 1.32 m high. It strikes the floor at a point 4.45 m horizontally away from the edge of the table. What was the ball’s speed at the moment that it left the tabletop? Round your answer to the nearest 0.1 m/s. Solution: The ball’s initial velocity is horizontal; so if the table’s height is H, then the time it takes for the ball to fall to the ground is s 2H t= g During that time, it traverses a horizontal distance of X; so its horizontal speed is q √ (4.45 m) 9.8 m/s2 X g X √ =√ v0 = = = 8.6 m/s t 2 · 1.32 m 2H
348 9.3.4
Special Case II: The Full Parabola
Note: For the case of the full-parabola the initial height is zero and the ball is fired at an angle of incline 0 < θ0 ≤ π2 . Setting h = 0 and θ0 6= 0, which leads to v0x 6= 0 (except if θ0 = π2 ) and v0y 6= 0, in the general equations reduces them to velocity x-component: position x-component:
vx = v0x x = v0x t
(9.5a) (9.5b)
If we take our y-axis pointing upward, then ay = −g and we have velocity y-component: position y-component:
vy = v0y − gt 1 y = v0y t − gt2 2
(9.6a) (9.6b)
Problem 1144. Your Civil War musket has a muzzle velocity of 290 m/s. If you fire it at an elevation of 6.2◦ , what is the maximum height reached by the bullet? Round your answer to the nearest meter. (a) 36 m (b) 41 m (c) 45 m *(d) 50 m (e) None of these Solution: We have the initial velocity, elevation, and gravitational acceleration; we want to know the maximum height. We have a formula for this: ymax =
(290 m/s)2 (sin 6.2◦ )2 1 v02 sin2 θ0 = = 50 m 2 g 2(9.8 m/s2 )
Problem 1145. A golfer hits a ball from ground level with an initial speed of 42 m/s at an elevation of 19◦ . How far from the initial location does the ball hit the ground? Round your answer to the nearest 10 m. (a) 100 m (c) 120 m (e) None of these
*(b) (d)
110 m 130 m
Solution: We have the initial velocity, the elevation, and the gravitational acceleration. We want to know the range. We have a formula for this: xmax =
(42 m/s)2 sin(2 · 19◦ ) v02 sin(2θ0 ) = = 110 m g 9.8 m/s2
349 Problem 1146. A golfer hits a ball from ground level with an initial speed of 57 m/s at an elevation of 34◦ . How far from the initial location does the ball hit the ground? Round your answer to the nearest 10 m. *(a) (c) (e)
310 m 370 m None of these
(b) (d)
340 m 410 m
Solution: We have the initial velocity, the elevation, and the gravitational acceleration. We want to know the range. We have a formula for this: xmax =
(57 m/s)2 sin(2 · 34◦ ) v02 sin(2θ0 ) = = 310 m g 9.8 m/s2
Problem 1147. You fire your potato gun from ground level at an angle of 36◦ above the horizontal. The potato strikes the ground 47 m away from you. What was the initial speed of the potato? Round your answer to the nearest m/s. *(a) (c) (e)
22 m/s 26 m/s None of these
(b) (d)
24 m/s 29 m/s
Solution: We know the elevation, the gravitational acceleration, and the range; we want to know the initial velocity. Use the formula for range: xmax
v2 = 0 sin(2θ0 ) g
⇒
r v0 =
" #1/2 gxmax (9.8 m/s2 )(47 m) = = 22 m/s sin(2θ0 ) sin(2 · 36◦ )
Problem 1148. You are holding the neck of a champagne bottle at an elevation of 41◦ when you pop the cork. The highest point in the cork’s flight is 2.6 m above its initial height. What was the cork’s initial speed? Round your answer to the nearest m/s. (a) 9 m/s *(c) 11 m/s (e) None of these
(b) 10 m/s (d) 12 m/s
Solution: We know the elevation, the gravitational acceleration, and the maximum height; we want to know the initial velocity. Use the formula for maximum height: q √ 2(9.8 m/s2 )(2.6 m) 1 v02 2gymax 2 ymax = sin θ0 ⇒ v0 = = = 11 m/s 2 g sin θ0 sin 41◦
350 Problem 1149. You are holding the neck of a champagne bottle at an elevation of 34◦ when you pop the cork. The highest point in the cork’s flight is 1.9 m above its initial height. What was the cork’s initial speed? Round your answer to the nearest m/s. (a) 10 m/s (c) 12 m/s (e) None of these
*(b) (d)
11 m/s 13 m/s
Solution: We know the elevation, the gravitational acceleration, and the maximum height; we want to know the initial velocity. Use the formula for maximum height: q √ 2(9.8 m/s2 )(1.9 m) 2gymax 1 v02 2 sin θ0 ⇒ v0 = = = 11 m/s ymax = 2 g sin θ0 sin 34◦ Problem 1150. A motorcyclist would like to ride up a short ramp on a rooftop, fly across a gap between two buildings, and land on a second rooftop at the same height as the first. The ramp is inclined at 32◦ ; the buildings are 18 m apart. How fast does the motorcyclist have to be going when he makes the jump? Round your answer to the nearest m/s. (a) 13 m/s (c) 15 m/s (e) None of these
*(b) (d)
14 m/s 16 m/s
Solution: We know the gravitational acceleration, the elevation, and the range; we want to know the initial velocity. Use the formula for range: xmax
v2 = 0 sin(2θ0 ) g
⇒
r v0 =
" #1/2 gxmax (9.8 m/s2 )(18 m) = = 14 m/s sin(2θ0 ) sin(2 · 32◦ )
Problem 1151. A crossbow fires its arrow at 105 m/s. If the arrow is aimed at an elevation of 12◦ above the horizontal, what is the horizontal distance that it will travel before striking the ground? Round your answer to the nearest 10 m. (a) 370 m *(c) 460 m (e) None of these
(b) 410 m (d) 500 m
Solution: We know the initial velocity, the gravitational acceleration, and the elevation; we want to know the range. We have a formula for this: xmax
(105 m/s)2 sin(2 · 12◦ ) v02 sin(2θ0 ) = = 460 m = g 9.8 m/s2
351 Problem 1152. A Napoleon cannon has a muzzle velocity of 450 m/s. If the cannon is fired at an elevation of 5.1◦ , how much time elapses before the ball strikes the ground? Round your answer to the nearest 0.1 s. *(a) (c) (e)
8.2 s 9.8 s None of these
(b) (d)
9.0 s 10.6 s
Solution: We know the elevation, initial velocity, and gravitational acceleration. We want to know the flight time. We have a formula for this: txmax =
2(450 m/s) sin 5.1◦ 2v0 sin θ0 = = 8.2 s g 9.8 m/s2
Problem 1153. Space pirates have kidnapped you from your physics lab and are holding you for ransom on another planet. Luckily, they have overlooked your cell phone and your spring cannon. The spring cannon has a muzzle velocity of 8.6 m/s; when you fire it at an elevation of 45◦ , the ball strikes the ground 41 m away. (The pirates are keeping you in a rather large room.) What is the surface gravity of the planet on which you’re being held? Round your answer to the nearest 0.1 m/s2 . (a) 1.6 m/s2 (c) 2.0 m/s2 (e) None of these
*(b) 1.8 m/s2 (d) 2.2 m/s2
Solution: We know the initial velocity, elevation, and range; we want to know the gravitational acceleration, which we’ll call gX . Use the formula for range: xmax =
v02 sin(2θ0 ) gX
⇒
gX =
v02 (8.6 m/s)2 sin(2 · 45◦ ) sin(2θ0 ) = = 1.8 m/s2 xmax 41 m
Problem 1154. Your potato gun has a muzzle velocity of 23 m/s. You want to shoot a potato and have it strike the ground 3.5 seconds later. At what elevation do you have to fire the gun? Round your answer to the nearest degree. (a) 43◦ (c) 53◦ (e) None of these
*(b) (d)
48◦ 58◦
Solution: We know the initial velocity, the gravitational acceleration, and the time of flight. We want to know the angle of elevation. Use the formula for flight time: txmax = ⇒
2v0 sin θ0 g
gtxmax 2v0 " # 2 gt (9.8 m/s )(3.5 s) xmax θ0 = sin−1 = sin−1 = 48◦ 2v0 2(23 m/s) ⇒
sin θ0 =
352 Problem 1155. You have designed a tennis-ball cannon with a muzzle velocity of 62 m/s. You want to land a tennis ball in a swimming pool 210 m away. At what angle to the horizontal do you need to aim the gun? Round your answer to the nearest degree. (a) 11◦ (c) 15◦ (e) None of these
(b) 13◦ *(d) 16◦
Solution: We know the initial velocity, the gravitational acceleration, and the range. We want the angle of elevation. Use the formula for range: v02 gxmax −1 gxmax xmax = sin(2θ0 ) ⇒ sin(2θ0 ) = ⇒ 2θ0 = sin g v02 v02 # " 1 −1 gxmax 1 −1 (9.8 m/s2 )(210 m) ⇒ θ0 = sin = = 16◦ sin 2 v02 2 (62 m/s)2 Problem 1156. You have designed a tennis-ball cannon with a muzzle velocity of 62 m/s. You want to land a tennis ball in a swimming pool 210 m away. At what angle to the horizontal do you need to aim the gun? Your answer should be less than π/4 radians. (a) 1.08 rad (b) 0.13 rad (c) 0.97 rad (d) 0.28 rad (e) None of these Solution: The horizontal range of a projectile is the product of the flight time and the horizontal component of velocity. The horizontal component of velocity is vx = v0 cos θ. For a projectile fired from ground level, the flight time is t = 2v0 sin θ/g. Then the horizontal range is 2v0 2 cos θ sin θ xmax = g We know x = 210 m, v0 = 62 m/s, and g = 9.8 m/s2 . We want to find θ. The presence of both sin θ and cos θ in the formula creates some difficulty. Happily, we remember the trigonometric identity: sin(2θ) = 2 sin θ cos θ Substituting this produces xmax = Now we can solve for θ: 1 −1 gxmax 1 −1 θ = sin = sin 2 v0 2 2
v0 2 sin(2θ) g
(9.8 m/s2 )(210 m) (62 m/s)2
! = 0.28 rad
353 Problem 1157. A dietitian is standing at the edge of a cliff. He throws a piece of cheese upward and outward, so that it reaches a maximum height of 12 m above him, then falls down into the canyon 65 m below him and 40 m from the edge of the cliff. Between the time he releases the cheese and the time that it strikes the bottom, where is the acceleration at a minimum? Solution: There is no maximum or minimum acceleration: during the entire time that the cheese is in the air, it is experiencing acceleration of 9.8 m/s2 downward. The various numbers describing the trajectory are irrelevant to the problem. Problem 1158. The surface gravity on Mars is 0.4 times the surface gravity on the Earth. A golfer can hit a ball so that it strikes the ground 200 m from him on Earth. If he hits the ball with the same speed and at the same angle on Mars, how far will it go? Solution: The range of a projectile is the product of the horizontal velocity and the time that it takes for the projectile to reach the ground. The horizontal velocity is vx = v0 cos θ. If the projectile is launched from ground level, the time is t = 2v0 sin θ/a, where a is the acceleration due to gravity. Hence the range is x=
2v0 2 cos θ sin θ a
On Earth, a = g; on Mars, a = 0.4g. The values of v0 and θ are the same on both planets. Hence g · xE = 0.4g · xM
so
xM =
xE 200 m = = 500 m 0.4 0.4
354 Problem 1159. Your physics teacher is trying to get a planeload of “prescription” drugs across the border. As an extra-credit project, he has assigned you and your lab partner to shoot down the radar blimp. The blimp flies at a height of 10,000 ft. You fire a cannon at an angle of θ to the horizontal, and the maximum height attained by the shell is 5000 ft. Your lab partner says that if you increase the angle to 2θ, the shell will go high enough to hit the blimp. Is he correct? Why or why not? Solution: Your lab partner is wrong. The obvious counterexample is: let θ = 90◦ . Then 2θ = 180◦ ; but a shell fired at 180◦ will never rise above ground level, so will reach a maximum height of zero. More generally, let’s look at ymax as a function of θ. The time from the initial release of the projectile to its maximum height is the time that it takes for its vertical velocity to reach zero: v0 sin θ t= g The maximum height is the distance that a body travels in that time if its initial velocity is v0 sin θ and it experiences an acceleration of −g: v0 sin θ v0 2 sin2 θ gt2 gv0 2 sin2 θ = v0 sin θ = ymax = v0 t sin θ − − 2 g 2g 2 2g Clearly, a simple relation like “2θ ⇒ 2ymax ” isn’t going to hold. Problem 1160. You are trying to put out a fire 150 m away with a water cannon. The manual informs you that if you aim the water cannon at an elevation of 23◦ above the horizontal, it will have a range of 150 m. Unfortunately, there is a building between you and the fire, and at an elevation of 23◦ the water will hit it. Is there another angle of elevation θ that you could use to get a range of 150 m? Give θ, or explain why it doesn’t exist. Solution: A projectile launched from ground level at an elevation of θ will have the same range as a similar projectile launched at an elevation of 90◦ − θ. In this case, since you can’t use an elevation of 23◦ , an elevation of 90◦ − 23◦ = 67◦ should work (provided, of course, that the obstructive building isn’t too high).
355 Problem 1161. If a champagne bottle is held with its neck elevated 78◦ above the horizontal, the popped cork will reach a maximum altitude of 4.8 m. Is there another angle θ between 0◦ and 90◦ for which the cork will reach the same maximum altitude? Give θ, or explain why it doesn’t exist. Solution:
There is no such angle θ.
Let’s calculate the maximum altitude ymax as a function of θ. The time from the popping of the cork to its reaching its maximum height is the time that it takes for its vertical velocity to reach zero: v0 sin θ t= g The maximum height is the distance that a body travels in that time if its initial velocity is v0 sin θ and it experiences an acceleration of −g: v0 sin θ gv0 2 sin2 θ v0 2 sin2 θ gt2 = v0 sin θ − = ymax = v0 t sin θ − 2 g 2g 2 2g Between 0 and 90◦ , sin θ is a nonnegative and strictly increasing function of θ. That means that within that range, there are no two values of θ that will produce the same value of ymax . In conclusion, for projectile motion there are two θ values that will give the same distance, but only one that will give the same height. Problem 1162. The Paris Gun had a muzzle velocity of 1600 m/s. If the gun was fired at an elevation of 45◦ above the horizontal, what was the maximum altitude reached by the shell? Give your answer in kilometers. Solution: The vertical component of the shell’s velocity as a function of time is vy = v0 sin θ − gt. The shell reaches its maximum altitude when the vertical component of velocity is zero; so at time t = v0 sin θ/g. The maximum altitude is the vertical distance travelled in that time t by a projectile with initial vertical speed v0 sin θ, subjected to an acceleration of −g: v0 2 sin2 θ v0 sin θ gv0 2 sin2 θ gt2 = v0 sin θ − = ymax = v0 t sin θ − 2 g 2g 2 2g Substituting our values for v0 , θ, and g, we get: ymax =
(1600 m/s)2 sin2 45◦ = 65, 000 m = 65 km 2(9.8 m/s2 )
356 Problem 1163. A crossbow fires its arrow at 345 ft/s. If the arrow is aimed at an elevation of 12◦ above the horizontal, what is the horizontal distance that it will travel before striking the ground? Solution: The horizontal range of a projectile is the product of the horizontal component of velocity and the time that the projectile is in the air. If the projectile is fired from ground level with a velocity of v0 and an elevation of θ, the time in the air is 2v0 sin θ/g. The horizontal component of velocity is vx = v0 cos θ. Hence the horizontal range is 2(345 ft/s)2 2v0 2 sin θ cos θ = sin 12◦ cos 12◦ = 1500 ft x= 2 g 32 ft/s ˇ akov´a won a gold medal in the 2008 Olympics by throwing Problem 1164. Barbora Spot´ ˇ akov´a threw the javelin at an angle of π/4 radians a javelin 71.42 m. Assuming that Spot´ to the horizontal, what was its speed when initially thrown? Solution: The horizontal range of a projectile is the product of the horizontal component of velocity and the time that the projectile is in the air. If the projectile is fired from ground level with a velocity of v0 and an elevation of θ, the time in the air is 2v0 sin θ/g. The horizontal component of velocity is vx = v0 cos θ. Hence the horizontal range is 2v0 2 sin θ cos θ x= g We know x, θ, and g. Solving for v0 , we get 1/2 gx v0 = = 2 sin θ cos θ
(9.8 m/s2 )(71.42 m) 2 sin π4 cos π4
!1/2 = 26 m/s
Problem 1165. The Napoleon cannon of the Civil War had a muzzle velocity of 1485 ft/s. If the cannon was fired at an elevation of 5.0◦ above the horizontal, how much time would elapse between the firing and the ball’s striking the ground? Solution: If a projectile is fired from ground level with initial velocity v0 and elevation θ, its flight time is 2v0 sin θ 2(1485 ft/s) sin 5◦ = = 8.1 s g 32 ft/s2
357 Problem 1166. Using surgical tubing and ingenuity, you have constructed a waterballoon launcher in a window 22 m above the ground. Your device launches a water balloon at a speed of 19 m/s, and at an elevation of 0.25 radians above the horizontal. At what horizontal distance from the building will the water balloon strike the ground? Solution: The horizontal range is the product of the flight time and the horizontal velocity. The horizontal velocity is vx = v0 cos θ. The flight time is the time that it takes for the projectile to reach the ground. The projectile’s height above ground level is: y = y0 + v0 t sin θ −
gt2 2
We need to find t for which y = 0. Using the quadratic formula yields p p −v0 sin θ ± (v0 sin θ)2 + 4(g/2)y0 −v0 sin θ ± v0 2 sin2 θ + 2gy0 = t= −2(g/2) −g q (−19 m/s)(sin 0.25 rad) ± (19 m/s)2 (sin 0.25 rad)2 + 2(9.8 m/s2 )(22 m) = −9.8 m/s2 We want to use the positive value, which is about 2.65 s. (You should not use this rounded value in subsequent calculations; we present it only because it’s a good idea to check intermediate results and make sure they seem reasonable.) The range is x = vx t = v0 t cos θ = (19 m/s)(cos 0.25 rad)t = 49 m
358 Problem 1167. A bicyclist unfamiliar with the terrain is riding along a level sidewalk when he goes over the edge of a staircase. He hits the staircase 3.7 m below the edge, and 3.7 m horizontally from the edge. How fast was he going when he went over the edge? Solution: We will use the bicyclist’s vertical travel to calculate his flight time; and his flight time and horizontal travel to calculate his initial horizontal speed. We take the height of the sidewalk as y0 = 0. We want to know how long it takes for the bicyclist to reach a height of y = −3.7 m. Since his initial velocity was horizontal, his height as a function of time is gt2 y=− 2 Solving for t: s r 2y 2(3.7 m) t= − = = 0.87 s g 9.8 m/s2 During this time, the bicyclist travels a horizontal distance of x = 3.7 m. Thus his horizontal velocity is q √ (3.7 m) 9.8 m/s2 x g x √ vx = = √ = 4.3 m/s = t −2y 2 · 3.7 m Problem 1168. You fire a cannon at an angle of θ to the horizontal, with a muzzle velocity of v0 . The ball lands at a horizontal distance of x from the cannon. For your second shot, you increase the powder charge in the cannon so that the muzzle velocity will be 2v0 . How far from the cannon will the ball land? Solution: The range of a projectile is the product of the horizontal velocity and the time that it takes for the projectile to reach the ground. The horizontal velocity is vx = v0 cos θ. If the projectile is fired from ground level, the time is t = 2v0 sin θ/g. Hence the range is 2v0 2 x= cos θ sin θ g If we increase the muzzle velocity v0 by a factor of 2, we increase the range by a factor of 22 = 4; so the new range is 4x.
359
9.4
Lab application problems: Two-Dimensional Kinematics
This is a mixture of problem types based on actual equipment found in instructional physic labs everywhere. 9.4.1
Spring-cannon apparatus
Problem 1169. (Lab Problema ) You fire a spring cannon horizontally from a fixed height hcannon = ymax . It strikes the ground at a distance xmax from the base of the mouth of the muzzle. Find a formula for the ball’s initial velocity as a function of the initial height ymax and range xmax . The initial velocity is taken to be the velocity of the ball as it leaves the mouth of the cannon. r √ 2ymax (b) xmax (a) xmax 2gymax r r g 2g g (c) xmax *(d) xmax ymax 2ymax (e) None of these a
This is a common technique for determining initial velocity, based on measured lengths rather than on measured time. It is good for estimating the initial velocity when the spring cannon is shot at small to moderate angles of incidence (θ0 ≤ 45◦ ).
Solution:
Since the ball is fired horizontally, v0x = v0 . Using fundamental kinematic r 2ymax is the time equation (1) in the x-direction we have xmax = v0 tfall , where tfall = g for the ball to fall a distance ymax . Thus, r xmax g ÷tfall xmax = v0 tfall −−−−−−→ v0 = = xmax tfall 2ymax
360 Problem 1170. (Lab Similarity Problem) You fire a spring cannon horizontally from a fixed height. The initial velocity of the ball is v0,1 ; it strikes the ground at a distance xmax,1 from the base of the mouth of the muzzle. If you want the ball to travel twice the horizontal distance, what must the initial velocity be? √ *(b) 2v0,1 (a) √2 v0,1 (c) 2 2 v0,1 (d) 4v0,1 (e) None of these Solution: If a projectile is fired horizontally, the horizontal range is the initial velocity times the time that it takes for the projectile to fall to the ground: xmax = v0 tfall . Since the cannon is fired horizontally, the fall time is the same as if we dropped the ball straight down from rest from the height of the cannon hcannon . The fall time is then s s 2hcannon 2hcannon ·v0 . −−−−−→ xmax = v0 tfall = g g This is our governing equation. Since g and hcannon are the same in both situations (i.e., these are our common parameters) we re-arrange the governing equation as s s xmax 2hcannon 2hcannon ÷v0 −−−−−→ . tfall = = g v0 g Using the standard form of the governing equation to equate the two situations through the common parameters gives s xmax,1 xmax,2 xmax,2 2hcannon = = ⇒ v0,2 = v0,1 = 2v0,1 . v0,1 g v0,2 xmax,1
361 Problem 1171. (Lab Similarity Problem) If you fire a spring cannon horizontally from a height of hcannon,1 , the ball travels a horizontal distance of xmax,1 before hitting the floor. How high must the spring cannon be to get the ball to travel twice the horizontal distance before hitting the ground? √ (b) 2hcannon,1 (a) √2 hcannon,1 (c) 2 2 hcannon,1 *(d) 4hcannon,1 (e) None of these Solution: If a projectile is launched horizontally, the horizontal distance that it travels is the initial p velocity times the time that it takes for it to fall to the ground: xmax = v0 tfall = v0 2hcannon /g. This is our governing equation. Re-arranging the terms to put the common factors on the right and the parameters that are situation dependent on the left gives: s r √ xmax 2hcannon 2 x2max 2v02 ( )2 ÷ hcannon √ −−−−−−→ xmax = v0 = v0 −−−→ = . g g hcannon g hcannon The last step is to use the governing equation in the similarity standard form to determine the relationship between the heights of the two similar problems: x2max,1 x2max,2 2v 2 = 0 = hcannon,1 g hcannon,2 hcannon,2 hcannon,1 flip equation = 2 −−−−−−−−−−→ 2 xmax,1 xmax,2 2 ·x2max,2 xmax,2 −−−−−−−→ hcannon,2 = hcannon,1 = 4 hcannon,1 xmax,1 Problem 1172. (Lab Problem) You fire a spring cannon from ground level at an elevation of 23◦ . The ball emerges from the cannon at a speed of 14.2 m/s. How long does it take before the ball strikes the ground? Round your answer to the nearest 0.1 s. (a) 1.0 s (c) 1.2 s (e) None of these
*(b) (d)
1.1 s 1.4 s
Solution: We know the initial velocity, the elevation, and the surface gravity; we want to know the flight time. We have a formula for this: txmax =
2(14.2 m/s) sin 23◦ 2v0 sin θ0 = = 1.1 s g 9.8 m/s2
362 Problem 1173. (Lab Problem) A high-powered spring cannon discharges its ball at 16.6 m/s. If the cannon is fired at an elevation of 77◦ , what is the maximum height reached by the ball? Round your answer to the nearest meter. (a) 12 m (c) 15 m (e) None of these
*(b) (d)
13 m 16 m
Solution: We have the initial velocity, elevation, and gravitational acceleration; we want to know the maximum height. We have a formula for this: ymax =
(16.6 m/s)2 (sin 77◦ )2 1 v02 sin2 θ0 = = 13 m 2 g 2(9.8 m/s2 )
Problem 1174. (Lab Problem) In your physics lab, you are using a new high-powered spring cannon with a muzzle velocity of 19 m/s. The ceiling of the lab is 2.8 m above the level of the cannon. What is the maximum elevation at which you can fire the cannon without hitting the ceiling? Round your answer to the nearest degree. (a) 16◦ (c) 21◦ (e) None of these
(b) 18◦ *(d) 23◦
Solution: We know the initial velocity, the maximum height, and the gravitational acceleration. We want to know the angle of elevation. Use the formula for maximum height: √ 2gymax 2gymax 1 v02 2 2 sin θ0 ⇒ sin θ0 = ⇒ sin θ0 = ymax = 2 2 g v0 v0 q √ 2(9.8 m/a2 )(2.8 m) 2gymax −1 −1 = 23◦ ⇒ θ0 = sin = sin v0 19 m/s
363 Problem 1175. (Lab Similarity Problem) You fire a spring cannon at a fixed elevation θ0 6= 0. The spring cannon uses a compressed spring to launch the ball. The cannon has three settings with the first being the lowest and the third being the strongest. Using the first setting (one click) you find that the ball emerges from the muzzle with a speed of v0,1 ; it strikes the ground at a distance xmax,1 from the cannon. You then increase the energy stored in the spring by using the second setting (two clicks). The muzzle velocity of the ball is now measured to be v0,2 = α v0,1 , where α is a positive constant. What is the new range of the ball? √ α xmax,1 (b) α xmax,1 (a) √ *(d) α2 xmax,1 (c) α α xmax,1 (e) None of these Solution:
Use the formula (the governing equation) for range:
xmax =
v02 sin(2θ0 ) g
÷v 2
−−−−0−→
xmax sin(2θ0 ) = 2 v0 g
(the common factors) ,
where we’ve re-arranged the governing equation to be in the standard form for a similarity equation with the common factors on the right-hand side and the situation-dependent variables on the left-hand side. If we increase the initial velocity from v0,1 to v0,2 = α v0,,1 , we get 2 2 ·v0,2 v0,2 sin(2θ0 ) xmax,1 xmax,2 = = 2 −−−−−−→ xmax,2 = xmax,1 = α2 xmax,1 2 v0,2 g v0,1 v0,1 An alternative approach is to make a direct substitution for v0,2 in terms of v0,1 in the original equation: 2 (α V )2 V 2 xmax,2 = sin(2θ0 ) = α sin(2θ0 ) = α2 xmax,1 g g Thus, if we double the initial speed we quadruple the distance traveled by the ball. Actually, in the real world the faster the ball travels, the larger the air drag.
364 Problem 1176. (Lab Similarity Problem) You fire a spring cannon at a fixed elevation θ0 6= 0. The ball emerges from the muzzle with a speed of v0,1 ; it reaches a maximum height of ymax . You would like to increase the muzzle velocity so that the ball can double its height. What must the new muzzle velocity be? √ (b) 2 v0,1 *(a) √2 v0,1 (c) 2 2 v0,1 (d) 4 v0,1 (e) None of these Solution: situations.
Use the formula for maximum height as the governing equation for the two
ymax =
1 v02 sin2 θ0 2 g
÷v 2
−−−−0−→
sin2 θ0 ymax = v02 2g
(common factors)
We want to find v0,2 such that the maximum height is ymax,2 = 2ymax,1 . sin2 θ0 ymax,1 ymax,2 = = 2 2 v0,2 2g v0,1 2 v0,1 = ymax,2 ymax,1 ymax,2 ·ymax,2 2 2 −−−−−−−→ v0,2 = v0,1 ymax,1 r √ ymax,2 √ −−−−−−→ v0,2 = v0,1 = 2 v0,1 ymax,1
flip equation
−−−−−−−−−−→
2 v0,2
Problem 1177. (Derive Problem) A ball rolls off the edge of a horizontal table of height H. It strikes the floor at a horizontal distance X away from the edge of the table. What was the ball’s speed at the moment that it left the tabletop? Derive a formula that involves the given data: H and X. Solution: The ball’s initial velocity is horizontal; so if the table’s height is H, then the time it takes for the ball to fall to the ground is s 2H tfall = g During that time, it traverses a horizontal distance of X; so its horizontal speed is r X g v0 = =X tfall 2H Problem 1178. (Derive Problem) A child whirls a stone in a horizontal circle at a height of H above the ground, on the end of a string with length L. The string breaks and the stone flies off horizontally, striking the ground at a horizontal distance of X from where it was when the string broke. What was the centripetal acceleration of the stone before the string broke? Your answer should be a formula of the form ac = ac (H, L, X, g),
365 where g is the acceleration due to gravity. Solution: To find the centripetal acceleration, we need to know the speed v of the stone while it was circling. If we knew the time t that it took for the stone to cover the horizontal distance X, we could calculate: v = X/t. We are not given t, but we know that it’s how long it takes for the stone to fall to the ground from an initial height of H. Thus we’re going to proceed in three steps: use H and g to determine t; use X and t to determine v; and use v and L to determine the centripetal acceleration ac . The time t is:
s t=
The speed v is:
2H g
√ X g X =√ v= t 2H
The centripetal acceleration ac is: ac =
X 2g v2 = L 2HL
Problem 1179. (Lab Derive Problem) A ball is fired horizontally out of a spring cannon with an initial velocity of v0 ; the cannon is a height h above the ground. The ball strikes the ground at a horizontal distance xmax from the muzzle of the cannon. Find v0 as a function of hcannon , xmax , and g. Note: The purpose of this problem is to find the initial velocity of the ball as it leaves the muzzle of the spring cannon. Using this initial velocity one can then point the cannon in any direction and predict the maximum height and the range of the projectile. Solution: Since the cannon is fired horizontally, the initial vertical velocity of the ball is zero; and we’re assuming that the horizontal component of the velocity remains v0 for the entire flight. We will use hcannon and g to determine the flight time (the time it takes the object to fall a distance hcannon ) tfall ; then use tfall and xmax to determine v0 . The time it takes for the ball to fall from a height of hcannon to the ground is s 2hcannon tfall = g The horizontal component of the ball’s velocity is r xmax g v0 = = xmax tfall 2hcannon
366
9.5
Advanced-Level Problems (Kinematics)
General algorithm for physics derivation problems Step 1: (Setup) Write down what you are given, what you want, and list any assumptions that you need to make to solve the problem: for example, “the pulley and string are massless”, “the surface is frictionless”, etc. Draw any relevant pictures and label them appropriately. Step 2: Starting from the general equations for projectile motion written on the formula sheet, substitute the values for the initial position, velocity, etc. That is, find the set of equations that are particular to this problem. Step 3: Solve the resulting equations. Derive problems will be graded out of 10 points. The setup step is worth between 3 and 4 points. The point breakdown for the setup is given below. Given: (1 point) Write down what you are given. Want: (1 point) Write down what you want (what you’re solving for). Assumptions: (1 point) Write down any assumptions and key physical laws/equations that you are going to use. Notice that I am including writing down the relevant equations and physical laws in this step. Here are some examples: (1.) Assume the surface is made of ice that can be treated as a frictionless surface. You should include this even if it is in the problem statement because it is a good habit to get into. Such an assumption of frictionless in a real problem could lead to dubious results. (2.) Use fundamental equation 2 and Newton’s second law. Picture: (1 point) Draw a picture, if one is not drawn for you. If you are resolving forces on a system, then draw a free-body diagram for each object in the system that you’re analyzing, based on the picture. This should include a coordinate system for each object.. (If a picture has been provided for you, then you will not receive credit for this step). How the rest of the problem is graded is too problem specific to provided an explicit algorithm. Here is a general guide line: • Use the free-body diagram/diagrams to determine the net force acting on each object in the system, in each component direction. • Apply the correct physical law/laws and solve the problem. This is the hard part!
367 Problem 1180. (Lab Derive Problem) Consider the situation shown in the figure 3 below. Take the mouth of the cannon to be at the point (x0 , y0 ) = (0, h). Assume that you know the speed v0 and direction θ of the initial velocity ~v0 = v0 hcos θ, sin θi. Derive a formula for the vertex of the trajectory (xymax , ymax ) and the range point (xmax , 0) in terms of the given information. Note: The purpose of this problem is to find the vertex and the range of a projectile.
Figure 3: Spring cannon Solution: See the file chapter4-projectile-motion-general-case under the physics lecture notes webpage.
368 Problem 1181. In one of your physics labs you are given the challenge of shooting a ball from a spring cannon into a bucket. Your job is to figure out where to place the center of the bucket in order to launch the ball into the bucket! Also, you are only allowed one attempt to put the ball in the bucket. You are given the following information (see figure 5): • The ball is released from the mouth of the cannon and is located at the initial position (x0 , y0 ) = (0, Hcannon ). • The bucket stands Hbucket units above the table with the center of the bucket located xmax units in the horizontal direction from the mouth of the cannon. Also, as the picture suggests, Hbucket < Hcannon . • The mouth of the bucket is just large enough to allow the ball to be captured in the bucket provided that the ball lands close enough to the point (xmax , Hbucket ). • The initial speed v0 = k~v0 k and direction θ0 that the ball leaves the mouth of the spring cannon.
Figure 4: Spring cannon and bucket
369 Problem 1182. (Lab Derive Problem [Put the ball in the bucket]) In one of your physics labs you are given the challenge of shooting a ball from a spring cannon into a bucket. Specifically, you need to work out the angle the cannon should be aimed in order to launch the ball into the bucket! You are only allowed one attempt to put the ball in the bucket and you cannot move the bucket. You are also told that the cannon has plenty enough power to launch the ball into the bucket. You are given the following information (see figure 5): • The bucket stands Hbucket units above the table with the center of the bucket located xmax units in the horizontal direction from the mouth of the cannon. • The mouth of the bucket is just large enough to allow the ball to be captured in the bucket provided that the ball lands close enough to the point (xmax , Hbucket ). • You are instructed to take the mouth of the cannon to be at the point (x0 , y0 ) = (0, Hcannon ). By using a swivel mount at the mouth of the cannon, the device is designed in such a way as to keep the height of the ball the same regardless of the direction θ that the cannon is aimed. Thus Hcannon is independent of θ. • The initial speed v0 = k~v0 k that the ball leaves the cannon. Derive a formula for θ in terms of the given information.
Figure 5: Spring cannon and bucket Solution: This is identical to the problem solved in class except that we must reduce the height of the cannon by the height of the bucket. Let h = Hcannon − Hbucket > 0.
370 Problem 1183. (Lab Derive Problem [Put the ball in the moving bucket]) The Idea: In one of your physics labs you are given the challenge of shooting a ball from a spring cannon into a bucket mounted on top of a glider moving at a constant velocity along an air track that is aligned in the direction of the cannon. The Apparatus: A horizontal air track is sitting atop a table. The track passes under a spring cannon that is attached to a mount that holds the cannon directly above the air track. From a top view the cannon is aimed along the track in the direction of motion of the glider. From a side view, the cannon makes an angle 0 ≤ θ0 ≤ π/2 with respect to the horizontal air track (see figure 6 below).
A bucket sits atop a glider moving down the air track at constant speed Vglider . When the bucket passes directly below the the cannon, the cannon is fired and the ball emerges at the muzzle of the cannon with a velocity of V0 . The apparatus is designed so that the center of mass of the ball, denoted by x, is directly over the center of the bucket xcenter as the ball leaves the muzzle. Take this time to be the initial time t0 = 0 and take the location along the air track to be x0 = 0.
Figure 6: Side view of the spring cannon firing ball into a bucket mounted on top of a glider The Setup: You are given the following information • You are instructed to take the mouth of the cannon to be at the point (x0 , y0 ) = (0, Hcannon ). By using a swivel mount at the mouth of the cannon, the device is designed in such a way as to keep the height of the ball the same regardless of the direction θ0 that the cannon is aimed. Thus Hcannon is independent of θ0 .
371 • The bucket stands Hbucket units above the table with the center of the bucket located a distance xcenter = Vglider t from the mouth of the cannon in the horizontal direction along the track. Thus the ordinates of the center of the bucket, the target, are (xcenter , Hbucket ). • The mouth of the bucket is large enough to offset the small effects of air drag so that the ball can be captured in the bucket provided that the ball lands close enough to the point (xcenter , Hbucket ). • Assume Hbucket < Hcannon . • The initial speed of the ball as it leaves the mouth of the cannon is k~v0 k = V0 > 0. • Assume that you know the speed of the glider Vglider and the initial speed of the ball as it leaves the mouth of the cannon V0 . The Questions: (a) Starting from the general equations for projectile motion with x0 = 0: ( x(t) = v0x t (the ball’s x-component:) (the ball’s y-component:) y(t) = h + v0y t − 12 gt2 substitute the values for the initial position (0, h), velocity ~v0 = V0 hcos(θ0 ), sin(θ0 )i, etc. That is, find the set of equations that are particular to this problem. (b) Taking the initial speed of the ball as leaves the mouth of the cannon V0 to be known from a previous experiment, what is the range of speeds for the glider Vglider > 0 for which there is a solution. That is, for what range of speeds will the ball land in the bucket, provided it is aimed properly. (c) In order for the ball to land in the bucket, what must the cannon’s angle of elevation θ0 be? (d) At what horizontal distance xmax from the mouth of the cannon will the ball land in the bucket? Potential future problem: What would happen if the air track were not level?
372
Part IV
Newton’s Laws
373
10 10.1
Introduction to Newton’s Laws Concept questions involving Newton’s three laws
Problem 1184. (Derived quantities) Determine the expression for the unit of force (the Newton N) in terms of the fundamental dimensions mass, length, and time from the equation expressing Newton’s 2nd Law F = ma. Solution: gives
Using our take-the-dimensions-of-the-quantity operator [ ] on the equation [F ] = [ma] = M
L m = kg 2 2 T s
⇒
1 Newton = kg
m s2
Problem 1185. An object is subjected to a nonzero constant net force F~ . Which of the following properties of the object will be constant? (a) Position (c) Velocity (e) None of these
(b) *(d)
Speed Acceleration
Problem 1186. (Similarity Problem) Consider the following two experiments. In the first experiment you push a block along a level surface at a constant velocity V . To overcome the friction between the block and the surface, you must apply a constant horizontal force Fpush,1 . In the second experiment you push the block at a constant speed 2V across the same surface. Assuming that the friction is the same as it was the first experiment, which of the following choices best describes the new push force Fpush,2 in terms of the first push force? *(a)A constant force Fpush,1 (b) A constant force 2Fpush,1 (c) A force that increases from Fpush,1 to 2Fpush,1 (d) A force of zero, since the block does not accelerate. Solution: For the block to move at constant speed, by Newton’s 1st law the net force on it must be zero. Doing a force balance in the horizontal direction we find: 0 = Fnet = Fpush,1 − f ⇒ Fpush,1 = f , where f is the frictional force. Thus when the block is moving at speed V , the pushing force Fpush,1 is equal to the frictional force. When the table is moving at 2V , the frictional force is the same; so the pushing force must again be Fpush,1 . This is a subtle problem because students generally choose (b). The key to understanding this problem is to realize that that the force that it takes to move the block at a constant speed is different that the force it takes to accelerate it from rest to a certain speed. Also the time that it would take. These other issues have to do with the concepts of work and power, concepts that we have not yet discussed.
374 Problem 1187. You are standing on a spring scale in an elevator that is moving at a constant speed V . Before the elevator started moving, the scale registered your weight as W . It now registers a weight of Wapp . What is the relationship between Wapp and W ? (a) Wapp < W *(b) Wapp = W (c) Wapp > W (d) Not enough information; it depends on the direction of V~ . Solution: The elevator is moving at constant speed, so it’s not accelerating upward or downward. Your apparent weight is the same as your true weight. Problem 1188. You come into your physics lab and find three books stacked on a table, as shown at right. What is the net force on the middle book? (a) 10 N downward (b) 20 N upward (c) 15 N downward *(d) 0 N
10 N 15 N 20 N
Solution: The book is motionless, so it’s not accelerating in any direction. Hence the net force on it is zero. Problem 1189. You are piloting your spaceship through interstellar space where there is no gravity and no air, when your timing belt breaks and your engine abruptly quits. This occurs at time T , when you are travelling at speed V . Which of the graphs below describes your motion starting at time T ? v
v
(a) 6
a
*(b) 6
V
a
(c) 6
(d) 6
V
T
-
t
T
-
t
T
-
t
T
-
t
Solution: There is no gravity, no friction, and no air resistance in this situation; so we assume that once the engine has stopped, there are no forces acting on the spaceship. Thus it keeps going at the same velocity forever. Graph (b) shows that. Graph (a) shows a changing speed; and graphs (c) and (d) show nonzero acceleration. Problem 1190. A piano with a weight of 4500 N is resting on a frictionless horizontal ice-covered parking lot. You push on the piano with a constant horizontal force of F . Which of the following statements is true? *(a) The piano will accelerate, no matter how small F is. (b) The piano will not accelerate unless F > 4500 N. (c) The piano will move with constant velocity, because F is constant. (d) The piano will move as long as you apply F ; once the force is removed, it will stop. Solution: Since the lot is level and there is no friction, the only horizontal force on the piano is your pushing force F . Hence as long as F > 0, there is a net force on the piano, so it will accelerate. Answer (d) is wrong because the piano, once moving, will continue to move at a constant velocity once there is no horizontal force on it.
375
Problem 1191. You come into your physics lab and find three weights suspended by thin strings from the ceiling, as shown at right. Which of the following statements is true about the tensions T1 , T2 , and T3 in the strings? (a) T1 = T2 = T3 *(b) T1 > T2 > T3 (c) T3 > T2 > T1 (d) T3 > T1 > T2
T1 4 kg T2 1 kg T3 6 kg
Solution: The tension in each string is equal to the weight of the masses below it. Hence T1 = 11 kg · g; T2 = 7 kg · g; and T3 = 6 kg · g. Problem 1192. Your physics instructor asks you to help him push an atomic bomb into the classroom. The bomb weighs 8000 N. You push on it with a horizontal force of 400 N. Which of the following statements is true? (a) If the atomic bomb moves across the floor, then you feel it pushing back on you with a force of less than 400 N. *(b) You feel the atomic bomb pushing back on you with a force of 400 N, whether it moves or not. (c) If the atomic bomb does not move, you feel it pushing back on you with a force of 8000 N. (d) You feel the atomic bomb pushing back on you with a force of 8000 N, whether it moves or not. Solution: This is Newton’s third law. If you’re pushing on the bomb with a force of 400 N, then the bomb is pushing on you with an equal force. Problem 1193. You are pushing two crates across a level frictionless floor, as shown at right. You apply a horizontal force of 300 N to crate A, which is heavier than crate B. Which of the following statements is true?
300 N -
A
B
(a)
Crate A pushes on crate B with a force of 300 N; crate B pushes on crate A with a force of 300 N. (b) Crate A pushes on crate B with more force than crate B pushes on crate A. *(c) Crate A and crate B push on one another with forces that are equal to one another and less than 300 N. (d) The two crates will not move if their weight is greater than 300 N. Solution: We can rule out answer (b) by Newton’s third law; and we can rule out (d), since the floor is level and frictionless: any horizontal force will move the crates. The two crates together experience a force of 300 N; so they accelerate at a = (300 N)/(MA + MB ). The force on crate B is the mass of crate B times this acceleration: MB FB = MB · a = (300 N) < 300N MA + MB
376 Problem 1194. A spaceship is very far away from stars, planets, or other sources of gravitational force. It is moving at high speed through the intergalactic void when its rockets stop firing. Which of the following describes the spaceship’s future course? (a) (b)
It will immediately stop, throwing the cargo and crew violently forward. It will begin slowing down, eventually coming to a complete stop in the vacuum of space. (c) It will move at constant speed for a while, but will then start slowing down. *(d) It will keep on going forever at constant speed. (e) None of these Problem 1195. You are idly playing with a ball on a level floor. You give the ball a light push, and it rolls about halfway across the floor before slowing down and coming to a stop. Why does the ball slow and stop? (a) Because you weren’t pushing it. (b) Because speed is proportional to force. *(c) Because there must have been some force on it opposing the direction of its motion. (d) Because the net force on the ball was zero. (e) None of these Problem 1196. Two objects with masses M and m, where M > m, are on a level frictionless surface. If a force F is applied to the smaller object, it will experience an acceleration of a. If that same force F is applied to the larger object, what will happen to that object? (a) It will experience an acceleration greater than a. (b) It will experience an acceleration of a. *(c) It will experience an acceleration less than a. (d) It will move only if the force F is greater than some minimum value. (e) None of these Problem 1197. An object is moving northward with increasing speed. What do we know about the forces on the object? (a) There is a single force on the object, directed northward. *(b) There is a net force on the object, directed northward. (c) There may be several forces on the object, but the largest is directed northward. (d) We know nothing about the forces on the object. (e) None of these
377 Problem 1198. An object is moving northward. What do we know about the forces on the object? (a) There is a single force on the object, directed northward. (b) There is a net force on the object, directed northward. (c) There may be several forces on the object, but the largest is directed northward. *(d) We know nothing about the forces on the object. (e) None of these Problem 1199. Which of the following objects is not experiencing a net force directed northward? (a) An object moving southward with decreasing speed. (b) An object moving northward with increasing speed. (c) An object that is instantaneously at rest, and that then begins moving northward. *(d) An object moving northward with constant speed. (e) None of these Problem 1200. A rock is at rest on a level surface. The force that the surface exerts on the rock has magnitude FSR ; the force that the rock exerts on the surface has magnitude FRS . What do we know about the two forces? (a) (d) (e)
FSR < FRS *(b) FSR = FRS (c) FSR > FRS We don’t know anything about the relative sizes of the forces. None of these
Problem 1201. An object with a mass of 20 kg is suspended from two ropes, as shown at right. The magnitude of the force exerted by each rope on the object is 139 N; the magnitude of the force of gravity on the object is 196 N. What is the magnitude of the net force on the object?
\ \ \ \ \ \ \ \
(a) 47.4 N (b) 33.5 N 20 kg (c) 13.9 N *(d) 0 N (e) None of these Solution: Since the system is static, the sum of the forces in each component direction is zero. Thus the net force is zero.
378 Problem 1202. An astronaut with a mass of 100 kg is trying to move a very small asteroid with a mass of 400 kg. The astronaut pushes on the asteroid with a force of 120 N. What force does the asteroid exert on the astronaut? Round your answer to the nearest newton. (a) 30 N (b) 96 N *(c) 120 N (d) 480 N (e) None of these Solution: This doesn’t require any computation: only Newton’s third law. The asteroid must exert the same magnitude of force, but in the opposite direction, as the astronaut by Newton’s 3rd Law. Since the astronaut exerts a force of 120 N on the asteroid, the asteroid exerts a force of 120 N on the astronaut. The masses of the astronaut and the asteroid have nothing to do with the problem. Problem 1203. You are pushing a crate with mass m across a floor. You are pushing forward with a force whose magnitude is Fpush ; the force of friction is directed backward, with magnitude f . If the crate is moving at a constant velocity, which of the following is true? *(a) Fpush = f (b) Fpush > f (c) Fpush = mg + f (d) Fpush > mg + f (e) None of these Solution: Since the crate is moving at a constant velocity, the net horizontal force on it must be zero. Hence Fpush = f .
Problem 1204. An object with a mass of 12 kg is suspended from the ceiling by two ropes of equal length, each making an angle of 74◦ to the horizontal. Each rope has a tension of 61 N. What is the magnitude of the net force on the object? Round your answer to the nearest newton. *(a) 0 N (b) 59 N (c) 118 N (d) 122 N (e) None of these
B B B
B
74◦ B
◦ 74
B
12 kg
Solution: You don’t have to do any calculations on this. Since the object isn’t moving, it has a constant velocity of zero; so the net force on it is zero.
379
10.2
Free-body diagrams
Problem 1205. A book is given a push along a frictionless horizontal tabletop. After it has been pushed, it slides along the tabletop and off the edge. (a) What are the forces acting on the book while it is being pushed along the table? What are the reaction forces to those forces? Solution: Sketch a free-body diagram showing action-reaction pairs. While the book is being pushed, it experiences a horizontal force from the hand doing the pushing; it pushes back on the hand with equal force. It experiences two vertical forces adding up to zero: the downward force of gravity, and the normal force of the table pushing upward. The reaction forces are the gravitational force pulling the Earth upward toward the book, and the force of the book pushing downward on the table. (b) What are the forces acting on the book while it is sliding along the table after being pushed? What are the reaction forces to those forces? Solution: As in part (a), only with no horizontal forces. (c) What are the forces acting on the book while it is falling from the table to the floor? What are the reaction forces to those forces? Solution: The only force acting on the book is the Earth’s gravity pulling it downward. The reaction force is the gravitational force pulling the Earth upward toward the book. Problem 1206. Two crates are connected by a thin horizontal rope and sitting on a horizontal floor. You are pulling them by a second horizontal rope attached to one of the crates. There is friction between the crates and the floor. Draw a free-body diagram for each of the crates. Solution: We isolate each crate as a separate system and draw a free-body diagram for that system. In this course, we will usually assume that the rope is massless and doesn’t stretch; so T1 = T2 .
m2
m1
N1 6
N2 6 T + f1
P
u -1-
m1 g ?
T2
u
f-2
m2 g ?
380
10.3
Pushing and pulling objects on horizontal frictionless surfaces
Problem 1207. Two blocks are connected by a thin rope and sitting on a horizontal frictionless surface. You are pulling them by a second rope attached to the smaller block. You pull on your rope with a horizontal force of P ; the tension in the rope connecting the blocks is T . Which of the following is true of P and T ? (a) P < T *(c) P > T
P
37 kg
T
89 kg
(b) P = T (d) Not enough information; it could be any of these
Solution: P is giving a certain acceleration to the combined mass of both crates. T is giving the same acceleration to the 89-kg crate alone. Thus T is smaller than P ; in fact, 89 kg ·P T = 37 kg + 89 kg Problem 1208. You are pushing two blocks across a horizontal frictionless surface, as shown at right. You apply a horizontal force of 240 N to the smaller block. How much force does the smaller block exert on the larger one? (a) 60 N *(c) 180 N
240 N-
20 kg
60 kg
(b) 120 N (d) 240 N
Solution: You are applying a net horizontal force of 240 N to a total mass of 20 + 60 = 80 kg. By Newton’s second law, the two crates together will accelerate at (240 N)/(80 kg) = 3 m/s2 . Again by Newton’s second law, the net horizontal force on the larger crate must be F = ma = (60 kg)(3 m/s2 ) = 180 N.
381
11
Applications of Newton’s Laws to Linear Motion
11.1
Newton’s laws in one-dimension
11.1.1
One-dimensional kinematics and Newton’s second law
The idea behind these problems is add one more equation to our three fundamental kinematic equations:
{Fnet
n Fund. Eq 1: v = v0 + at = ma} + Fund. Eqn 2: v 2 = v02 + 2a(x − x0 ) Fund. Eqn 3: x = x0 + v0 t + 21 at2
This builds our system of possible fundamental equations to 4 equations: Fund. Eqn 1: v = v0 + at Fund. Eqn 2: v 2 = v 2 + 2a(x − x ) 0 0 1 n 2 Fund. Eq 3: x = x0 + v0 t + 2 at Fund. Eqn 4: Fnet = ma with the catch that when mass or force is involved in a problem then fundamental equation 4 is always used together with one of the other three equations. That is, we are always solving a system of equations: F = ma plus one of the fundamental kinematic equations.
382 Problem 1209. A block with a mass of 132 kg is pulled with a horizontal force of F~ across a rough floor. The coefficient of friction between the floor and the block is 0.51. If the block is moving at a constant velocity, what is the magnitude of F~ ? Round your answer to the nearest newton. (a) 67 N (b) 259 N *(c) 660 N (d) 2536 N (e) None of these Solution: Since the block’s velocity is constant, the net force on it is zero. The two forces acting in the horizontal direction are the pull, directed forward with magnitude F ; and the frictional force fk , directed backward with magnitude µmg. Hence: F = fk = µmg = (0.51)(132 kg)(9.8 m/s2 ) = 660 N
Problem 1210. A 5.5-kg block is initially at rest on a frictionless horizontal surface. It is pulled with with a constant horizontal force of 3.8 N. How long must it be pulled before its speed is 5.2 m/s? (a) 2.5 s (b) 5.0 s *(c) 7.5 s (d) 10.0 s (e) None of these Solution: Start by using Newton’s 2nd law for 1-d motion to determine a. Once a is known, this becomes a kinematics problem and we can apply one of our 3 fundamental equations. 3.8 kg m/s2 F ÷m = = 0.69 m/s2 . F = ma −−→ a = m 5.5 kg We now know the acceleration a; so write down what information you are given, what you want, and which fundamental equation you need to use. vi = 0 m/s v = 5.2 m/s f Want: t Which Equation: Fundamental equation (1) Given: 2 a = .69 m/s ti = 0 Solving for t in equation 1: t =
vf − vi 5.2 m/s = = 7.5 s. a .69 m/s2
Problem 1211. A 523-kg experimental rocket sled can be accelerated from rest to 1620 km/h in 1.82 s. What is the net force required to produce this motion? Assume the applied force is constant. *(a) 1.29 × 105 N (b) 6.34 × 103 N (c) 2.74 × 106 N (d) 1.29 × 104 N (e) None of these
383 Solution: Start by down what we’re given and what we want: m = 5 kg vi = 0 m/s 1000 m 1h Given: vf = 1620 km Want: F = ? = 450m/s h 3600 s 1 km ti = 0 t = 1.82 s f We can find the net force using Newton’s 2nd law for 1-d motion F = ma, but we’ll need to use 1-d kinematics to determine a. To find a we can apply one of our 3 fundamental equations. Since we know both velocities and time, but there is no x in the problem we should apply equation (1). Put another way, since the acceleration is constant,
×m
−−−−−→
a = aave ∆v = ∆t vf − vi = tf − ti 450 m/s = = 247 m/s2 1.82 s F = ma = (523 kg)(247 m/s2 ) = 1.29 × 105 N .
Problem 1212. A box has a mass of 160 kg. It is sitting on an icy sidewalk, which constitutes a horizontal frictionless surface. A horizontal force of 32 N is applied to the box. What is the box’s acceleration? (a) 5 m/s2 *(b) 0.2 m/s2 2 (c) 6.4 m/s (d) 0.16 m/s2 (e) None of these Solution: Let +x be the direction of the force and acceleration. The net force on the box is 32 N. Since F = ma, we can solve for acceleration: a=
Fnet 32 N = = 0.2 m/s2 m 160 kg
384 Problem 1213. A box has a mass of 60 kg. It is sitting on an icy sidewalk, which constitutes a horizontal frictionless surface. A horizontal force of 30 N is applied to the box. What is the box’s acceleration? (a) 15 m/s2 (b) 5 m/s2 2 (c) 2 m/s *(d) 0.5 m/s2 (e) None of these Solution:
The only horizontal force on the box is the push of 30 N. Since F = ma, a=
30 N F = = 0.5 m/s2 m 60 kg
Problem 1214. You are trying to move a 150-lb crate across a floor by pulling on a rope inclined at 17◦ to the horizontal. The coefficient of static friction is 0.52; the coefficient of kinetic friction is 0.35. You pull on the rope until the crate starts moving. What is the crate’s initial acceleration? Solution:
See the solution for Problem 25 in Chapter 5 Exercises on my website.
Problem 1215. A 10-kg block is initially at rest on a frictionless horizontal surface. It is pulled with with a constant horizontal force of 10 N until it reaches a final velocity of 10 m/s. How far does it move in this time? (a) 5 m (b) 10 m (c) 100 m *(d) 50 m (e) None of these
Solution: Start by using Newton’s 2nd law for 1-d motion to determine a. Once a is known, this becomes a kinematics problem and we can apply one of our 3 fundamental equations. 10 kg m/s2 F ÷m F = ma −−→ a = = = 1.0 m/s2 . m 10 kg We now know the acceleration a; so write down what information you are given, what you want, and which fundamental equation you need to use. Notice that time does not appear explicitly in the problem, and we are not asked to find the time. vi = 0 m/s v = 10 m/s f Want: ∆x Which Equation: Fundamental equation (2) Given: 2 a = 1.0 m/s ti = 0 Solving for ∆x in equation 2: ∆x =
vf2 − vi2 102 (m/s)2 = = 50 m. 2a 2(1.0) m/s2
385 Problem 1216. A 1.0-kg block is initially at rest on a frictionless horizontal surface. It is pulled with with a constant horizontal force of 1.0 N. How long must it be pulled before its speed is 1.0 m/s? (a) 0.25 s (b) 0.5 s (c) .75 s *(d) 1.0 s (e) None of these Solution: Start by using Newton’s 2nd law for 1-d motion to determine a. Once a is known, this becomes a kinematics problem and we can apply one of our 3 fundamental equations. 1.0 kg m/s2 F ÷m = = 1.0 m/s2 . F = ma −−−−−→ a = m 1.0 kg We now know the acceleration a; so write down what information you are given, what you want, and which fundamental equation you need to use. vi = 0 m/s v = 1.0 m/s f Want: t Which Equation: Fundamental equation (1) Given: 2 a = 1.0 m/s ti = 0 Solving for t in equation 1: t =
vf − vi 1.0 m/s = = 1.0 s. a 1.0 m/s2
386 Problem 1217. A crate of auto parts is sitting on a horizontal frictionless frozen lake. A worker applies a constant horizontal force of 750 N to the crate. Starting from rest, the crate moves 20 m in the first 4 s. What is the mass of the crate? (a) 250 kg *(b) 300 kg (c) 350 kg (d) 400 kg (e) None of these Solution: If you sketch a free-body diagram, you’ll see that the only horizontal force is the constant 750 N. That is therefore the net horizontal force. There’s no net vertical force. Given: Fnet = 750 N; vi = 0 m/s; vf = ? ; ∆x = 20 m; and ∆t = 4 s. Want: m = ? If we try to solve for m directly using Newton’s second law, we get stuck: Fnet = ma
÷a
−−−−−→
m=
Fnet 750 N = a a
(11.1)
We need to use one of our three fundamental kinetic equations to find a. Since we know ∆x, vi = 0, and t, we can use 1 2 at2 ∆x = vi t + at = 2 2
· 2/t2
−−−−→
a=
2∆x t2
(11.2)
Substituting equation (11.5) into equation (11.4) gives us m=
Fnet t2 (750 N)(4 s)2 = = 300 kg 2∆x 2(20 m)
Problem 1218. A ball with a mass of 0.4 kg is at rest on top of a tee when it is hit with a bat moving horizontally. The bat is in contact with the ball for 20 ms; when the ball leaves the bat, it is moving horizontally at 50 m/s. If the bat applies a constant force Fnet to the ball during the whole time that they are in contact, what is the magnitude of Fnet ? (a) 400 N *(b) 1000 N (c) 2500 N (d) 4000 N (e) None of these Solution: We know the ball’s mass m, its initial speed v0 and final speed vf , and the time over which it accelerates ∆t. We want to know the force Fnet . We can use Newton’s second law: Fnet = ma. We don’t know a, but we can find it from the equation vf = vi + a∆t, in which we know everything but a: vf − vi ∆t m(vf − vi ) (0.4 kg)(50 m/s − 0 m/s) = ma = = = 1000 N ∆t 0.02 s vf = vi + a∆t
Fnet
⇒
a=
387 Problem 1219. A 10.0-kg block is initially at rest on a frictionless horizontal surface. It is pulled with with a constant horizontal force of 10.0 N. (a) How long must it be pulled before its speed is 100 m/s? Round your answer to the nearest second (b) How far does it move in this time? Round your answer to the nearest meter. Solution: Since F = ma, the acceleration of the block is a=
10.0 N F = = 1 m/s2 m 10.0 kg
(a) Since the block starts at v0 = 0, speed as a function of time is v = at; so t=
v 100 m/s = 100 s = a 1 m/s2
(b) Since v0 = 0 and x0 = 0, position as a function of time is 1 2 (1 m/s2 )(100 s)2 x = at = = 5000 m 2 2 Problem 1220. A 1-kg ketchup bottle is slid across a smooth counter at a local diner. You notice that the bottle slid 1.5 meters and took 2 seconds to come to rest. Assuming a constant net force owing to the friction on the table, what was the initial velocity of the bottle? Round your answer to the nearest 0.1 m/s. (a) 5 m/s (b) 0.5 m/s *(c) 1.5 m/s (d) 0.15 m/s (e) None of these Solution: If the force on the bottle was constant, then so was the acceleration a. If we let x0 = 0, then position as a function of time is: x = v0 t + 21 at2 . Since the acceleration is constant and slows the bottle from v0 to v = 0 in time t = 2 s, we can write: a = −v0 /t. Substituting that expression for a into the equation of position, we get x = v0 t −
1 v0 2 v0 t t = 2 t 2
Solving for v0 yields 2x 2(1.5 m) = = 1.5 m/s t 2s [Notice that we didn’t need the mass of the bottle.] v0 =
388 Problem 1221. A block of mass M is pulled along a horizontal frictionless surface by a bar of mass m that is attached to the block and horizontal to the surface . A horizontal force P~ (P for pull) is applied to the free end of the bar (see figure below). (a) Find the acceleration of the bar-block system (i.e., the bar and the block taken together). (b) What is the force that the bar exerts on the block? Solution: (a) The bar-block system has mass M + m. The acceleration of the system is the force applied to it divided by its mass: a=
P M +m
(b) The force on the block is its mass times its acceleration. Its mass is M and its acceleration is that of the whole bar-block system, so F = Ma =
11.1.2
MP M +m
Weight
Remember weight is defined as w = mg, where m is mass and g is the magnitude of the acceleration due to gravity. g is always positive. Problem 1222. The acceleration due to gravity at the surface of Neptune’s moon Triton is 2.41 m/s2 . If a rock weighs 106 N on Triton, what is its mass? Round your answer to the nearest kilogram. (a) 26 kg (c) 255 kg (e) None of these Solution:
w = mgT
*(b) 44 kg (d) 431 kg ÷g
T −−−→
m=
106 N w = = 44 kg gT 2.41 m/s2
Problem 1223. Triton’s gravity is 2.41 m/s2 . If a rock weighs 482 N on Triton, what is its mass? Round your answer to the nearest kilogram. (a) 206 kg (c) 225 kg (e) None of these Solution:
w = mgT
*(b) (d) ÷g
T −−−−− −→
200 kg 22 kg m=
w 482 N = = 200 kg. gT 2.41 m/s2
389 Problem 1224. The acceleration due to gravity at the surface of Neptune is 11.15 m/s2 . If a rock weighs 266 N on Neptune, what is its mass? Round your answer to the nearest 0.1 kg. (a) 27.1 kg (c) 0.4 kg (e) None of these Solution:
(b) *(d)
2965.9 kg 23.9 kg
On Neptune, WN = mgN
⇒
m=
266 N WN = 23.9 kg = gN 11.15 m/s2
Problem 1225. The acceleration due to gravity at the surface of Planet X is 12.2 m/s2 . If a rock weighs 307 N on Planet X, what is its mass? Round your answer to the nearest kilogram. *(a) (c) (e)
25 kg 382 kg None of these
Solution:
(b) 247 kg (d) 3745 kg
On Planet X, WX = mgX
⇒
m=
307 N WX = 25 kg = gX 12.2 m/s2
Problem 1226. An astronaut weighs 790 N on Earth. On Planet X, he weighs 640 N. What is astronaut’s mass on Planet X? Round your answer to the nearest kilogram. *(a) 81 kg (b) 89 kg (c) 98 kg (d) 107 kg (e) None of these Solution: Remember that the mass of an object is the same wherever it may be. The only way to change mass is by adding material to or taking it away from the object. We know that on Earth, wE = mgE ; and we know both wE and gE . Hence wE = mgE
÷g
E −−→
m=
wE 790 N = = 81 kg gE 9.8 m/s2
390 Problem 1227. An astronaut weighs 790 N on Earth. On Planet X, he weighs 640 N. What is the astronaut’s mass on Earth? Round your answer to the nearest kilogram. (a) 73 kg *(b) 81 kg (c) 89 kg (d) 98 kg (e) None of these Solution: Remember that the mass of an object is the same wherever it may be. The only way to change mass is by adding material to or taking it away from the object. We know that on Earth, wE = mgE ; and we know both wE and gE . Hence wE = mgE
÷g
E −−→
m=
790 N wE = 81 kg = gE 9.8 m/s2
Problem 1228. (Similarity Problem) An astronaut weighs 790 N on Earth. On Planet X, he weighs 640 N. What is the gravitational acceleration on Planet X? Round your answer to the nearest 0.1 m/s2 . (a) 5.8 m/s2 (b) 6.4 m/s2 (c) 7.1 m/s2 *(d) 7.9 m/s2 (e) None of these Solution: Step 1: Write down what you’re given, what you want, and any assumptions: Given: wE = mgE = 790 N; wX = mgX = 640 N In addition we know gE = 9.8 m/s2 , but don’t know m Want: gX = ? Step 2: Solve. We can eliminate the common parameter m by solving both equations for m and setting the expressions equal (m is our common link):
×
gX gE wE
−−−−→
wX wE =m= gE gX wX gE (640 N)(9.8 m/s2 ) gX = = = 7.9 m/s2 wE 790 N
391 Problem 1229. A uniform chain is 2.0 m long and has a mass of 20 kg. It hangs from the ceiling and supports a lamp with a mass of 60 kg. What is the tension in the top link of the chain? For ease of calculation, assume that g = 10 m/s2 . (a) 200 N (c) 700 N (e) None of these
(b) 600 N *(d) 800 N
u u u
Ttop 1 w 2 c
1 w 2 c
wL
6 u
wc ? wL ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the top link. The top link supports the entire weight of the chain wc , plus the weight of the lamp wL . Since the system is in static equilibrium, there is no acceleration, so no net force; so Ttop = wc + wL = (mc + mL )g = (20 kg + 60 kg)(10 m/s2 ) = 800 N Problem 1230. A uniform chain is 2.0 m long and has a mass of 20 kg. It hangs from the ceiling and supports a lamp with a mass of 60 kg. What is the tension in the middle link of the chain? For ease of calculation, assume that g = 10 m/s2 . (a) 100 N *(c) 700 N (e) None of these
(b) 600 N (d) 800 N
u u u
Tmiddle 1 w 2 c
1 w 2 c
wL
6 u 1 w 2 c
?
wL ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the middle link. The middle link supports half of the weight of the chain, 21 wc , plus the weight of the lamp wL . Since the system is in static equilibrium, there is no acceleration, so no net force; so 1 1 Ttop = wc + wL = ( mc + mL )g = (10 kg + 60 kg)(10 m/s2 ) = 700 N 2 2
392 Problem 1231. A uniform chain is 2.0 m long and has a mass of 20 kg. It hangs from the ceiling and supports a lamp with a mass of 60 kg. What is the tension in the bottom link of the chain? For ease of calculation, assume that g = 10 m/s2 . (a) 200 N (c) 700 N (e) None of these
*(b) (d)
u u u
600 N 800 N
Tbottom 1 w 2 c
1 w 2 c
wL
6 u
wL ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the bottom link. The bottom link supports the weight of the lamp wL , but none of the weight of the chain. Since the system is in static equilibrium, there is no acceleration, so no net force; so Tbottom = wL = mL g = (60 kg)(10 m/s2 ) = 600 N Problem 1232. (Force-to-weight ratio) A softball is at rest on top of a tee when it is hit with a bat moving horizontally. The bat is in contact with the ball for 20 ms; when the ball leaves the bat, it is moving horizontally at 50 m/s. If the bat applies a constant force Fnet to the ball during the whole time that they are in contact, what is the force-to-weight ratio Fnet /wball for the ball? For ease of calculation, assume that g = 10 m/s2 . *(a) 250 (b) 500 (c) 1250 (d) 2500 (e) None of these Solution: We know the ball’s initial speed v0 and final speed vf , and the time over which it accelerates ∆t. We want to know the ratio of net force to weight Fnet /w. Since we don’t know the ball’s mass, we can’t know the net force or the weight; but in calculating the ratio, the mass cancels out: ma a Fnet = = wball mg g We know enough to calculate the acceleration: vf − vi t a vf − vi 50 m/s − 0 m/s = = = = 250 g gt (10 m/s2 )(0.02 s) vf = vi + at
Fnet wball
⇒
a=
393 Problem 1233. (Determining the weight of an object without a scale) You have just landed on Planet X and realized that you’ve left your scale behind. However, you have a 5.0 kg ball, and a spring cannon that can launch the ball straight upward at a speed of 15 m/s. You launch the ball upward from ground level; it strikes the ground again 3.0 s after launching. What is the magnitude of gravity on Planet X? Use this to determine the weight of the object. Round your answer to the nearest newton. (a) 40 N (b) 45 N *(c) 50 N (d) 55 N (e) None of these Solution: We know m = 5.0 kg; vi = 15 m/s; and ∆t = 3.0 s. We want wX = mgX , where gX is the gravitational acceleration on Planet X. We are assuming that there is no air resistance, so the ball returns to the ground at the same speed at which it was launched, and it takes as much time going up as it does falling back down: vf = −vi = −15 m/s, and ∆t = 2tymax , where tymax is the time that the ball takes to reach its maximum height after being launched. The velocity as a function of time is v(t) = vi − gX t. At the top of the trajectory (i.e. at t = tymax ), the velocity is zero. Hence v(tymax ) = 0 = vi − gX tymax ⇒ wX = mgX =
⇒
gX =
vi tymax
2(5.0 kg)(15 m/s) 2mvi = = 50 N ∆t 3.0 s
∆t=2tymax
−−−−−−→
gX =
2vi ∆t
394 Problem 1234. You are standing on a 100 m tall cliff on Planet X. You drop a 5 kg stone off the cliff, and notice that it takes 10 seconds for it to hit the ground below. What is the magnitude of gravity gX on planet X? You may assume that you can ignore air drag. (a) 1.0 m/s2 *(b) 2.0 m/s2 (c) 5.0 m/s2 (d) 10.0 m/s2 (e) None of these Solution: This is just a kinematics problem! The weight of the stone is irrelevant. By Newton’s 2nd law for 1-d motion with weight as our force: mgX = w = F = ma
÷m
−−−−−→
gX = a .
We can now apply one of our 3 fundamental equations. Start by writing down what information you are given, what you want, and which fundamental equation you need to use. We’ll take our initial position to be the origin yi = 0, and the final position at the bottom of the cliff as our final position yf = 100 m. Since this is ”drop problem” the natural direction for the y-axis is in the direction of motion, so we’ll take the y-axis pointing downward. yi = 0 m yf = 100 m Want: a = gX Which Equation: Equation (3) Given: vi = 0 m/s ti = 0 s t = 10 s f Solving for a in equation 3 with v0 = 0 and y0 = 0 gives: 1 yf = gX t2fall 2
⇒
gX =
2yf 2 (100) m = = 2.0 m/s2 . 2 2 2 tfall 10 s
395 11.1.3
Apparent weight problems (The Elevator Equation)
The apparent weight of a mass that is accelerating (upward or downward) with the y-axis taken pointing upward is ay wapp = w 1 + g Problem 1235. A block is hanging from a spring scale that is attached to the ceiling of an elevator. Under which of the following circumstances will the reading on the scale be less than the weight of the block? (a) The elevator is moving downward and slowing down. *(b) The elevator is moving upward and slowing down. (c) The elevator is moving downward at constant speed. (d) The elevator is moving upward at constant speed.
Tu 6 u
wb ?
Solution: Sketch a free-body diagram for the block, as at right. There are two forces acting on the block: the weight wb pulling down, and the tension in the scale Tu pulling upward. If Tu < wb , then the net force is downward; so the block is experiencing downward acceleration. This can happen if the elevator is moving upward and slowing down, or if it is moving downward and speeding up. Only the former option is in the answers. Problem 1236. A block is hanging from a spring scale that is attached to the ceiling of an elevator. What will the reading on the scale be if the elevator cable breaks? Solution: The vertical forces on the block are the weight pulling downward, and the tension in the scale pulling up. If the cable breaks and the elevator is in free fall, then the net force on the block is equal to its weight; so the tension in the scale is zero. Hence the scale will register a reading of zero. Problem 1237. An object is hung from a spring scale attached to the ceiling of an elevator. The scale reads 100 N when the elevator is sitting still. What is the reading on the scale if the elevator is moving downward with a constant speed of 10 m/s? Approximate g as 10 m/s2 . (a) 0 N (b) 50 N *(c) 100 N (d) 200 N (e) None of these Solution: Since the elevator is moving at constant speed, the acceleration a is 0 m/s2 . From Newton’s 2nd law it follows that the net force is just the force due to gravity, which was present when the elevator was sitting still. Thus the reading on the scale is 100 N.
396 Problem 1238. What is the apparent weight of a 150 N block if it is accelerating upward at 1 m/s2 ? For ease of calculation, assume that g = 10 m/s2 . (a) 20 N (b) 135 N (c) 150 N *(d) 165 N (e) None of these Solution:
Use the apparent weight formula.
Problem 1239. What is the apparent weight of a 150 N block if it is accelerating downward at 1 m/s2 ? For ease of calculation, assume that g = 10 m/s2 . (a) 20 N (b) 135 N (c) 150 N (d) 165 N (e) None of these Solution:
Use the apparent weight formula.
Problem 1240. What is the apparent weight of a 150 N block if it is moving upward at a constant velocity of 1 m/s? For ease of calculation, assume that g = 10 m/s2 . (a) 20 N (b) 135 N *(c) 150 N (d) 165 N (e) None of these Solution:
Constant velocity ⇒ No acceleration ⇒ wapp = w.
Problem 1241. What is the weight of a 150 N block if it is accelerating downward at 1 m/s2 ? For ease of calculation, assume that g = 10 m/s2 . (a) 20 N (b) 135 N *(c) 150 N (d) 165 N (e) None of these Solution: The weight of an object is the force exerted on it by gravity, and that’s not affected by its upward or downward acceleration.
397 Problem 1242. An elevator weighing 6200 lb is pulled upward by a cable, with an upward acceleration of 3.2 ft/s2 . What is the tension in the cable? Round your answer to the nearest 100 lb. (a) 218,200 lb (b) 19,800 lb *(c) 6800 lb (d) 600 lb (e) None of these Solution: A free-body diagram shows the force of weight pointing straight downward with magnitude W = mg, and the tension T pointing straight upward. The sum of the two is the net force, which produces an upward acceleration of a. Hence Newton’s second law becomes: T − W = T − mg = ma. We solve for tension: T = m(g + a). We know that W = mg, so m = W/g. Hence ! 2 3.2 ft/s a W (g + a) =W 1+ = 6800 lb = (6200 lb) 1 + T = m(g + a) = g g 32 ft/s2
Problem 1243. A bucket weighing 500 N is lifted with an upward acceleration of 2 m/s2 by a uniform chain weighing 200 N. What is the tension in the top link of the chain? For ease of calculation, assume that g = 10 m/s2 . (a) 360 N (b) 720 N *(c) 840 N (d) 1080 N (e) None of these
u u u
Ttop 1 w 2 c
1 w 2 c
wb
6 u
wc ? wb ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the top link. The top link supports the weight of the bucket wb plus the entire weight of the chain wc . Since the system is accelerating upward, it experiences a net upward force. We will use Newton’s second law: Ttop − wb − wc = Fnet = msystem a = (mb + mc )a ⇒ Ttop = wb + wc + (mb + mc )a (wb + wc )a w=mg −−−→ = wb + wc + g g+a 10 m/s2 + 2 m/s2 = (wb + wc ) = (500 N + 200 N) = 840 N g 10 m/s2
398 Problem 1244. A bucket weighing 500 N is lifted with an upward acceleration of 2 m/s2 by a uniform chain weighing 200 N. What is the tension in the middle link of the chain? For ease of calculation, assume that g = 10 m/s2 . (a) 600 N *(b) 720 N (c) 840 N (d) 960 N (e) None of these
u u u
Tmid 1 w 2 c
1 w 2 c
wb
6 u 1 w 2 c
?
wb ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the middle link. The middle link supports the weight of the bucket wb plus half the weight of the chain, 21 wc . Since the system is accelerating upward, it experiences a net upward force. We will use Newton’s second law: wc mc Tmid − wb − = Fnet = msystem a = mb + a 2 2 wc mc ⇒ Tmid = wb + + mb + a 2 2 wc (wb + wc /2)a w=mg + −−−→ = wb + 2 g wc g + a 200 N 10 m/s2 + 2 m/s2 = wb + = 500 N + = 720 N 2 g 2 10 m/s2 Problem 1245. A bucket weighing 500 N is lifted with an upward acceleration of 2 m/s2 by a uniform chain weighing 200 N. What is the tension in the bottom link of the chain? For ease of calculation, assume that g = 10 m/s2 . *(a) 600 N (b) 660 N (c) 720 N (d) 840 N (e) None of these
u u u
Tbottom 1 w 2 c
1 w 2 c
wb
6 u
wb ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system consisting of the bottom link. The bottom link supports the weight of the bucket wb , but none of the weight of the chain wc . Since the system is accelerating upward, it experiences a net upward force. We will use Newton’s second law: Tbottom − wb = Fnet = msystem a = mb a ⇒ Tbottom = wb + mb a wb a w=mg −−−→ = wb + g g+a 10 m/s2 + 2 m/s2 = wb = (500 N) = 600 N g 10 m/s2
399 Problem 1246. You are standing on a spring scale on an elevator. When the elevator is not moving, the scale registers a weight of 600 N. When the elevator starts moving, it has an upward acceleration of 2 m/s2 . What weight does the scale register while the elevator is accelerating? For ease of calculation, assume that g = 10 m/s2 . (a) 360 N (b) 500 N *(c) 720 N (d) 1200 Nv (e) None of these
Ns 6 u
w?
Solution: Sketch a free-body diagram, as at right. The forces working on the system (you) are the normal force of the scale Ns , which is the weight registered on the scale; and your weight. Since you are accelerating, there is a nonzero net force. Use Newton’s second law: Fnet = Ns − w = ma ⇒ Ns = w + ma a wa w=mg = 1+ w= −−−→ Ns = w + g g
2 m/s2 1+ 10 m/s2
! (600 N) = 720 N
Problem 1247. You are standing on a spring scale on an elevator. When the elevator is not moving, the scale registers a weight of 730 N. When the elevator starts moving, it has an upward acceleration of 1.6 m/s2 . What weight does the scale register while the elevator is accelerating? Round your answer to the nearest 10 N. (a) 760 N *(b) 850 N (c) 930 N (d) 1030 N (e) None of these Solution: There are two vertical forces acting on you: your weight w = mg = 730 N, pulling down; and the normal force of the scale, Nscale = wapp , your apparent weight, pushing upward. The combination of these two is the net force, which gives you an upward acceleration of a = 1.6 m/s2 . Taking the upward direction as positive, this gives us Nscale − w = Fnet = ma ⇒ Nscale = w + ma We solve w = mg to get: m = w/g. Substituting this gives Nscale
! a 1.6 m/s2 wa =w 1+ = (730 N) 1 + = 850 N = w + ma = w + g g 9.8 m/s2
400 Problem 1248. Someone has left a spring scale on an elevator. While the elevator is not moving, you get on the scale and see that your weight is 810 N. When the elevator starts moving, you see that the scale registers 890 N. What is the acceleration of the elevator? Round your answer to the nearest 0.1 m/s2 ; use a positive sign if the acceleration is upward, a negative sign if the acceleration is downward. *(a) 1.0 m/s2 (c) −1.0 m/s2 (e) None of these
(b) 10.8 m/s2 (d) −10.8 m/s2
Solution: If your mass is m, then gravity is pulling you downward with a force of your weight: −w = −mg = −810 N. When the elevator is accelerating, the scale reading is the normal force of the floor pushing up on you. This upward force is Fscale = 890 N. The net vertical force on you is Fnet = Fscale − mg = 80 N. Since Fnet = ma, your acceleration is a = Fnet /m. We can get m from your weight: m = w/g. Hence Fnet g (80 N)(9.8 m/s2 ) Fnet = = = 1.0 m/s2 a= m w 810 N Problem 1249. Someone has left a spring scale in an express elevator. While the elevator is standing still, you get on the scale and see that you weigh 142 lb. You ride the elevator nonstop to the top floor of the building. In the middle of the ride, you are going upward at at a constant speed of 44 ft/s. What weight does the scale register at this point? Round your answer to the nearest pound. (a) 103 lb (c) 154 lb (e) None of these
*(b) (d)
142 lb 195 lb
Solution: Since the elevator is moving at a constant speed, it is experiencing no acceleration. The weight registered by the scale is the same as it would be if the elevator were standing still: 142 lb.
Problem 1250. A lamp with a mass of 7.0 kg is hanging from a thin cord attached to the ceiling of an elevator. When the elevator is accelerating upward at 3 m/s2 , what is the tension in the cord? For ease of calculation, assume that g = 10 m/s2 . (a) 21 N (b) 49 N (c) 70 N *(d) 91 N (e) None of these
T6 u
w = mg ?
Solution: Sketch a free-body diagram for the lamp, as at right. The forces working on the system are the tension of the cord T , and the lamp’s weight w = mg. Since the system is accelerating, there is a nonzero net force. Use Newton’s second law: Fnet = T − mg = ma ⇒
T = mg + ma = m(g + a) = (7.0 kg)(10 m/s2 + 3 m/s2 ) = 91 N
401 Problem 1251. A block with weight w is hanging from a spring scale that is attached to the ceiling of an elevator. If the elevator has an upward acceleration of a > 0 and the reading on the scale is T , what is w as a function of a and T ? T (g − a) Tg (b) w = (a) w = g−a g T (g + a) Tg (d) w = *(c) w = g+a g (e)
T6
u
w?
None of these
Solution: Sketch a free-body diagram of the block, as at right. There are two vertical forces acting on the block: the weight w pulling down, and the tension in the scale T pulling up. The block is experiencing an upward acceleration of a. Using Newton’s second law and the fact that w = mg: wa T − w = ma = g a g+a ⇒ T =w 1+ =w g g Tg ⇒ w= g+a Problem 1252. A block with weight w is hanging from a spring scale that is attached to the ceiling of an elevator. If the elevator has an upward acceleration of a > 0, what is the reading on the scale T ? g a (b) T = w (a) T = w g a g a (c) T = w 1 + (d) T = w 1 + a g (e)
T6
u
w?
None of these
Solution: Sketch a free-body diagram of the block, as at right. There are two vertical forces acting on the block: the weight w pulling down, and the tension in the scale T pulling up. The block is experiencing an upward acceleration of a. Using Newton’s second law and the fact that w = mg: wa T − w = ma = g a ⇒ T =w 1+ g Quick way: Notice the the tension in the string is the force that the scale reads. But that is just the apparent weight. Now apply the apparent weight formula.
402 11.1.4
Spring Problems
Problem 1253. (Derived quantities) Determine the units of the spring constant k from the equation for the ideal spring force: Fspring = −kx, where F is force, k is the spring constant, and x is the displacement from the equilibrium position. Write the answer in terms of Newtons. Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expression for the spring force gives [F ] = [−kx] = [k][x]
÷[x]
−−−−−−→
[k] =
[F ] N = [x] m
There is no special name for a Newton divided by a meter. Problem 1254. A rubber band has an unstretched length of 8 cm and a force constant of 4 N/cm. What is the force required to stretch it to a length of 12 cm? (a) 3 N *(b) 16 N (c) 32 N (d) 48 N (e) None of these Solution: The rubber band is being stretched by ∆x = 12 cm − 8 cm = 4 cm. The force required to stretch it is F = k∆x = (4 N/cm)(4 cm) = 16 N. Problem 1255. A spring is hanging from the ceiling. Its unstretched length is 12 cm. When a block with a mass of 6 kg is hanging from the end, it stretches to a length of 15 cm. What is the spring constant of the spring? For ease of calculation, assume that g = 10 m/s2 . (a) 4 N/m (b) 20 N/m (c) 400 N/m *(d) 2000 N/m (e) None of these Solution: Don’t forget to convert centimeters to meters. When a mass of m is hanging from the spring, the force on the spring is w = mg. By Hooke’s law, F = k∆y; so mg (6 kg)(10 m/s2 ) F = k= = = 2000 N/m ∆y yf − y0 0.15 m − 0.12 m
403 Problem 1256. A spring is hanging from the ceiling. Its unstretched length is 10 cm. When a block with a mass of 8 kg is attached to the end, it stretches to a length of 20 cm. If you want to stretch the spring to a length of 30 cm, how much mass must you hang from it? (a) 8 kg (b) 12 kg *(c) 16 kg (d) 24 kg (e) None of these Solution: When a mass of m is hanging from the spring, the force on the spring is w = mg. By Hooke’s law, F = mg = k∆y. When we change the mass from m1 to m2 , we change from ∆y1 to ∆y2 ; k and g don’t change. Hence
⇒
m m1 m2 k = = = g ∆y ∆y1 ∆y2 m1 ∆y2 (8 kg)(0.30 m − 0.10 m) m2 = = = 16 kg ∆y1 0.20 m − 0.10 m
Problem 1257. A horizontal spring with a spring constant of 500 N/m is used to pull a sled across a frictionless frozen lake. The sled has a mass of 40 kg, and is undergoing an acceleration of 0.5 m/s2 . How much does the spring stretch while pulling the sled? For ease of calculation, use g = 10 m/s2 . (a) 2.5 cm *(b) 4 cm (c) 10 cm (d) 25 cm (e) None of these Solution: Since the ice is horizontal and frictionless, we need only concern ourselves with the horizontal forces. There is only one: the tension in the spring, pulling on the sled. Using Newton’s second law gives us the value of the force: F = ma. Using Hooke’s law gives us the amount by which the spring stretches. F = ma = k∆x
⇒
∆x =
(40 kg)(0.5 m/s2 ) ma = = 0.04 m = 4 cm k 500 N/m
404 Problem 1258. An unstretched spring has a length of 22.8 cm. When an object with a mass of 12.0 kg is hung from it, it stretches to a length of 35.5 cm. What is the spring constant of the spring? Round your answer to the nearest N/m. (a) 34 N/m (c) 331 N/m (e) None of these
(b) *(d)
94 N/m 926 N/m
Solution: The force applied to the spring is the weight of the object w = mg. Since there is no acceleration in the mass-spring system, applying Newton’s 2nd law in the vertical direction gives |Fspring | = w = mg. The spring stretches by x = 35.5 cm − 22.8 cm = 0.127 m. (Since the problem asks for the spring constant in N/m, you must convert centimeters to meters.) Hooke’s law gives us |Fspring | = kx = mg
⇒
(12.0 kg)(9.8 m/s2 ) mg = = 926 N/m k= x 0.127 m
Problem 1259. An unstretched spring is 62 cm long. When you hang an object with a mass of 4.6 kg from it, it stretches to a length of 71 cm. What is the mass of the object you would have to hang from the spring to get it to stretch to a length of 90 cm? Round your answer to the nearest 0.1 kg. (a) 1.8 kg *(c) 14.3 kg (e) None of these
(b) 5.8 kg (d) 450.8 kg
Solution: Let m1 = 4.6 kg be the first mass, and let x1 = 71 cm − 62 cm = 9 cm be the distance by which it stretches the spring. We can calculate the force constant of the spring: m1 g m1 g = kx1 ⇒ k = x1 We want to know what mass m2 it takes to stretch the spring by a distance of x2 = 90 cm − 62 cm = 28 cm. m2 g = kx2
⇒
m2 =
kx2 m 1 x2 (4.6 kg)(28 cm) = = = 14.3 kg g x1 9 cm
405 Problem 1260. A bungee cord obeys Hooke’s law, with a spring constant of 850 N/m. It is attached to a crate with a mass of 170 kg and pulled at an angle of 19◦ to the horizontal, in order to move the crate across a horizontal frictionless floor. If the crate accelerates at 0.81 m/s2 , how much does the bungee cord stretch? Round your answer to the nearest centimeter. (a) 16 cm *(b) 17 cm (c) 159 cm (d) 168 cm (e) None of these Solution: Let T be the tension in the bungee cord. Since it makes an angle of 19◦ to the horizontal, the horizontal component of the force is Fh = T cos 19◦ . This is the only horizontal force acting on the crate, and it is accelerating the crate at a = 0.81 m/s2 . By Newton’s second law, T cos 19◦ = ma. By Hooke’s law, T = kx, where x is the amount by which the bungee cord stretches. Hence kx cos 19◦ = ma
⇒
x=
(170 kg)(0.81 m/s2 ) ma = = 0.17 m = 17 cm k cos 19◦ (850 N/m) cos 19◦
Problem 1261. A bungee cord obeys Hooke’s law, with a spring constant of 620 N/m. It is attached to a crate with a mass of 220 kg and pulled at an angle of 24◦ to the horizontal, in order to move the crate across a horizontal frictionless floor. If the crate accelerates at 0.72 m/s2 , how much does the bungee cord stretch? Round your answer to the nearest centimeter. (a) 13 cm (b) 14 cm (c) 26 cm *(d) 28 cm (e) None of these Solution: Let T be the tension in the bungee cord. Since it makes an angle of 19◦ to the horizontal, the horizontal component of the force is Fh = T cos 24◦ . This is the only horizontal force acting on the crate, and it is accelerating the crate at a = 0.72 m/s2 . By Newton’s second law, T cos 24◦ = ma. By Hooke’s law, T = kx, where x is the amount by which the bungee cord stretches. Hence ◦
kx cos 24 = ma
⇒
(220 kg)(0.72 m/s2 ) ma = = 0.28 m = 28 cm x= k cos 24◦ (620 N/m) cos 24◦
406 Problem 1262. (Lab Problem) A spring is hanging from the ceiling in the physics lab. When there is no weight hanging from it, the spring is 42 cm long; it has a spring constant of 8300 N/m. You hang a lead block with a mass of 87 kg from the spring, stretching it. What is the length of the spring with the weight hanging from it? Round your answer to the nearest centimeter. (a) 10 cm (b) 21 cm (c) 32 cm *(d) 52 cm (e) None of these Solution: We know the unstretched length of the spring Lu = 42 cm = 0.42 m; the spring constant k = 8300 N/m; and the mass m = 87 kg. If y is the change in length of the spring, we want the stretched length: Ls = Lu + y. Use the equation |Fspring | = ky to find y. We assume that the system is static, so there is no acceleration; so the two forces on the weight match: mg = |Fspring | = ky. Divide by k: y=
mg k
⇒
Ls = Lu +
mg (87 kg)(9.8 m/s2 ) = 0.42 m + = 0.52 m = 52 cm k 8300 N/m
Problem 1263. (Lab Problem) An unstretched spring is 62 cm long. When you hang an object with a mass of 4.6 kg from it, it stretches to a length of 71 cm. What is the mass of the object you would have to hang from the spring to get it to stretch to a length of 90 cm? Round your answer to the nearest 0.1 kg. (a) 1.8 kg *(c) 14.3 kg (e) None of these
(b) 5.8 kg (d) 450.8 kg
Solution: Let m1 = 4.6 kg be the first mass, and let x1 = 71 cm − 62 cm = 9 cm be the distance by which it stretches the spring. We can calculate the force constant of the spring: m1 g m1 g = kx1 ⇒ k = x1 We want to know what mass m2 it takes to stretch the spring by a distance of x2 = 90 cm − 62 cm = 28 cm. m2 g = kx2
⇒
m2 =
kx2 m 1 x2 (4.6 kg)(28 cm) = = = 14.3 kg g x1 9 cm
Problem 1264. (Derive Problem) Two air-track gliders of mass m1 and m2 with m1 < m2 are connected by a spring whose mass is negligible compared to the mass of the gliders. The gliders sit on a horizontal frictionless air-track and are constrained to one-dimensional motion. Treat the problem as a one-dimensional problem by assuming that the force of gravity is exactly cancelled by the upward blowing air of the air track. The system: Take the pair of gliders together with the spring to be our system under investigation. There are no external forces acting on the system in the direction of the track.
407 The initial condition: The system is initially at rest with the spring compressed and held in place by a thin massless string connecting the two gliders. At time t = 0, the string is cut setting the mass-spring system in motion. The final condition: (a) If at time tf > 0 the acceleration of mass m2 is a2 , then what is the force on m2 by the spring, denoted Fm2 s ? (b) What is the acceleration of mass m1 at time tf ? Comment: After the string is cut, the energy stored in the spring sets the system in motion. The motion comes from internal forces, there are no external forces acting on the mass-spring system.
Solution:
See solutions to chapter 3 homework on my website (Exercises #11 ).
408
11.2
Newton’s laws in two-dimensions
11.2.1
Static equilibrium problems
Problem 1265. Two nails are driven into a wall at r different heights, and a weight is hung from two thin θ strings fastened to the nails, as shown at right. Both r ZZθ strings make the same angle θ to the horizontal, but Z T1 T 2 Z the first string is longer than the second. If T1 is the tension in the longer string and T2 is the tension in M the shorter, what is the relationship between T1 and T2 ? (a) T1 < T2 *(b) T1 = T2 (c) T1 > T2 (d) Not enough information: could be any of these Solution: Since the weight isn’t moving, the net force on it must be zero. In particular, the net horizontal force must be zero. There are two horizontal forces acting on it: the horizontal component of T2 pulling to the left, and the horizontal component of T1 pulling to the right. These horizontal forces must be equal to one another: T1x = T2x
⇒
T1 cos θ = T2 cos θ
⇒
T1 = T2
The difference in the lengths of the strings doesn’t matter; what’s important is that their angles are the same.
409 Problem 1266. A rope is stretched between two poles. The rope is initially horizontal. After an 82 N block is hung from the middle of the rope each side of the rope makes an angle of 9.5◦ with the horizontal. What is the tension in the rope? Round your answer to the nearest 10 N. (a) 40 N (b) 50 N (c) 80 N *(d) 250 N (e) None of these Solution: Begin by sketching a free-body diagram for the block, as at right. There are two horizontal forces acting on the block: the horizontal components of the tension in the two ropes, T cos θ. These forces are equal and opposite.
9.5◦
9.5◦ h hhh hhhh
T
( R((( @ ( h((((
T
82 N
T PP i PP
T sin θ T sin θ
T
1 66 PP θ PP u θ -
T cos θ
T cos θ w?
There are three vertical forces acting on the block: the force due to gravity (i.e., the weight w) pulling downward, and the vertical components of the tension in the two ropes pulling upward, each with magnitude Ty = T sin θ. Since the crate is in static equilibrium, the net force on it is zero; so 2T sin θ = w
⇒
T =
82 N w = = 250 N 2 sin θ 2 sin 9.5◦
410 Problem 1267. In an attempt to keep bears from eating his food, a hiker stretches a rope between two trees and hangs the food from the middle of the rope. When the food is hung from the rope, each side of the rope makes an angle of 11.4◦ with the horizontal. The rope will break if its tension exceeds 220 N. What is the maximum weight of food wmax that the hiker can hang without breaking the rope? Round your answer to the nearest newton. (a) 43 N (c) 216 N (e) None of these
*(b) (d)
87 N 432 N
11.4◦
11.4◦ h hhh hhhh
T
h((((
( R((( @ (
T
wmax
T
T
T sin θ T sin θ
PP i PP
1 66 PP θ PP u θ -
Solution: Begin by sketching a free-body diT cos θ T cos θ agram for the food, as at right. There are two horizontal forces acting on the food: the horwmax ? izontal components of the tension in the two ropes, T cos θ. These forces are equal and opposite. There are three vertical forces acting on the food: the weight wmax pulling downward, and the vertical components of the tension in the two ropes pulling upward, each with magnitude Ty = T sin θ. Since the food is in static equilibrium, the net force on it is zero; so wmax = 2T sin θ = 2(220 N) sin 11.4◦ = 87 N
Problem 1268. In an attempt to keep bears from eating his food, a hiker stretches a rope between two trees and hangs the food from the rope. The food weighs 82 N; each side of the rope makes an angle of 9.5◦ with the horizontal. What is the tension in the rope? Round your answer to the nearest 10 N. (a) 40 N (b) 50 N (c) 80 N *(d) 250 N (e) None of these
θ = 9.5◦ h hhh h
T
θ = 9.5◦
( ( R @ hhh (((( h(((
T
82 N
Solution: The food is not accelerating in any direction, so the net force on it is zero. In particular, the net vertical force on it is zero. The vertical forces on it are the weight, w = 82 N, pulling downward; and the vertical component of the tension of the rope on either side, pulling upward. If the tension is T , its vertical component on either side of the food is Ty = T sin θ. If we take upward as the positive direction, the net vertical force
411 is: 2T sin θ − w = 0 +w
−−−−−→ 2T sin θ = w w ÷2 sin θ −−−−−−−→ T = 2 sin θ 82 N evaluate −−−−−−−−→ T = = 250 N 2 sin 9.5◦
(11.3)
Notice that from equation (11.3), that as θ → 0, sin θ → 0, and so T → ∞. Problem 1269. Two ropes are attached to the ceiling, and their free ends are tied together. A block with weight w is hung from the point where the ropes are joined. Rope A makes an angle of 51◦ to the horizontal; rope B makes an angle of 68◦ to the horizontal. The tension in rope B is TB = 138 N. What is the weight w of the block? Round your answer to the nearest newton.
\ 51◦ 68◦ \ \ \ B A\ \ \\
w
Solution: Since the weight isn’t accelerating to the left or to the right, we know that the net horizontal force on it is zero. The only horizontal forces acting on it are the horizontal components of the tensions in the ropes. Thus these two must be equal: ◦
◦
TA cos 51 = TB sin 68
⇒
TB cos 68◦ TA = cos 51◦
Since the weight isn’t accelerating up or down, the net vertical force on it is zero. The vertical forces are the weight w pulling down, and the vertical components of TA and TB pulling up. Hence: w = TA sin 51◦ + TB sin 68◦ =
TB cos 68◦ sin 51◦ + TB sin 68◦ cos 51◦
Since TB = 138 N, w = (138 N) [cos 68◦ tan 51◦ + sin 68◦ ] = 192 N
412 Problem 1270. A hanging mass with weight w is suspended by two thin ropes, labelled A and B. Rope A makes an angle of θA with the horizontal; rope B makes an angle of θB with the horizontal. The tension in rope A is TA ; the tension in rope B is TB . Which of the following equations represents the condition for static equilibrium of the hanging-mass system in the horizontal direction? *(a) TA cos θA = TB cos θB (c) TA cos θB = TB cos θA (e) None of these
Z θA Z Z
θB Z
Z TA Z Z
TB
Z Z
Z
w
(b) TA sin θA = TB sin θB (d) TA sin θB = TB sin θA
Solution: The hanging mass is not moving, so the net force on it must be zero. The net horizontal force is the sum of the horizontal components of the tensions in the two ropes: TA cos θA , pulling to the left; and TB cos θB , pulling to the right. Since the net horizontal force must be zero, TA cos θA = TB cos θB . Problem 1271. A hanging mass with weight w is suspended by two thin ropes, labelled A and B. Rope A makes an angle of θA with the horizontal; rope B makes an angle of θB with the horizontal. The tension in rope A is TA ; the tension in rope B is TB . Which of the following equations represents the condition for static equilibrium of the hanging-mass system in the vertical direction? (a)
w = TA cos θA + TB cos θB
*(c) w = TA sin θA + TB sin θB (e)
θB
Z θA Z Z Z
TB
Z TA Z Z
Z Z
Z
w
TA TB + cos θA cos θB TA TB (d) w = + sin θA sin θB
(b) w =
None of these
Solution: The hanging mass is not moving, so the net force on it must be zero. The vertical forces acting on it are the weight w, pulling downward; and the vertical components of the tensions in the two ropes, TA sin θA and TB sin θB , both pulling upward. Since the net vertical force must be zero, w = TA sin θA + TB sin θB .
413 Problem 1272. A block with weight w = 130 N is suspended by two thin ropes, labelled A and B. Rope A makes an angle of θA = 37◦ with the horizontal; rope B makes an angle of θB = 54◦ with the horizontal. The tension in rope A is TA ; the tension in rope B is TB . Find TA ; round your answer to the nearest newton. (a) 62 N *(c) 76 N (e) None of these
Z θA Z Z
θB Z
Z TA Z Z
TB
Z Z
Z
w
(b) 69 N (d) 85 N
Solution: The block is not moving, so the net force on it must be zero. The net force in the horizontal direction is the sum of the horizontal components of the two tensions: TA cos θA pulling leftward; and TB cos θB pulling rightward. The net force in the vertical direction is the sum of the weight w, pulling downward; and the vertical components of the two tensions, TA sin θA and TB sin θB , pulling upward. Thus we get two equations that we can solve for the two variables TA and TB : TA cos θA = TB cos θB TA sin θA + TB sin θB = w We want to solve for TA , so we will begin by solving the first equation for TB , then substitute into the second equation. TA cos θA = TB cos θB substitute
−−−−−→ TA sin θA +
÷ cos θ
B −−−−→
TB =
TA cos θA cos θB
TA cos θA sin θB =w cos θB
× cos θ
B −−−−→ TA sin θA cos θB + TA cos θA sin θB = w cos θB
factor
−−−→ TA (sin θA cos θB + cos θA sin θB ) = w cos θB w cos θB divide −−−→ TA = sin θA cos θB + cos θA sin θB w cos θB trig identity −−−−−−→ TA = sin(θA + θB ) (130 N) cos 54◦ = sin 37◦ cos 54◦ + cos 37◦ sin 54◦ = 76 N
414 Problem 1273. A block with weight w = 130 N is suspended by two thin ropes, labelled A and B. Rope A makes an angle of θA = 37◦ with the horizontal; rope B makes an angle of θB = 54◦ with the horizontal. The tension in rope A is TA ; the tension in rope B is TB . Find TB ; round your answer to the nearest newton. (a) 93 N (c) 115 N (e) None of these
*(b) (d)
Z θA Z Z
θB Z
Z TA Z Z
TB
Z Z
Z
w
104 N 128 N
Solution: The block is not moving, so the net force on it must be zero. The net force in the horizontal direction is the sum of the horizontal components of the two tensions: TA cos θA pulling leftward; and TB cos θB pulling rightward. The net force in the vertical direction is the sum of the weight w, pulling downward; and the vertical components of the two tensions, TA sin θA and TB sin θB , pulling upward. Thus we get two equations that we can solve for the two variables TA and TB : TA cos θA = TB cos θB TA sin θA + TB sin θB = w We want to solve for TB , so we will begin by solving the first equation for TA , then substitute into the second equation. TA cos θA = TB cos θB substitute
−−−−−→
÷ cos θ
A −−−−→
TA =
TB cos θB cos θA
TB cos θB sin θA + TB sin θB = w cos θA
× cos θ
A −−−−→ TB cos θB sin θA + TB sin θB cos θA = w cos θA
factor
−−−→ TB (cos θB sin θA + sin θB cos θA ) = w cos θA w cos θA divide −−−→ TB = cos θB sin θA + sin θB cos θA w cos θA trig identity −−−−−−→ TA = sin(θA + θB ) (130 N) cos 37◦ = cos 54◦ sin 37◦ + sin 54◦ cos 37◦ = 104 N
415 Problem 1274. (Lab Derive Problem) You walk into the lab and find a block hanging from Z θL θR Z the ceiling suspended by two threads as shown Z in the figure to the right. Using a scale, you Z TR Z TL find the weight of the block wblock . Using a proZ Z tractor, you find the angle between the ceiling Z Z and the left-hand string to be θL . Similarly, θR Z is the angle between the ceiling and the rightwblock hand thread. From this information determine the tensions in the threads: TL and TR . Solution: The block is not moving, so the net force on it must be zero. The net force in the horizontal direction is the sum of the horizontal components of the two tensions: TL cos θL pulling leftward; and TR cos θR pulling rightward. The net force in the vertical direction is the sum of the weight of the block wblock , pulling downward; and the vertical components of the two tensions, TL sin θL and TR sin θR , pulling upward. Thus we get two equations that we can solve for the two variables TL and TR : TL cos θL = TR cos θR TL sin θL + TR sin θR = wblock The order in which we solve for the tensions is arbitrary, so we’ll solve for TR . We will begin by solving the first equation for TL , then substitute this result into the second equation. cos θR ÷ cos θL TR TL cos θL = TR cos θR −−−−−−−−→ TL = cos θL cos θR substitute −−−−−→ TR sin θL + TR sin θR = wblock cos θL × cos θ
L −−−−→ TR sin θL cos θR + TR sin θR cos θL = wblock cos θL
factor
−−−→ TR (sin θL cos θR + sin θR cos θL ) = wblock cos θL trig identity
−−−−−−→ TR sin(θL + θR ) = wblock cos θL cos θL divide wblock −−−→ TR = sin(θL + θR ) We can now substitute this result back into our equation for TL and TR to find TL . There is a faster and more insightful way to solve for TL . By symmetry, we can reverse the roles of the left and right angles, just call the left angle right and the right angle left (i.e., just switch the left and right subscripts) to get cos θR cos θR TL = wblock = wblock . sin(θR + θL ) sin(θL + θR )
416 Problem 1275. Two blocks with weights wA and wB are held in place on a frictionless inclined plane by two thin ropes, as shown at right. One rope connects block A to a wall at the top of the plane; a second rope connects block B to block A. Both ropes are parallel to the inclined plane, which makes an angle of α with the horizontal. What is the tension TAB in the rope that connects blocks A and B? (a) wB cos α (c) (wA + wB ) cos α (e) None of these
y ! TAw !! !! L !! ! L wA L L !! ! TAB! L ! ! ! ! L !! L!! ! ! L wB L L !! L ! ! ! !L! ! ! LL ! L!!
!
x
α
*(b) wB sin α (d) (wA + wB ) sin α
Solution: Begin by drawing a free-body diagram for each block. We will take the xaxis as pointing up the slope, and the y-axis as perpendicular to the slope pointing upward: see the axes sketched on the upper figure. There are only two forces acting in the xdirection on block B: the tension TAB in the positive direction, and the x-component of the weight wBx = wB sin α in the negative direction. Since the system is in static equilibrium, the net force on block B is zero; so TAB = wB sin α.
NB
NA 6
wBx u -TAB α ?
wB
wBy
6 wAx u T -Aw TAB α ?
wA
wAy
417 Problem 1276. Two blocks with weights wA and wB are held in place on a frictionless inclined plane by two thin ropes, as shown at right. One rope connects block A to a wall at the top of the plane; a second rope connects block B to block A. Both ropes are parallel to the inclined plane, which makes an angle of α with the horizontal. What is the tension TAwall in the rope that connects block A to the wall? (a) wA cos α (c) (wA + wB ) cos α (e) None of these
y
!
T
Awall! ! ! !! L !! L ! L wA L !! ! TAB! L ! ! ! ! L !! L!! ! ! L wB L L !! L ! ! ! !L! ! !
LL ! L!!
x
α
(b) wA sin α *(d) (wA + wB ) sin α
Solution: Begin by drawing a free-body diagram for each block. We will take the xaxis as pointing up the slope, and the y-axis NB NA as perpendicular to the slope pointing upward: 6 6 see the axes sketched on the upper figure. wBx u -TAB Awall wAx u TThere are only two forces acting in the x TAB direction on block B: the tension TAB in the α α positive direction, and the x-component of the ? ? wA wAy wB wBy weight wBx = wB sin α in the negative direction. Since the system is in static equilibrium, the net force on block B is zero; so TAB = wB sin α. Now we can look at the forces acting on block A. There are three forces in the x-direction: TAB and the x-component of the weight, wAx = wA sin α, both pulling in the negative x-direction; and TAwall in the positive x-direction. (Since the ropes are thin, we assume that TAB doesn’t change over the length of the rope.) Since the system is in static equilibrium, the net force on block A must be zero; so TAwall = wAx + TAB = wA sin α + wB sin α = (wA + wB ) sin α
418 Problem 1277. Two blocks with weights wA and wB are held in place on a frictionless inclined plane by two thin ropes, as shown at right. One rope connects block A to a wall at the top of the plane; a second rope connects block B to block A. Both ropes are parallel to the inclined plane, which makes an angle of α with the horizontal. What is the magnitude of the force that the plane exerts on block B? *(a) wB cos α (c) (wA + wB ) cos α (e) None of these
y LL ! L!!
!
x
! !! !! L !! ! L wA L L !! ! L ! ! ! ! ! L !! L!! ! ! L wB L L !! L ! ! ! !L! ! !
α
(b) wB sin α (d) (wA + wB ) sin α
Solution: Begin by drawing a free-body diagram for each block. We will take the xaxis as pointing up the slope, and the y-axis as perpendicular to the slope pointing upward: see the axes sketched on the upper figure. There are only two forces acting in the ydirection on block B: the normal force NB in the positive y-direction, and the y-component of the block’s weight, wBy = wB cos α. Since the block is in static equilibrium, the net force on it is zero; so NB = wB cos α.
NB
NA 6
wBx u -TAB α ?
wB
wBy
6 wAx u T -Aw TAB α ?
wA
wAy
419 Problem 1278. Two blocks with weights wA and wB are held in place on a frictionless inclined plane by two thin ropes, as shown at right. One rope connects block A to a wall at the top of the plane; a second rope connects block B to block A. Both ropes are parallel to the inclined plane, which makes an angle of α with the horizontal. What is magnitude of the force that the plane exerts on block A? *(a) wA cos α (c) (wA + wB ) cos α (e) None of these
y LL ! L!!
!
x
! !! !! L !! ! L wA L L !! ! L ! ! ! ! ! L !! L!! ! ! L wB L L !! L ! ! ! !L! ! !
α
(b) wA sin α (d) (wA + wB ) sin α
Solution: Begin by drawing a free-body diagram for each block. We will take the xaxis as pointing up the slope, and the y-axis as perpendicular to the slope pointing upward: see the axes sketched on the upper figure. There are only two forces acting in the ydirection on block A: the normal force NA in the positive y-direction, and the y-component of the block’s weight, wAy = wA cos α. Since the block is in static equilibrium, the net force on it is zero; so NA = wA cos α.
NB
NA 6
6
wBx u -TAB α ?
wB
wBy
wAx u T -Aw TAB α ?
wA
wAy
Problem 1279. Two forces with magnitudes F1 and F2 are acting on an object. Which of the following inequalities always hold for the magnitude of the net force Fnet on the object? (a) F1 ≤ Fnet ≤ F2 *(c) |F1 − F2 | ≤ Fnet ≤ F1 + F2 (e) None of these
(b) (F1 − F2 )/2 ≤ Fnet ≤ (F1 + F2 )/2 2 (d) F12 − F22 ≤ Fnet ≤ F12 + F22
Problem 1280. An object with a mass of 20 kg is suspended from two ropes, as shown at right. The magnitude of the force exerted by each rope on the object is 139 N; the magnitude of the force of gravity on the object is 196 N. What is the magnitude of the net force on the object? (a) 47.4 N (b) 33.5 N (c) 13.9 N *(d) 0 N (e) None of these Solution: The system is static, so the net force is zero.
\ \ \ \ \ \ \\
20 kg
420 11.2.2
Block and pulley frictionless systems
Key Idea: Frictionless block-and-pulley systems are just one-dimensional systems in disguise! The system sits in two dimensions, but can be converted to a one-dimensional system. The background space is known as the ambient space and the system can be shown to be one dimensional by mathematical arguments from a field known as Topology. We will just take these topological arguments as fact. Problem 1281. A rope, one end of which is attached to a ceiling, passes through a pulley and then goes straight upward where it is held in place by a man that is standing on the ground. A weight of 400 N is attached to the pulley. Assuming that the weight of the pulley can be neglected in comparison to the 400 N weight and that there is no friction between the pulley and the rope, what upward force must the man exert to hold the weight in place? (a) 800 N (c) 400 N (e) None of these
(b) 640 N *(d) 200 N
Solution: Let W = 400 N be the downward force of the weight, and let T be the tension of the rope. If you sketch a free-body diagram, it will show a downward force of W on the pulley, an upward force of T exerted on the pulley by the rope to the ceiling, and another upward force of T exerted on the pulley by the man pulling upward on the other end of the rope. Since the pulley isn’t moving, the net force must be zero: so 2T = W ; so T = W/2 = 200 N.
Problem 1282. Two crates are connected by a thin rope and sitting on a frictionless horizontal floor. You are pulling them by a second rope attached to the smaller crate. You pull on your rope with a horizontal force of P ; the tension in the rope connecting the crates is T . Which of the following is true of P and T ? (a) P < T *(c) P > T
P
37 kg
T
89 kg
(b) P = T (d) Not enough information; it could be any of these
Solution: P is giving a certain acceleration to the combined mass of both crates. T is giving the same acceleration to the 89-kg crate alone. Thus T is smaller than P ; in fact, 89 kg T = ·P 37 kg + 89 kg
421 Problem 1283. In your physics lab, you find the apparatus shown at right. A block with a mass of m1 is resting on a horizontal table. A light string runs horizontally from the block, over a massless frictionless pulley, and down to a hanging bucket. The coefficient of static friction between the block and the table is µs , and the coefficient of kinetic friction between the block and the table is µk . Initially, the system does not move. You slowly add water to the bucket until the system starts moving. At this point, what is the weight w of the bucket? (a) m1 (b) m1 g (c) µs m1 g (d) µk m1 g
m1 given: µs given: µk w
Solution: The system is initially at rest, so you must add water to the bucket until its weight is sufficient to overcome the static friction of the block on the table. This occurs when W = µs N = µs M g, where N is the normal force on the block from the table. Problem 1284. (Lab Problem) In your physics lab, you find the apparatus shown at right. A block with a weight of 60 N is resting on a horizontal table. A light string runs horizontally from the block, over a massless frictionless pulley, and down to a hanging bucket. The coefficient of static friction between the block and the table is µs = 0.6, and the coefficient of kinetic friction between the block and the table is µk = 0.2. Initially, the system does not move. You slowly add water to the bucket until the system starts moving. At this point, what is the weight w of the bucket? (a) 12 N *(b) 36 N (c) 48 N (d) 60 N
60 N µs = 0.6 µk = 0.2 w
Solution: The system is initially at rest, so you must add water to the bucket until its weight is sufficient to overcome the static friction of the block on the table. This occurs when W = µs · 60 N = (0.6)(60 N) = 36 N.
422 Problem 1285. In your physics lab, you find the apparatus shown at right. A glider with a weight of m1 is resting on a horizontal air-track with the air off. A light string runs horizontally from the glider, over a massless frictionless pulley, and down to a hanging block with a weight of m2 . The coefficient of static friction between the glider and the air-track is µs , and the coefficient of kinetic friction between the glider and the air-track is µk . The coefficients of friction are assumed known. After the glider-stringblock system have been released to move freely, and before the hanging block has hit the floor, what is the tension T in the string?
m1 (a) (1 + µk )g m1 + m2 m1 m2 (1 − µk )g (c) m1 + m2
m1 µs > µk
m2
m1 m2 *(b) (1 + µk )g m1 + m2 m1 + m2 (d) (1 + µk )g m1 m2
Solution: First, you should determine a condition between m1 and m2 that assures us that the system moves when the glider-block system is released. The maximal static friction of the glider on the air-track is fk = µs N1 = µs m1 g. That is, we need m2 g > µs m1 g ⇒ m2 > µs m1 . This is our condition for motion. Assuming the system will move, we can determine the forces on it. We will first analyze the forces on the two separate subsystems: mass m1 and m2 , and then we will connect them by examining the combined system: glider, massless string, and hanging-mass. We will connect the forces on the systems by observing that because the string is taunt and assumed not to stretch, that the velocity and the acceleration of the two masses must be the same.
423 The glider m1 : If we take the x1 -axis for m1 to point in the direction of motion and the y1 -axis to point upward, then applying Newton’s second law we arrive at the system of equations:
N1 6
T1
Fnet1,x = T1 − fk = m1 a1,x , Fnet1,y = N1 − m1 g = m1 a1,y = 0 ,
f u- k m1 g ?
where we’ve used the fact that the glider does not move in the vertical direct, so ay,1 = 0 (see free-body diagram).
free-body diagram for mass m1
From the second equation we have N1 = m1 g
⇒
fk = µk N1 = µk m1 g .
Upon substituting this equation into the first equation we have T1 = fk + m1 a1,x = µk m1 g + m1 a1,x . The hanging mass m2 : Since the motion of m2 is pure vertical, we only need a y2 axis to describe the motion. We take the y2 -axis pointing downward in the direction of motion. Applying Newton’s second law to mass m2 , we arrive at the system of equations: Fnet1,y = m2 g − T2 = m2 a2,y
⇒
T2 = m2 g − m2 a2,y .
By hypothesis, the pulley is massless and frictionless, so it only serves to redirect the forces. Thus, T1 = T2 and we have T1 = T = µk m1 g + m1 a1,x , T2 = T = m2 g − m2 a2,y . Equating these equations through T and using the fact that since the glider and hanging mass are connected by a string, they move as one and therefore have the same magnitude of acceleration (i.e., a1,x = a2,y = a) gives the following equation for a µk m1 g + m1 a = T = m2 g − m2 a re-arraging terms
−−−−−−−−−→ (m1 + m2 )a = (m2 − µk m1 )g m2 − µk m1 ÷ (m1 +m2 ) g −−−−−−→ a = m1 + m2
424 We can now use a to determine T . T = m2 g − m2 a m 2 − µk m 1 = m2 g − m2 g m1 + m2 (m1 m2 + m22 ) − (m22 − µk m1 m2 ) = g m1 + m2 m1 m2 = (1 + µk )g m1 + m2
As a check on our formulas that we found for acceleration and tension, let’s ask what happens if the mass of the glider goes to zero (i.e., m1 → 0+ )? Then a = g and T = 0 as would be expected since the hanging mass would then be in free fall. Notice that we cannot let m2 go to zero because of the constraint m2 > µs m1 > 0.
425 Problem 1286. (Lab Problem) This problem has a fundamental mistake in it, can you find it? A glider is sitting at rest on a horizontal frictionless air track. A light string is attached to a light string that runs over a massless frictionless pulley and attaches to a hanging mass of 1 kg. The system is held in place and released from rest. Using photo gates that are placed 1 m apart the time is measured to be 2 s. Assuming that the string remains horizontal and taunt throughout the entire run, what is the mass of the glider? For ease of computation take g to be 10 m/s2 . Round your answer to the nearest kilogram. (a) 1 kg (b) 2 kg (c) 4 kg (d) 19 kg (e) None of these Solution: The massless, frictionless pulley justifies the assumption that the tension is the same throughout the string, from which it follows that there is a constant horizontal force on the glider of mg = 10 N from the hanging weight. This force has been transmitted through the string. Recall that a pulley is a machine that redirects force. The light string means that we can ignore the mass of the string. The mistake is highlighted in italics: as the hanging weight falls, it accelerates, so it’s apparent weight increases since it’s in an accelerating frame of reference. This leads to an increase in tension. If you sketch a free-body diagram of the glider, you’ll see that the only horizontal force is the constant 1 N. There’s no net vertical force. Given: Fnet = 1 N; vi = 0 m/s; ∆x = 1 m; and t = 2s; Want: m = ? Note: We don’t know, nor do we want vf . If we try to solve for m directly using Newton’s second law, we get stuck: Fnet = ma
÷a
−−−−−→
m=
1N Fnet = a a
(11.4)
We need to use one of our three fundamental kinetic equations to find a. Since we know ∆x, vi = 0, and t, we can use 1 at2 ∆x = vi t + at2 = 2 2
· 2/t2
−−−−→
a=
Substituting equation (11.5) into equation (11.4) gives us m=
Fnet t2 (1 N)(2 s)2 = = 2 kg 2∆x 2(1 m)
2∆x t2
(11.5)
426
Problem 1287. In your physics lab, you find the apparatus shown at right. A tank of water is resting on a horizontal frictionless table. A light string runs horizontally from the tank, over a massless frictionless pulley, and down to a hanging block with a mass of m1 . You add increasing amounts of water to the tank, each time releasing the system and observing its motion or lack thereof. How does the system behave as you increase the mass m2 of the tank?
m2
m1
(a)
The acceleration does not change, since there is no friction and m1 remains constant. (b) The system accelerates as long as m2 < m1 ; when m2 = m1 , the system no longer accelerates. (c) The system moves as long as m2 < m1 ; when m2 = m1 , the system no longer moves. *(d) The system always accelerates; but the acceleration gets smaller as m2 becomes larger. Solution: Since there is no friction, the system always accelerates, no matter how large m2 is. However, since the force remains the same (F = m1 · g), and since the mass of the system is m1 + m2 , increasing m2 increases the mass of the system and decreases a. In fact, m1 · g a= m1 + m2
427 Problem 1288. (Derive Lab Problem) In your physics lab, you find the apparatus shown at right. A block with mass m1 is resting on a horizontal air-track with the air turned off. A light string runs horizontally from the block, over a massless frictionless pulley, and down to a hanging bucket. The coefficient of static friction between the block and the track is µs , and the coefficient of kinetic friction between the block and the table is µk < µs . Initially, the system does not move. You slowly add water to the bucket until the system just starts to move. Call this moment in time t = t0 = 0; at this time, call the mass of the bucket of water m2 .
m1
µk < µs m2 Note: Not drawn to scale.
Setup: For the block, take the x1 -axis in the direction of motion along the table and the y1 -axis going upward. Take the y2 -axis pointing downward, in the direction of motion. Let the origin along the x1 -axis be the initial position of the block with mass m1 (x1,i = 0). The initial velocity of mass m1 is zero (at t = 0, v1,i = 0). (a) Find the value of m2 : that is, find m2 such that the force of gravity on the hanging bucket is just enough to overcome the force of static friction on the block with mass m1 . (b) Find a formula for the initial acceleration of the system, denoted asys . (c) Assuming that the initial position and velocity of the block with mass m1 is zero, find an equation for its position as a function of time: x1 = x1 (t). Ignore the fact that the block will eventually crash into the pulley. Solution:
Work it out!
428 Problem 1289. (Derive Problem) In your physics lab, you find the apparatus shown below. A glider with mass m1 is resting on a horizontal air track. A light string runs horizontally from the glider, over a massless frictionless pulley, and down to a hanging bucket. The air track is specially designed so that the coefficient of static friction between the glider and the air track is µs > 0 (a constant), and the coefficient of kinetic friction between the block and the table is µk = µs e−x (a function of position) provided that the front of the glider is initially at rest at a location marked as the origin on the air track. The coefficient of static friction between the block and the table is µs (a constant), and the coefficient of kinetic friction between the block and the table is µk = µs e−x (a function of position). Initially, the system does not move. You slowly add water to the bucket until the system just starts to move. Call this moment the initial time t0 = 0; at this time, call the mass of the bucket of water m2 . Define the initial position of the center of mass for block m1 to be the origin and take the initial velocity to be zero, since the block starts from rest.
m1
µk = µs e−x m2
The Coordinate System: For the glider, take the x1 -axis in the direction of motion along the table and the y1 -axis going upward. Take the y2 -axis pointing downward, in the direction of motion. Let the origin along the x1 -axis be the initial position of the glider with mass m1 (x1,i = 0). Since the direction of motion is one dimensional, you can simplify things by letting x = x1 = y2 . The Setup and Initial Conditions: Take the initial position of the glider to be at the origin defined above (i.e., x1,i = 0) and the initial velocity of the two-mass-plus-string system to be zero at time t0 = 0. Take the point of zero potential to be y2 = initial height of the hanging mass m2 . This is equivalent to taking x = 0 as the point of zero potential. The Problem: Find an equation for the mechanical energy of the system as a function of time and position: K.E.sys (x, t) + P.E.sys (x, t) = D(x, t) ,
429 where D(x, t) is a dissipation term. Ignore the fact that the glider will eventually crash into the pulley, or the hanging mass will hit the ground. Your model is only designed to describe the motion in the interim time. Comments, Advice, and Hints: Comment (1): The first step is to determine if the system will even move! To do this you will need to find the force exerted by the hanging weight m2 that over comes the force of static friction. That is, find m2 such that the force of gravity on the hanging bucket plus water is just enough to overcome the force of static friction fs,max on the block with mass m1 . Comment (2): Notice that the coefficient of kinetic friction µk is a function of position! We have never solved such a problem in class! However, we have solved a similar equation in the new lab 6. Comment (3): Be sure to clearly label all of the intermediate steps in comments (1)-(3). If I have to go searching for them, I will take points off. Advice: To find the kinetic and potential energy, and any possible dissipation term for the system, as a function of position and time, you will need to first find a formula for the initial acceleration of the system asys as a function of time t and position x and integrate the system using the initial conditions. When you get your result, ask yourself what is the source of the kinetic energy of the system, and what is the source of the potential energy of the system. Also, ask yourself: is the system mechanically conservative? If not, then what are the sources of energy loss? Can you identify these terms in the equation? Hint: To perform one integration on a differential equation of the form: d2 x dF (x(t)) + = constant , dt2 dx The standard trick is to multiple the equation by the velocity term v = x˙ and integrate the equation using the chain rule together with the Fundamental Theorem of Calculus (FTC). With this hint, you can solve the resulting equation with the techniques that you learn in Calculus 1 (a prerequisite for the course). Solution:
Work it out!
430 Problem 1290. Two identical blocks, each with mass m, are attached to opposite ends of a thin rope that passes over a massless frictionless pulley. The pulley is suspended from the ceiling by a thin chain. Draw a free-body diagram for each block, and one for the block-pulley system. Solution: The diagrams are drawn to the right of the illustration. The diagram on the left is for an individual block; the one on the right is for the block-pulley system.
6
m
u
wb = mg ?
wsys = 2mg
m
?
6
Tchain
r
u
m
w = 2mg
m
Problem 1292. Two identical blocks, each with mass m, are attached to opposite ends of a thin rope that passes over a massless frictionless pulley. The pulley is suspended from the ceiling by a thin chain. What is the tension in the rope? (a) 0 (b) mg/2 *(c) mg (d) 2mg Solution: A free-body diagram for one of the blocks is sketched at right. There are only two forces acting on the block: the weight w = mg pulling downward, and the tension Trope pulling upward. Since the system is in static equilibrium, Fnet = 0; so Trope = mg.
Trope
u
Problem 1291. Two identical blocks, each with mass m, are attached to opposite ends of a thin rope that passes over a massless frictionless pulley. The pulley is suspended from the ceiling by a thin chain. What is the tension in the chain holding up the pulley-block system? (a) 0 (b) mg/2 (c) mg *(d) 2mg Solution: A free-body diagram for the pulley-blocks system is sketched at right. There are two forces acting on the system: the weight w = 2mg pulling down; and the tension Tchain pulling up. Since the system is in static equilibrium, Fnet = 0; so Tchain = 2mg.
Tchain
6
r
?
6
Trope
r
u
m m
w = mg ?
431
Problem 1293. In your physics lab, you find the apparatus at right. A block with mass M1 is on a frictionless slope. A thin string from the block runs parallel to the slope, up to and over a massless frictionless pulley, then down to a second block with mass M2 . Initially, the system is in static equilibrium. If it is shaken slightly, what will the system do?
u@ @ @ @ @ @ @ @ @ M1 @ @ @ @ M2 @ @ @ @
(a) The first block will accelerate up the slope; the second will accelerate downward. (b) The first block will accelerate down the slope; the second will accelerate upward. *(c) The blocks will not accelerate. Solution: Since the system is in static equilibrium and there is no friction, the downward force of gravity on M2 equals the component of the gravitational force on M1 parallel to the slope. Vibrating the system doesn’t change this; the net force on the system will still be zero afterward, so it won’t accelerate in either direction. Problem 1294. A block with mass m1 is on a frictionless ramp that makes an angle of θ to the horizontal. A thin string runs from the block, parallel to the ramp, up and over a massless frictionless pulley, and down to a hanging block with mass m2 . The system is held in place, then released. If the blocks do not move after the release, which of the following must be true? *(a) m1 sin θ = m2 (b) m2 sin θ = m1 (c) m1 cos θ = m2 (d) m2 cos θ = m1 (e) None of these Solution: Sketch a free-body diagram for each of the blocks, as at right. Since the rope is thin and the pulley is massless and frictionless, we assume that the tension in the rope is constant over its entire length. In the diagram for the sliding block, we take the x-direction as parallel to the slope, and the y-direction as perpendicular to it.
u A A m1 A A m2
θ
N
T KA A A
A *T A u m1 A w1x A θA A AU ? w1y
w1 = m1 g
6
um2
?
w2 = m2 g
If the system does not move, then the net force on each block must be zero. For the hanging block, this means that T = w2 = m2 g. For the sliding block, the net force in the x-direction must be zero; so T = w1x = m1 g sin θ. Combining these, we get T = m1 g sin θ = m2 g
⇒
m1 sin θ = m2
432 Problem 1295. A block with mass m1 is on a frictionless ramp that makes an angle of θ to the horizontal. A thin string runs from the block, u parallel to the ramp, up and over a massless A frictionless pulley, and down to a hanging block A m1 A with mass m2 . The system is held in place, A m2 then released. What is the acceleration of the θ hanging block? Use downward as the positive direction. (m2 − m1 cos θ)g (m2 − m1 sin θ)g (a) *(b) m1 + m2 m1 + m2 (m2 − m1 sin θ)g (m2 − m1 cos θ)g (d) (c) m2 m2 (e)
None of these
Solution: Sketch a free-body diagram for each of the blocks, as at right. Since the rope is thin and the pulley is massless and frictionless, we assume that the tension in the rope is constant over its entire length. In the diagram for the sliding block, we take the x-direction as parallel to the slope, and the y-direction as perpendicular to it. We know the masses m1 and m2 , the angle θ, and the gravitational acceleration g. We don’t know the tension in the rope T . We want to know the acceleration of the hanging block; and since the two blocks are connected by a taut rope, the acceleration of the sliding block must have equal magnitude.
N
T KA A A
6
T
A
u m1A
um2
*
A w1x A θA A AU ?
w1 = m1 g
w1y
?
w2 = m2 g
The net force acting on the hanging block in the positive (downward) direction is F2 = m2 g − T . The net force acting on the sliding block in the same direction (up the slope) is F1 = T − m1 g sin θ. The net force acting on the whole system is Fnet = F1 + F2 = T − m1 g sin θ + m2 g − T = m2 g − m1 g sin θ The mass of the system is m1 + m2 ; so by Newton’s second law, the acceleration of the system is (m2 − m1 sin θ)g Fnet a= = msystem m1 + m2
433 Problem 1296. A block with a mass of 50 kg is on a frictionless ramp that makes an angle of 30◦ to the horizontal. A thin string runs parallel to the ramp from the block to a massless frictionless pulley, then down to a second block with a mass of 30 kg. The system is initially held in place, then released. What is the acceleration of the hanging (30 kg) block? Round your answer to the nearest 0.1 m/s2 . [The picture is not drawn to scale.] (a) 0.6 m/s2 upward *(b) 0.6 m/s2 downward (c) 1.6 m/s2 upward (d) 1.6 m/s2 downward
u@ @ @ @ @ @ @ @ @ 50 kg@ @ @ @ 30 kg @ @ @ @
Solution: Since the pulley is massless and frictionless and since the string is massless, the tension is the same throughout the length of the string: T . Sketch a free-body diagram of each block. The hanging block is easy: the weight w1 = m1 g pulling straight down, and the tension T pulling straight up. (We’ll use the subscripts 1 for the hanging block and 2 for the sliding block.) The sliding block is more complicated. There are N three forces: the weight w2 = m2 g pulling straight T YH H down; the tension T pulling upward parallel to the H HHr slope; and the normal force N pushing upward perH H j w2k H pendicular to the slope. We decompose w2 into two components: w2k = w2 sin 30◦ , parallel to the slope; 30◦ ◦ and w2⊥ = w2 cos 30 , perpendicular to the slope. A w2⊥ free-body sketch is shown at right. ?w2 We can ignore forces on the sliding block perpendicular to the slope, since there’s no friction. We will take forces as positive if they tend to move the system so that the hanging block moves downward and the sliding block moves upward. The net force on the hanging block is F1 = w1 − T = m1 g − T ; the net force parallel to the slope on the sliding block is F2 = T − w2k = T − m2 g sin 30◦ . Then the net force on the whole system is Fnet, system = F1 + F2 = m1 g − T + T − m2 g sin 30◦ = m1 g − m2 g sin 30◦ By Newton’s second law, the acceleration of the system is the net force on the system divided by the total mass: a= =
Fnet, system m1 g − m2 g sin 30◦ g(m1 − m2 sin 30◦ ) = = msystem m1 + m2 m1 + m2 (9.8 m/s2 )[30 kg − (50 kg) sin 30◦ ] = 0.6 m/s2 30 kg + 50 kg
Since the sign is positive, the hanging block accelerates downward.
434 11.2.3
Accelerometers
Problem 1297. (Accelerometer) A thin string is attached to the ceiling of a car, and a lead sinker with mass m is attached to the free end of the string. The car is going at highway speed when the driver hits the brakes. While the car is slowing down, the string makes an angle of θ to the vertical. What is the magnitude of the car’s acceleration? Take the direction of motion to be the positive x-axis and the y-axis to point upward. *(a) −g tan θ (b) −mg tan θ mg g (d) (c) − tan θ tan θ (e)
θ
L L L L Lu
None of these
Solution: Sketch a free-body diagram of the object on the string. There are two forces acting on it: gravity, pulling straight downward with a force of w = mg, where m is the mass of the object; and the tension T in the string, pulling upward and to the left at an angle of θ to the vertical. The latter can be decomposed into its x- and y-components: Tx = T sin θ and Ty = T cos θ. Since there is no vertical acceleration, the net force in the ydirection must be zero: thus 0 = may = Fnet,y = Ty − w
⇒
Ty
T AK
6 A A A
Tx
θ A Au
T cos θ = mg
The only force in the x-direction is Tx ; so it supplies all the horizontal acceleration. Hence max = Fnet,x = −Tx
⇒
T sin θ = −max
From these, we can calculate θ: tan θ =
Tx −max −ax = = Ty mg g
⇒
ax = −g tan θ
Note: If we take the x-axis pointing in the initial direction of motion, then ax < 0.
? w
435 Problem 1298. (Accelerometer) A thin string is attached to the ceiling of a car, and a weight is attached to the free end of the string. The car is going at 43 m/s when the driver hits the brakes. The car’s acceleration has a magnitude of 2.3 m/s2 . While the car is slowing down, what angle θ does the string make to the vertical? Take the direction of motion to be the positive x-axis and the y-axis to point upward. Round your answer to the nearest degree. (a) 11◦ (b) 12◦ ◦ *(c) 13 (d) 15◦
θ
Solution: Sketch a free-body diagram of the object on the string. There are two forces acting on it: gravity, pulling straight downward with a force of w = mg, where m is the mass of the object; and the tension T in the string, pulling upward and to the left at an angle of θ to the vertical. The latter can be decomposed into its x- and y-components: Tx = T sin θ and Ty = T cos θ. Since there is no vertical acceleration, the net force in the ydirection must be zero: thus 0 = may = Fnet,y = Ty − w
⇒
L L L L Lu
⇒
6 A A A
Tx
θ A Au
T cos θ = mg ? w
The only force in the x-direction is Tx ; so it supplies all the horizontal acceleration. Hence max = Fnet,x = −Tx
Ty
T AK
T sin θ = −max
From these, we can calculate θ: Tx −max −ax tan θ = = = Ty mg g
⇒
θ = tan−1
−ax g
= tan−1
2.3 m/s2 9.8 m/s2
Note: If we take the x-axis pointing in the initial direction of motion, then ax < 0
⇒
−ax > 0
⇒
θ > 0.
! = 13◦
436 Problem 1299. (Accelerometer) A thin string is attached to the ceiling of a car, and a lead sinker with mass m is attached to the free end of the string. The car is going at highway speed when the driver hits the brakes, producing an acceleration with magnitude ax . While the car is slowing down, the string makes an angle of θ to the vertical. What is θ? Take the direction of motion to be the positive x-axis and the y-axis to point upward. g g −1 (b) θ = tan (a) θ = tan −ax −ax −ax −ax −1 *(d) θ = tan (c) θ = tan g g (e)
None of these
L L L L Lu m
θ
T AK
Ty = T cos θ 6
A A Aθ A Au
Tx = T sin θ Solution: Sketch a free-body diagram for the sinker, as at right. There are two forces acting on it: the tension in the string T , pulling ? w = mg upward and to the left; and the weight w = mg, pulling straight down. While the car is slowing down, the sinker has a horizontal acceleration of ax and no vertical acceleration. Since there is no vertical acceleration, the two vertical forces must be equal and opposite: Ty = T cos θ = mg. The only horizontal force is Tx = T sin θ; so max = Fnet,x = −T sin θ. We can eliminate the unknowns T and m by taking the ratio of these two equations: T cos θ = mg and T sin θ = −max sin θ −max −ax T sin θ ⇒ = = tan θ = = T cos θ cos θ mg g −a a x x ⇒ θ = tan−1 = − tan−1 g g In the last step we used the fact that the arctan is an odd function to factor out the negative sign.
437 11.2.4
Static and kinetic friction problems
Problem 1300. (Similarity Problem) You are pushing a crate of physics books across a concrete floor, with friction present. If you push with a horizontal force of F1 , the crate experiences an acceleration a1 . If you increase the horizontal force to F2 = 2F1 , the acceleration is a2 . What is the relationship between a2 and a1 ? (a) a2 = a1 (b) a1 < a2 < 2a1 (c) a2 = 2a1 *(d) a2 > 2a1 Solution: If you push the crate hard enough to move it, then the net horizontal force on it is your pushing force minus the force of kinetic friction. In the first case, that’s F1 − fk . If you push with twice the horizontal force, it doesn’t affect the friction; so in the second case, the net horizontal force on the crate is 2F1 − fk > 2(F1 − fk ). Since the mass of the crate doesn’t change from the first case to the second, the acceleration is proportional to the net horizontal force; so a2 > 2a1 . Problem 1301. Two blocks, one with mass m1 and one with mass m2 , are connected by a thin horizontal rope. A second horizontal rope is attached to the block with mass m1 and pulled to the left with a force whose magnitude is P . The coefficient of kinetic friction between each block and the floor is µk . If the blocks move at a constant velocity, what is the tension T in the rope connecting the two blocks? (a) m1 gµk *(c) m2 gµk (e) None of these
(b) (d)
(m2 + m1 )gµk (m2 − m1 )µk
P
m2
T
m1
N2
N1
6 6
u -
P
f1k ?
-
T
T
u -
f2k
? Solution: Sketch a free-body diagram for each w1 = m1 g w2 = m2 g block, as at right. Let T be the tension in the rope connecting the two boxes. Since the floor is horizontal, N1 = m1 g and N2 = m2 g. Hence f1k = µk m1 g and f2k = µk m2 g. Since the blocks are moving at constant velocity, the net force on each is zero. For block 2, this means that
T = f2k = m2 gµk
438 Problem 1302. Two blocks, one with mass m1 and one with mass m2 , are connected by a thin horizontal rope. A second horizontal rope is attached to the block with mass m1 and pulled to the left with a force whose magnitude is P . The coefficient of kinetic friction between each block and the floor is µk . If the blocks move at a constant velocity, what is P ?
P
m2
T
m1
N2
N1
6 6
(a) 0 (c) m1 µk (e) None of these
*(b) (m1 + m2 )gµk (d) (m2 − m1 )µk
u -
P
-
T
f1k
u -
T
f2k
? Solution: Sketch a free-body diagram for each ? w = m1 g 1 block, as at right. Let T be the tension in the rope w2 = m2 g connecting the two boxes. Since the floor is horizontal, N1 = m1 g and N2 = m2 g. Hence f1k = µk m1 g and f2k = µk m2 g. Since the blocks are moving at constant velocity, the net force on each is zero. For block 2, this means that T = f2k = µk m2 g. For block 1, it means that
P = f1k + T = µk m1 g + µk m2 g = (m1 + m2 )gµk Problem 1303. Two blocks, one with mass m1 = 3 kg and one with mass m2 = 7 kg, are connected by a thin horizontal rope. A second horizontal rope is attached to the block with mass m1 and pulled to the left with a force whose magnitude is P . The coefficient of kinetic friction between each block and the floor is µk = 0.4. If the blocks move at a constant velocity, what is P ? For ease of calculation, assume that g = 10 m/s2 . (a) 12 N (c) 28 N (e) None of these
(b) 16 N *(d) 40 N
P
m2
T
m1
N2
N1
6 6
u -
P
f1k
-
T
T
u -
f2k
?
? Solution: Sketch a free-body diagram for each w1 = m1 g w2 = m2 g block, as at right. Let T be the tension in the rope connecting the two boxes. Since the floor is horizontal, N1 = m1 g and N2 = m2 g. Hence f1k = µk m1 g and f2k = µk m2 g. Since the blocks are moving at constant velocity, the net force on each is zero. For block 2, this means that T = f2k = µk m2 g. For block 1, it means that
P = f1k + T = µk m1 g + µk m2 g = (m1 + m2 )gµk = (3 kg + 7 kg)(10 m/s2 )(0.4) = 40 N
439 Problem 1304. Two blocks, one with mass m1 = 3 kg and one with mass m2 = 5 kg, are connected by a thin horizontal rope. A second horizontal rope is attached to the block with mass m1 and pulled to the left with a force whose magnitude is P . The coefficient of kinetic friction between each block and the floor is µk = 0.2. If the blocks move at a constant velocity, what is the tension T in the rope connecting the two blocks? For ease of calculation, assume that g = 10 m/s2 . (a) 6 N *(c) 10 N (e) None of these
(b) 8 N (d) 16 N
P
m2
T
m1
N2
N1
6 6
u -
P
f1k
-
T
u -
T
f2k
?
w1 = m1 g
?
w2 = m2 g Solution: Sketch a free-body diagram for each block, as at right. Let T be the tension in the rope connecting the two boxes. Since the floor is horizontal, N1 = m1 g and N2 = m2 g. Hence f1k = µk m1 g and f2k = µk m2 g. Since the blocks are moving at constant velocity, the net force on each is zero. For block 2, this means that T = f2k = m2 gµk = (5 kg)(10 m/s2 )(0.2) = 10 N
440 Problem 1305. You are towing a block on a horizontal surface, using a rope that makes an angle of θ to the horizontal. The block has a mass of m; the coefficient of kinetic friction between the block and the floor is µk . As you pull the block, you are giving the block a forward acceleration of a. What is the tension T in the rope, expressed in terms of m, a, θ, µk , and the gravitation g? (a)
(mgµk + a) cos θ
*(b)
(c)
m(gµk + a) cos θ
m(gµk + a) cos θ + µk sin θ
(d)
mgµk + a
(e)
None of these T m
* θ
Solution: The first step is draw a picture and a free-body diagram. The forces on the block are the normal force, the weight, the kinetic frictional force, and the tension that must be broken into its x and y components, where the x-axis points to the right and the y-axis pointing upward. See the diagram. The horizontal force from the rope is Tx = T cos θ, directed forward and the vertical force is Ty = T sin θ. The frictional force is given by fk = µk N . Notice that because of the vertical component of the tension the normal force is not equal to the weight. Applying Newton’s second law in each of the component directions: ( T cos θ − µk N = Fnet,x = max = ma T sin θ + N − mg = Fnet,x = may = 0 This is a system of two equations and two unknowns: N and T . Solving the second equation for N gives N = mg − T sin θ, and substituting this equation into the first equation yields T cos θ − µk (mg − T sin θ) = ma
Isolate T
−−−−−→
T (cos θ + µk sin θ) = m(gµk + a) m(gµk + a) Solve for T −−−−−−→ T = cos θ + µk sin θ
Note 1: A common mistake is to take N = mg, giving an incorrect force of friction fk = −µk mg, which would only be the case if there was no y component of tension. Note 2: If the force is horizontal, then θ = 0, and the problem is straight forward.
441 Problem 1306. A large stone block has been loaded in the bed of a dump truck. The bed is slowly raised until it makes an angle of θs with the horizontal; at this point, the block starts to slide downward. What is the coefficient of static friction µs between the block and the truck bed? (a) sin θs *(b) tan θs (c) cos θs (d) θs (e) None of these Solution: Sketch a picture and then a free-body diagram, as at right. For the free-body diagram, we choose the x-axis parallel to the truck bed and the y-axis perpendicular to it.
N AK A fs * A AA r y AA A A AK x A A* ?w θs
N 6
wx = w sin θs
Since the block does not sink into the truck bed or rise above it, the net force in the y-direction must be zero. Hence N = wy = w cos θs . At the point where the block begins to slide, the downslope force in the x-direction is equal to the force of static friction. Hence w sin θs = fs = µs N = µs w cos θs
⇒
µs =
w
u
θs ? w
y
f s = µs N
= w cos θs
w sin θs sin θs = = tan θs w cos θs cos θs
Problem 1307. A brick is sitting on a roof, which makes an angle of θk with the horizontal. If the brick is given a slight nudge, it slides down the rooftop at a constant speed. What is the coefficient of kinetic friction µk between the brick and the rooftop? (a) sin θk *(b) tan θk (c) cos θk (d) θk (e) None of these Solution: Sketch a picture and then a free-body diagram, as at right. For the free-body diagram, we choose the x-axis parallel to the roof top and the y-axis perpendicular to it.
N AK A y KA x A* θ k
A fk * A AA r AA A A A ?w
N 6
wx = w sin θk
f k = µk N u Since the brick does not sink into the roof or rise above it, the net force in the y-direction must be zero. Hence θk N = wy = w cos θk . When the brick slides down the ? wy = w cos θk w roof without accelerating, the downslope force in the xdirection exactly matches the force of kinetic friction. Hence w sin θk sin θk w sin θk = fk = µk N = µk w cos θk ⇒ µk = = = tan θk w cos θk cos θk
442 Problem 1308. A crate whose weight is w is being pulled up a ramp that makes an angle of θ with the horizontal, using a thin rope that runs parallel to the ramp. The coefficient of kinetic friction between the crate and the ramp is µk . If the crate is moving at a constant speed of v0 , what is the tension in the rope? *(a) w(sin θ + µk cos θ) (c) w(cos θ + µk sin θ) (e) None of these
! ! ! !! ! ! ! L ! L !! ! L !! L L w L ! x ! ! L! L !L! ! ! L! !! ! ! !!
y
!
θ
(b) w(sin θ − µk cos θ) (d) w(cos θ − µk sin θ)
Solution: Draw a free-body diagram with the x-axis parallel to the ramp and the y-axis perpendicular to it. Since the crate does not sink into or rise above the ramp, the net force in the y-direction is zero; so N = w cos θ. Hence fk = µk N = µk w cos θ. Since the crate is moving at constant speed, the net force on it is zero. Hence the net force in the x-direction is zero; so
N wx = 6 w sin θ
fk = µk N w
T
u
θ ? w
-
y
= w cos θ
T = wx + fk = w sin θ + µk w cos θ = w(sin θ + µk cos θ) Problem 1309. A crate whose mass is mc is sliding down a ramp that makes an angle of θ with the horizontal. The coefficient of kinetic friction between the crate and the ramp is µk . What is the magnitude of the crate’s acceleration? (a) g(sin θ + µk cos θ) (c) g(cos θ + µk sin θ) (e) None of these
y
! ! !! LL ! L !! LL mc L ! x ! L!! L !L! ! ! L! !! ! !! !! θ
*(b) g(sin θ − µk cos θ) (d) g(cos θ − µk sin θ)
Solution: Draw a free-body diagram with the x-axis parallel to the ramp and the y-axis perpendicular to it. N 6 Since the crate does not sink into or rise above the ramp, wx = the net force in the y-direction is zero; so N = w cos θ. w sin θ f = µk N u- k Hence fk = µk N = µk w cos θ. The net force on the crate in the downslope x-direction θ is Fx = wx − fk = w sin θ − µk w cos θ. We can use New ? wy = w cos θ w ton’s law, plus the fact that w = mc g, to calculate the acceleration. Fx mc g sin θ − µk mc g cos θ a= = = g(sin θ − µk cos θ) mc mc
443 Problem 1310. Two crates, one full of avocados and the other full of beets, are connected by a light horizontal rope. A second horizontal rope is attached to the crate of avocados and used to pull both crates across the floor at a constant speed of 1.4 m/s. The avocados have a mass of 72 kg; the beets have a mass of 93 kg. The coefficient of kinetic friction between the crates and the floor is 0.21. What is the tension TAB on the rope connecting the two crates? Round your answer to the nearest newton. (See the illustration following the answers.) (a) 148 N (c) 219 N (e) None of these
*(b) (d)
191 N 340 N
B: 93 kg
TAB
A: 72 kg
-
Solution: In this situation, the properties of the crate of avocados have nothing to do with TAB . The exact speed of the two crates is also unimportant; all that matters is that they’re moving at a constant velocity. The horizontal forces acting on the beets are TAB pulling forward, and the friction mB gµ pulling backward. Since the crate is moving at a constant velocity, the net horizontal force must be zero: so TAB = mB gµ = (93 kg)(9.8 m/s2 )(0.21) = 191 N Problem 1311. A block with a mass of 9.7 kg is sliding down a plane at a constant speed of 4.4 m/s. The plane is angled at 22.4◦ to the horizontal. What is the coefficient of kinetic friction µk of the block on the plane? Round your answer to the nearest 0.01. (a) 0.37 *(c) 0.41 (e) None of these
(b) 0.39 (d) 0.43
Solution: Take the x-axis to be down the plane and the y-axis to be the upward normal to the plane. Since the speed is constant, ax = 0 ⇒ Fnet,x = 0 ⇒ the force of friction must match the component of the gravitational force parallel to the plane. This component is: Wx = mg sin θ. The component of the gravitational force normal to the plane is: N = mg cos θ; so the frictional force is: fk = µk N = µk mg cos θ. Equating the two forces gives: mg sin θ = µk mg cos θ
⇒
µk =
sin θ = tan θ = tan 22.4◦ = 0.41 cos θ
The mass of the block and the value of its constant speed are irrelevant to the problem.
444 Problem 1312. A block with a mass of 132 kg is pulled with a horizontal force of F~ across a rough floor. The coefficient of friction between the floor and the block is 0.51. If the block is moving at a constant velocity, what is the magnitude of F~ ? Round your answer to the nearest newton. (a) 67 N (b) 259 N *(c) 660 N (d) 2536 N (e) None of these Solution: Since the block’s velocity is constant, the net force on it is zero. The two forces acting in the horizontal direction are the pull, directed forward with magnitude F ; and the frictional force fk , directed backward with magnitude µmg. Hence: F = fk = µmg = (0.51)(132 kg)(9.8 m/s2 ) = 660 N
Problem 1313. A block with a mass of 84 kg is pulled with a horizontal force of F~ across a rough floor. The coefficient of friction between the floor and the block is 0.64. If the block is moving at a constant velocity, what is the magnitude of F~ ? Round your answer to the nearest newton. (a) 54 N (b) 131 N *(c) 527 N (d) 1286 N (e) None of these Solution: Since the block is moving at constant velocity, the net force on it must be zero. The horizontal forces on it are the pull, directed forward with a magnitude of F ; and the frictional force, directed backward with a magnitude of µmg. Hence F = µmg = (0.64)(84 kg)(9.8 m/s2 ) = 527 N
Problem 1314. You trying to move a crate across a floor by pulling horizontally on a rope. The crate has a mass of 114 kg. The coefficient of static friction between the crate and the floor is 0.76; the coefficient of kinetic friction is 0.51. You gradually increase your pull on the rope until the crate starts to move. What is the crate’s initial acceleration? Round your answer to the nearest 0.01 m/s2 . Solution: To get the crate to start moving, you must exert a horizontal force equal to the crate’s weight times the coefficient of static friction: Fpull = µs mg. Once the crate begins to move, the force pulling backward on it is kinetic friction: Fk = µk mg. At that point, the net force on it is: F = Fpull − Fk = (µs − µk )mg. The acceleration is a=
F = (µs − µk )g = (0.76 − 0.51)9.8 m/s2 = 2.45 m/s2 m
445 Problem 1315. A block has mass m. It is sliding at a constant speed v0 down a plane that makes an angle of θk to the horizontal. If the coefficient of kinetic friction of the block on the plane is µk , which equation gives the value of µk ? 2∆x 1 (a) µk = mg sin θk (b) µk = µs − g cos θs (∆t)2 mg (c) µk = *(d) µk = tan θk sin θk (e)
None of these
Solution: The component of the gravitational force parallel to the surface of the plane is: Fp = mg sin θk . The frictional force is µk mg cos θk . Since the block is sliding at constant velocity, the net force on it is zero; so mg sin θk = µk mg cos θk
⇒
µk =
sin θk = tan θk cos θk
446 11.2.5
Stopping distance problems
Problem 1316. The coefficient of kinetic friction between a set of tires and the road surface is µk . If you are driving on level pavement at a speed of v0 , and you lock up the brakes and skid to a halt, how far do you travel before you stop? v0 v02 (b) *(a) 2µk g µk g v0 µ2k v0 g 2 (c) (d) g2 µ2k (e)
N 6
f k = µk N
w = mg
u
?
None of these
Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Since the road surface is level, N = w = mg, so fk = µk N = µk mg. Use Newton’s law to find your acceleration: a = fk /m = µk g. Then we know v0 , vf = 0, and a = µk g; we want to know the distance ∆x. We have a kinetic equation that we can use in this situation. Remember that the signs of a and of ∆x are opposite; since we want ∆x to be positive, we will use a = −µk g. vf2
=
v02
+ 2a∆x
⇒
vf2 − v02 −v02 v2 = = 0 ∆x = 2a −2µk g 2µk g
447 Problem 1317. The coefficient of kinetic friction between a set of tires and the road surface is µk,dry when the road is dry, and µk,wet when the road is wet, with 2µk,wet = µk,dry . If you are driving on level wet pavement at a speed of vwet and hit the brakes, you will travel a distance of ∆x before you come to a stop. How fast can you drive on dry pavement and be able to stop in the same distance? √ 2 vwet (a) vwet *(b) (c) 2vwet (d) 4vwet (e) None of these
N 6
f k = µk N
w = mg
u
?
Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Since the road surface is level, N = w = mg, so fk = µk N = µk mg. Use Newton’s law to find your acceleration: a = fk /m = µk g. Then we know ∆x, vf = 0, and a = µk g; we want to know the initial speed v0 . We have a kinetic equation that we can use in this situation. Remember that the signs of a and of ∆x are opposite. vf2 = v02 + 2a∆x
⇒
v02 = vf2 − 2a∆x = 2µk g∆x
In going from the wet- to the dry-pavement situation, the only things that change are µk and v0 ; 2g∆x remains the same. Thus we can write:
⇒ ⇒
2 vdry v02 v2 = 2g∆x = = wet µk µk,dry µk,wet µk,dry 2 2µk,wet 2 2 2 vdry = vwet = v = 2vwet µk,wet µk,wet wet √ vdry = 2 vwet
448 Problem 1318. The coefficient of kinetic friction between a set of tires and the road surface is µk,dry when the road is dry, and µk,ice = µk,dry /4 when the road is icy. If you are driving on level dry pavement at a speed of v0 and hit the brakes, you will travel a distance of ∆xdry before you come to a stop. What will your stopping distance be if you are driving at v0 on an icy road? (a) ∆xdry (b) 2∆xdry *(c) 4∆xdry (d) 16∆xdry (e) None of these
N 6
f k = µk N
w = mg
u
?
Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Since the road surface is level, N = w = mg, so fk = µk N = µk mg. Use Newton’s law to find your acceleration: a = fk /m = µk g. Then we know v0 , vf = 0, and a = µk g; we want to know the distance ∆x. We have a kinetic equation that we can use in this situation. Remember that the signs of a and of ∆x are opposite; since we want ∆x to be positive, we will use a = −µk g. vf2
=
v02
+ 2a∆x
⇒
vf2 − v02 −v02 v2 = = 0 ∆x = 2a −2µk g 2µk g
In switching from a dry to an icy road, the only things that change are the coefficient of friction µk and the stopping distance ∆x; v0 and g do not change. Thus we can write v02 = µk,ice ∆xice = µk,dry ∆xdry 2g µk,dry µk,dry = ∆xdry = ∆xdry = 4∆xdry µk,ice µk,dry /4
µk ∆x = ⇒ ∆xice
449
12
Applications of Newton’s Laws to Uniform Circular Motion
12.1
Centripetal force
12.1.1
Conceptual questions: centripetal force
Problem 1319. A satellite is in circular orbit around the Earth. What is the source of the centripetal force? (a) Momentum *(c) Gravity (e) None of these
(b) Kinetic energy (d) There is no centripetal force
Solution: Look for a force directed toward the center of the circle. In this case, it’s gravity. Momentum and kinetic energy aren’t forces at all; and there must be a centripetal force, since the satellite is in circular motion. Problem 1320. A car goes around a banked curve on an icy highway. What is the source of the centripetal force? (a) Kinetic energy *(c) Normal force (e) None of these
(b) The car’s engine (d) There is no centripetal force
Solution: The highway is described as icy, which suggests that there’s no friction. The curve is banked, so there’s a normal force directed toward the inside of the curve. Kinetic energy is not a force at all; and since the car is going around a curve, there must be some centripetal force. Problem 1321. A car goes around an unbanked curve. What is the source of the centripetal force? (a) Momentum *(c) Friction (e) None of these
(b) Normal force (d) There is no centripetal force
Solution: Momentum is not a force at all. Since the curve is unbanked, the normal force points vertically upward, with no component pointing toward the inside of the curve. Friction is the only force that might work; there must be a centripetal force, since the car is going around a curve.
450 Problem 1322. A model airplane is attached to a wire so that it flies in horizontal circles around a central pylon. What is the source of the centripetal force? *(a) (c) (e)
Tension in the wire Inertia None of these
(b) Air resistance (d) There is no centripetal force
Solution: Look for a force directed toward the center of the circle. Tension in the wire is such a force. Air resistance is directed backward against the plane’s motion. Inertia is not a force at all. There must be a centripetal force, since the plane is moving in a circle. Problem 1323. A bullet is fired straight up into the air; it slows down, comes to a momentary halt, then falls straight down again. What is the source of the centripetal force? (a) Inertia (b) Air resistance (c) Normal force *(d) There is no centripetal force (e) None of these Solution: The bullet is moving in a straight line: first straight up, then straight down. Since it is not moving in a curve, there is no centripetal force. Problem 1324. You have left your coffee mug on the flat roof of your car. As you drive around an unbanked curve, what is the source of the centripetal force on the mug? (a) Normal force *(c) Static friction (e) None of these
(b) Kinetic energy (d) There is no centripetal force
Solution: The mug is moving in a curve, so there is a centripetal force. Since the roof and the road are level, the normal force is directed straight upward. Kinetic energy is not a force at all. Static friction is the source of the centripetal force: if there were no static friction, the mug would slide off on the curve. Problem 1325. Your keys are attached to the end of a thin chain. You are holding the end of the chain and twirling the keys in a vertical circle. What is the source of the centripetal force on the keys at the top of the circle? (a) Gravity *(c) Both gravity and tension (e) None of these
(b) Tension in the chain (d) There is no centripetal force
Solution: The keys are moving in a circle, so there is a centripetal force. At the top of the circle, both gravity and the tension in the chain are directed straight downward, toward the center of the circle. Hence the combination of the two produces the centripetal force.
451 Problem 1326. A car is moving down a straight and level road. The driver comes to a stop sign, at which he slows down but does not stop completely, then speeds up again. What is the source of the centripetal force? (a) Kinetic friction (c) Normal force (e) None of these
(b) *(d)
Air resistance There is no centripetal force
Solution: The car is moving in a straight line. Since it is not moving in a curve, there is no centripetal force. Problem 1327. A dead cockroach is lying on its back on the flat blade of a ceiling fan, which is slowly turning in a horizontal circle. What is the source of the centripetal force on the roach? *(a) Static friction (b) Kinetic friction (c) Gravity (d) There is no centripetal force (e) None of these Solution: The late roach is moving in a circle, so there must be a centripetal force. Gravity is pulling straight downward, so has no component toward the center of the circle. There is no kinetic friction, since the roach is not moving with respect to the fan blade on which it is resting. Static friction is the force that is holding it in place as the fan turns, so is the source of the centripetal acceleration. Problem 1328. Which of the following statements is correct? (a) Uniform circular motion causes a constant force toward the center. (b) Uniform circular motion is caused by a constant force toward the center. *(c) Uniform circular motion is caused by a net force toward the center with a constant magnitude. (d) Uniform circular motion is caused by a net force away from the center with a constant magnitude. (e) None of these
452 Problem 1329. Let F~net be the net force on an object moving in a circle (not necessarily with constant speed). Which of the following statements best describes the component of the net force in the radial direction? *(a) The radial component of the net force is the centripetal force. (b) The radial component of the net force is the centrifugal force. (c) The radial component of the net force causes the object to spiral inwards towards the center of the circular path. (d) The radial component of the net force causes the object to spiral outwards away from the center of the circular path. (e) None of these Solution: Recall: The word Centripetal originates from the Greek word for center seeking. By assumption, the object is moving in a circle, so (c) and (d) are wrong. The object is traveling in a circle because the radial component of the net force points towards the center of the circle. The tangential component of the net force will cause the object to accelerate in the θ-direction (i.e., it will cause the object to speed up or slow down, but it will continue to move in a circle). There is no such thing as centrifugal force; it is an apparent force, not a real force (i.e., it doesn’t originate from a push or a pull). The effect of centrifugal force comes from linear inertial. For example, suppose you are a passenger in a car that is initially traveling in a straight line and then makes a hard left. Why are you thrown outward against the car door? There is no magical force that is throwing you outward. The explanation is simple. You were initially traveling in a straight line, and by Newton’s first law (the law of linear inertial) you want to remain moving in a straight line, but the car turned out from under you. You went straight and the car didn’t! How did the car mange to do this? The friction between the tires and the road allowed the car to move in a circular arc. It should also be mentioned that the friction between the tires and the road is the source of the very real centripetal force. Problem 1330. Determine whether the two claims are true or false. Claim 1: Centripetal force is a real force and its existence is brought about when objects move in circles. Claim 2: Centrifugal force is a real force that points radially outwards when objects move in circles. (a) Claim 1 is true; Claim 2 is true (b) Claim 1 is true; Claim 2 is false (c) Claim 1 is false; Claim 2 is true *(d) Claim 1 is false; Claim 2 is false Solution: Claim 1 is false. The centripetal force comes from a real push or pull between two or more objects, it does not just appear magically because you are moving in a circle. To the contrary, it is a real force that causes you to move in a circle. Claim 2 is also false. There is no such thing as a centrifugal force.
453 12.1.2
Horizontal Motion
Problem 1331. A car is driving at a constant speed counter-clockwise around the horizontal track shown at right. At each of the labelled points on the track, sketch a vector representing the net force on the car. Ignore gravity, which points downward into the page. Remember to make the lengths of your vectors roughly proportional to the magnitude of the force: greater forces should be indicated by longer vectors. Solution: Since the car never changes speed, there is no acceleration in the direction tangent to the track. The acceleration vector and thus the force must therefore always be perpendicular to the track. At points A, B, and C, there is a centripetal force keeping the car on the curves. The curves at A and B have about the same radius; the curve at C has a radius about twice as big.
r-
C F~net
Dr = ~0
r
B 6 r
A
Since arad = v 2 /r, and since v is the same at all points, the centripetal force at C should have about half the magnitude that it has at A and B. At all three points, the force vector should point toward the inside of the curve (the concave side of the curve). Note: You don’t need to be fussy about making the vector at C exactly half as long as the vectors at A and B; but it should clearly be shorter on your drawing. The car is moving in a straight line at D, so it is experiencing no acceleration, hence no force. Indicate the zero vector at that point.
454 Problem 1332. A car is travelling at a constant speed clockwise around a horizontal track. Which of the diagrams below best shows the force vectors on the car at two points on the track? 6
(a)
(b)
?
6
*(c)
(d) ?
-
-
Solution: Since the car never changes speed, there is no acceleration in the direction tangent to the track. The acceleration vector and thus the force must always be perpendicular to the track. That doesn’t help us: it doesn’t allow us to rule out any of the four diagrams. At each of the two points shown, the car is on a curve. There must be a centripetal force keeping the car on the curve, and that centripetal force must be directed toward the concave side of the curve. That lets us rule out (b) and (d), which show the force directed toward the convex side of the curves. The magnitude of the centripetal force is Frad = v 2 /r. Since the speed of the car is constant, Frad should be proportional to 1/r. The curve on the left of the picture is wide (large r); the curve on the top of the picture is tight (small r). Hence Frad should be smaller on the left and larger at the top. We see this in (c) but not in (a). Hence the correct answer is (c). Problem 1333. A model gas airplane is flying in a horizontal circle, attached to the central point by a string 5 m long. The airplane has a mass of 1 kg. The string will break if subjected to a tension of more than 100 N. How fast can the airplane fly without breaking the string? Round your answer to the nearest 0.1 m/s. Solution: If the airplane’s speed is v, then the centripetal acceleration necessary to keep it moving in a circle of radius r is: arad =
v2 r
We assume that the tension T in the string is supplying all of the centripetal force; so if the airplane has mass m, then T = marad . We want to find the value of v for which T = 100 N. Thus: r 1/2 mv 2 rT (5 m)(100 N) T = marad = ⇒ v= = = 22.4 m/s r m 1 kg
455 Problem 1334. A circular curve on a highway is designed for traffic moving at 15 m/s. The radius of the unbanked curve is 100 m. What is the minimum coefficient of friction between tires and the highway necessary to keep cars from sliding off the curve? Round your answer to the nearest 0.01. (a) 0.15 *(c) 0.23 (e) None of these
(b) 0.19 (d) 2.25
Solution: In order for a car to remain on the curve, it must experience a centripetal acceleration of arad = v 2 /r. For this acceleration to come from friction, it must be the case that gµs ≥ arad =
v2 r
⇒
µs ≥
v2 (15 m/s)2 = 0.23 = rg (100 m)(9.8 m/s2 )
Problem 1335. You are designing a highway interchange that includes an unbanked circular curve. Assuming that the coefficient of static friction between tires and pavement is µs = 0.51 (pointing in the direction transverse to the motion), and that cars will need to negotiate the curve at 20 m/s, what must the radius of the curve be? Round your answer to the nearest meter. (a) 39 m (c) 204 m (e) None of these
*(b) (d)
80 m 784 m
Solution: For a car with mass m to follow the curve, it must experience a centripetal force of mv 2 /r. The maximum frictional force on such a car is mgµr . Hence mv 2 = mgµr r
⇒
r=
v2 (20 m/s)2 = 80 m = gµr (9.8 m/s2 )(0.51)
456 Problem 1336. A highway curve in northern Minnesota has a radius of 140 m. The curve is banked so that a car travelling at 25 m/s will not skid sideways, even if the curve is coated with a frictionless glaze of ice. Sketch a free-body diagram of a car under these circumstances. Solution: See figure at right. The vector N is the normal force, so it’s perpendicular to the highway surface. The y-component of N must match the weight w, since the car doesn’t move up or down. The x-component of N is an unbalanced force. By Newton’s second law, this force must lead to an acceleration. It is this force that provides the centripetal acceleration arad . Problem 1337. A highway curve in northern Minnesota has a radius of 160 m. The curve is banked so that a car travelling at 25 m/s will not skid sideways, even if the curve is coated with a frictionless glaze of ice. At what angle to the horizontal is the curve banked? Round your answer to the nearest degree. (a) 18◦ (b) 20◦ *(c) 22◦ (d) 24◦ (e) None of these
N AK
Ny = mg 6 A A Aθ A Ar Nx = marad θ ? w = mg
N AK
Ny = mg 6 A A Aθ A Ar Nx = marad θ ? w = mg
Solution: Sketch a free-body diagram. There are two forces acting on the car: the weight w = mg, pulling straight down; and the normal force of the highway N , pushing perpendicular to the road surface. If the banking angle of the highway is θ = θbank , then N decomposes into two components: an x-component Nx = N sin θ; and a y-component Ny = N cos θ. Since the car doesn’t move up or down, the total vertical force must be zero; so Ny = mg. Since Nx is the only horizontal force, it must account for the centripetal acceleration; so if the speed of the car is V and the radius of the curve is R, then Nx = marad = mV 2 /R. Hence mV 2 /R V2 Nx = = Ny mg Rg 2 V (25 m/s)2 −1 −1 = tan = tan = 22◦ 2 Rg (160 m)(9.8 m/s )
tan θ = ⇒ θbank
457 Problem 1338. (Derive Problem) You are a civil engineer working in the wilds of North Dakota. You are designing a highway with a circular curve of radius r. You know that at times, the highway will be glazed with ice so that the coefficient of friction between the highway and a car’s tires is essentially zero. In order for the cars to safely negotiate the icy curve at a speed of v, the road must be banked at an angle of θbank above the horizontal. Find a formula for θbank in terms of r, v, and g. rg (b) tan−1 v 2 rg (a) tan−1 2 v 2 v v −1 −1 *(c) tan (d) tan rg r2 g (e) None of these Solution: The highway must exert two components of force on a car: a vertical force equal to the car’s weight; and the centripetal force, directed horizontally. If there is no friction, then the sum of these two forces must be normal to the surface of the highway. The vertical weight force is mg; the horizontal centripetal force is mv 2 /r. Hence 2 v2 v mv 2 /r −1 = ⇒ θbank = tan tan θbank = mg rg rg Problem 1339. A highway curve in northern Minnesota has a radius of 90 m. The curve is banked so that a car travelling at 15 m/s will not skid sideways, even if the curve is coated with a frictionless glaze of ice. At what angle to the horizontal is the curve banked? Round your answer to the nearest degree. *(a) 14◦ (b) 16◦ ◦ (c) 17 (d) 19◦ (e) None of these
N AK
Ny = mg 6 A A Aθ A Ar Nx = marad θ ? w = mg
Solution: Sketch a free-body diagram. There are two forces acting on the car: the weight w = mg, pulling straight down; and the normal force of the highway N , pushing perpendicular to the road surface. If the banking angle of the highway is θ = θbank , then N decomposes into two components: an x-component Nx = N sin θ; and a y-component Ny = N cos θ. Since the car doesn’t move up or down, the total vertical force must be zero; so Ny = mg. Since Nx is the only horizontal force, it must account for the centripetal acceleration; so if the speed of the car is V and the radius of the curve is R, then Nx = marad = mV 2 /R. Hence Nx mV 2 /R V2 = = Ny mg Rg 2 V (15 m/s)2 −1 −1 = tan = tan = 14◦ 2 Rg (90 m)(9.8 m/s )
tan θ = ⇒ θbank
458 Problem 1340. A circular curve on a highway is designed for traffic moving at 15 m/s. The radius of the curve is 100 m. What is the correct angle of banking for the road? Round your answer to the nearest degree. *(a) 13◦ (c) 9◦ (e) None of these
(b) 11◦ (d) 7◦
Solution: The centripetal acceleration is horizontal, and has magnitude arad = v 2 /r. The acceleration due to gravity is vertical and has magnitude g. The surface of the highway at the correct angle of banking is normal to the sum of these vectors; so 2 v2 v (15 m/s)2 arad −1 −1 = ⇒ θbank = tan = tan = 13◦ tan θbank = 2 g rg rg (100 m)(9.8 m/s )
Problem 1341. (Similarity Problem) A car travelling at a speed of V goes around an unbanked curve with a radius of R. The centripetal force that keeps it on the road is the frictional force F between the road and the tires. A few miles later, the car comes to another curve, this one with radius R/2, and tries to go around the curve at the same speed. What must the frictional force fs be in order to keep the car on the road? √ (a) F/2 (b) F/ 2 √ (c) 2F *(d) 2F (e) None of these Solution: On the first curve, F = M V 2 /R, where M is the mass of the car. On the second curve, 2V 2 V2 =M = 2F . fs = M R/2 R Since this is a similarity problem, we could also solve this problem by finding the parameters that are in common (held fixed) in both experiments. Notice that the mass of the car M and the speed of the car V are the same in both situations (experiments). 2 ·R R 2 F R = M V = fs −−−− −→ fs = 2F . 2 Notice that the result is independent of the mass of the car.
459 Problem 1342. (Similarity Problem) A car with mass M and a pickup with mass 3M both go around an unbanked curve with radius R. Each vehicle is moving at a constant speed V . While the car is going around the curve, it experiences a centripetal force of F . What is the centripetal force fs experienced by the pickup on the curve? √ 3F (a) F (b) *(c) 3F (d) 9F (e) None of these Solution: The two vehicles go around the same curve at the same speed, so they experience the same centripetal acceleration. The centripetal force is the mass times the acceleration. Since the pickup has three times the mass of the car, it experiences three times the centripetal force: fs = 3F . Problem 1343. (Similarity Problem) A car with mass M and a pickup with mass 2M both go around a curve with radius R. Each vehicle is moving at a constant speed V . While the car is going around the curve, it experiences a centripetal force of F . What is the centripetal force Ftruck experienced by the pickup on the curve? (a) F/2 (b) F *(c) 2F (d) 4F (e) None of these Solution: The two vehicles go around the same curve at the same speed, so they experience the same centripetal acceleration. The centripetal force is the mass times the acceleration. Since the pickup has twice the mass of the car, it experiences twice the centripetal force: Ftruck = 2F . Problem 1344. (Similarity Problem) An object with a mass of m is on a frictionless horizontal surface. The object is attached to a horizontal string with length r whose other end is attached to a pivot fixed in the surface; the object is then set to moving in a horizontal circle at a constant speed v1 . Under these circumstances, the tension in the string is T1 . If the speed of the object were doubled to a constant v2 = 2v1 , what would be the tension in the string T2 ? √ 2 T1 (a) T1 (b) (c) 2T1 *(d) 4T1 (e) None of these Solution: The tension in the string is the sole source of centripetal acceleration; so the tension equals mac . When the speed of the object is v1 , the tension is T1 = mv12 /r. When the speed is doubled, it becomes T2 =
mv22 m(2v1 )2 4mv12 = = = 4T1 r r r
460 Problem 1345. (Similarity Problem) An object with a mass of m is on a frictionless horizontal surface. The object is attached to a horizontal string with length r1 whose other end is attached to a pivot fixed in the surface; the object is then set to moving in a horizontal circle at a constant speed v1 . Under these circumstances, the tension in the string is T1 . If the speed of the object is doubled to a constant v2 = 2v1 , and the length of the string is increased to r2 = 4r1 , what will be the tension in the string T2 ? √ 2 T1 *(a) T1 (b) (c) 2T1 (d) 4T1 (e) None of these Solution: The tension in the string is the sole source of centripetal acceleration; so the tension equals mac . When the speed of the object is v1 and the radius is r1 , the tension is T1 = mv12 /r1 . When the speed is doubled and the radius quadrupled, the tension is unchanged: m(2v1 )2 mv12 mv22 = = = T1 . T2 = r2 4r1 r1
461 12.1.3
Vertical Motion
Problem 1346. A block with mass M is whirled on the end of a thin rigid rod that is attached to the axel of a motor so that it moves at a constant speed in a vertical circle with radius R. At the top of the circle, the tension in the rod is twice the weight of the block. What is the speed of the block? √ √ *(b) √3gR (a) √2gR (d) 4 gR (c) 2 gR (e) None of these Solution: At the top of the circle, two forces are pulling straight down on the block: gravity, pulling downward with the weight of the block W = M g; and tension in the rod at the top of the circle Ttop , pulling downward with twice the weight of the block. The sum of these two is the centripetal force: Frad = W + Ttop = 3W = 3M g. Since Frad = M arad , it follows that the centripetal acceleration is 3g. Hence arad = 3g =
v2 R
⇒
v=
p 3gR
Problem 1347. A brass sphere with weight W is attached to the end of a thin rigid rod; the other end of the rod is attached to the axle of a motor, which is set to turning so that the sphere moves in a vertical circle at a constant speed. When the sphere is at the lowest point on the circle, the tension in the rod is 2W . What is the tension in the rod when the sphere is at the top of the circle? *(a) 0 (b) W/2 (c) W (d) 2W (e) None of these Solution: At the bottom of the circle, the weight is pulling downward and the tension is pulling upward; so Frad = Tbottom − W = 2W − W = W
⇒
Tbottom = W + Frad .
Notice that the tension on the rod when the sphere is at the bottom of the circle is the sum of the centripetal force, needed to overcome the sphere’s inertia and keep the sphere moving in a circle, and the gravitational force. This should be the maximum tension in the rod. At the top of the circle, the radial force is the same Frad because the object is travelling with constant speed, but now the weight and the tension are pulling in the same direction. Thus Frad = W = Ttop + W ⇒ Ttop = 0 . Notice that the tension on the rod when the sphere is at the top of the circle is the difference between the centripetal force and the gravitational force. This should be the minimum tension in the rod. Since Ttop < Tbottom , this result agrees with our intuition.
462 Problem 1348. A brass sphere with weight W is attached to the end of a thin rigid rod; the other end of the rod is attached to an axle, which is set to turning so that the sphere moves in a vertical circle at a constant speed. When the sphere is at the lowest point on the circle, the tension in the rod is 3W . What is the tension in the rod when the sphere is at the top of the circle? (a) 0 *(b) W (c) 2W (d) 4W (e) None of these Solution: At the bottom of the circle, the weight is pulling downward and the tension is pulling upward; so Frad = Tbottom − W = 3W − W = 2W
+W
−−−−−−→
Tbottom = W + Frad .
Notice that the tension on the rod when the sphere is at the bottom of the circle is the sum of the centripetal force, needed to overcome the sphere’s inertia and keep the sphere moving in a circle, and the gravitational force. This should be the maximum tension in the rod. At the top of the circle, the radial force is the same Frad because the object is travelling with constant speed, but now the weight and the tension are pulling in the same direction. Thus Frad = 2W = Ttop + W ⇒ Ttop = W . Notice that the tension on the rod when the sphere is at the top of the circle is the difference between the centripetal force and the gravitational force. This should be the minimum tension in the rod. Since Ttop < Tbottom , this result agrees with our intuition.
463 Problem 1349. A pendulum in the physics lab consists of a metal sphere with a mass of 220 g attached to the end of a thin string 104 cm long. The sphere is pulled away from the center with the string taut and then released to swing freely. When the sphere reaches the lowest point in its arc, it is moving at 83 cm/s. At that point, what is the magnitude of the sphere’s acceleration? Round your answer to the nearest 0.01 m/s2 . *(a) 0.66 m/s2 (b) 0.73 m/s2 (c) 0.80 m/s2 (d) 0.88 m/s2 (e) None of these Solution: At the pendulum’s lowest point, there is no acceleration due to gravity, since gravity is pulling directly against the string (see the comments below). Hence the only acceleration to consider is the centripetal acceleration. Not forgetting to convert centimeters to meters, we get arad =
(0.83 m/s)2 v2 = = 0.66 m/s2 r 1.04 m
The mass of the sphere is irrelevant to this problem. Comment (1): When the pendulum is at its lowest point (the bottom of the circular arc), there is no component of gravity in the tangential direction. The tension in the string counters the gravitational force and must also bear the burden of having to change the direction of the mass (the sphere), which possesses inertia. Remember that by Newton’s 1st Law, the sphere wants to travel in a straight line. If the string did not pull on the mass, then it would not go in a circle! Comment (2): If the sphere were brought to rest (no motion), then the sphere would hang straight down and a simple free-body diagram would show that the tension in the string would equal the weight (no motion ⇒ no acceleration ⇒ net force is zero). The only difference between the static situation and the one involving the motion of the pendulum through the lowest point is that a centripetal force is needed to change the pendulum’s direction. This additional force is the centripetal force. Problem 1350. A Ferris wheel has a radius of 79 m. At its maximum speed, it makes one revolution every 38 s. If you are riding the Ferris wheel at this speed, how fast are you moving? Round your answer to the nearest m/s. *(a) 13 m/s (b) 14 m/s (c) 16 m/s (d) 17 m/s (e) None of these Solution: Since you are moving at a constant speed, your average velocity is equal to your instantaneous velocity: v = v. In travelling a complete revolution, you go a distance of one circumference: 2πr. If you travel this distance in a time period Tperiod , your speed is 2πr 2π(79 m) v= = = 13 m/s Tperiod 38 s
464 Problem 1351. A Ferris wheel has a radius of 65 m. At its maximum speed, it makes one revolution every 31 s. A dog is sitting on a spring scale on one of the seats on the wheel. When the wheel is not moving, the scale registers 124 N. When the wheel is turning at its maximum speed, what weight does the scale register at the highest point in the circle? Round your answer to the nearest newton. (a) 81 N *(b) 90 N (c) 99 N (d) 109 N (e) None of these Solution: At the top of the wheel, the dog is experiencing two forces: the force of gravity pulling downward with a magnitude of w = 124 N, and the normal force of the scale pushing upward with a magnitude of Nscale = wapp , which will be the weight registered by the scale. This additional force on an object, resulting from the fact that the object is accelerating, together with the force of gravity on the object (its weight) is known as the object’s apparent weight, which we will denote by wapp and use from this point forward. The sum of these two forces will produce a net centripetal force, which will give rise to a downward-directed centripetal acceleration arad . Hence if the dog’s mass is m, then marad = w − wapp ; so wapp = w − marad . Since we know the dog’s weight on the ground, w = 124 N, we can calculate the mass: w = mg ⇒ m = w/g. Calculating arad requires the dog’s speed on the wheel. If the radius of the wheel is R and the dog makes a complete circle in time T , then the speed is V = 2πR/T . Then the centripetal acceleration is 4π 2 R V2 = arad = R T2 Putting all of these equations together gives us wapp = w − marad warad =w− g arad =w 1− (the elevator equation with y-axis pointing up) g 4π 2 R =w 1− gT 2 4π 2 (65 m) = (124 N) 1 − (9.8 m/s2 )(31 s)2 = 90 N Notice that in the elevator equation, we have ay = −arad , since ~arad always points towards the center of the Ferris wheel.
465 Problem 1352. A Ferris wheel has a radius of 79 m. At its maximum speed, it makes one revolution every 44 s. A dog is sitting on a spring scale on one of the seats on the wheel. When the wheel is not moving, the scale registers 227 N. When the wheel is turning at its maximum speed, what weight does the scale register at the lowest point in the circle? Round your answer to the nearest newton. (a) 193 N (b) 214 N (c) 238 N *(d) 264 N (e) None of these Solution: At the bottom of the wheel, the dog is experiencing two forces: the force of gravity pulling downward with a magnitude of w = 227 N, and the normal force of the scale pushing upward with a magnitude of Nscale = wapp , which will be the apparent weight registered by the scale. The sum of these two forces will produce the centripetal acceleration arad , which is directed upward. Hence if the dog’s mass is m, then marad = wapp − w; so wapp = w + marad . Since we know the dog’s weight on the ground, w = 227 N, we can calculate the mass: w = mg ⇒ m = w/g. Calculating arad requires the dog’s speed on the wheel. If the radius of the wheel is R and the dog makes a complete circle in time T , then the speed is V = 2πR/T . Then the centripetal acceleration is 4π 2 R V2 = arad = R T2 Putting all of these equations together gives us wapp = w + marad warad =w+ g arad =w 1+ (the elevator equation with y-axis pointing up) g 4π 2 R =w 1+ gT 2 4π 2 (79 m) = (227 N) 1 + (9.8 m/s2 )(44 s)2 = 264 N Notice that in the elevator equation we have ay = −arad since ~arad always points towards the center of the Ferris wheel.
466 Problem 1353. A Ferris wheel has a radius of 88 m. A dog is sitting on a spring scale on one of the seats on the wheel. When the wheel is not moving, the scale registers 265 N. When the wheel is turning at its maximum speed, the scale registers zero at the top of the circle. At this speed, how long does it take for the wheel to make one revolution? Round your answer to the nearest second. (a) 14 s (b) 15 s (c) 17 s *(d) 19 s (e) None of these Solution: At the top of the wheel, the dog is experiencing two forces: the force of gravity pulling downward with a magnitude of w = mg = 265 N, and the normal force of the scale pushing upward with a magnitude of Nscale = wapp , which will be the apparent weight registered by the scale. The sum of these two forces will produce the centripetal acceleration arad , which is directed downward. Hence if the dog’s mass is m, then marad = w − wapp . Since the scale registers zero, wapp = 0, and marad = w = mg. We can factor out the dog’s mass (its weight is irrelevant to this problem): arad = g. If the radius of the wheel is R and the time it takes to make one revolution is T , then the dog is going around at a speed of V = 2πR/T . Then g = arad = ·
T2 g
−−−−−−→ √
−−−−−−→ evaluate
−−−−−−−−→
4π 2 R g s R T = 2π g s
4π 2 R V2 = R T2
T2 =
T = 2π
88 m = 19 s 9.8 m/s2
467 Problem 1354. A Ferris wheel has a radius of 81 m. At its maximum speed, it makes one revolution every 56 s. A dog is sitting on a spring scale on one of the seats on the wheel. When the wheel is not moving, the scale registers 119 N. When the wheel is turning at its maximum speed, what weight does the scale register at the lowest point in the circle? Round your answer to the nearest newton. (a) 96 N (b) 107 N *(c) 131 N (d) 145 N (e) None of these Solution: At the bottom of the wheel, the dog is experiencing two forces: the force of gravity pulling downward with a magnitude of w = 119 N, and the normal force of the scale pushing upward with a magnitude of Nscale = wapp , which will be the apparent weight registered by the scale. The sum of these two forces will produce the centripetal acceleration arad , which is directed upward. Hence if the dog’s mass is m, then marad = wapp − w; so wapp = w + marad . Since we know the dog’s weight on the ground, w = 119 N, we can calculate the mass: w = mg ⇒ m = w/g. Calculating arad requires the dog’s speed on the wheel. If the radius of the wheel is R and the dog makes a complete circle in time T , then the speed is V = 2πR/T . Then the centripetal acceleration is 4π 2 R V2 = arad = R T2 Putting all of these equations together gives us wapp = w + marad warad =w+ g arad =w 1+ (the elevator equation with y-axis pointing up) g 4π 2 R =w 1+ gT 2 4π 2 (81 m) = (119 N) 1 + (9.8 m/s2 )(56 s)2 = 131 N Notice that in the elevator equation we have ay = −arad , since ~arad always points towards the center of the Ferris wheel.
468 Problem 1355. A Ferris wheel has a radius of 57 m. At its maximum speed, the seats are moving at 12.2 m/s. A dog is sitting on a spring scale on one of the seats on the wheel. When the wheel is not moving, the scale registers 303 N. When the wheel is turning at its maximum speed, what weight does the scale register at the lowest point in the circle? Round your answer to the nearest newton. (a) 345 N *(b) 384 N (c) 422 N (d) 464 N (e) None of these Solution: At the bottom of the wheel, the dog is experiencing two forces: the force of gravity pulling downward with a magnitude of w = 303 N, and the normal force of the scale pushing upward with a magnitude of Nscale = wapp , which will be the apparent weight registered by the scale. The sum of these two forces will produce the centripetal acceleration arad , which is directed upward. Hence if the dog’s mass is m, then marad = wapp − w; so wapp = w + marad . Since we know that the dog’s true weight is w = 303 N, we can calculate the mass: w = mg ⇒ m = w/g. We know V = 12.2 m/s and R = 57 m for the Ferris wheel, so we can calculate the centripetal acceleration: arad = V 2 /R. Putting all of these equations together gives us wapp = w + marad warad =w+ g arad =w 1+ g V2 =w 1+ gR = (303 N) 1 +
(the elevator equation with y-axis pointing up)
(12.2 m/s)2 (9.8 m/s2 )(57 m)
= 384 N Notice that in the elevator equation we have ay = −arad , since ~arad always points towards the center of the Ferris wheel.
469 Problem 1356. An airplane pilot is flying downward. He wants to pull out of the dive and start flying upward again by making a vertical semicircle. At the bottom of the semicircle, his speed will be 125 m/s. If he experiences more than 8.2 g’s, he will lose consciousness and crash the plane. What is the minimum radius of the semicircle? Round your answer to two significant figures. (a) 200 m *(b) 220 m (c) 2100 m (d) 2200 m (e) None of these Solution: If the pilot’s centripetal acceleration at the bottom of the circle is arad , then the apparent gravity, denoted gapp , he will experience is gapp = arad + g. We want to find the radius of the semicircle such that arad + g = 8.2g, so arad = 7.2g. If V is his speed and R is the radius of the semicircle, then arad =
V2 = 7.2g R
⇒
R=
V2 (125 m/s)2 = = 220 m 7.2g (7.2)(9.8 m/s2 )
Comment: The apparent weight and apparent gravity for a mass m are related by the equation wapp = mgapp . To derive this relationship, we start by finding the centripetal force. The centripetal force is Frad = N − w, where N is the normal force of the pilot’s seat. If the pilot were sitting on a scale, then the reading on the scale would be N , usually denoted Nscale , and Nscale = wapp . From the equation Frad = N − w we get wapp = w + Frad = mg + marad = m(g + arad ) = mgapp . Problem 1357. A skateboard park includes a U-shaped half-pipe with a semicircular cross section whose radius is 3.4 m. A skater who starts at the top will be moving at 7.4 m/s at the lowest point. If a skater weighs 160 lb, what is his apparent weight at the lowest point? Round your answer to the nearest 10 lb. (a) 260 lb (b) 330 lb *(c) 420 lb (d) 540 lb (e) None of these Solution: At the lowest point on the circle, the apparent weight of the skateboarder will be wapp = w + marad , where m is the mass of the skateboarder. We don’t know the mass of the skater, and it would be tedious to calculate it in kg from the weight in pounds. Happily, we don’t need to do so. The true weight is w = mg; so m = w/g. Thus arad w(g + arad ) =w 1+ (The elevator equation) wapp = mgapp = m(g + arad ) = g g Now we need to calculate arad . We know the radius of the circle R and the speed of the skater V . Hence arad = V 2 /R. Then the apparent weight is V2 (7.4 m/s)2 wapp = w 1 + = (160 lb) 1 + = 420 lb Rg (3.4 m)(9.8 m/s2 )
470 Problem 1358. You have tied a short length of rope to the handle of a bucket full of water, and are swinging the bucket in a vertical circle with radius 83 cm. What is the minimum speed that the bucket must be moving in order for no water to spill out at the top of the circle? Round your answer to two significant figures. (a) 2.6 m/s *(b) 2.9 m/s (c) 26 m/s (d) 29 m/s (e) None of these Solution: When the bucket is moving just fast enough to keep the water in at the top of the circle, then arad = g. To see this, notice that at the top of the circular arc Frad = Ttop + w, where Ttop is the tension in the rope at the top of the circle, and w is the weight of the bucket of water. Now, since the speed of the bucket V is related to the centripetal acceleration via the equation arad = V 2 /R, and since we are assuming that the radius R is fixed, it follows that if we want to minimize V , then we must minimize arad . But to minimize arad we must minimize Frad , since the mass is assumed to be a fixed constant. Finally, to minimize Frad = Ttop + w, we need to minimize the tension in the rope, since the weight of the water is a fixed constant. Since tension is always greater than or equal to zero, it follows that to minimize V , we should take the tension at the top of the rope to be zero. That is, to minimize V = Vmin , we must set Ttop = 0. Hence if the minimum speed of the bucket is Vmin and the radius of the circle is R (and if we don’t forget to convert R from centimeters to meters), then 2 Vmin =g R
⇒
Vmin
q p = Rg = (0.83 m)(9.8 m/s2 ) = 2.9 m/s
Problem 1359. A block with a mass of M slides down a frictionless loop-the-loop track. The loop has a radius of R. At the top of the loop, the block is moving at a speed of V . What is the net force acting on the block at that point? Solution: The net force is the centripetal force: the centripetal acceleration times the mass of the block. MV 2 Fnet = M arad = R
471 Problem 1360. A sack of potatoes has a mass of 4.4 kg. It is hanging from a spring scale that is hanging from the ceiling of a car. The car is driven around a vertical loopthe-loop track with a radius of 7.9 m. At the top of the track, the car is going at a speed of 9.6 m/s. At this point, what apparent weight does the scale show for the potatoes? Round your answer to the nearest 0.1 N. *(a) 8.2 N (b) 9.0 N (c) 9.9 N (d) 10.9 N (e) None of these Solution: At the top of the circle, the potatoes are experiencing two forces: gravity, pulling downward with a force of w = mg, where m = 4.4 kg is the mass of the potatoes; and the tension in the scale Tscale = wapp , the apparent weight registered by the scale. Since the car is upside down at the top of the loop, with the potatoes “hanging” upward from the ceiling, Tscale is directed downward. The sum of these two forces will produce the centripetal acceleration arad , which is directed downward toward the center of the loop. Hence mg + wapp = marad ; so wapp = marad − mg = m(arad − g) We know the car’s speed V = 9.6 m/s and the loop’s radius R = 7.9 m. Thus we can calculate arad = V 2 /R. Substituting that into the equation for wapp gives us 2 (9.6 m/s)2 V 2 − g = (4.4 kg) − 9.8 m/s = 8.2 N wapp = m(arad − g) = m R 7.9 m
472 Problem 1361. The college’s Wellness Committee has decreed that overweight faculty members will not receive pay raises. To get his raise, a certain physics instructor must get his weight from 200 lbs down to 180 lbs. Rather than giving up pizza and beer, he decides to employ trickery. He stands on a spring scale in the back of a moving van and has it driven around a vertical loop-the-loop track with a radius of 8.7 m. At the top of the loop, he has a friend take a picture with the scale showing an apparent weight of 180 lbs. At what speed does the van need to be going at the top of the loop? Round your answer to the nearest 0.1 m/s. *(a) 12.7 m/s (b) 14.0 m/s (c) 15.4 m/s (d) 16.9 m/s (e) None of these Solution: At the top of the loop, the pudgy instructor will experience two forces: gravity, pulling downward with a force of W = mg = 200 lbs, where m is his mass; and the normal force of the scale, pushing downward with a magnitude Nscale = Wapp = 180 lbs, the apparent weight. (Since the van is upside down at the top of the loop, the scale is pushing the instructor downward.) The sum of these two forces produces the centripetal acceleration arad , which is directed downward toward the center of the loop. Hence W + Wapp = marad . It would be nice if we could avoid having to calculate m. Happily, we can. Notice that m = W/g. Substituting this expression into the equation W + Wapp = marad gives W arad = W + Wapp g g ·W (W + Wapp )g Wapp −−−−−→ arad = = 1+ g W W Using our given information, we can now compute arad . 180 lbs Wapp = = 0.9 W 200 lbs
⇒
arad = 1.9g .
Now that we know arad we can use the formula for centripetal acceleration. We know the radius R = 8.7 m of the loop; we want to know the speed v. Setting the two expressions for arad equal to one another will give us v v2 Wapp = arad = 1 + g R W Wapp ·R 2 −−−−−→ v = 1 + gR W s √ Wapp −−−−−→ v = 1+ gR W 1/2 evaluate −−−−−−−−→v = (1.9)(9.8 m/s2 )(8.7 m) = 12.7 m/s
473
12.2
Gravitation and circular orbits
12.2.1
Gravation
Problem 1362. A wrench has a mass of 1.43 kg. It has been left outside of a spaceship at a distance of 1.61 × 107 m from the center of a planet whose mass is 5.98 × 1024 kg. What is the gravitational force on the wrench? Round your answer to the nearest 0.01 N. (a) 1.78 N (b) 1.98 N *(c) 2.20 N (d) 2.42 N Solution: Fg =
Use the formula for gravitational force: (6.674 × 10−11 N m2 /kg2 )(1.43 kg)(5.98 × 1024 kg) Gm1 m2 = = 2.20 N r2 (1.61 × 107 m)2
Problem 1363. A very small asteroid with a mass of 32.5 kg is at a distance of 7.71×106 m from the center of a planet whose mass is 2.12 × 1021 kg. What is the gravitational force on the asteroid? Round your answer to three significant figures. (a) 6.96 × 10−2 N *(b) 7.74 × 10−2 N −2 (c) 8.51 × 10 N (d) 9.36 × 10−2 N Solution: Fg =
Use the formula for gravitational force:
Gm1 m2 (6.674 × 10−11 N m2 /kg2 )(32.5 kg)(2.12 × 1021 kg) = = 7.74 × 10−2 N 2 6 2 r (7.71 × 10 m)
Problem 1364. An astronaut with a mass of 73.3 kg is standing on the surface of a planet whose radius is 1.15 × 106 m and whose mass is 1.31 × 1022 kg. What is the astronaut’s weight on the planet? Round your answer to the nearest 0.1 N. (a) 39.3 N (b) 43.6 N *(c) 48.5 N (d) 53.3 N Solution: The weight of an object is the gravitational force acting on it. We can therefore use the formula for gravitational force: Fg =
(6.674 × 10−11 N m2 /kg2 )(73.3 kg)(1.31 × 1022 kg) Gm1 m2 = = 48.5 N r2 (1.15 × 106 m)2
474 Problem 1365. An unmanned space probe has a mass of 129 kg. It is sent to the surface of a planet with a radius of 2.48 × 107 m and a mass of 1.02 × 1026 kg. What is the probe’s weight on the planet? Round your answer to the nearest 10 N. (a) 1040 N (b) 1160 N (c) 1290 N *(d) 1430 N Solution: The weight of an object is the gravitational force acting on it. We can therefore use the formula for gravitational force: Fg =
Gm1 m2 (6.674 × 10−11 N m2 /kg2 )(129 kg)(1.02 × 1026 kg) = = 1430 N r2 (2.48 × 107 m)2
Problem 1366. A space probe has a mass of 31.2 kg. It is sent to the surface of Planet X, whose radius is 2.11 × 106 m. On the planet’s surface, the probe’s weight is 109 N. What is the mass of Planet X? Round your answer to three significant figures. (a) 1.70 × 1023 kg (b) 1.89 × 1023 kg 23 (c) 2.10 × 10 kg *(d) 2.33 × 1023 kg Solution: The weight of an object is the force of gravity on it. We know the gravitational force, the mass of the probe mp , and the radius of the planet; we want to know the mass of the planet mX . We can use the formula for gravitational force and solve for mX . wr2 Gmp mX ·(r2 /Gmp ) − − − − − − → m = X r2 Gmp 6 2 (109 N)(2.11 × 10 m) = 2.33 × 1023 kg mX = (6.674 × 10−11 N m2 /kg2 )(31.2 kg) Fg = w =
Problem 1367. An astronaut has a mass of 67.3 kg. On the surface of Planet X, his weight is 418 N. The radius of Planet X is 4.87 × 106 m. What is the mass of Planet X? Round your answer to three significant figures. (a) 1.99 × 1024 kg *(b) 2.21 × 1024 kg (c) 2.43 × 1024 kg (d) 2.67 × 1024 kg Solution: The weight of an object is the force of gravity on it. We know the gravitational force, the mass of the astronaut ma , and the radius of the planet; we want to know the mass of the planet mX . We can use the formula for gravitational force and solve for mX . Gma mX wr2 ·(r2 /Gma ) Fg = w = −−−−−−→ mX = r2 Gma (418 N)(4.87 × 106 m)2 mX = = 2.21 × 1024 kg 2 2 −11 (6.674 × 10 N m /kg )(67.3 kg)
475 12.2.2
Orbits
Problem 1368. You want to calculate the period of a satellite in a circular orbit around a planet. Which of the following do you not need to know for your calculation? (a) The mass of the planet (c) The radius of the orbit (e) None of these Solution:
*(b) (d)
The radius of the planet The gravitational constant G
The formula for orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the planet). The radius of the planet does not matter. Problem 1369. Astronomers discover a small planet in a circular orbit around a distant star. They are able to determine the distance of the planet from the star, and the period of the planet’s orbit. Which of the following can they calculate based on these? (a) The mass of the planet *(c) The mass of the star (e) None of these Solution:
(b) (d)
The radius of the planet The radius of the star
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the star). We know T , r, and G; the only other variable in the formula is M . We can solve the formula for M and calculate it. The mass of the planet and the radius of the planet and the star do not come into the formula at all.
476 Problem 1370. A satellite in circular orbit around the Earth has a period of 90 minutes. A second satellite in circular orbit around Planet X has a period of 180 minutes. If the mass of the Earth is ME and the mass of Planet X is MX , which planet’s mass is greater? (a) ME < MX (c) ME > MX (e) None of these Solution:
*(b) (d)
Not enough information: need radii of orbits Not enough information: need masses of satellites
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary. We can solve this for M : 2πr3/2 T =√ GM ( )2
−−→ ·M/T 2
−−−−→
T2 =
4π 2 r3 GM
M=
4π 2 r3 GT 2
We see that M depends on r as well as on T . Since we don’t know anything about r, we can’t conclude anything about M . Problem 1371. A satellite with a mass of 24 kg in circular orbit at a distance of 10,000 km from the center of Planet X has a period of 2 hours. Which of the following satellites circling Planet X would also have a period of 2 hours? (a) A satellite with *(b) A satellite with (c) A satellite with (d) A satellite with (e) None of these Solution:
a a a a
mass mass mass mass
of of of of
24 kg at a distance of 20,000 km 96 kg at a distance of 10,000 km 3 kg at a distance of 20,000 km 3 kg at a distance of 5,000 km
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, of Planet X). Hence T depends on r and M alone. Since both satellites are circling the same planet, M is the same; so if r is the same, then T is the same as well. The mass of the satellite doesn’t enter into the formula at all.
477 Problem 1372. Russell’s celestial teapot was observed to be in a circular orbit around the Earth at a distance of 1.48 × 107 m from the center of the planet. The mass of the Earth is 5.97 × 1024 kg. What is the teapot’s speed? Round your answer to the nearest 10 m/s. (a) 3780 m/s (b) 4200 m/s (c) 4670 m/s *(d) 5190 m/s Solution: For an object in circular orbit, gravitation provides the centripetal force. Thus we can use the formula for gravitational force and the one for centripetal force: r √ GmE GmE mt v 2 Gmt mE ·(r/mt ) 2 − − − − → v = − − → v = = Fg = Frad = r2 r r r 1/2 (6.674 × 10−11 N m2 /kg2 )(5.97 × 1024 kg) v= = 5190 m/s 1.48 × 107 m Problem 1373. An asteroid is in circular orbit around the sun, at a distance of 1.50×1011 m from the sun’s center. The mass of the sun is 1.99 × 1030 kg. What is the asteroid’s speed? Round your answer to three significant figures. (a) 2.41 × 104 m/s (b) 2.68 × 104 m/s 4 *(c) 2.98 × 10 m/s (d) 3.27 × 104 m/s Solution: For an object in circular orbit, gravitation provides the centripetal force. Thus we can use the formula for gravitational force and the one for centripetal force: r √ Gma ms ma v 2 Gm Gms ·(r/ma ) s Fg = Frad = = −−−−→ v 2 = −−→ v = 2 r r r r 1/2 −11 2 2 30 (6.674 × 10 N m /kg )(1.99 × 10 kg) v= = 2.98 × 104 m/s 1.50 × 1011 m Problem 1374. A small moon is in circular orbit around Planet X, at a distance of 1.68 × 107 m from the planet’s center. The moon’s orbital speed is 1120 m/s. What is the mass of Planet X? Round your answer to three significant figures. (a) 2.84 × 1023 kg *(b) 3.16 × 1023 kg 23 (c) 3.47 × 10 kg (d) 3.82 × 1023 kg Solution: For an object in circular orbit, gravitation provides the centripetal force. Thus we can use the formula for gravitational force and the one for centripetal force: mm v 2 Gmm mX v2r ·(r2 /Gmm ) = = − − − − − − → m = X r2 r G 2 7 (1120 m/s) (1.68 × 10 m) 23 mX = kg 2 2 = 3.16 × 10 −11 (6.674 × 10 N m /kg )
Fg = Frad =
478 Problem 1375. A small planet is in circular orbit around a star, at a distance of 3.78 × 1011 m from the star’s center. The orbital speed of the planet is 3.44 × 104 m/s. What is the star’s mass? Round your answer to three significant figures. (a) 5.43 × 1030 kg (b) 6.03 × 1030 kg 30 *(c) 6.70 × 10 kg (d) 7.37 × 1030 kg Solution: For an object in circular orbit, gravitation provides the centripetal force. Thus we can use the formula for gravitational force and the one for centripetal force: v2r Gmp ms mp v 2 ·(r2 /Gmp ) = − − − − − − → m = = s r2 r G (3.44 × 104 m/s)2 (3.78 × 1011 m) = 6.70 × 1030 kg ms = (6.674 × 10−11 N m2 /kg2 ) Fg = Frad =
479 Problem 1376. A satellite has a mass of 6 kg. It is in circular orbit around a planet with a mass of 6 × 1024 kg. The radius of the planet is 106 m; the radius of the orbit is 108 m. What is the satellite’s orbital period? For ease of calculation, assume that G = 23 × 10−10 N m2 /kg2 . (a) π × 102 s (c) π × 1014 s (e) None of these Solution:
*(b) (d)
π × 105 s π × 1017 s
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the planet). The mass of the satellite and the radius of the planet don’t enter into the formula at all, and have nothing to do with this problem. Hence: 2π(108 m)3/2 2πr3/2 = T =√ 1/2 GM ( 23 × 10−10 N m2 /kg2 )(6 × 1024 kg) We’ll look at the numbers first, then check the units and make sure that they work. T =
2π × 1012 2π × 1012 = = π × 105 [4 × 1014 ]1/2 2 × 107
Now, we need to check the units. In the numerator, the units are m3/2 . In the denominator, remember that 1 N = 1 kg m/s2 . Hence the units of the denominator are:
N m2 kg · 1 kg2
1/2
kg2 m3 = kg2 s2
1/2 =
m3/2 s
Thus the units of the whole formula (numerator over denominator) are: m3/2 =s m3/2 /s Our solution is: T = π × 105 s.
480 Problem 1377. A satellite with a mass of 150 kg is in circular orbit around a planet with a mass of 1.5 × 1026 kg. The radius of the planet is 108 m; the radius of the orbit is 1010 m. What is the satellite’s orbital period? For ease of calculation, assume that G = 23 × 10−10 N m2 /kg2 . (a) 2π × 104 s (c) 2π × 106 s (e) None of these Solution:
(b) 2π × 105 s *(d) 2π × 107 s
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the planet). The mass of the satellite and the radius of the planet don’t enter into the formula at all, and have nothing to do with this problem. Hence: 2π(1010 m)3/2 2πr3/2 = T =√ 1/2 GM ( 23 × 10−10 N m2 /kg2 )(1.5 × 1026 kg) We’ll look at the numbers first, then check the units and make sure that they work. T =
2π × 1015 2π × 1015 = = 2π × 107 [1 × 1016 ]1/2 108
Now, we need to check the units. In the numerator, the units are m3/2 . In the denominator, remember that 1 N = 1 kg m/s2 . Hence the units of the denominator are:
N m2 kg · 1 kg2
1/2
kg2 m3 = kg2 s2
1/2 =
m3/2 s
Thus the units of the whole formula (numerator over denominator) are: m3/2 =s m3/2 /s Our solution is: T = 2π × 107 s.
481 Problem 1378. A star has a mass of 2.4 × 1031 kg and a radius of 109 m. A planet in circular orbit around the star has a mass of 2.4 × 1025 kg and a radius of 8 × 106 m. The radius of the planet’s orbit is 4 × 1012 m. What is the planet’s orbital period? For ease of calculation, assume that G = 32 × 10−10 N m2 /kg2 . *(a) (c) (e)
4π × 108 s 4π × 1010 s None of these
Solution:
(b) 4π × 109 s (d) 4π × 1011 s
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the star). The mass of the planet and the radii of the planet and of the star don’t enter into the formula at all, and have nothing to do with this problem. Hence: 2πr3/2 2π(4 × 1012 m)3/2 T =√ = 1/2 GM ( 23 × 10−10 N m2 /kg2 )(2.4 × 1031 kg) We’ll look at the numbers first, then check the units and make sure that they work. T =
2π(8 × 1018 ) 16π × 1018 16π × 1018 = = = 4π × 108 [1.6 × 1021 ]1/2 [16 × 1020 ]1/2 4 × 1010
Now, we need to check the units. In the numerator, the units are m3/2 . In the denominator, remember that 1 N = 1 kg m/s2 . Hence the units of the denominator are:
N m2 kg · 1 kg2
1/2
kg2 m3 = kg2 s2
1/2 =
m3/2 s
Thus the units of the whole formula (numerator over denominator) are: m3/2 =s m3/2 /s Our solution is: T = 4π × 108 s.
482 Problem 1379. (Similarity Problem) A satellite with mass m1 is in circular orbit at a distance of r0 from the center of a planet with mass Mp . A second satellite with mass m2 = 2m1 is in circular orbit around the same planet, at the same distance from the center. If the period of the first satellite is T1 , what is the period of the second satellite? (a) T1 /4 *(c) T1 (e) None of these
(b) T √1 /2 2 T1 (d)
Solution: The orbital period only depends on the orbital radius r0 and the mass of the primary (in this case, the planet) Mp . The mass of the satellite does not enter into the calculation. Since both satellites orbit the same planet at the same distance, they have the same period. Problem 1380. (Similarity Problem) A satellite with mass m1 is in circular orbit around the Earth at a distance of r1 from the center of the planet; it has a period of T1 . Scientists would like to launch a second satellite into circular orbit with a period of T2 = 2T1 . What should they do to give the second satellite this period? (a) Orbit the second satellite at a distance of r2 = 64r1 *(b) Orbit the second satellite at a distance of r2 = 41/3 · r1 (c) Give the second satellite a mass of m2 = 4m1 (d) Give the second satellite a mass of m2 = m1 /4 (e) None of these Solution:
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, the Earth). The mass of the satellite does not appear in the formula, so we can rule out answers (c) and (d). M and G are the same from one experiment to the other; so the only things that change are r and T . Separating the changing from the unchanging variables, we get: 2πr3/2 T =√ GM ⇒
T2 =
4π 2 r3 GM
T2 T12 T22 (2T1 )2 4T12 = = = = r3 r13 r23 r23 r23
·r3 r3 /T 2
1 −−1−2−−→ r23 = 4r13
( )1/3
( )2
−−→
−−−→ r2 = 41/3 r1
÷r3
−−→
T2 4π 2 = r3 GM
483 Problem 1381. (Similarity Problem) Planet X has mass MX . A satellite with mass mX is in circular orbit around it at a distance of rX from the center of the planet. Planet Y has mass MY ; a satellite with mass mY is in circular orbit around it at a distance of rY = 2rX from the planet’s center. If the two satellites have the same period, which of the following is true? (a) MY = MX /8 (c) mY = mX /8 (e) None of these Solution:
*(b) MY = 8MX (d) mY = 8mx
The formula for the orbital period is 2πr3/2 T =√ GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the primary (in this case, Planet X). The mass of the satellite does not appear in the formula, and has no bearing on the problem; thus we can rule out answers (c) and (d). In the two experiments, M and r are different; G and T are the same. Separating the changing from the unchanging variables, we get: 2πr3/2 T =√ GM ⇒ ·MX MY /r3
( )2
−−→
T2 =
4π 2 r3 GM
3 3 r3 rX rY3 (2rX )3 8rX = = = = M MX MY MY MY
X −−−−−−−→ MY = 8MX
·G/4π 2
−−−−→
T 2G r3 = 4π 2 M
484
13
13.1
Applications of Newton’s Laws to Rotating Bodies Torque
Problem 1382. A diving board extends 2 m horizontally from the side of a pool. A swimmer weighing 600 N stands on the very end of the board. What torque does the swimmer exert about the base of the board? Round your answer to two significant figures. (a) 300 N·m *(c) 1200 N·m (e) None of these
(b) 600 N·m (d) 2400 N·m
Solution: Since the board is horizontal and gravity is vertical, the angle θ between the board and the force on it is 90◦ . Hence τ = F r sin θ = F r sin 90◦ = F r = (600 N)(2 m) = 1200 N·m Problem 1383. A fisherman is holding a fishing rod horizontally over the edge of a bridge. The rod is 150 cm long; from the end of it hangs a catfish with a mass of 6.0 kg. What torque does the fish exert about the base of the road? Round your answer to two significant figures. (a) 0.92 N·m *(c) 88 N·m (e) None of these
(b) 9.0 N·m (d) 900 N·m
Solution: Don’t forget to convert centimeters to meters, and to convert the catfish’s mass to its weight. The rod is horizontal and the fish’s weight pulls vertically, so the angle θ between the rod and the force is 90◦ . Hence τ = F r sin θ = mgr sin 90◦ = mgr = (6.0 kg)(9.8 m/s2 )(1.50 m) = 88 N·m Problem rusty, and to pull on significant
1384. A cord wraps around a pulley with a radius of 8 cm. The pulley is requires a torque of 120 N·m to make it turn. How much force do you have the cord with in order to make the pulley turn? Round your answer to two figures.
(a) 15 N *(c) 1500 N (e) None of these
(b) 96 N (d) 9600 N
Solution: When we’re applying force to a cord or a belt wrapped around a wheel, we can assume that the force is perpendicular to the radius; so τ = F r sin 90◦ = F r. We know τ and r; we want to know F . Don’t forget to convert centimeters to meters. τ = Fr
⇒
F =
120 N·m τ = = 1500 N r 0.08 m
485 Problem 1385. On a construction site, a plank is fastened at one end and projects 2.0 m horizontally from the building. The fastening will break if subjected to a torque of more than 1500 N·m. What is the mass of the heaviest piece of equipment that can safely be placed on the end of the plank? Round your answer to two significant figures. *(a) (c) (e)
77 kg 750 kg None of these
(b) 310 kg (d) 3000 kg
Solution: The plank is horizontal and the force of weight is vertical; so τ = ◦ F r sin 90 = F r. We know r and τ ; we want to know m, for which we have to use the fact that F = mg. ⇒
τ = F r = mgr
m=
1500 Nm τ = 77 kg = gr (9.8 m/s2 )(2.0 m)
Problem 1386. You are out in the middle of the desert with a flat tire. Your lug nuts are rusty, and require a torque of 300 N·m to loosen them. You are strong enough to exert a force of 500 N on the end of a wrench. How long does the wrench have to be in order for you to get the lug nuts off? Round your answer to two significant figures. (a) 15 cm *(c) 60 cm (e) None of these
(b) 17 cm (d) 67 cm
Solution: Since no angle is mentioned, assume that you can apply the force at 90◦ to the wrench handle; so τ = F r. We know τ and F ; we want to know r. τ = Fr
⇒
r=
300 N·m τ = = 0.6 m = 60 cm F 500 N
Problem 1387. An action-movie hero who weighs 70 kg is hanging from a flagpole that projects horizontally from the side of a building. The flagpole will break off if subjected to a torque of more than 1200 N·m. How far out on the flagpole can the hero go? Round your answer to two significant figures. (a) 0.84 m (c) 1.4 m (e) None of these
(b) 1.1 m *(d) 1.7 m
Solution: Since the flagpole is horizontal and the hero’s weight is vertical, θ = 90◦ and τ = F r. We know τ and m (which we can convert to F using F = mg); we want to know r. τ 1200 N·m τ = F r = mgr ⇒ r = = = 1.7 m mg (70 kg)(9.8 m/s2 )
486 Problem 1388. You are trying to loosen a rusty nut with a wrench 30 cm long. Because the nut is in an awkward place, you can only pull at an angle of 70◦ to the wrench handle. If you can pull with a force of 600 N, how much torque do you exert on the nut? Round your answer to two significant figures. (a) 62 N·m *(c) 170 N·m (e) None of these Solution:
(b) 100 N·m (d) 280 N·m
Don’t forget to convert centimeters to meters. τ = F r sin θ = (600 N)(0.30 m) sin 70◦ = 170 N·m
Problem 1389. A fisherman has a bass on the end of his line. His rod is 1.2 m long; the line makes an angle of 50◦ with the rod; and the bass is pulling on the line with a force of 40 N. How much torque does the bass produce about the handle of the rod? Round your answer to two significant figures. (a) 31 N·m (c) 35 N·m (e) None of these Solution:
(b) *(d)
33 N·m 37 N·m
τ = F r sin θ = (40 N)(1.2 m) sin 50◦ = 37 Nm
Problem 1390. A telephone pole is supported by a guy wire that is attached 3.0 m up on the pole and makes an angle of 60◦ with the ground. If the tension in the wire is 6000 N, how much torque does it exert about the base of the pole? Round your answer to two significant figures. *(a) 9000 N·m (c) 13,000 N·m (e) None of these
(b) 11,000 N·m (d) 16,000 N·m
◦
r
30 J ) J JF J 60◦JJ
Solution: Careful! The θ in the formula τ = F r sin θ is the angle between the force and the radius. In this case, the radius is vertical (from the base of the pole to the point where the wire joins the pole), so θ = 90◦ − 60◦ = 30◦ . τ = F r sin θ = (6000 N)(3.0 m) sin 30◦ = 9000 N·m
487 Problem 1391. A fisherman is holding a rod 120 cm long at an angle of 55◦ above the horizontal. A carp weighing 60 N is hanging from the end of the rod. How much torque does the fish exert about the handle of the rod? Round your answer to two significant figures. (a) 29 N·m (c) 59 N·m (e) None of these
*(b) (d)
41 N·m 72 N·m
θ = 35◦
r
? ◦ F 55
Solution: Careful! The rod’s angle is 55◦ above the horizontal; the force of the carp’s weight is vertical. Thus the angle θ between the rod and the force is 90◦ − 55◦ = 35◦ . It’s a good idea to sketch a diagram, as at right. Also, don’t forget to convert centimeters to meters. τ = F r sin θ = (60 N)(1.2 m) sin 35◦ = 41 N·m F Problem 1392. A flagpole with a length of 40 m has been installed carelessly, and makes an angle of 12◦ to the vertical. The wind exerts a horizontal force of 300 N on the flag at the top of the pole. How much torque does this produce about the base of the pole? Round your answer to two significant figures. (a) 2500 N·m (c) 7000 N·m (e) None of these
(b) *(d)
4200 N·m 12,000 N·m
θ
3 = 78◦ r 12◦ -
Solution: Careful! The pole is at 12◦ to the vertical; the force of the wind is horizontal. Thus the angle θ between the force and the flagpole is 90◦ − 12◦ = 78◦ . It never hurts to sketch a diagram, as at right. τ = F r sin θ = (300 N)(40 m) sin 78◦ = 12, 000 N·m Problem 1393. A bird with a mass of 4.2 kg lands on top of a pole 3.5 m long that makes an angle of 15◦ to the vertical. How much torque does the bird exert about the base of the pole? Round your answer to two significant figures. (a) 19 N·m (c) 72 N·m (e) None of these Solution:
*(b) (d)
37 N·m 140 N·m
Don’t forget to calculate the bird’s weight from its mass.
τ = F r sin θ = mgr sin θ = (4.2 kg)(9.8 m/s2 )(3.5 m) sin 15◦ = 37 N·m
488 Problem 1394. A fisherman has a marlin on the end of his line. The line makes an angle of 70◦ with the rod, which is 2.2 m long. The fisherman will lose his grip on the rod if it is subjected to a torque of more than 1200 N·m. How much force must the marlin exert on the line in order to break the fisherman’s grip? Round your answer to two significant figures. *(a) (c) (e)
580 N 1600 N None of these
Solution:
τ = F r sin θ
(b) 900 N (d) 2500 N
⇒
F =
1200 N·m τ = = 580 N r sin θ (2.2 m) sin 70◦
Problem 1395. A flagpole 2.5 m long projects from the side of a building, at an angle of 35◦ to the vertical. A bird lands on the end of the flagpole, producing a torque of 80 N·m about the flagpole’s base. What is the bird’s mass? Round your answer to two significant figures. (a) 4.2 kg (c) 5.1 kg (e) None of these Solution:
(b) 4.6 kg *(d) 5.7 kg
Use torque to calculate weight, then use that to calculate mass.
τ = F r sin θ = mgr sin θ
⇒
m=
80 N·m τ = = 5.7 kg 2 gr sin θ (9.8 m/s )(2.5 m) sin 35◦
Problem 1396. Rebels have overthrown the President-for-life of Tyrannia, and are now attempting to pull down a large statue of him. They have tied a cable around the neck of the statue, 12 m above the ground, and attached the other end to a pickup. The rope makes an angle of 40◦ to the vertical. In order to topple the statue, it is necessary to produce a torque of 80,000 N·m about its base. How much force does the pickup need to apply to the rope? Round your answer to two significant figures. (a) 8700 N (c) 12,000 N (e) None of these Solution:
τ = F r sin θ
*(b) 10,000 N (d) 15,000 N
⇒
F =
τ 80, 000 N·m = = 10, 000 N r sin θ (12 m) sin 40◦
489 Problem 1397. A flagpole is supported by a guy wire that is attached to the pole 5.0 m above the ground, and that makes an angle of 65◦ to the horizontal. To keep the flagpole from blowing over in high winds, the wire must be able to produce a torque of 6000 N·m about the base of the pole. How much tension must the wire be able to withstand? Round your answer to two significant figures. (a) 1300 N (c) 6100 N (e) None of these
*(b) (d)
2800 N 13,000 N
Solution: Be careful with your angle. The wire makes an angle of 65◦ with the horizontal, so it makes an angle of θ = 90◦ − 65◦ = 25◦ with the vertical flagpole. Sketch a diagram if you’re not sure about this. τ = F r sin θ
⇒
F =
6000 N·m τ = = 2800 N r sin θ (5.0 m) sin 25◦
Problem 1398. A branch projects diagonally upward from the trunk of a tree, making an angle of 70◦ with the vertical trunk. A monkey weighing 240 N climbs out along the branch. The branch will break off if it is subjected to a torque of more than 300 N·m about its base. How far along the branch can the monkey go without breaking it? Round your answer to two significant figures. *(a) (c) (e)
1.3 m 2.6 m None of these
Solution:
τ = F r sin θ
(b) (d)
⇒
1.9 m 3.7 m
r=
300 N·m τ = = 1.3 m F sin θ (240 N) sin 70◦
Problem 1399. A partly open drawbridge makes an angle of 15◦ to the horizontal. A truck weighing 1.2 × 105 N drives slowly up the slope. The drawbridge will collapse if subjected to a torque of more than 2.0 × 106 N. How far along the drawbridge can the truck go? Round your answer to two significant figures. (a) 11 m (c) 33 m (e) None of these
*(b) (d)
17 m 64 m
Solution: Be careful with your angle. The bridge makes an angle of 15◦ to the horizontal; but the force of the truck’s weight is vertical, so it makes an angle of 90◦ −15◦ = 75◦ with the bridge. Sketch a diagram if you’re unclear on this. τ = F r sin θ
⇒
r=
τ 2.0 × 106 N·m = = 17 m F sin θ (1.2 × 105 N) sin 75◦
490 Problem 1400. The boom of a crane is 20 m long. A cement truck weighing 300,000 N is hanging from the end of the crane. The crane will topple over if it is subjected to a torque about its base greater than 5.0 × 106 N·m. What is the largest angle with the vertical that the boom can safely make? Round your answer to the nearest degree. (a) 34◦ (c) 49◦ (e) None of these Solution:
τ = F r sin θ
(b) 41◦ *(d) 56◦
⇒
−1
θ = sin
h τ i 5.0 × 106 N·m −1 = sin = 56◦ Fr (300, 000 N)(20 m)
Problem 1401. Rebels have overthrown the Dear Leader of Annexia, and are attempting to pull down his statue. They have attached a cable around the statue’s neck, 11 m above ground level, and have attached the other end to a truck. If the truck can pull with a force of 50,000 N, and if it will require a torque of 400,000 N·m about the statue’s base to pull it over, what angle must the cable make with the horizontal? Round your answer to the nearest degree. *(a) 43◦ (c) 50◦ (e) None of these
(b) 47◦ (d) 53◦
Solution: Careful! The problem asks for the cable’s angle with the horizontal; but the radius (from the statue’s base to the neck) is vertical. Thus the problem is asking for φ = 90◦ − θ, where θ is the angle between the cable and the vertical. Sketch a diagram if you’re not clear on this. h τ i τ = F r sin θ ⇒ θ = sin−1 Fr 400, 000 N·m ◦ −1 ⇒ φ = 90 − sin = 43◦ (50, 000 N)(11 m) Problem 1402. Two people are standing on a horizontal diving board. One person weighs 400 N, and is standing 3 m from the base of the board. The other weighs 800 N, and is standing 2 m from the base. What is the magnitude of the net torque that the two people produce about the base of the board? Round your answer to two significant figures. (a) 400 N·m *(c) 2800 N·m (e) None of these
(b) 1200 N·m (d) 3000 N·m
Solution: Both people are standing on the same side of the base; so their torques point in the same direction, and therefore have the same sign. Since the board is horizontal and the forces (weight) are vertical, θ = 90◦ and F r sin θ = F r. τ = τ1 + τ2 = F1 r1 + F2 r2 = (400 N)(3 m) + (800 N)(2 m) = 2800 N·m
491 Problem 1403. A teeter-totter consists of a massless board pivoting on a fulcrum. With the board held horizontal, a physics instructor weighing w1 = 800 N sits 1 m from the fulcrum. A physics student weighing w2 = 500 N sits on the opposite side, 2 m from the fulcrum. What is the net torque about the fulcrum? Assume that the instructor produces a positive torque. Round your answer to two significant figures. (a) 200 N·m (c) 300 N·m (e) None of these
w1
w2 T
TT
*(b) −200 N·m (d) −300 N·m
Solution: We will use ri = 1 m and rs = −2 m, so that the instructor’s torque τi will be positive and the student’s torque τs will be negative. Since the board is horizontal and the forces are vertical, θ = 90◦ and F r sin θ = F r. τ = τi + τs = Fi ri + Fs rs = (800 N)(1 m) + (500 N)(−2 m) = −200 N·m Problem 1404. Two pulleys are fixed on a shaft, as shown at right. One pulley has a radius of 15 cm; a weight of w1 = 6 N hangs from that pulley on the left side of the shaft. The second pulley has a radius of 10 cm; a weight of w2 = 4 N hangs from it, on the right side of the shaft. What is the net torque on the shaft? Assume that w1 produces a positive torque. Round your answer to two significant figures. *(a) (c) (e)
0.50 N·m 1.3 N·m None of these
(b) −0.50 N·m (d) −1.3 N·m
u
w1
w2
Solution: Don’t forget to convert centimeters to meters. We will use r1 = 0.15 m and r2 = −0.10 m, so that the directions of the torques will be right. τ = τ1 + τ2 = w1 r1 + w2 r2 = (6 N)(0.15 m) + (4 N)(−0.10 m) = 0.50 N·m
492 Problem 1405. You are trying to close a door; your dog is on the opposite side, trying to push it open. You apply a force of 200 N at a distance of 80 cm from the hinges. The dog applies a force of 400 N at a distance of 60 cm from the hinges. What is the net torque on the door? Assume that you are applying a positive torque. Round your answer to two significant figures. (a) −160 N·m (c) 120 N·m (e) None of these
*(b) −80 N·m (d) 400 N·m
Solution: We will use Fy = 200 N and Fd = −400 N so that the torques will have the right signs. Since no angles are mentioned, we’ll assume that you and the dog are both pushing at right angles to the door, so sin θ = 1 for both. Don’t forget to convert centimeters to meters. τ = τy + τd = Fy ry + Fd rd = (200 N)(0.80 m) + (−400 N)(0.60 m) = −80 N·m Problem 1406. Two civil engineering students are attempting to loosen a nut on a bridge, using a very large wrench. Both students are pulling in the same direction. One student is applying a force of 400 N at an angle of 90◦ to the handle of the wrench, 1.2 m from the nut; the other student is applying a force of 500 N at an angle of 80◦ to the handle, 1.6 m from the nut. What is the net torque applied to the nut? Round your answer to two significant figures. (a) 310 N·m *(c) 1300 N·m (e) None of these
(b) 900 N·m (d) 1400 N·m
Solution: Since both students are producing torque in the same direction, we don’t have to worry about signs. τ = τ1 + τ2 = F1 r1 sin θ1 + F2 r2 sin θ2 = (400 N)(1.2 m) sin 90◦ + (500 N)(1.6 m) sin 80◦ = 1300 N·m
493 Problem 1407. Rebels have overthrown the Tsar of Fanatistan, and are now trying to pull down his statue. However, the two rebel organizations are in disagreement on how to do it. The People’s Front has attached a rope 6.0 m up on the statue, and is pulling northward with a force of 20,000 N at an angle of 40◦ to the horizontal. The Popular Front has attached their rope 7.5 m up on the statue, and is pulling southward with a force of 15,000 N at an angle of 35◦ to the horizontal. What is the net torque about the base of the statue? Assume that the People’s Front is applying positive torque. Round your answer to the nearest 1000 N·m. *(a) (c) (e)
0 N·m 38,000 N·m None of these
(b) −13, 000 N·m (d) 180,000 N·m
Solution: Careful! Notice that we are given angles to the horizontal, but that to calculate the torque we need angles measured to the vertical. Also, to make the signs of the torques work out correctly, we’ll give the Popular Front a negative force: F2 = −15, 000 N. τ = τ1 + τ2 = F1 r1 sin θ1 + F2 r2 sin θ2 = (20, 000 N)(6.0 m) sin 50◦ + (−15, 000 N)(7.5 m) sin 55◦ = 0 N·m Problem 1408. A massless plank is lying on the ground and extending partly out over the edge of the Grand Canyon. A tourist weighing 800 N wants to walk to the end of the plank, 2.5 m beyond the cliff edge, and have his wife take his picture. If the wife weighs 500 N, how far back from the cliff edge must she stand on the plank to keep her husband from falling into the canyon? Round your answer to two significant figures. (a) 1.6 m *(c) 4.0 m (e) None of these
(b) 2.5 m (d) 6.4 m
Solution: If the wife really doesn’t want her husband to fall in, then she must produce a torque about the cliff edge that matches his. We assume that the plank is horizontal, so F r sin θ = F r. Then Fw rw = Fh rh
⇒
rw =
(800 N)(2.5 m) Fh rh = = 4.0 m Fw 500 N
494 Problem 1409. A teeter-totter consists of a massless board pivoting on a fulcrum. A physics instructor weighing w1 = 800 N sits 1.2 m from the fulcrum. A physics student weighing w2 = 500 N sits on the opposite side. How far from the fulcrum must the student sit to balance the instructor’s torque? Round your answer to two significant figures. (a) 0.75 m (c) 1.6 m (e) None of these
w1
w2 T TT
(b) 1.4 m *(d) 1.9 m
Solution: Since no angle is mentioned, we can assume that the board is horizontal; so F r sin θ = F r. To balance, the torques produced by instructor and student must match. w1 r1 = w2 r2
⇒
r2 =
(800 N)(1.2 m) w 1 r1 = 1.9 m = w2 500 N
Problem 1410. A vertical pulley has a radius of 5 cm. A thin string is wound around the outside of the pulley, and a weight of 8 N is suspended from it. The pulley has a brake, located 3 cm from the pulley’s axle. How much force must the brake exert to keep the pulley from turning? Round your answer to two significant figures. (a) 1.9 N *(c) 13 N (e) None of these
(b) 4.8 N (d) 120 N
Solution: To keep the pulley from turning, the torques produced by the weight and by the brake must match. F1 r1 = F2 r2
⇒
F2 =
(8 N)(0.05 m) F1 r1 = 13 N = r2 0.03 m
Problem 1411. A massless horizontal pole projects 1.6 m from the side of a building. It is held up by a cable running from the side of the building to the end of the pole, making an angle of 35◦ to the horizontal. A sign weighing w = 30 N hangs from the pole 1.5 m out from the building. What is the tension in the cable? Round your answer to two significant figures. (a) 28 N (c) 39 N (e) None of these
HH HH
(b) 34 N *(d) 49 N
HH H
w
Solution: Since the system is not moving, the torques produced by the weight and by the tension must match. Let T be the tension. T rT sin θT = wrw
⇒
T =
wrw (30 N)(1.5 m) = = 49 N rT sin θ (1.6 m) sin 35◦
495 Problem 1412. A flagpole 2.4 m long projects from the side of a building, making an angle of 40◦ to the vertical. The flagpole has a mass of 80 kg; its center of mass is at its midpoint. It is supported by a horizontal cable running from the side of the building to the end of the pole. What is the tension in the cable? Round your answer to two significant figures. (a) 34 N *(c) 330 N (e) None of these
(b) 40 N (d) 390 N
T r 50◦ r 40◦ ? w=
mg
Solution: We need to balance two torques about the base of the pole. The first is produced by the weight of the flagpole: F1 = w = mg, r1 = 1.2 m (the midpoint of the pole), and θ1 = 40◦ . The second is produced by the tension in the cable: F2 = T , r2 = 2.4 m (the end of the pole), and θ2 = 50◦ . mgr1 sin θ1 = T r2 sin θ2 mgr1 sin θ1 (80 kg)(9.8 m/s2 )(1.2 m) sin 40◦ ⇒ T = = = 330 N r2 sin θ2 (2.4 m) sin 50◦ Problem 1413. A flagpole 7 m high is supported by a guy wire attached 5 m high on the north side of the pole and meeting the ground at an angle of θ = 50◦ . The wind blows southward, exerting a force of 200 N on the flag at the top of the pole. What is the tension in the wire? Round your answer to two significant figures. (a) 180 N (c) 370 N (e) None of these
(b) 210 N *(d) 440 N
J J J J θJJ
Solution: Since the pole doesn’t move, the torque produced by the wire must balance the torque produced by the wind on the flag. Be careful with the angles; the angle between the wire and the flagpole is φ = 90◦ − 50◦ = 40◦ . Let T be the tension; then F1 r1 = T r2 sin φ
⇒
T =
F 1 r1 (200 N)(7 m) = = 440 N r2 sin φ (5 m) sin 40◦
496 Problem 1414. A pendulum consists of a lead sinker with weight W attached to the end of a thin string with length L; the other end of the string is attached to a hook in the ceiling. The pendulum is pulled back and released. When the string makes an angle of θ with the vertical, it has a tension of T . At that point, what is the net torque about the hook in the ceiling? (a) 0 *(c) W L sin θ (e) None of these
(b) (T + W ) sin θ (d) (T − W )L cos θ
A A
θA
L
A A A Au
θ W?
Solution: The tension is pulling straight away from the hook, so it makes no contribution to the torque: r = 0. The weight W is pulling straight downward, at an angle of 180◦ − θ to the string and at a distance of L from the hook. Hence τ = W L sin(180◦ − θ) = W L sin θ
13.2
Angular acceleration
Problem 1415. A windmill has a moment of inertia of 5.0 kg·m2 . When the wind begins to blow, it accelerates at 0.40 rad/s2 . What is the torque on the windmill? Round your answer to two significant figures. (a) 0.80 N·m (c) 5.0 N·m (e) None of these Solution:
*(b) 2.0 N·m (d) 13 N·m
τ = Iα = (5.0 kg·m2 )(0.40 rad/s2 ) = 2.0 N·m
Problem 1416. A shaft’s moment of inertia is 24 kg·m2 . How much torque does it take to accelerate the shaft at 3.0 rad/s2 ? Round your answer to two significant figures. (a) 8.0 N·m (c) 35 N·m (e) None of these Solution:
(b) *(d)
17 N·m 72 N·m
τ = Iα = (24 kg·m2 )(3.0 rad/s2 ) = 72 N·m
497 Problem 1417. A flywheel has a mass of 1200 kg, a radius of 0.75 m, and a moment of inertia of 480 kg·m2 . When a motor connected to the flywheel is switched on, the flywheel accelerates at 0.80 rad/s2 . How much torque does the motor apply to the flywheel? Round your answer to two significant figures. (a) 290 N·m *(c) 380 N·m (e) None of these Solution:
(b) 340 N·m (d) 900 N·m
We don’t need the mass or the radius for this problem. τ = Iα = (480 kg·m2 )(0.80 rad/s2 ) = 380 N·m
Problem 1418. A bicycle wheel’s moment of inertia is 0.30 kg·m2 . If you apply a torque of 20 N·m to it, what is its angular acceleration? Round your answer to two significant figures. (a) 1.8 rad/s2 (c) 20 rad/s2 (e) None of these Solution:
τ = Iα
(b) *(d)
⇒
6.0 rad/s2 67 rad/s2 20 N·m τ = 67 rad/s2 = I 0.30 kg·m2
α=
Problem 1419. A revolving door has moment of inertia 120 kg·m2 . You apply a torque of 50 N·m to it. What is its angular acceleration? Round your answer to two significant figures. (a) 0.12 rad/s2 (c) 1.4 rad/s2 (e) None of these Solution:
τ = Iα
*(b) 0.42 rad/s2 (d) 2.4 rad/s2
⇒
τ 50 N·m = = 0.42 rad/s2 I 120 kg·m2
α=
Problem 1420. You are trying to determine the moment of inertia of a roast turkey on a massless turntable. Unfortunately, your physics book does not include a formula for the moment of inertia of a turkey, roasted or otherwise. You apply a torque of 0.060 N·m to the turntable, and it accelerates at 1.4 rad/s2 . What is the turkey’s moment of inertia? Round your answer to two significant figures. *(a) 0.043 kg·m2 (c) 0.067 kg·m2 (e) None of these Solution:
τ = Iα
(b) (d)
⇒
I=
0.054 kg·m2 0.084 kg·m2
τ 0.60 N·m = = 0.043 kg·m2 α 1.4 rad/s2
498 Problem 1421. A motor supplies a torque of 160,000 N·m to a merry-go-round, which accelerates at 0.3 rad/s2 . What is the merry-go-round’s moment of inertia? Round your answer to two significant figures. (a) 48,000 kg·m2 (c) 240,000 kg·m2 (e) None of these Solution:
τ = Iα
(b) *(d)
⇒
I=
110,000 kg·m2 530,000 kg·m2
τ 160, 000 N·m = = 530, 000 kg·m2 2 α 0.3 rad/s
Problem 1422. A flywheel has a radius of 60 cm and a moment of inertia of 150 kg·m2 . A belt around the rim of the wheel imparts a force of 80 N. What is the flywheel’s angular acceleration? Round your answer to two significant figures. (a) 0.29 rad/s2 (c) 0.35 rad/s2 (e) None of these Solution:
τ = F r = Iα
*(b) (d)
⇒
0.32 rad/s2 0.39 rad/s2
α=
(80 N)(0.60 m) Fr = = 0.32 rad/s2 I 150 kg·m2
Problem 1423. A gate is 2 m long, and has a moment of inertia of 200 kg·m2 . A cow pushes on the gate 1.5 m from the hinges, with a force of 400 N. What is the gate’s angular acceleration? Round your answer to two significant figures. (a) 1.5 rad/s2 *(c) 3.0 rad/s2 (e) None of these
(b) (d)
2.3 rad/s2 4.0 rad/s2
Solution: Since no angle is mentioned, we assume that the cow pushes at a right angle; so τ = F r. τ = F r = Iα
⇒
α=
Fr (400 N)(1.5 m) = = 3.0 rad/s2 I 200 kg·m2
Notice that we don’t need the length of the gate. Problem 1424. A shaft is 10 m long, with a radius of 10 cm and a moment of inertia of 12 kg·m2 . A belt around the shaft imparts a force of 50 N. What is the shaft’s angular acceleration? Round your answer to two significant figures. (a) 0.30 rad/s2 (c) 0.38 rad/s2 (e) None of these Solution:
(b) *(d)
0.34 rad/s2 0.42 rad/s2
The length of the shaft is irrelevant to this problem. τ = F r = Iα
⇒
α=
FR (50 N)(0.10 m) = = 0.42 rad/s2 I 12 kg·m2
499 Problem 1425. A large gear has a radius of 1.2 m, a mass of 4100 kg, and a moment of inertia of 4000 kg·m2 . A second gear engages the teeth on the outer edge of the first one and imparts a force of 200 N. What is the angular acceleration of the first gear? Round your answer to two significant figures. (a) 0.050 rad/s2 (c) 0.067 rad/s2 (e) None of these Solution:
*(b) (d)
0.060 rad/s2 0.075 rad/s2
The mass of the gear doesn’t matter. τ = F r = Iα
⇒
α=
Fr (200 N)(1.2 m) = = 0.060 rad/s2 2 I 4000 kg·m
Problem 1426. A merry-go-round has a radius of 2.0 m. A physics professor pushes with a force of 120 N tangent to the outer edge of the merry-go-round, giving it an angular acceleration of 1.8 rad/s2 . What is the merry-go-round’s moment of inertia? Round your answer to two significant figures. (a) 110 kg·m2 *(c) 130 kg·m2 (e) None of these Solution:
τ = F r = Iα
(b) (d)
⇒
120 kg·m2 150 kg·m2
I=
Fr (120 N)(2.0 m) = = 130 kg·m2 α 1.8 rad/s2
Problem 1427. A belt runs over a pulley with a radius of 12 cm. When the belt is pulled with a force of 60 N, the pulley experiences a radial acceleration of 2.5 m/s2 . What is the pulley’s moment of inertia? Round your answer to two significant figures. (a) 2.1 kg·m2 (c) 2.6 kg·m2 (e) None of these Solution:
τ = F r = Iα
(b) 2.3 kg·m2 *(d) 2.9 kg·m2
⇒
I=
Fr (60 N)(0.12 m) = = 2.9 kg·m2 α 2.5 m/s2
Problem 1428. A fountain features a stone sphere with a radius of 2.8 m, suspended frictionlessly on a film of water in a hemispherical basin. The sphere has a moment of inertia of 2800 kg·m2 . How much force do you have to apply along the sphere’s equator in order to produce an acceleration of 0.04 m/s2 ? Round your answer to two significant figures. *(a) (c) (e)
40 N 48 N None of these
Solution:
τ = F r = Iα
(b) (d)
⇒
44 N 53 N Iα (2800 kg·m2 )(0.04 m/s2 ) F = = = 40 N r 2.8 m
500 Problem 1429. A gate is 2.1 m long, and has a moment of inertia of 260 kg·m2 . How hard do you have to push on the outer edge of the gate to induce an acceleration of 1.5 rad/s2 ? Round your answer to two significant figures. (a) 140 N (c) 170 N (e) None of these Solution: the gate.
(b) 150 N *(d) 190 N
Since we don’t mention an angle, assume that you push at right angles to
τ = F r = Iα
⇒
F =
(260 kg·m2 )(1.5 rad/s2 ) Iα = = 190 N r 2.1 m
Problem 1430. A windmill’s moment of inertia is 20 kg·m2 . It is initially at rest; when the wind starts to blow, it begins turning, reaching a speed of 4.5 rad/s in 8 seconds. What is the magnitude of the torque that the wind applied to the windmill? Round your answer to two significant figures. (a) 10 N·m (c) 12 N·m (e) None of these
*(b) (d)
11 N·m 14 N·m
Solution: We can’t use τ = Iα directly, since we don’t know α. However, we know ω0 = 0, ω, and t; so we can calculate α. ⇒
ω = ω0 + αt = αt τ = Iα =
α=
ω t
Iω (20 kg·m2 )(4.5 rad/s) = = 11 N·m t 8s
Problem 1431. A shaft has a radius of 8 cm and a moment of inertia of 15 kg·m2 . It is driven by a belt wrapped around it. If the shaft is initially at rest and the belt is pulled with a force of 20 N, how long does it take for the shaft to turn 10 radians? Round your answer to two significant figures. (a) 12 s (c) 15 s (e) None of these
*(b) (d)
14 s 17 s
Solution: Since we know r and F , we can calculate τ ; since we know I, we can calculate α. Then we know α, ω0 = 0, θ0 = 0, and θ; we want to know t. Fr I 1 1 Fr 2 θ = θ0 + ω0 t + αt2 = αt2 = t 2 2 2I r 1/2 2Iθ 2(15 kg·m2 )(10 rad) t= = = 14 s Fr (20 N)(0.08 m) τ = F r = Iα
⇒
⇒
α=
501 Problem 1432. A wheel’s moment of inertia is 30 kg·m2 . It is spinning at 25 rad/s when a brake is applied at a distance of 30 cm from the wheel’s center. The wheel turns through 15 radians before coming to a halt. How much force was produced by the brake? Round your answer to two significant figures. *(a) (c) (e)
2100 N 2500 N None of these
(b) (d)
2300 N 2800 N
Solution: We know r and we want to know F . If we knew τ , we could calculate it directly. We know I; if we knew α, we could calculate τ . We know ω0 , ∆θ, and ω = 0, from which we can calculate α. ⇒
tau = Iα = F r 2α∆θ = ω 2 − ω02 = −ω02
F = ⇒
Iα r
α=−
ω02 2∆θ
Since we only care about the magnitude of the force applied by the brake, we’ll ignore the negative sign. (30 kg·m2 )(25 rad/s)2 Iω02 = = 2100 N F = 2r∆θ 2(0.30 m)(15 rad) Problem 1433. An evil physics professor decides to stop the rotation of the Earth. He embeds a large rocket in a mountain on the equator, pointing westward. The Earth has a radius of 6.4 × 106 m and a moment of inertia of 9.8 × 1037 kg·m2 ; it is initially turning at 7.3 × 10−5 rad/s. If the professor wants to stop the rotation in 3.2 × 107 s (approximately one year), how much thrust must his rocket produce? Round your answer to two significant figures. (a) 3.2 × 1018 N (c) 3.9 × 1020 N (e) None of these
*(b) 3.5 × 1019 N (d) 4.2 × 1021 N
Solution: We know r. If we knew τ , we could calculate F . We know I; if we knew α, we could calculate τ . We know ω0 , ω = 0, and t; from these, we can calculate α. τ = F r = Iα ω = 0 = ω0 + αt
⇒ ⇒
Iα r ω0 α=− t
F =
Since we want the magnitude of the force, we will ignore the negative sign. F =
Iω0 (9.8 × 1037 kg·m2 )(7.3 × 10−5 rad/s) = = 3.5 × 1019 N rt (6.4 × 106 m)(3.2 × 107 s)
502 Problem 1434. A large gear’s radius is 30 cm, and its moment of inertia is 40 kg·m2 . A second gear engages the teeth on the outer edge of the first one and applies a force of 20 N. The gears are initially at rest. After the first gear has turned through 5 rad, how fast will it be turning? Round your answer to two significant figures. (a) 1.0 rad/s *(c) 1.2 rad/s (e) None of these
(b) 1.1 rad/s (d) 1.3 rad/s
Solution: We know ω0 = 0 and ∆θ; we want to know ω. If we knew α, we could calculate ω. We know r and F , so we can calculate τ ; and we know I, so we can calculate α. τ = F r = Iα
⇒
⇒
α=
Fr I
2α∆θ = ω 2 − ω02 = ω 2 1/2 1/2 √ 2F r∆θ 2(20 N)(0.30 m)(5 rad) = = 1.2 rad/s ω = 2α∆θ = I 40 kg·m2
Problem 1435. A flywheel with a radius of 0.50 m and a moment of inertia of 120 kg·m2 is driven by a belt around the rim. The flywheel is initially at rest. You would like it to turn through 20 rad in 6 seconds. What force do you need to apply to the belt? Round your answer to two significant figures. (a) 190 N (c) 240 N (e) None of these
(b) 220 N *(d) 270 N
Solution: We know r and I; we want to know F . If we knew α, we could calculate τ , and from that we could calculate F . We know θ0 = 0, ω0 = 0, θ, and t. From these, we can calculate α. 1 1 θ = θ0 + ω0 t + αt2 = αt2 2 2 τ = F r = Iα
⇒
F =
⇒
α=
2θ t2
Iα 2Iθ 2(120 kg·m2 )(20 rad) = 2 = = 270 N r rt (0.50 m)(6 s)2
503
13.3
Moment of inertia
Problem 1436. A physics instructor has a mass of 90 kg. His wife has a mass of 60 kg. The two are riding together on a merry-go-round with two concentric circles of wooden horses. In which of the following situations is the system’s moment of inertia the smallest? (a) Both sit together one one of the outer horses (b) Both sit on outer horses, on opposite sides of the merry-go-round *(c) She sits on one of the outer horses; he sits on the inner horse next to her (d) He sits on one of the outer horses; she sits on an inner horse on the opposite side Solution: In all of the possible solutions, the mass is the same. We can reduce the moment of inertia by moving mass closer to the axis of rotation. We don’t have the option of putting both people on inner horses (making both rp and rw small). Since mp > mw , our best option is to put the professor on an inner horse (making rp small). Problem 1437. A flywheel consists of a solid iron disc with two holes drilled through it, turning on a shaft through its center. In which of the following arrangements of holes is the moment of inertia the largest? (b)
*(a) ku k
(c) k u
k
(d) k k
u
u kk
Solution: To make the moment of inertia large, we want as much mass as possible to be far away from the axis of rotation. In situation (a), we are drilling our holes near the axis, where the mass removed contributed little to I. In the other situations, the mass removed contributed more to I, so drilling holes there makes greater reductions in I. Problem 1438. A lead block with mass mb is attached to a massless turntable, at a distance of r1 from the center. The system’s moment of inertia is I1 . The block is then moved to a distance of 2r1 from the center. What is the new moment of inertia? √ 2 I1 (a) I1 (b) (c) 2I1 *(d) 4I1 (e) None of these We assume that the block is small enough to be treated as a point mass. Then I1 = mb r12
and I2 = mb (2r1 )2 = 4mb r12 = 4I1
504 Problem 1439. A lead block with mass mb is attached to a massless turntable, at a distance of r1 from the center. The system’s moment of inertia is I1 . A second block with the same mass is then attached at the same distance from the center, on the opposite side. What is the new moment of inertia? √ 2 I1 (a) I1 (b) *(c) 2I1 (d) 4I1 (e) None of these Solution: Then
We assume that the block is small enough to be treated as a point mass. I1 = mb r12
and I2 = mb r12 + mb r12 = 2mb r12 = 2I1
Problem 1440. A system consists of a massless turntable with a lead block with mass m1 attached at a distance of r1 from the center. A second system has the same moment of inertia, but uses a block with mass m2 = 2m1 . At what distance r2 is this block attached? (a) r1 /4√ (b) r1 /2 (d) r1 *(c) r1 / 2 (e) None of these Solution:
We assume that the blocks can be treated as point masses.
I1 = m1 r12
and I2 = I1 = m2 r22 = (2m1 )r22
⇒
r12 = 2r22
⇒
r1 r2 = √ 2
Problem 1441. A gate can be regarded as a thin rectangular plate hinged along one side, with a mass of 80 kg and a length of 3.0 m. The gate is pushed with a force of 50 N, 2.0 m from its hinges. What angular acceleration does it experience? Round your answer to two significant figures. (a) 0.30 rad/s2 (c) 0.38 rad/s2 (e) None of these
(b) 0.34 rad/s2 *(d) 0.42 rad/s2
Solution: We’ll use L = 3.0 m to denote the length of the gate, and r = 2.0 m to denote the distance at which force is applied. We know r and F , so we can calculate τ . To calculate α, we need to know I. We can calculate I from the mass and dimensions of the gate. mL2 I= 3 τ = F r = Iα
⇒
α=
Fr 3F r 3(50 N)(2.0 m) = = = 0.42 rad/s2 I mL2 (80 kg)(3.0 m)2
505 Problem 1442. A wind turbine has three blades. Each blade has a mass of 1200 kg and a length of 20 m. Each blade can be regarded as a thin rod connected at one end to the axis. How much torque must be applied to the turbine to make it accelerate at 0.10 rad/s2 ? Round your answer to two significant figures. (a) 35,000 N·m (c) 43,000 N·m (e) None of these
(b) 39,000 N·m *(d) 48,000 N·m
Solution: We know α. We want to know τ . We can calculate I from the description of the turbine. We’ll use L = 20 m to denote the length of a blade. M L2 = M L2 Iturbine = 3Iblade = 3 3 τ = Iα = M L2 α = (1200 kg)(20 m)2 (0.10 rad/s2 ) = 48, 000 N·m Problem 1443. A flywheel consists of a solid cylinder with a radius of 70 cm and a thickness of z = 8 cm; it is made of steel with a density of ρ = 7.5 g/cm3 . The flywheel is driven by a belt that wraps around its rim. If you want it to go from rest to a speed of 12 rad/s in 5 seconds, how much force do you need to apply to the belt? Round your answer to two significant figures. *(a) (c) (e)
780 N 940 N None of these
(b) (d)
850 N 1000 N
Solution: We know the dimensions and density of the flywheel. From these, we can calculate m, and then I. We know ω0 = 0, ω, and t; from these, we can calculate α. With that information, we can calculate τ and then F . m = ρV = πr2 zρ πr4 zρ mr2 = I= 2 2 ω ω = ω0 + αt = αt ⇒ α = t τ = F r = Iα ⇒
F =
Iα πr3 zρω π(0.70 m)3 (0.08 m)(7500 kg/m3 )(12 rad/s) = = = 780 N r 2t 2(5 s)
506 Problem 1444. A flywheel can be regarded as a cylindrical shell with an inside radius of 50 cm, an outside radius of 60 cm, a thickness (length) of z = 7.5 cm, and a density of ρ = 7.5 g/cm3 . It is mounted on a shaft with radius 1.5 cm. (The moment of inertia of the shaft, and of the wheel’s spokes, can be ignored.) A belt around the shaft drives the flywheel. How much force do you need to apply to the belt if you want the wheel, starting from rest, to reach a speed of 1.2 rad/s by the time it has made one complete revolution? Round your answer to two significant figures. *(a) (c) (e)
450 N 550 N None of these
(b) (d)
500 N 600 N
Solution: We know the dimensions and density of the flywheel; from these we can calculate m and then I. We know ω0 = 0, ω, and ∆θ; from these, we can calculate α. That will allow us to calculate τ ; and since we know the radius r of the shaft, we can calculate F . We’ll use Ri and Ro for the inner and outer radii of the flywheel. m = ρV = πρz(Ro2 − Ri2 ) m(Ro2 + Ri2 ) πρz(Ro4 − Ri4 ) = 2 2 ω2 2α∆θ = ω 2 − ω02 = ω 2 ⇒ α = 2∆θ I=
τ = F r = Iα =
⇒
F =
Iα πρzω 2 (Ro4 − Ri4 ) = r 4r∆θ
π(7500 kg/m3 )(0.075 m)(1.2 rad/s)2 [(0.60 m)4 − (0.50 m)4 ] = 450 N 4(0.015 m)(2π rad)
507 Problem 1445. (Similarity Problem) A solid sphere, solid cylinder, thin-walled hollow cylinder, and thin-walled hollow sphere all having the same mass M and radius R are released from rest from the same height H on a ramp. Assuming that the ramp is smooth and each of the four objects rolls without slipping, which of these objects will be the first one to reach the bottom of the hill? *(a) solid sphere (b) solid cylinder (c) thin-walled hollow cylinder (d) thin-walled hollow sphere (e) None of these Solution: Take the y-axis pointing upward and the origin y = 0 to be at the bottom of the ramp. Since each object has the same mass M and initial height y = H, they have the same initial potential energy Ui = M gH. Since they also start from rest, they all have the same initial kinetic energy Ki = 0. Thus, the initial mechanical energy is MEi = M gH. At each point along the ramp they all have the same potential energy. When the object is y units above ground level its potential energy is Uf (y) = M gy. Notice that when they reach ground level that the potential energy is zero. The final kinetic energy at height y above the ground is the sum of the kinetic energy due to translation and rotation: 2 + 21 Icm ω 2 . When an object with radius R has rotated through one Kf (y) = 12 M vcm complete revolution (2π radians), it has rolled a distance equal to its circumference (2πR). Thus the distance travelled during any time interval is R times the angular velocity. It follows that vcm = Rω. Substituting ω = vcm /R in the expression for kinetic energy gives 2 vcm 1 1 Icm 1 2 2 . Kf (y) = M vcm + Icm 2 = M vcm 1 + 2 2 R 2 M R2 We now make an observation about the moment of inertia about the center of mass of each of the four objects. They all have the structure: Icm = βM R2 , where 0 < β ≤ 1. Thus the final mechanical energy can be expressed as 1 2 MEf = M gy + M vcm (1 + β) . 2 Ignoring any loss of energy due to rolling friction we can apply the conservation of mechanical energy to arrive at the equation 1 2 M gH = MEi = MEf = M gy + M vcm (1 + β) 2 1 −M gy 2 −−−−−−−→ M g(H − y) = M vcm (1 + β) 2 ×
2
(1+β)−1
2 −−−−M−−−−−−−→ 2g(H − y) (1 + β)−1 = vcm p √ 2g(H − y) −−−−−−→ vcm = √ 1+β
(13.1)
Notice that the final velocity depends on the parameters H, g, y, and β. This is an amazing result; it says that the final velocity is independent of the mass M and the radius R!
508 If we compare any two objects at a time, then we can treat them like two similar experiments with common parameters: H, g, y. We wish to find the similarity between the two different velocities. Treating the above equation as the governing equation and put all of the common parameters on the right-hand side of the equation and all of the experiment-dependent parameters on the left we arrive at √ × 1+β
−−−−−−−−→ vcm
p p 1 + β = 2g(H − y)
Comparing two similar experiments (objects) with β1 < β2 and using the above equation with the common parameters as our link we have p p p vcm,1 1 + β1 = 2g(H − y) = vcm,2 1 + β2 √ vcm,1 1 + β2 =√ (13.2) ⇒ vcm,2 1 + β1 √ √ Since β1 < β2 ⇒ 1 + β1 < 1 + β2 , it follows from equation (13.2) that vcm,1 > vcm,2 . Define the following subscripts: ss = solid sphere sc = solid cylinder hc = thin-walled hollow cylinder hs = hollow sphere. Then βss = 25 < βsc = 12 < βhs = 32 < βhc = 1
⇒
vcm,ss > vcm,sc > vcm,hs > vcm,hc .
Thus, the solid sphere reaches the bottom of the hill first.
509
Part V
Work, Energy, and Momentum
510
14 14.1
Work and mechanical energy Work
Problem 1446. A rocket is fired upward. As it rises, the work done by gravity on the rocket is: (a) positive
*(b) negative
(c) zero
Solution: The rocket is moving upward; gravity is pulling downward. Since the force is in the opposite direction of the motion, the work done by the force is negative. Problem 1447. A piano is dropped from a high window. As it falls, the work done by gravity on the piano is: *(a) positive
(b) negative
(c) zero
Solution: The piano moves downward; gravity is pulling downward. Since the force and the motion are in the same direction, the work done by the force is positive. Problem 1448. An ice skater is gliding across the frictionless surface of a frozen pond. As he crosses the pond, the work done by gravity on the skater is: (a) positive
(b) negative
*(c) zero
Solution: The force of gravity is pulling straight down. The skater is moving horizontally. Since the force is perpendicular to the motion, the work done by the force is zero. Problem 1449. A pickup is towing a car along a level road. The two are connected by a horizontal tow-strap, which has a tension of 4000 N. How much work does the pickup do in towing the car a distance of 100 m in a time of 10 s? (a) 2 × 104 J *(c) 4 × 105 J (e) None of these
(b) 2 × 105 J (d) Not enough information: need car’s mass
Solution: The force and the motion are in exactly the same direction, so the angle θ between force and motion is zero. Hence W = F s cos θ = F s = (4000 N)(100 m) = 4 × 105 J The time it takes to move the car is irrelevant to the problem.
511 Problem 1450. You are pushing a shopping cart across a level parking lot. The cart has a mass of 10 kg. You push it with a horizontal force of 100 N to move it at a constant speed of 4 m/s. How much work do you do on the cart? (a) 400 J (c) 1000 J (e) None of these
(b) *(d)
800 J Not enough information: need the distance
Solution: The force and the motion are in exactly the same direction, so the angle θ between force and motion is zero. Hence W = F s cos θ = F s. We know F = 100 N, but we don’t know the distance s, and we have no way of finding it out from the information we’re given in the problem. Hence: (d) Problem 1451. A filing cabinet has a mass of 69 kg. You push on it horizontally in order to move it at a constant velocity. The coefficient of kinetic friction between the cabinet and the floor is 0.77. How much work do you do in moving the cabinet 0.85 m? Round your answer to two significant figures. (a) 45 J *(c) 440 J (e) None of these
(b) 59 J (d) 520 J
Solution: To move the cabinet at a constant velocity, the force with which you push it has to equal the force of kinetic friction. The work done is this force, times the distance that you push the cabinet: W = fk · s = mgµk s = (69 kg)(9.8m/s2 )(0.77)(0.85 m) = 440 J Problem 1452. A crate full of pineapples has a mass of 170 kg. A worker pushes horizontally on the crate to move it 12 m across the floor at a constant speed. The coefficient of kinetic friction between the sliding crate and the floor is 1.12. How much work does the worker do on the crate? Round your answer to the nearest 100 J. (a) 900 J (c) 20,000 J (e) None of these
(b) 2300 J *(d) 22,400 J
Solution: To overcome friction, the worker has to push with a horizontal force of F = mgµk . Exerting this force over a horizontal distance of s does work: W = F s = mgµk s = (170 kg)(9.8 m/s2 )(1.12)(12 m) = 22, 400 J
512 Problem 1453. A sidewalk runs parallel to a street. A skateboarder is being towed down the sidewalk by a rope attached to the back of a car on the street. The rope makes an angle of 23◦ to the direction in which the car and the skateboarder are moving. The tension in the rope is 185 N. How much work does the car do in towing the skateboarder 78 m down the sidewalk? Round your answer to the nearest 100 J. (a) 5600 J (c) 14,400 J (e) None of these
*(b) 13,300 J (d) 15,700 J
Solution: The component of the tension in the direction of the sidewalk is T cos 23◦ . The work is this parallel component of force, times the distance through which it acts: W = (T cos 23◦ )s = (185 N)(78 m) cos 23◦ = 13, 300 J Problem 1454. A road runs beside and parallel to a straight stretch of railroad track. A train robber attaches his pickup truck to a boxcar, using a horizontal rope that makes an angle of 17◦ to the tracks, and tows the boxcar 1200 m down the tracks. If the tension in the rope was 7400 N, how much work was done on the boxcar? Give your answer in scientific notation, rounded to two significant figures. (a) 2.4 × 106 J *(c) 8.5 × 106 J (e) None of these
(b) 2.6 × 106 J (d) 9.3 × 106 J
◦ 17
Solution: The component of force parallel to the track is T cos θ = (7400 N) cos 17◦ . The work done is this component of force times the distance that the boxcar was moved: W = T x cos θ = (7400 N)(1200 m) cos 17◦ = 8.5 × 106 N·m = 8.5 × 106 J Problem 1455. You are taking your German shepherd for a walk in Reid Park when she decides to chase a Black-crowned Night-Heron by the pond. You pull back with a force of T on her leash, which makes an angle of θ with the horizontal. However, the dog is stronger than you are, and pulls you forward a distance of x. How much work do you do on the dog? (a) T x cos θ (c) T x sin θ (e) None of these
*(b) −T x cos θ (d) −T x sin θ
Solution: The dog is moving horizontally forward; the force is directed backward at an angle of θ to the horizontal. Hence the component of force parallel to the dog’s displacement is −T cos θ. Since the dog moves a distance of x against this force, the work done is −T x cos θ.
513 Problem 1456. A case of ball bearings has a mass of 145 kg. A worker pushes horizontally on the case to move it 4.2 m across the floor at a constant speed. The coefficient of friction between the sliding case and the floor is 0.58. How much work does the worker do on the case? Round your answer to two significant figures. (a) 180 J (c) 1700 J (e) None of these
(b) *(d)
350 J 3500 J
Solution: To overcome friction, the worker has to push with a horizontal force of F = mgµ. Exerting this force over a horizontal distance of s does work: W = F s = mgµs = (145 kg)(9.8 m/s2 )(0.58)(4.2 m) = 3500 J
14.2
Power
Problem 1457. You are pushing a shopping cart with a mass of 23 kg across a parking lot. The forces of resistance total 52 N. How much power is necessary to push the cart at 1.8 m/s? Round your answer to two significant figures. (a) 47 W (c) 460 W (e) None of these
*(b) (d)
94 W 920 W
Solution: Use the equation for power as a product of force times velocity: P = F v. Then P = Fpush · vcart = (52N)(1.8m/s) = 94 W. To find Fpush , draw a free-body diagram and apply Newton’s second law, along with the fact that the cart is moving at a constant speed (so it’s not accelerating). That will give you: Fpush = fk . Problem 1458. (Power) How much power, in kilowatts, must be developed by the electric motor of a 1600-kg car moving at 26 m/s on a level road if the forces of resistance total 720 N? Round your answer to the nearest tenth of a kilowatt. (a) 95.2 kW (c) 133.2 kW (e) None of these
(b) 36.3 kW *(d) 18.7 kW
Solution: Use the equation for power as a product of force times velocity: P = F v. Then P = Fmotor · vcar = (720N)(26m/s) = 18.7 kW. To find Fmotor , draw a free-body diagram and apply Newton’s second law together with the fact that the car is moving at a constant speed (so it’s not accelerating) to arrive at Fmotor = fk . See my solutions on the web for the details.
514
14.3
Kinetic energy (K.E.)
Problem 1459. A model airplane with a mass of 6 kg is flying at a speed of 10 m/s. What is the airplane’s kinetic energy? (a) 30 J (b) 60 J *(c) 300 J (d) 600 J (e) None of these Solution:
K = 12 mv 2 = 12 (6 kg)(10m/s)2 = 300 J
Problem 1460. A skier with a mass of 80 kg is sliding down a slope at 20 m/s. What is the skier’s kinetic energy? (a) 3.2 × 103 J (b) 6.4 × 103 J 4 *(c) 1.6 × 10 J (d) 3.2 × 104 J (e) None of these Solution:
K = 12 mv 2 = 12 (80 kg)(20 m/s)2 = 1.6 × 104 J
Problem 1461. A suitcase with a mass of 20 kg falls out of an airplane flying at a height of 500 m. Just before it hits the ground, the suitcase is falling with a speed of 80 m/s. What is the suitcase’s kinetic energy at this point? (a) 1.6 × 103 J (b) 5 × 104 J *(c) 6.4 × 104 J (d) 1 × 105 J (e) None of these Solution:
K = 12 mv 2 = 12 (20 kg)(80 m/s)2 = 6.4 × 104 J
The height from which the suitcase falls is irrelevant to the problem. If you’re feeling advanced, you can calculate the potential energy of the suitcase at that height. It’s less than the kinetic energy of the suitcase hitting the ground, so the suitcase lost mechanical energy (probably to air resistance) as it fell. Problem 1462. (Similarity Problem) A car with a mass of M and a truck with a mass of 9M are moving at the same speed V . If K1 is the kinetic energy of the car and K2 is the kinetic energy of the truck, which of the following equations is true? (a) K2 = K1 (b) K2 = 3K1 *(c) K2 = 9K1 (d) K2 = 81K1 (e) None of these Solution:
Since K = 21 mv 2 : K1 =
MV 2 2
K2 =
(9M )V 2 = 9K1 2
515 Problem 1463. (Similarity Problem) When a car is moving at speed v1 , its kinetic energy is K1 . When it is moving at speed v2 , its kinetic energy is K2 = 2K1 . What is the relationship between v2 and v1 ? √ (b) v2 = 2v1 *(a) v2 = √2 v1 (c) v2 = 2 2 v1 (d) v2 = 4v1 (e) None of these Solution: Use K = 12 mv 2 . In this problem, the mass doesn’t change; the speed and kinetic energy do. We separate the variables that don’t change from the ones that do: K K2 m K1 = 2 = 2 = 2 2 v v1 v2
⇒
v22 =
K2 2 v = 2v12 K1 1
⇒
v2 =
√ 2 v1
Problem 1464. (Similarity Problem) When a crate is pushed at speed vc across a smooth floor with a low coefficient of friction, it has kinetic energy K1 . When the same crate is pushed at the same speed across a rough floor with a higher coefficient of friction, its kinetic energy is K2 . What is the relationship between K1 and K2 ? (a) K1 > K2 *(b) K1 = K2 (c) K1 < K2 (d) Not enough information: need mass of crate (e) None of these Solution: The kinetic energy depends only on the mass of the crate and its speed. Since these are the same in both situations, the kinetic energy is the same. Problem 1465. You are testing your new potato gun at the ballistics lab. A potato fired at a speed of v has kinetic energy of KE. What is the mass of the potato? r r 2KE KE (b) (a) r 2m r m m KE (d) (c) 2KE 2m (e) None of these Solution:
Use the formula for kinetic energy: r 1/2 1 2 2K 2K 2 · 60 J 2 K = mv ⇒ v = ⇒ v= = = 18 m/s 2 m m 0.37 kg
516 Problem 1466. You have just acquired a Civil War cannon, and you want to know its muzzle velocity. You fire a cannonball with a mass of 7.3 kg into a ballistic pendulum, and discover that the kinetic energy of the cannonball was 370,000 J. What was the speed of the cannonball? Round your answer to the nearest 10 m/s. (a) 70 m/s *(c) 320 m/s (e) None of these Solution:
(b) 230 m/s (d) 5170 m/s
Since K = 12 mv 2 , solving for v gives: r v=
2K = m
s
2 · 370, 000 J = 320 m/s 7.3 kg
Problem 1467. (Impulse and K.E.) Two railroad cars are at rest on two horizontal frictionless tracks. One car is three times as heavy as the other. Each car is pushed with the same horizontal force of F for the same time T . At the end of this time, the kinetic energy of the lighter car is K. What is the kinetic energy of the heavier car? √ (a) 3K√ (b) 3K (c) K/ 3 *(d) K/3 (e) None of these Solution: Let the mass of the light car be M . We’ll construct a table with the kinematic properties of the two cars: Mass Acceleration = force/mass Speed = acceleration × time Kinetic energy =
1 2
× mass × speed2
Light car M F/M F T /M F 2T 2 K= 2M
Heavy car 3M F/3M F T /3M F 2T 2 K = 6M 3
517
14.4
Potential energy (P.E.)
Problem 1468. Which of the following statements is true? (a) If a force is directed downward, it does negative work. (b) The work done by a force on an object equals the increase in the object’s potential energy. (c) If a force has moved an object around a circle and back to its starting point, then the force has done no work on the object. *(d) If an object is moving in the positive direction, and if a force is applied to it in the negative direction, then the force does negative work on the object. Solution: We can rule out (a): what matters is not whether the force is directed upward or downward or sideways, but whether it has a positive or negative component in the direction of the object’s motion. We can rule out (b), since the work done on an object might have the effect of increasing its kinetic energy without changing the potential energy (e.g. accelerating a glider along a horizontal air track). We can rule out (c) for much the same reason: if the object is moving faster when it returns to its starting point than it was when it started, it has greater kinetic energy, so work has been done on it. Only (d) is true. Problem 1469. You move 56 kg of potatoes from the floor to a shelf 1.9 m high. By how much did you increase the gravitational potential energy of the potatoes? Round your answer to the nearest joule. (a) 101 J (c) 202 J (e) None of these Solution:
(b) 106 J *(d) 1043 J
The gravitational potential of the potatoes is Ugrav = mgy = (56 kg)(9.8 m/s2 )(1.9 m) = 1043 J
Problem 1470. A case of rat food has a mass of 10.3 kg. To keep rats from eating it, you move it from the floor to a shelf 2.2 m high. By how much did you increase the potential energy of the case? Round your answer to the nearest 10 J. (a) 160 J (c) 200 J (e) None of these
(b) 180 J *(d) 220 J
Solution: Let y = 0 be the level of the floor. Then the initial potential energy of the case is zero. The potential energy of the case on the shelf is Ugrav = mgy = (10.3 kg)(9.8m/s2 )(2.2 m) = 220 J
518 Problem 1471. A cat has a mass of 4.3 kg. It jumps up from the floor onto a countertop 1.2 m high. How much did the cat’s gravitational potential energy increase in going from the floor to the countertop? Round your answer to the nearest joule. (a) 5 J (c) 30 J (e) None of these
(b) 25 J *(d) 51 J
Solution: The gravitational potential energy of an object with mass m at a height h above the floor is Ugrav = mgh = (4.3 kg)(9.8 m/s2 )(1.2 m) = 51 J Problem 1472. A spring can store 36 J of energy when compressed by 0.12 m. What is the spring constant of the spring? (a) 300 N/m (c) 2500 N/m (e) None of these Solution:
(b) *(d)
600 N/m 5000 N/m
Since the energy stored in a spring is U = 12 kx2 , k=
2 · 36 J 2U = = 5000 N/m x2 (0.12 m)2
We were given x in meters, so we didn’t have to convert it. Problem 1473. You move 41 kg of physics books from the floor to a shelf 2.1 m high. By how much did you increase the gravitational potential energy of the books? Round your answer to the nearest joule. (a) 86 J (c) 422 J (e) None of these Solution:
(b) *(d)
181 J 844 J
The gravitational potential of the books is Ugrav = mgy = (41 kg)(9.8 m/s2 )(2.1 m) = 844 J
519
14.5
Work-Mechanical-Energy Theorem
Work-Mechanical-Energy Theorem: Wext = ∆M E. 14.5.1
Work-Kinetic-Energy Theorem
Special case 1: Work-Kinetic-Energy Theorem: Wext = ∆KE. Problem 1474. A bartender slides a glass of beer with a mass of 480 g along the top of a level bar. The coefficient of kinetic friction between the glass and the bar is µk = 0.13. When the glass leaves the bartender’s hand, it is moving at a speed of v0 = 2.1 m/s. How far does the glass travel before coming to a stop? Round your answer to the nearest 10 cm. (a) 140 cm (b) 160 cm *(c) 170 cm (d) 190 cm (e) None of these Solution: We know the mass m, the coefficient of friction µk , the initial velocity vi , and the final velocity vf = 0. We want the stopping distance ∆x. Apply the work-energy principle: Wnet = ∆K = Kf − Ki . The net force after the glass is released is friction, so Wnet = Fnet · (distance) = −µk N ∆x Since the top of the bar is level, N = mg. Hence Wnet = −µk mg∆x. Since vf = 0, Kf = 12 mvf2 = 0. Hence Wnet = −Ki = − 12 mv02 . Hence 1 − µk mg∆x = − mv02 2 2 v ÷(−m) −−−−−−−→ µk g∆x = 0 2 v02 (2.1 m/s)2 ÷µk g −−−−−−→ ∆x = = = 1.7 m = 170 cm 2µk g 2(0.13)(9.8 m/s2 )
520 Problem 1475. A skier slides down a slope and onto a frozen lake, where he slows down and comes to a halt. When he first reaches the lake, he is moving at 24 m/s. The coefficient of kinetic friction between his skis and the ice is 0.13. How far does he travel across the lake before stopping? Round your answer to the nearest 10 m. *(a) 230 m (b) 250 m (c) 270 m (d) 300 m (e) None of these Solution: Use the work-energy principle. If the skier’s mass is m and his initial speed is vi , then his initial kinetic energy is Ki = 21 mvi2 . He will slide until the negative work done by friction equals Ki . Since he is on a level surface, N = mg; so the force of kinetic friction is Fk = µk mg. If he slides a total distance of s, then the negative work done by friction is Fk s = µk mgs. Hence: 1 µk mgs = Ki = mvi2 2
⇒
s=
mvi2 v2 (24 m/s)2 = i = = 230 m 2µk mg 2µk g 2(0.13)(9.8 m/s2 )
Problem 1476. A skier slides down a small slope and onto a frozen lake, where he slows down and comes to a halt. When he first reaches the lake, he is moving at 10 m/s. He travels 50 meters across the lake before coming to a stop. What is the coefficient of kinetic friction between his skis and the ice? Round your answer to the nearest hundredth and approximate g by 10 m/s2 . (a) .15 *(b) .10 (c) .29 (d) .19 (e) None of these Solution: Use the work-energy principle. If the skier’s mass is m and his initial speed is vi , then his initial kinetic energy is Ki = 21 mvi2 . He will slide until the negative work done by friction equals Ki . Since he is on a level surface, the normal force is equal to weight N = mg; so the force of kinetic friction is fk = µk mg. If he slides a total distance of s, then the negative work done by friction is fk s = µk mgs. Hence: 1 µk mgs = Wfriction = Ki = mvi2 2 2 2 mvi v ÷ mgs −−−−−−−→ µk = = i 2smg 2sg (10 m/s)2 evaluate −−−−−−−−→ µk = = .10 2(50 m)(10 m/s2 )
521 Problem 1477. A car has a weight of 13,500 N. The coefficient of kinetic friction between its wheels and the wet highway is 0.47. The car is travelling at 18.4 m/s when the driver locks up the brakes. How far does the car travel before it comes to a complete stop? Round your answer to the nearest meter. *(a) (c) (e)
37 m 339 m None of these
(b) 159 m (d) 647 m
Solution: Step 1: Draw a picture and a free-body diagram (not shown here). Step 2: Write down what you’re given and what you want.
given:
want:
w = mg = 13, 500N (weight = mass × gravity) µ = 0.47 (coefficient of kinetic friction-car is moving) k vi = 18.4m/s (initial velocity) vf = 0m/s (final velocity) ∆x (the stopping distance)
Note: Time is not given, nor is it needed in this problem. Step 3: Use the appropriate fundamental kinematic equation and solve for the unknown. Alternatively, we can use the work-kinetic-energy theorem, which is what we’ll do. We can determine the mass m of the car from its weight w = mg, where g = 9.8 m/s2 . We can determine the kinetic energy K from the mass and the speed v. We can determine the force of kinetic friction fk from the weight and the coefficient of friction µk . Then the force of friction times the stopping distance x equals the kinetic energy. m= 1 1 w 2 wvi2 Ki = mvi2 = v = 2 2g i 2g
13, 500 N w = g 9.8 m/s2 and Kf = 0 (final K.E. - no motion)
fk = µw = (0.47)(13, 500 N) ∆K = Kf −Ki = −fk ·∆x
⇒
∆x =
Ki wv 2 /2g v2 (18.4 m/s)2 = = = = 37 m fk µw 2µg 2(4.7)(9.8 m/s2 )
Notice that the weight of the car cancels out. A heavy car would require more force to stop it than a light one; but it would have a greater frictional force.
522 Problem 1478. (Similarity Problem) You are driving at a speed of V on a straight and level road when you see a deer in front of you and lock up the brakes. You travel a distance D before coming to a complete stop. If your original speed had been 2V when you hit the brakes, what distance would you have travelled before stopping? √ 2D (a) D (b) (c) 2D *(d) 4D (e) None of these Solution: We will use the work-energy principle. Your car will move until the negative work done by friction equals its initial kinetic energy. If m is the mass of the car, v is its initial speed, Fk is the force of kinetic friction, and d is the stopping distance, then 1 2 mv = Fk d 2
⇒
d=
mv 2 2Fk
The mass of the car and the force of friction don’t change between trials. In the first trial, your initial speed is V , so mV 2 D= 2Fk In the second trial, your initial speed is 2V , so d=
14.5.2
m(2V )2 4mV 2 = = 4D 2Fk 2Fk
Work-Potential-Energy Theorem
Special case 2: Work-Potential-Energy Theorem: Wext = ∆P E. Problem 1479. (The potential energy in a spring) A giant spring with spring constant 50 N/m is 10 m long in its unstretched position. Suppose a force acts on the spring and stretches it to 12 m long. How much energy is stored in the spring? (a) 50 J (b) 100 J (c) 150 J (d) 200 J (e) None of these Solution: Step 1: Write down what you’re given and what you want. Given: Lu = 10 m, s = 12 m, and k = 50 N/m. Want: Uspring The equation for the potential energy stored in a spring is: 1 Uspring (x) = kx2 , where x = Ls − Lu is the displacement. Here x = 12 m − 10 m = 2 m. 2 1 Uspring (x = 2) = 50 22 J = 100 J 2
523 Problem 1480. A spring is hanging from the ceiling in the physics lab. When there is no weight hanging from it, the spring is 42 cm long; it has a spring constant of 8300 N/m. You hang a lead block with a mass of 87 kg from the spring, stretching it. What is the length of the spring with the weight hanging from it? Round your answer to the nearest centimeter. (a) 10 cm (b) 21 cm (c) 32 cm *(d) 52 cm (e) None of these Solution: We know the unstretched length of the spring Lu = 42 cm = 0.42 m; the spring constant k = 8300 N/m; and the mass m = 87 kg. If y is the change in length of the spring, we want the stretched length: Ls = Lu + y. Use the equation |Fspring | = ky to find y. We assume that the system is static, so there is no acceleration; so the two forces on the weight match: |Fspring | = mg = ky. Divide by k: mg y= k
⇒
mg (87 kg)(9.8 m/s2 ) Ls = Lu + = 0.42 m + = 0.52 m = 52 cm k 8300 N/m
Problem 1481. An unstretched spring is 15.0 cm long. If you hang an object with a mass of 8.8 kg from it, it stretches to a length of 19.2 cm. What is the spring constant of the spring? Round your answer to two significant figures. (a) 4.5 N/m (c) 210 N/m (e) None of these
(b) *(d)
22 N/m 2100 N/m
Solution: From Hooke’s law, F = kx. The force acting on the spring is F = mg, where m = 8.8 kg. The value of x is the difference between the stretched and the unstretched length of the spring: x = 0.192 m − 0.150 m = 0.042 m. (The answers are given in N/m, so we have to convert centimeters to meters.) Hence F mg (8.8 kg)(9.8 m/s2 ) = = = 2100 N/m k= x x 0.042 m
524 Problem 1482. An unstretched spring has a length of 22.8 cm. When an object with a mass of 12.0 kg is hung from it, it stretches to a length of 35.5 cm. What is the spring constant of the spring? Round your answer to the nearest N/m. (a) 34 N/m (c) 331 N/m (e) None of these
(b) *(d)
94 N/m 926 N/m
Solution: The force applied to the spring is the weight of the object w = mg. Since there is no acceleration in the mass-spring system, applying Newton’s 2nd law in the vertical direction gives |Fspring | = w = mg. The spring stretches by x = 35.5 cm − 22.8 cm = 0.127 m. (Since the problem asks for the spring constant in N/m, you must convert centimeters to meters.) Hooke’s law gives us |Fspring | = kx = mg
⇒
(12.0 kg)(9.8 m/s2 ) mg = = 926 N/m k= x 0.127 m
Problem 1483. A box has a mass of 22.2 kg. It is sitting on an icy sidewalk, which constitutes a horizontal frictionless surface. A spring with a spring constant of 103 N/m is attached to the box and pulled horizontally, so that the box accelerates at 1.31 m/s2 . How much does the spring stretch? Round your answer to the nearest 0.01 m. *(a) (c) (e)
0.28 m 2.11 m None of these
(b) (d)
1.38 m 2.77 m
Solution: The spring is pulled horizontally and there is no friction, so the force exerted by the spring is the mass of the box times its acceleration: F = ma. By Hooke’s law, F = kx. Hence ma = kx
⇒
x=
(22.2 kg)(1.31 m/s2 ) ma = = 0.28 m k 103 N/m
Problem 1484. A spring has a spring constant of 1700 N/m. You pull on it with a force that increases to 120 N until it stops stretching. How much work have you done on the spring? Round your answer to two significant digits. *(a) (c) (e)
4.2 J 14 J None of these
(b) 8.5 J (d) 60 J
Solution: When you pull on a spring with a force of magnitude F , the amount by which it stretches is given by: F = kx. Solving this for x gives: x = F/k. The work that it takes to stretch a spring by x is: 2 1 2 k F F2 (120 N)2 W = kx = = = = 4.2 J 2 2 k 2k 2 · 1700N/m
525 Problem 1485. A spring has a spring constant of 680 N/m. You pull on it with a force that increases to 170 N until it stops stretching. How much work have you done on the spring? Round your answer to two significant digits. *(a) (c) (e)
21 J 5400 J None of these
(b) 43 J (d) 11000 J
Solution: When you pull on a spring with a force of magnitude F , the amount by which it stretches is given by: F = kx. Solving this for x gives: x = F/k. The work that it takes to stretch a spring by x is: 2 F2 (170 N)2 1 2 k F = = = 21 J W = kx = 2 2 k 2k 2 · 680N/m Problem 1486. With no force applied to it, a spring is 24 cm long. You pull on it with a force of 330 N, stretching it to a length of 30 cm. How much work have you done in stretching the spring? *(a) 9.9 J (b) 49.5 J (c) 990 J (d) 4950 J (e) None of these Solution: The work that you have done is equal to the potential energy stored in the spring. We can find that using the formula: U = 21 kx2 . Here x is the change in length from equilibrium: x = 30 cm − 24 cm = 6 cm = 0.06 m. We can calculate k using the spring formula: F = kx ⇒ k = F/x. Then the potential energy is: U=
F x2 Fx (330 N)(0.06 m) kx2 = = = = 9.9 J 2 2x 2 2
Problem 1487. An unstretched spring is 0.24 m long. When a force that increases to 190 N is applied, it stretches until its total length is 0.30 m. How much work does it take to stretch the spring to this length? *(a) (c) (e)
5.7 J 28.5 J None of these
(b) 11.4 J (d) 57 J
Solution: The change in the spring’s length is x = 0.30 m − 0.24 m = 0.06 m. By Hooke’s law, F = kx, so k = F/x. The work that it takes to stretch the spring is kx2 Fx (190 N)(0.06 m) W = = = = 5.7 J 2 2 2 The spring’s lengths were given in meters, so we didn’t have to convert.
526 Problem 1488. An unstretched spring has a length of 23.8 cm. When an object with a mass of 18.2 kg is hung from it, it stretches to a length of 32.2 cm. What is the spring constant of the spring? Round your answer to the nearest N/m. (a) 217 N/m *(c) 2123 N/m (e) None of these
(b) 554 N/m (d) 9435 N/m
Solution: The force applied to the spring is the weight of the object w = mg. Since there is no acceleration in the mass-spring system, applying Newton’s 2nd law in the vertical direction gives |Fspring | = w = mg. The spring stretches by x = 32.2 cm − 23.8 cm = 0.084 m. (Since the problem asks for the spring constant in N/m, you must convert centimeters to meters.) Hooke’s law gives us |Fspring | = kx = mg
⇒
(18.2 kg)(9.8 m/s2 ) mg = = 2123 N/m k= x 0.084 m
Problem 1489. With no force applied to it, a spring hanging from the ceiling is 24 cm long. You hang a weight of 330 N from it, stretching it to a length of 30 cm. How much work was done by gravity in stretching the spring? (a) 9.9 J (b) 49.5 J (c) 990 J (d) 4950 J (e) None of these Solution: By applying a force (the weight of the object) through a distance x (the displacement) gravity has done work on the system. The work done on the springmass system is equal to the potential energy stored in the spring. We can find that using the formula: U = 12 kx2 . Here x is the change in length from equilibrium: x = Ls − Lu = 30 cm − 24 cm = 6 cm = 0.06 m. We can calculate k using the spring formula: F = kx ⇒ k = F/x. Then the potential energy is: U=
kx2 F x2 Fx (330 N)(0.06 m) = = = = 9.9 J 2 2x 2 2
527 Problem 1490. (Derive Problem-Requires Calculus) [Approximating the strength of gravity as a function of height] The magnitude of the gravitational attraction between two particles of mass m1 and m2 is given by |Fg (r)| = G
m1 m2 , r2
where r is the distance between the two particles and the force that each particle exerts on the other is directed along the radial direction between the two particles, and G is the universal gravitational constant. (a) What is the potential energy function U = Ug (r), if we take U0 = lim Ug (r) = 0? r→∞ That is, we take our reference point out at r = ∞. (b) How much work Wg must be done against the conservative gravitational force to increase the separation distance from r = R to r = R + h? (c) Let m1 = Me the mass of the earth and m2 = m the mass of a small point particle on the earth, where we assume that m Me . Let R = Re be the mean radius of the earth and h be the height that the particle is above the ground, where h Re . Compute the work done against gravity in moving the object from the earth’s surface at r = Re to a height h above the ground r = Re + h. Denote this work by Wg . Express your answer h GMe , and the variable . Since Re , Me , m, and G are in terms of the parameter g = 2 Re Re fixed, the work will be a function of h. However, it will turn out to be advantageous to h think of the work as a function of . Re (d) We are now going to repeat the calculation in part (c) except this time we are going to approximate the force of gravity as a constant force Fgrav = −mg, where g = GMe is the constant magnitude of gravity that is assumed to be independent of the Re2 separation distance between the two attracting masses. This is nothing new, when we apply Newton’s second law to trajectory problems involving gravity on the earth’s surface, such as computing the path of a cannon ball launched from the ground, we take the force of gravity as a constant. Compute the work Wgrav done in moving the same object of mass m in part (c) from the ground at y = 0 to height y = h against the constant force of gravity Fgrav = −mg. (e) We now want to compare the exact result for the work found in part (c) with the approximate value of work found in part (d). What is the size of the relative error between the work found in part (c) and (d)? Hint: W.L.O.G. we can take Wgrav as our reference measure. Show that the relative error is h Wg − Wgrav =O . Wgrav Re (f) The mean radius of the earth is approximately 6370 km. Use the relative error that you find to predict how high you can go above the earth’s surface and safely ignore the fact that gravity is changing with height when you compute the work done in lifting an object from the earth’s surface.
528
Figure 7: Earth-mass system. Solution:
See hand-written solutions.
529
14.6
Conservation of mechanical energy (M.E.)
Problem 1491. Planet X has no air, hence no air resistance. An astronaut standing atop a cliff on Planet X drops a rock down into the lava lake below. If the gravitational potential energy of the rock just before being dropped is Ugrav , and its kinetic energy just as it hits the lava is K, what is the relationship between Ugrav and K? Use the level of the lava lake as y = 0. (a) Ugrav < K *(b) Ugrav = K (c) Ugrav > K (d) Not enough information: it depends on Planet X’s gravity gX . Solution: Since there is no air resistance, we assume that mechanical energy is conserved: Ui + Ki = Uf + Kf . Initially, just as the astronaut drops the rock, it has zero velocity; so Ki = 0. Just as the rock hits the lava, it is at y = 0, so Uf = 0. Hence Ugrav = Ui = Kf = K Problem 1492. Two identical stone blocks are transported from the bottom to the top of a cliff. The first block is fastened to a rope and pulled straight up through the air to the top. The second block is pulled up a frictionless ramp to the top of the cliff. If the work done on the first block is W1 and the work done on the second block is W2 , what is the relationship between W1 and W2 ? (a) W1 < W2 *(b) W1 = W2 (c) W1 > W2 (d) Not enough information: any of these could be true Solution: Since there is no friction, conservation of energy applies; so the work done on each block is equal to the change in its potential energy. Since the blocks are identical, they have the same mass; and they are raised by the same height. Hence the potential energy change, and therefore the work, is the same for both blocks. Problem 1493. Two frictionless inclined planes have the same vertical height Y . Plane 1 makes an angle of 60◦ to the horizontal; plane 2 makes an angle of 30◦ to the horizontal. Two objects with the same mass M are released from the tops of the planes to slide down. The object that slides down plane 1 has speed v1 when it reaches the bottom; the object that slides down plane 2 has speed v2 when it reaches the bottom. What is the relationship between v1 and v2 ? *(a) v1 = v2 (b) v1 = 2v2 ◦ ◦ (c) v1 sin 30 = v2 sin 60 (d) v1 cos 60◦ = v2 cos 30◦ (e) None of these Solution: The key word is “frictionless”. Since there are no nonconservative forces, mechanical energy must be conserved. Initially, both objects have zero kinetic energy and the same potential energy (since they have the same mass M and they’re at the same height Y ). When they reach the bottom of their planes, they still have the same mechanical energy; since they have height zero, their potential energy is zero; so their mechanical energy is all kinetic, and it’s the same for both. Since they have the same kinetic energy and the same mass, their speeds must be the same.
530 Problem 1494. Ricky, the 5 kg snowboarding raccoon, is on his snowboard atop a frictionless ice-covered quarter-pipe with radius 3 meters (see figure at right). If he starts from rest at the top of the quarter-pipe, what is his speed at the bottom of the quarter-pipe? For ease of calculation, assume that g = 10 m/s2 . Round your answer to the nearest 0.1 m/s. (a) 7.0 m/s (c) 8.5 m/s (e) None of these
*(b) (d)
x
Raccoon: 5 kg 3m 3m
7.7 m/s 9.3 m/s
Solution: We are given: m = 5 Kg, R = 3 m, vi = 0 m/s, and g = 10 m/s2 . We want: vf =? We assume: no friction, hence the mechanical energy of the system is conserved. Since the system, Ricky and the earth, has no external forces on it, such as friction, we can apply the conservation of mechanical energy: MEi = MEf . Now, the initial potential due to gravity is Ugrav = mgh = mgR and the initial kinetic energy is Ki = 0 since Ricky is at rest. The final P.E. is zero, since Ricky is now at ground level h = 0. The final K.E. is Kf = 12 mvf2 . Using the definition of mechanical energy: ME = PE + KE gives 1 mgR = PEi + KEi = MEi = MEf = PEf + KEf = mvf2 2 ÷1m
2 −−−− −−→ vf2 = 2Rg √ p 1/2 −−−−−−→ vf = 2Rg = (2)(3 m)(10 m/s2 ) = 7.7 m/s
Notice that since R = h, this is the same speed that Ricky would have if he had jumped straight down (go look back at our 1-D kinematic problems). The only difference is the direction of his speed would have been different. If he had jumped straight down, then his velocity would also have been straight down, and into the ground would go Ricky; but the quarter pipe redirected his motion to being tangent to the ground, while preserving his speed. So instead of busting his head, and winding up dead; Ricky sped off to work, where he was a clerk. Were it not for conservation of mechanical energy he would have been late for his job, and replaced by his competitor Rob.
531 Problem 1495. The surface gravity on Planet X is gX . An astronaut standing on the surface of the planet fires a bullet with mass m straight upward with kinetic energy K. What is the maximum height hmax reached by the bullet? s 2K K (a) *(b) mgX mgX r 2K K2 (c) (d) m mgX (e)
None of these
Solution: Apply the conservation of mechanical energy. Take the initial position to be the ground at the moment the potato is launched, and the final position to be the highest point of the trajectory. Let y = 0 correspond to the initial position. At the initial and final points: ( Ui = mgy = 0 (y = 0) Ki = K and ( Uf = mgX hmax Kf = 0 (ball is momentarily at rest at the top of the trajectory) Notice that at its highest point, the potato will have zero kinetic energy, so all of its energy will be potential. At this height hmax , K = mgX hmax
⇒
hmax =
K mgX
532 Problem 1496. Planet X has a surface gravity of 13.6 m/s2 and no atmosphere. An astronaut on Planet X fires a potato gun straight upward; the potato has a mass of 0.73 kg, and leaves the gun with a speed of 34.3 m/s. What is the maximum height that the potato reaches? Round your answer to the nearest meter. (a) 7 m (c) 34 m (e) None of these
(b) 24 m *(d) 43 m
Solution: Apply the conservation of mechanical energy. Take the initial position to be the ground at the moment the potato is launched, and with the final position to be the highest point of the trajectory. Let y = 0 correspond to the initial position. At the initial and final points: ( Ui = mgy = 0 (y = 0) Ki = 12 mvi2 = 430J and ( Uf = mgX hmax Kf = 0 (ball is momentarily at rest at the top of the trajectory) Notice that at its highest point, the potato will have zero kinetic energy, so all of its energy will be potential. At this height hmax , Ki = mgX hmax
⇒
hmax =
Ki 430 J = 43 m = mgX (0.73 kg)(13.6 m/s2 )
533 Problem 1497. (Lab Problem) A spring cannon shoots a ball with a mass of 28 g vertically into the air from ground level. In this situation, air resistance can be ignored. If the spring is compressed by a distance of 1.2 cm, the ball reaches a maximum height of 2.4 m. What is the spring constant of the spring in the cannon? Round your answer to the nearest 100 N/m. (a) 6700 N/m (b) 7400 N/m (c) 8200 N/m *(d) 9100 N/m (e) None of these Solution: We know the mass of the ball m, the compression distance y of the spring, the maximum height hmax of the ball, and the value of g. We want to know the spring constant k. Since there’s no air resistance and no friction, we can assume that mechanical energy is conserved: Ki + Ui = Kf + Uf . We will take the initial position (subscript i) as that of the cannon with its spring compressed, just before it is fired; and the final position (subscript f ) as the system when the ball has reached its maximum height. Since the ball isn’t moving in either case, Ki = Kf = 0. Hence Ui = Uf . The initial potential energy is all in the spring: Ui = 12 ky 2 . The final potential energy is all gravitational: Uf = mghmax . Hence 1 2 ky = mghmax 2 2mghmax 2(0.028 kg)(9.8 m/s2 )(2.4 m) · (2/y 2 ) −−−−−→ k = = = 9100 N/m y2 (0.012 m)2 Problem 1498. (Lab Problem) A spring cannon shoots a ball with a mass of 9.6 g vertically into the air from ground level. In this situation, air resistance can be ignored. The spring is compressed by a distance of 3.4 cm below the mouth of the cannon. If the ball reaches a maximum height of 96 cm above the mouth of the cannon, then what is the spring constant of the spring in the cannon? Round your answer to the nearest 10 N/m. Take y = 0 to be at the mouth of the cannon. (a) 140 N/m (b) 150 N/m *(c) 160 N/m (d) 170 N/m (e) None of these Solution: We know the mass of the ball m, the compression distance ∆y of the spring, the maximum height ymax of the ball, and the value of g. We want to know the spring constant k. Since there’s no air resistance and no friction, we can assume that mechanical energy is conserved: Ki + Ui = Kf + Uf . We will take the initial position (subscript i) as that of the cannon with its spring compressed, just before it is fired; and the final position (subscript f ) as the system when the ball has reached its maximum height ymax . Since the ball isn’t moving in either case, Ki = Kf = 0. Hence Ui = Uf . The initial potential energy is the energy stored in the spring plus the negative gravitational potential energy (because the ball is ∆y below the zero gravitational reference
534 potential line at y = 0): Ui = Ugrav + Uspring = −mg∆y + 21 k(∆y)2 . The final potential energy is all gravitational: Uf = mgymax . Hence Ui = Uf becomes
+ mg∆y −−−−−−→
1 − mg∆y + k(∆y)2 = mgymax 2 1 k(∆y)2 = mgymax + mg∆y 2
2 · (∆y)2 2mg (ymax + ∆y) −−−−−−→ k = (∆y)2 evaluate
−−−−−−−−→ k =
2(0.0096 kg)(9.8 m/s2 ) (0.034 m)2
! (0.96 + 0.034) m = 162 N/m ≈ 160 N/m
Note: We could have taken the zero reference line to be the top of the ball when it is in its initial position sitting on the compressed spring. In this case the maximum height would be htextmax = ymax + ∆y. Problem 1499. (Lab Similarity Problem) A spring cannon is used to launch two balls straight upward: first, a ball with mass M ; then a ball with mass 2M . The spring in the cannon is compressed by the same amount in both launchings. Which of the following statements is true? (a) The two balls reach the same maximum height. (b) The larger ball will leave the spring cannon at a higher speed than the smaller ball. *(c) The two balls leave the spring cannon with the same kinetic energy. (d) At their maximum heights, the smaller ball has greater potential energy than the larger ball. Solution: In both trials, the spring is compressed by the same amount; so in both trials, it has the same potential energy. When the balls are launched, the potential energy of the spring is entirely converted into kinetic energy of the ball. Thus both balls leave the spring cannon with the same kinetic energy. Since the balls have different masses, they leave the spring cannon with different speeds: the small ball at a greater speed than the large ball. That lets us rule out (b). At their maximum heights, each ball stops moving; so its initial kinetic energy is completely converted into gravitational potential energy. Since they started with the same kinetic energy, they must have the same potential energy at their maximum heights; so we can rule out (d). Since they have different masses, they must reach different maximum heights in order to have the same potential energy there; that rules out (a).
535 Problem 1500. A baseball with mass M and a cannonball with mass 9M are thrown vertically upward from ground level at the same initial speed V , with no air resistance. The initial kinetic energy of the baseball is Kb ; it reaches a maximum height of Yb ; and at this height, its gravitational potential energy is Ub . The corresponding values for the cannonball are Kc , Yc , and Uc . Which of the following is true? (a) Yb = 3Yc *(b) Uc = 9Ub (c) Kb = Kc (d) Kc = 81Kb (e) None of these Solution: Since there is no air resistance, we assume that mechanical energy is conserved. Initially at ground level, the potential energy is zero and the mechanical energy consists entirely of the ball’s kinetic energy: Kb = M V 2 /2 and Kc = (9M )V 2 /2 = 9Kb . At the high point, the ball is motionless, so its kinetic energy is zero. At this point, the potential energy of each ball equals its initial kinetic energy; so Uc = Kc = 9Kb = 9Ub . Problem 1501. (Lab Problem) A frictionless horizontal air track has a horizontal spring at one end, with a spring constant of 800 N/m. A glider with a mass of 500 g is pressed against the spring, compressing it by 4 cm. When the glider is released, the spring pushes it down the track. How fast is the glider moving when it leaves the spring? (a) 20 cm/s (b) 40 cm/s (c) 80 cm/s *(d) 160 cm/s (e) None of these Solution: We assume that energy is conserved; so the potential energy of the spring before the glider is released equals the kinetic energy of the glider after it leaves the spring. Don’t forget to convert units to kilograms and meters. 1 1 Uel = kx2 = K = mv 2 2 2 ⇒ kx2 = mv 2 kx2 ⇒ v2 = rm 1/2 k 800 N/m ⇒ v= x= (0.04 m) = 1.6 m/s = 160 cm/s m 0.5 kg
536 Problem 1502. (Lab Problem) A horizontal spring with spring constant k1 is at one end of a frictionless horizontal air track. A glider is pushed against the spring, compressing it by a distance of x1 , so that the energy stored in the spring is 100 J. After the glider is released, it reaches a speed of v1 and a kinetic energy of K1 . The procedure is then repeated, with the spring replaced by one with spring constant k2 > k1 ; the new spring is compressed by a distance of x2 , so that the energy stored in it is once again 100 J. When the glider is released this time, it reaches a speed of v2 and a kinetic energy of K2 . Which of the following is true? (a) x1 = x2 (b) K1 < K2 *(c) v1 = v2 (d) k1 x21 < k2 x22 (e) None of these Solution: The air track is frictionless, so we assume that there are no nonconservative forces, which means that mechanical energy is conserved. Initially, the spring is compressed and the glider is not moving; so the kinetic energy is zero and the mechanical energy equals the potential energy of the spring. After the glider has been released and has reached its maximum speed, there is no potential energy left in the spring, and the mechanical energy equals the kinetic energy of the glider. In both trials, the initial potential energy is U1 = U2 = 100 J. Hence in both trials, the final kinetic energy is K1 = K2 = 100 J. Since the mass of the glider doesn’t change from the first trial to the second, and since its kinetic energy is the same in both trials, its velocity must be the same: v1 = v2 .
537 Problem 1503. (Lab Problem) A frictionless horizontal air track has a spring at either end. The spring on the left has a spring constant of kL ; the one on the right has a spring constant of kR . A glider with a mass of m is pressed against the left-hand spring, compressing it by a distance of xL from its equilibrium position. The glider is then released, so that the spring propels it rightward. It slides along the track and into the right-hand spring. Find a formula for the maximum compression of the right-hand spring xR as a function of the given parameters: xL , kL , kR , m. (The diagram following the answers shows the situation before the glider is released, when the left-hand spring is compressed and the glider is not moving.) q q (b) xR = kkRL xL *(a) xR = kkRL xL (c) (e)
xR = kkRL xL None of these
(d) xR = xL
m E E E E
C C C C C C
Solution: We know the mass m of the glider, the spring constants KL and KR of the left- and right-hand springs, and the compression xL of the left-hand spring. We want to know the compression xR of the right-hand spring. There is no friction, so we can assume that mechanical energy is conserved: Ui + Ki = Uf + Kf . We will take the initial situation to be that in which the left-hand spring is fully compressed, before the glider is released to move; and the final situation to be that in which the glider has pushed the right-hand spring to its maximum compression. Since the glider isn’t moving in either case, Ki = Kf = 0; so Ui = Uf . In both situations, the potential energy is all in a spring: Ui = 12 kL x2L = 12 kR x2R = Uf . We can solve this for xR : ·2
−−−−→ kR x2R = kL x2L kL x2L ÷kR −−−−− −→ x2R = k sR r √ kL x2L kL −−−−−→ xR = = xL kR kR Notice that we never needed the mass of the glider. Its only function in the problem was to carry all the energy from one spring to the other.
538 Problem 1504. (Lab Problem) A frictionless horizontal air track has a spring at either end. The spring on the left has a spring constant of 1800 N/m; the one on the right has a spring constant of 1300 N/m. A glider with a mass of 4.30 kg is pressed against the left-hand spring, compressing it by 3.9 cm. The glider is then released, so that the spring propels it rightward. It slides along the track and into the right-hand spring. What is the maximum compression of the right-hand spring? Round your answer to the nearest 0.1 cm. (The diagram following the answers shows the situation before the glider is released, when the left-hand spring is compressed and the glider is not moving.) (a) 2.0 cm *(c) 4.6 cm (e) None of these
(b) 2.8 cm (d) 5.4 cm
4.30 kg E E E E
C C C C C C
Solution: We know the mass m of the glider, the spring constants KL and KR of the left- and right-hand springs, and the compression xL of the left-hand spring. We want to know the compression xR of the right-hand spring. There is no friction, so we can assume that mechanical energy is conserved: Ui + Ki = Uf + Kf . We will take the initial situation to be that in which the left-hand spring is fully compressed, before the glider is released to move; and the final situation to be that in which the glider has pushed the right-hand spring to its maximum compression. Since the glider isn’t moving in either case, Ki = Kf = 0; so Ui = Uf . In both situations, the potential energy is all in a spring: Ui = 12 kL x2L = 12 kR x2R = Uf . We can solve this for xR : ·2
−−−−→ kR x2R = kL x2L kL x2L ÷kR −−−−− −→ x2R = k sR s r √ kL x2L kL 1800 N/m −−−−−→ xR = = xL = (0.039 m) = 0.046 m = 4.6 cm kR kR 1300 N/m Notice that we never needed the mass of the glider. Its only function in the problem was to carry all the energy from one spring to the other.
539 Problem 1505. You drop a rock off the Navajo Bridge and let it fall 467 feet to the Colorado River. This is a situation in which air resistance is too large to be ignored. If the gravitational potential energy of the rock just before being dropped is Ugrav , and its kinetic energy just as it hits the water is K, what is the relationship between Ugrav and K? Use the level of the river as y = 0. (a) Ugrav < K (b) Ugrav = K *(c) Ugrav > K (d) Not enough information: any of these could be true Solution: If there were no air resistance, then the mechanical energy (M.E.) of the rock-earth system would be conserved: Ui +Ki = Uf +Kf . However, air resistance means that the rock losses some of its kinetic energy (K.E.) to the molecules in the atmosphere (the atmosphere is not considered part of the rock earth system). The atmosphere acts like a K.E. sink, it converts K.E. into heat! Thus, the final M.E. will be less than the initial M.E. In particular, Ui + Ki = Uf + Kf + Heat > Uf + Kf . Initially, the rock has no K.E. since it starts from rest (zero velocity); so Ki = 0. Just as the rock hits the river, it is at y = 0, so Uf = 0. Hence Ugrav = Ui > Kf = K Problem 1506. A frozen pond acts as a frictionless horizontal surface. A spring with a spring constant of 32,000 N/m is attached to a wall at the edge of the pond, and a piano with a mass of 450 kg is pushed against the spring, compressing it by 0.18 m. When the piano is released, how fast does it slide across the ice? Round your answer to the nearest 0.1 m/s. *(a) (c) (e)
1.5 m/s 4.6 m/s None of these
(b) (d)
2.3 m/s 12.8 m/s
Solution: Before the piano is released, it is stationary; so the initial kinetic energy of the spring-piano system is Ki = 0. The potential energy of the system is that stored in the spring: Ui = 12 kx2 . Hence the total mechanical energy of the system is Ui +Ki = 12 kx2 . After the piano is released, this energy is completely transferred from the spring to the piano. The final kinetic energy of the system is Kf = 12 mv 2 , and the final potential energy is Uf = 0. Thus the total mechanical energy of the system is Uf + Kf = 21 mv 2 . By conservation of energy, the initial and final mechanical energies of the system must be the same. Hence: s r 2 2 2 kx mv kx (32, 000 N/m)(0.18 m)2 = ⇒ v= = = 1.5 m/s 2 2 m 450 kg
540 Problem 1507. (Derive Problem [Ball in the bucket]) You want to shoot a ball of mass m out of a spring cannon into a bucket. The spring cannon is aimed to fire the ball in the horizontal direction. The bucket’s radius is just a little larger than the ball’s radius. The bucket is designed to capture the ball so long as the ball’s center of mass hits the center of the top of the bucket. The cannon’s manufacturer gives the spring constant as k. You measure the following distances: H, the height from the top of the bucket to the center of mass of the ball as it sits in the cannon; and R, the distance from the end of the muzzle of the cannon (where the ball exits the cannon) to the center of the top of the bucket. Neglecting friction and air drag, and treating the ball as a point particle, derive a formula for the distance d that the spring must be compressed in order to shoot the ball into the bucket (see figure below). Your final answer should be an expression for d as a function of m, k, g, H, and R. Justify your answer. r r mg mg *(b) R (a) 2Hk 2Hk s r Hk mg (d) R (c) R mg Hk (e) None of these
Figure 8: Ball in the bucket. Solution:
See hand-written solutions
541 Problem 1508. (Derive Problem) A small block of mass m slides along a frictionless loop-the-loop track. The radius of the loop is R. At what height H above the bottom of the track should the block be released from so that it just makes it through the loop-theloop without losing contact with the track? Hint: We want the maximum height such that the normal force exerted by the track on the block at the top of the loop is zero. That is, if we release the block from a height of H + , where is arbitrarily small, then the normal force at the top of the loop will not be zero. In order to receive any credit for this problem, you must show all the work necessary to arrive at the answer. Simply writing down the final formula will earn you a grade of zero. 3 2 R (b) R (a) 3 4 3 5 R (d) *(c) 2 2 (e) None of these
Figure 9: The loop-the-loop. Solution:
See hand-written solutions
542 Problem 1509. (Derive Problem) A small block of mass m sits at rest at the top of a frictionless hemispherical mound of ice of radius R. The block is given a slight tap by a very small bug and begins to slide down the side of the mound. Find the height H above the ground where the block leaves the ice. Express this height as a function of the radius R. Hint: Look for the position where the normal force first vanishes. This will be the point where the block leaves the ice. 3 2 R (b) R *(a) 3 4 5 3 (c) R (d) 2 2 (e) None of these
Figure 10: Block sliding off a hemispherical piece of ice. Solution:
See hand-written solutions
543
15 15.1
Conservation of Linear Momentum Computing Linear Momentum p = mv
Problem 1510. A skateboarder with a mass of 60 kg is moving at a speed of 3 m/s. What is his momentum? *(a) 180 kg m/s (b) 270 kg m/s (c) 360 kg m/s (d) 540 kg m/s (e) None of these Solution:
p = mv = (60 kg)(3 m/s) = 180 kg m/s.
Problem 1511. A shopping cart with a mass of 13.4 kg is rolling across a parking lot at 4.0 m/s. What is the cart’s momentum? (a) 26.8 kg m/s (c) 107.2 kg m/s (e) None of these Solution:
*(b) (d)
53.6 kg m/s 214.4 kg m/s
p = mv = (13.4 kg)(4.0 m/s) = 53.6 kg m/s.
Problem 1512. Your potato gun launches a potato with a mass of 370 g at a speed of 21 m/s. What is the magnitude of the potato’s momentum? Round your answer to two significant digits. (a) 3.9 kg m/s (c) 82 kg m/s (e) None of these
*(b) 7.8 kg m/s (d) 160 kg m/s
Solution: p = mv = (0.37 kg)(21 m/s) = 7.8 kg m/s. Don’t forget to convert grams to kilograms. Problem 1513. A flying duck with a mass of 2 kg has a momentum of 40 kg m/s. What is the duck’s speed? (a) 10 m/s (c) 40 m/s (e) None of these Solution:
p = mv
*(b) (d)
⇒
v=
20 m/s 80 m/s 40 kg m/s p = = 20 m/s m 2 kg
544 Problem 1514. An artillery shell has a mass of 5 kg and a momentum of 4000 kg m/s. What is the shell’s speed? (a) 40 m/s (c) 160 m/s (e) None of these Solution:
15.2
p = mv
(b) *(d)
⇒
80 m/s 800 m/s 4000 kg m/s p = = 800 m/s m 5 kg
v=
Applying conservation of momentum to isolated systems
Problem 1515. You drop an object from rest on a planet with no atmosphere. As the object falls, which of the quantities of the object listed below is conserved? (a) Momentum *(b) Mechanical energy (c) Potential energy (d) Kinetic energy Solution: The object is accelerating in a straight line due to the force of gravity. Its speed increases as it falls; so its momentum and its kinetic energy are both increasing. Thus they are not conserved. The potential energy decreases as the object falls, so it is not conserved. Since the planet has no atmosphere, there is no air drag or other dissipative force. Thus the object-planet system is an isolated system and conserves mechanical energy. Problem 1516. You drop a wrench from the top of a bridge. Which property of the wrench is conserved as it falls? Assume that there is no air resistance. (a) Momentum (b) Kinetic energy (c) Potential energy *(d) Mechanical energy (e) None of these Problem 1517. A cannon has a mass of 1100 kg. It fires a cannonball with a mass of 5.3 kg at a muzzle velocity of 390 m/s. How fast does the cannon move backward when the ball is fired? Round your answer to two significant digits. *(a) (c) (e)
1.9 m/s 18 m/s None of these
(b) (d)
2.8 m/s 27 m/s
Solution: Assuming that the cannon-cannonball system is initially at rest, the total momentum is zero. Hence the backward momentum of the cannon must equal the forward momentum of the cannonball: mc vc = mb vb
⇒
vc =
(5.3 kg)(390 m/s) mb vb = = 1.9 m/s mc 1100 kg
545 Problem 1518. A cannon has a mass of 1500 kg. It fires a cannonball with a mass of 5.5 kg at a muzzle velocity of 370 m/s. How fast does the cannon move backward when the ball is fired? Round your answer to two significant digits. *(a) (c) (e)
-1.4 m/s -11 m/s None of these
(b) (d)
-2.7 m/s -22 m/s
Solution: Assuming that the cannon-cannonball system is an isolated system that is initially at rest, the total momentum is zero. Hence by conservation of momentum, the magnitude of the backward momentum of the cannon must equal the forward momentum of the cannonball: Psys,f = mc vc + mb vb = 0
⇒
vc = −
(5.5 kg)(370 m/s) mb vb =− = −1.4 m/s mc 1500 kg
Problem 1519. An astronaut with a mass of 80 kg is adrift in space outside of his space station. Fortunately, he is holding a space wrench with a mass of 2 kg. How fast must he throw the wrench to give himself a recoil speed of 20 cm/s? *(a) (c) (e)
8 m/s 20 m/s None of these
(b) 16 m/s (d) 40 m/s
Solution: The initial momentum of the system is zero; so the backward momentum of the astronaut must equal the forward momentum of the wrench. Don’t forget to convert centimeters to meters. Starting from ∆Pastronaunt = −∆Pwrench and ignoring the negative sign, since we know the two objects move in opposite directions, we have ma va = mw vw
⇒
vw =
(80 kg)(0.2 m/s) ma va = = 8 m/s mw 2 kg
546 Problem 1520. A cannon with a mass of mc is sitting on a frictionless frozen lake. It fires a cannonball with a mass of mb horizontally at a speed of vb . After it has fired the ball, what is the cannon’s backward recoil speed? s mb vb2 mb vb (a) (b) mb + mc mb + mc s mb vb2 mb vb *(d) (c) mc mc (e)
None of these
Solution: We know the masses and initial velocities (zero) of the two bodies, and the final velocity of the cannonball. We want to know the final velocity vc of the cannon. We can use conservation of momentum. The initial momentum is zero, so after the cannon is fired, its backward momentum must equal the forward momentum of the ball: mc vc = mb vb
÷m
c −−−→
vc =
mb vb mc
Problem 1521. (Lab Problem: The spring-cannon glider - a discrete rocket) A spring cannon is fastened to a glider at rest on a horizontal frictionless air track. The spring-cannon-glider assembly has a mass of mc . It has been previously determined that the cannon can launch a ball with a mass of mb horizontally at a speed of vb . Use the conservation of momentum to determine the recoil speed of the cannon after it launches the ball. Take the positive x-axis in the direction of the ball. r mb mb vb (b) − vb (a) − mg mb + mg *(c) − (e)
mb vb mg
(d) −vb
None of these
Solution: We take our system to be the ball-spring-cannon-glider system. Since we are only interested in the motion of the spring-cannon glider, we can ignore any effects of gravity on the ball. We can then treat the system as isolated and apply the conservation of momentum to the system, which gives psys,i = psys,f . Since the system is initially at rest, psys,i = 0. It follows that psys,f = mg vg + mb vb = 0
⇒
vg = −
mb vb . mg
Note: This problem simulates, in a simple discrete way, the idea behind a rocket. The main difference is that a rocket expels gas continuously, where as the spring cannon only does it once. This type of problem is known as an explosion problem and it acts like a perfectly inelastic collision in reverse (if we ran the film backwards).
547 Problem 1522. (Lab Problem) A spring cannon is fastened to a glider at rest on a horizontal frictionless air track. The combined mass of the cannon and the glider is 400 g. The cannon fires a ball with a mass of 20 g horizontally at a speed of 8 m/s. How fast does the cannon-glider assembly move backward after the cannon is fired? (a) 20 cm/s *(c) 40 cm/s (e) None of these
(b) 25 cm/s (d) 50 cm/s
Solution: Use conservation of momentum: psys,i = psys,f . Since the system is initially at rest, its momentum is zero. After the cannon is fired, the forward momentum of the ball must equal the backward momentum of the cannon. Don’t forget to convert units. mb vb = mc vc
⇒
vc =
(0.02 kg)(8 m/s) mb vb = = 0.4 m/s = 40 cm/s mc 0.4 kg
548
15.3
Impulse
Problem 1523. An irate constituent throws a pie at a congressman. The pie has a mass of 1.2 kg and hits the congressman with a horizontal speed of 14 m/s. How large is the impulse that the pie gives to the congressman? Round your answer to two significant figures. (a) 8.4 kg m/s (c) 120 kg m/s (e) None of these
*(b) 17 kg m/s (d) 140 kg m/s
Solution: The impulse given by the pie is equal to the change ∆p in its momentum. We assume that the politician is large and heavy compared to the pie, so that the pie comes to a complete stop when it hits him. Then the final momentum of the pie is zero; so the impulse equals the original momentum of the pie: J = ppie = mv = (1.2 kg)(14 m/s) = 17 kg m/s Problem 1524. An irate constituent throws a pie at a congressman. The pie has a mass of 1.6 kg and hits the congressman with a horizontal speed of 14 m/s. How large is the impulse that the pie gives to the congressman? Round your answer to two significant figures. *(a) (c) (e)
22 kg m/s 160 kg m/s None of these
(b) 45 kg m/s (d) 310 kg m/s
Solution: The impulse given by the pie is equal to the change ∆p in its momentum. We assume that the politician is large and heavy compared to the pie, so that the pie comes to a complete stop when it hits him. Then the final momentum of the pie is zero; so the impulse equals the original momentum of the pie: J = ppie = mv = (1.6 kg)(14 m/s) = 22 kg m/s Problem 1525. A golf ball has a mass of 0.071 kg. It is placed on a tee and hit, giving it a speed of 42 m/s. Golf scientists determine that the club is in contact with the ball for 0.046 s. Assuming constant acceleration, what is the average force exerted on the ball by the club? Round your answer to the nearest newton. (a) 63 N (c) 125 N (e) None of these
*(b) (d)
65 N 1361 N
Solution: The golf ball starts at rest, so the impulse given it by the club is equal to its final momentum. This is J = p = mv. Since we’re assuming constant acceleration, the force is the impulse divided by the time: F =
(0.071 kg)(42 m/s) mv = = 65 N t 0.046 s
549 Problem 1526. A ball weighing 120 g and moving horizontally at 9.2 m/s strikes a wall and rebounds horizontally at 8.4 m/s. What is the magnitude of the impulse that the wall gives to the ball? Round your answer to two significant digits. (a) 0.096 kg m/s *(c) 2.1 kg m/s (e) None of these
(b) (d)
0.94 kg m/s 21 kg m/s
Solution: The impulse is the ball’s change in momentum. If we take the direction away from the wall as positive, then the ball’s initial momentum is pi = (0.12 kg)(−9.2 m/s); its final momentum is pf = (0.12 kg)(8.4 m/s). The change in momentum is ∆p = pf − pi = (0.12 kg)(8.4 m/s) − (0.12 kg)(−9.2 m/s) = (0.12 kg)(17.6 m/s) = 2.1 kg m/s Problem 1527. A ball weighing 290 g and moving horizontally at 14 m/s strikes a wall and rebounds horizontally at 10 m/s. What is the magnitude of the impulse that the wall gives to the ball? Round your answer to two significant digits. (a) 1.2 kg m/s (c) 14 kg m/s (e) None of these
*(b) 7.0 kg m/s (d) 43 kg m/s
Solution: The impulse is the ball’s change in momentum. If we take the direction away from the wall as positive, then the ball’s initial momentum is pi = (0.29 kg)(−14 m/s); its final momentum is pf = (0.29 kg)(10 m/s). The change in momentum is ∆p = pf − pi = (0.29 kg)(10 m/s) − (0.29 kg)(−14 m/s) = (0.29 kg)(24 m/s) = 7.0 kg m/s Problem 1528. A large truck travelling down the highway collides with a small stationary deer. Which of the following statements is true? (a) The average force acting on the deer during the collision is greater than the average force acting on the car. (b) After the collision, the magnitude of the deer’s velocity relative to the car is the same as it was before the collision. (c) The change in the car’s momentum is greater than the change in the deer’s momentum. *(d) The magnitude of the impulse given to the car by the deer is the same as the magnitude of the impulse given to the deer by the car. Solution: We are not told whether this collision is elastic or inelastic; and judging by experience and common sense, it’s probably not perfectly elastic. Thus we can rule out (b). We know that momentum is conserved, so we can rule out (a) and (c). That leaves us with (d), which is true: the changes in momentum of the car and the deer must be equal and opposite, which means that the impulses acting on them must be equal and opposite.
550 Problem 1529. A large freight train traveling down the tracks collides with a small, cute, cuddly stationary squirrel burying a nut. Which of the following statements is true? (a) The average force acting on the squirrel during the collision is greater than the average force acting on the train. (b) After the collision, the magnitude of the squirrel’s velocity relative to the train is the same as it was before the collision. (c) The change in the train’s momentum is greater than the change in the squirrel’s momentum. *(d) The magnitude of the impulse given to the train by the squirrel is the same as the magnitude of the impulse given to the squirrel by the train. Solution: We are not told whether this collision is elastic or inelastic; and judging by experience and common sense, it’s probably not perfectly elastic. Thus we can rule out (b). We know that momentum is conserved, so we can rule out (a) and (c). That leaves us with (d), which is true: the changes in momentum of the train and the squirrel must be equal and opposite, which means that the impulses acting on them must be equal and opposite. Problem 1530. (Impulse over long-time period) A crate with a mass of 30 kg is sliding down a ramp. Starting from rest, it reaches a speed of 3 m/s after 10 s. What is the magnitude of the net force on the crate? Assume the net force on the crate is constant. (a) 3 N *(b) 9 N (c) 27 N (d) 30 N (e) None of these Solution: The crate begins at rest, so the impulse given to it is equal to its final momentum. Thus J = ∆p = F ∆t = mv
⇒
F =
(30 kg)(3 m/s) mv = =9N ∆t 10 s
Problem 1531. A golf ball has a mass of 0.071 kg. It is placed on a tee and hit, giving it a speed of 42 m/s. Golf scientists determine that the club is in contact with the ball for 0.046 s. Assuming constant acceleration, what is the average force exerted on the ball by the club? Round your answer to the nearest newton. (a) 63 N (c) 125 N (e) None of these
*(b) (d)
65 N 1361 N
Solution: The golf ball starts at rest, so the impulse given it by the club is equal to its final momentum. This is J = p = mv. Since we’re assuming constant acceleration, the force is the impulse divided by the time: F =
mv (0.071 kg)(42 m/s) = = 65 N t 0.046 s
551 Problem 1532. (Similarity Problem) Two railroad cars are standing on two horizontal frictionless tracks. The first car has mass M ; the second car has mass 3M . Each car is pulled along its track with a force of F for a time of T . At the end of this time, the momentum of the first car is P . What is the momentum of the second car? *(a) P (b) P/3 √ 3P (c) 3P (d) (e) None of these Solution: Since both cars are subjected to the same force F for the same time T , they have the same final momentum P . The second car only experiences one-third the acceleration of the first, so it only ends up going at one-third the velocity; but since it has three times the mass, its momentum is the same. Problem 1533. (Similarity Lab Problem) Two gliders are at rest on parallel horizontal frictionless air-tracks. Glider 1 has mass m1 ; glider 2 has mass m2 . Each glider is pushed with a horizontal force of Fpush for the same total time T . At the end of this time, glider 1 has kinetic energy K1 . What is the relationship between the kinetic energy of glider 2 in terms of the kinetic energy of glider 1? *(a) K/2 (b) K √ 2K (d) 4K (c) (e) None of these Solution: This is a similarity problem. We have two experiments with certain parameters that are in common to both. Since the same force acts on both gliders for the same amount of time, both gliders experience the same impulse. This is our common parameter. Thus p1,f − p1,i = ∆p1 = J = ∆p2 = p2,f − p2,i . Since the gliders both start from rest we have v1,i = v2,i = 0. Substituting this result into the previous expression gives m1 v1 = p1,f = p2,f = m2 v2
÷m
−−−−−2−→
v2 =
m1 v1 . m2
The kinetic energy of glider 1 is K1 = 12 m1 v12 ; the kinetic energy of glider 2 is 1 K2 = m2 v22 2 2 1 m1 = m2 v1 2 m2 m1 1 2 = m1 v1 2 m2 m1 = K1 m2 Comments:
(substitute for v2 for the above equation)
552 1. Notice that the ratio of the kinetic energies is K2 M1 = K1 M2 This follows directly from the fact that the momentums are the same (p1 = p2 ), so K2 p2 v2 v2 m1 = = = K1 p1 v1 v1 m2 Problem 1534. (Similarity Problem) Two railroad cars are at rest on two horizontal frictionless tracks. One car is three times as heavy as the other. Each car is pushed with the same horizontal force of F for the same time T . At the end of this time, the kinetic energy of the lighter car is K. What is the kinetic energy of the heavier car? √ 3K (a) 3K√ (b) (c) K/ 3 *(d) K/3 (e) None of these Solution: Let the mass of the light car be M . We’ll construct a table with the kinematic properties of the two cars: Mass Acceleration = force/mass Speed = acceleration × time Kinetic energy =
1 2
× mass × speed2
Light car M F/M F T /M F 2T 2 K= 2M
Heavy car 3M F/3M F T /3M F 2T 2 K = 6M 3
Problem 1535. (Similarity Problem) Two carts are at rest on parallel horizontal frictionless tracks. One cart has mass M ; the second has mass 2M . Each cart is pushed with a horizontal force of F for a time T . At the end of this time, the magnitude of the first cart’s momentum is P . What is the magnitude of the second cart’s momentum? (a) P/2 *(b) P (c) 2P (d) 4P (e) None of these Solution: Newton’s second law allows us to calculate the accelerations of both carts. For the first, A = F/M ; for the second, A2 = F/2M = A/2. Both carts are pushed for time T ; so the speed of the first is V = AT and the speed of the second is V2 = A2 T = AT /2 = V /2. Then the momentum of the first cart is P = M V ; the momentum of the second is P2 = M2 V2 = (2M )(V /2) = M V = P .
553
15.4
Collisions
Problem 1536. (Lab Problem) Two gliders collide on a horizontal frictionless air track. Which of the following statements must be true regardless of whether the collision is elastic or inelastic? (a) Neither momentum nor kinetic energy is necessarily conserved. (b) Kinetic energy is conserved; momentum is not necessarily conserved. *(c) Momentum is conserved; kinetic energy is not necessarily conserved. (d) Momentum and kinetic energy are both conserved. Solution: In all collisions in an isolated system (no external forces), momentum is conserved. In elastic collisions, kinetic energy is conserved; in inelastic collisions, it is not. Problem 1537. (Lab Problem) Two gliders collide on a frictionless horizontal air track. If the collision is elastic, which of the following statements must be true? *(a) Momentum and kinetic energy are both conserved. (b) Momentum is conserved; kinetic energy is not conserved. (c) Kinetic energy is conserved; momentum is not conserved (d) Neither momentum nor kinetic energy is conserved. Solution: In all collisions in an isolated system (no external forces), momentum is conserved. In elastic collisions, kinetic energy is conserved (by definition). Problem 1538. (Lab Problem) Two gliders collide on a horizontal frictionless air track. If the collision is inelastic, which of the following statements must be true? (a) Momentum and kinetic energy are both conserved. *(b) Momentum is conserved; kinetic energy is not conserved. (c) Kinetic energy is conserved; momentum is not conserved (d) Neither momentum nor kinetic energy is conserved. Solution: In all collisions in an isolated system (no external forces), momentum is conserved. In inelastic collisions, kinetic energy is not conserved.
554 Problem 1539. Two railroad cars are rolling toward one another on a frictionless horizontal track. The two cars have the same mass and are moving at the same speed. When they collide, their couplings engage so that they remain attached together. Which of the following statements is true? (a) The kinetic energy of the system is conserved during the collision. (b) The momentum of each car is the same before and after the collision. *(c) The two cars lose all of their kinetic energy during the collision. (d) The momentum of the system decreases during the collision. Solution: Since the two cars have the same mass and the same initial speed in opposite directions, the initial momentum of the system is zero. Thus when the cars are coupled together, their speed must be zero, which means their total kinetic energy is zero. Thus (c) is true. We can rule out (a) because kinetic energy is lost; (b), because each car has nonzero momentum before the collision and zero momentum afterward; and (d), because the total momentum of the system is conserved. Problem 1540. (Lab Problem) A glider with a mass of 500 g is at rest on a frictionless horizontal air track. A second glider with a mass of 250 g is launched toward it at a speed of V . The second glider has a spring attached, which is compressed as it collides with the first glider. When the spring is at its maximum compression, which of the following statements is true?
V250 g CC CC C C
(a) The second glider is at rest relative to the ground. *(b) Both gliders have the same velocity. (c) Both gliders have the same momentum. (d) Both gliders have the same kinetic energy.
500 g
555 Problem 1541. (Lab Problem) A glider with a mass of M is at rest on a horizontal frictionless air track. A second glider with a mass of m, where m < M , is launched toward it at a speed of V . The second glider has a spring attached, which is compressed as it collides with the first glider. When the spring is at its maximum compression, which of the following statements is true?
Vm
C C CC CC
M
(a) The second glider is at rest relative to the ground. *(b) Both gliders have the same velocity. (c) Both gliders have the same momentum. (d) Both gliders have the same kinetic energy. Problem 1542. (Lab Problem) Two gliders collide on a frictionless horizontal air track. The first glider has a mass of 2.0 kg; before the collision, it is moving rightward at 5.0 m/s. The second glider has a mass of 3.0 kg; before the collision, it is at rest. After the collision, the second glider is moving rightward at 4.0 m/s. What is the velocity of the first glider after the collision? *(a) (c) (e)
1.0 m/s leftward 3.0 m/s leftward None of these
(b) (d)
1.0 m/s rightward 3.0 m/s rightward
Solution: We know the mass and initial velocity of each glider; and we know the final velocity of the second glider. We want to know the final velocity of the first glider. We can use conservation of momentum. Taking rightward as the positive direction, we get: P = m1 v1i + m2 v2i = m1 v1f + m2 v2f v =0
−−2i−−→ m1 v1i = m1 v1f + m2 v2f −m2 v2f
−−−−−→ m1 v1f = m1 v1i − m2 v2f (2.0 kg)(5.0 m/s) − (3.0 kg)(4.0 m/s) m1 v1i − m2 v2f ÷m1 = = −1.0 m/s −−−→ v1f = m1 2.0 kg Since the sign is negative, the glider is moving leftward.
556 Problem 1543. (Lab Problem) A glider with a mass of 2.0 kg is moving rightward at a speed of 3.0 m/s on a frictionless horizontal air track. It collides with a stationary glider whose mass is 4.0 kg. If the collision is elastic, what is the speed of the heavier glider after the collision? Round your answer to the nearest 0.1 m/s. (a) 0.0 m/s *(c) 2.0 m/s (e) None of these
(b) 1.0 m/s (d) 3.0 m/s
Solution: Since the collision is elastic, both momentum and kinetic energy are conserved. Since the heavier glider is stationary, the initial momentum and kinetic energy are only those of the first glider. Taking the positive direction as rightward, and using the subscript 1 for the lighter glider and 2 for the heavier, we find: pi = m1 v1i = (2.0 kg)(3.0 m/s) = 6.0 kg m/s 1 1 2 = (2.0 kg)(3.0 m/s)2 = 9.0 J Ki = m1 v1i 2 2 By conservation of momentum and kinetic energy, pf = pi and Kf = Ki . For brevity’s sake, we will drop the ”f” subscript and refer to the post-collision velocities as v1 and v2 . m1 v1 + m2 v2 = pi
and
1 1 m1 v12 + m2 v22 = Ki 2 2
Substituting numbers, we get 2v1 + 4v2 = 6 and 1v12 + 2v22 = 9 We want to know the speed of the heavier glider after the collision: v2 . We’ll solve the first equation for v1 , then substitute that into the second equation; that will give us an equation in v2 alone, which we will solve for v2 . Solving the first equation for v1 : 2v1 + 4v2 = 6
⇒
v1 =
6 − 4v2 = 3 − 2v2 2
Substitute this expression for v1 into the kinetic-energy equation: v12 + 2v22 = 9
⇒
(3 − v2 )2 + 2v22 = 9
Expand to eliminate the parentheses: 9 − 6v2 + v22 + 2v22 = 9
⇒
3v22 − 6v2 = 0
Factor: 3v2 (v2 − 2) = 0
⇒
v2 = 0 or v2 = 2
The solution v2 = 0 refers to the velocity of the glider before the collision; so the velocity after the collision is 2.0 m/s.
557 Problem 1544. (Lab Problem) Two gliders collide on a frictionless horizontal air track. Glider 1 has a mass of 0.88 kg; glider 2 has a mass of 1.13 kg. Before the collision, glider 1 is moving rightward at 3.3 m/s; glider 2 is at rest. After the collision, glider 1 is moving rightward at 1.1 m/s. How fast is glider 2 moving? Round your answer to the nearest 0.1 m/s. (a) 1.4 m/s (c) 2.7 m/s (e) None of these
*(b) 1.7 m/s (d) 4.4 m/s
Solution: We don’t know if this collision is elastic or inelastic. We know the masses of both gliders, and the velocities of both before the collision, so we can calculate the initial momentum. We know the velocity of glider 1 after the collision, so we can use conservation of momentum to calculate the post-collision velocity of glider 2. Since the initial velocity of glider 2 is zero, the initial momentum of the system is: pi = p1,i + p2,i = m1 v1,i + m2 v2,i = m1 v1,i After the collision, the momentum of the system is pf = p1,f + p2,f = m1 v1,f + m2 v2,f Since momentum is conserved, pf = pi ; so m1 v1,f + m2 v2,f = m1 v1,i Solve for v2,f : v2,f =
(0.88 kg)(3.3 m/s − 1.1 m/s) m1 (v1,i − v1,f ) = = 1.7 m/s m2 1.13 kg
Problem 1545. A block of wood with a mass of 2300 g is sitting on a frictionless horizontal tabletop. A rifle bullet with a mass of 44 g is fired horizontally into the block at a speed of 270 m/s and stops inside the block. After the collision, what is the speed of the combined block and bullet? Round your answer to the nearest 0.1 m/s. (a) 3.7 m/s (c) 4.6 m/s (e) None of these
(b) 4.1 m/s *(d) 5.1 m/s
Solution: Since the bullet stops inside the block, we have an inelastic collision. We use conservation of momentum. If mb and vbi are the mass and initial speed of the bullet, and mw and vwi = 0 are the mass and initial speed of the wood block, then P = mb vbi . If vf is the speed of block plus bullet after the collision, then P = (mb + mw )vf mb vbi (0.044 kg)(270 m/s) P ⇒ vf = = = = 5.1 m/s mb + mw mb + mw 0.044 kg + 2.3 kg
558 Problem 1546. A block of wood with a mass of 1300 g is sitting on a horizontal tabletop. The coefficient of kinetic friction between the block and the table is µk = 0.74. A rifle bullet with a mass of 39 g is fired horizontally into the block at a speed of 290 m/s and stops inside the block. How far does the block with the bullet embedded in it slide before coming to a stop? Round your answer to the nearest 0.1 m. (a) 4.4 m (c) 5.4 m (e) None of these
*(b) (d)
4.9 m 6.0 m
Solution: There are two processes going on here: an inelastic collision between the bullet and the block, followed by the block’s sliding to a halt as friction dissipates its kinetic energy. Since the bullet-wood collision is inelastic, we can’t use conservation of energy there. We have to use conservation of momentum. We know the masses and initial velocities of the bullet and the block: mb , vbi , mw , and vwi = 0. (We’ll use the subscript “w” for “wood”, since “block” and “bullet” both begin with the same letter.) We want to know the kinetic energy Ks of the block-bullet system immediately after the collision, and we can calculate that if we know the velocity vs at that time. By conservation of momentum, mb vbi + mw vwi = mb vbi = (mb + mw )vs mb vbi ÷(mb +mw ) −−−−−−−→ vs = mb + mw Now we can calculate the energy of the system immediately after the collision. There is no potential energy involved. 2 (mb + mw )(mb vbi )2 m2b vbi 1 = Ks = ms vx2 = 2 2(mb + mw )2 2(mb + mw )
As the block-bullet system slides to a halt, the only force working on it is friction. It will move until the work done against the frictional force equals its initial energy. Since the tabletop is horizontal, the normal force equals the weight; so the force of kinetic friction is Fk = µk N = µk (mb + mw )g The work done against friction is the force times the distance x that the block slides: W = Fk x. We want to know x; and we can find it by setting W = Ks and then solving for x. 2 m2b vbi = Ks 2(mb + mw ) 2 m2b vbi (0.039 g)2 (290 m/s)2 ÷µk (mb +mw )g −−−−−−−−−→ x = = = 4.9 m 2µk g(mb + mw )2 2(0.74)(9.8 m/s2 )(0.039 g + 1300 g)2
W = Fk x = µk (mb + mw )gx =
559 Problem 1547. (Derive Problem) A bullet of mass mb with an initial velocity of vb,i strikes a wooden block of mass mw at rest on a frictionless horizontal surface. The bullet emerges with its speed reduced to vb,f . Derive a formula for the resulting velocity vw,f of the block. Assume there is no loss of mass in the wooden block when the bullet passes through the block. Solution: This is a collision problem between the bullet and the wooden block. If we take the system to be the wooden block plus the bullet, then we would like to apply the conservation of momentum to the system before and after the collision. This requires demonstrating that we can ignore all of the external forces on the system in the direction tangent to the surface. Take this to be the x-direction. Since we have a horizontal frictionless surface, there is no forces due to gravity and friction on the system in the x-direction, and ignoring any effects due to air drag over this short time period, we are justified in applying conservation of momentum. We now proceed in steps. Step 1: Choose the coordinate system. Choose a fixed coordinate system with the x-axis running along the frictionless surface and the y-direction normal to the surface opposite the direction of gravity. Step 2: Draw a before and after picture.
Figure 11: Bullet-wooden-block collision.
560 Step 3: Write down what you are given and what you want. Also include any assumptions you are making to simplify the problem. mb = mass of the bullet mw = mass of the wooden block given: vb,i = initial speed of the bullet (before impact) vb,i = final speed of the bullet (after impact) v = 0 (initial speed of the wooden block (after impact)) w,i want: assumptions:
vb,i = final speed of the wooden block (after impact) This is a 1-D motion problem; no external forces on the system
Step 4: Apply conservation of momentum to the system: psys,i = psys,f . mb vb,i + mw vw,i = psys,i = psys,f = mb vb,f + mw vw,f set vw,i =0
−−−−−→ −mb vb,f
−−−−−→ ÷m
w −−−→
mb vb,i + 0 = mb vb,f + mw vw,f mw vw,f = mb vb,i − mb vb,f mb vw,f = − ∆vb mw
A faster way would be to use ∆pw = −∆pb with pw,i = 0. Problem 1548. (Lab Problem: Explosion problem) Two gliders, each with mass M , are at rest on a frictionless horizontal air track. The gliders are tied together with a horizontal spring between them; the spring has been compressed so that it stores a potential energy of U . When the string connecting the two gliders is cut, the spring pushes them apart. After the spring has fully expanded, what is the speed of each glider? U U (b) (a) 2M M r r 2U U (c) *(d) M M (e)
None of these
Solution: There is no friction in the system, so we will use conservation of energy. Since the gliders are initially at rest, the momentum of the system is zero. Therefore, since the gliders have the same mass M , they will move at the same speed v in opposite directions after the string is cut. Hence the kinetic energy of each glider will be Kg = 12 M v 2 ; so the kinetic energy of the whole system will be Ks = 2Kg = M v 2 . By conservation of energy, r U U 2 2 U = Ks = M v ⇒ v = ⇒ v= M M
561 Problem 1549. Two identical gliders have an elastic collision on a horizontal frictionless air track. Assume there are no external forces on the two-glider system, and the motion is one-dimensional. Before the collision, the initial velocity of first glider is v1,i 6= 0, and the initial velocity of the second glider is v2,i = 0. Suppose that after the collision the first glider is observed to be at rest. What is the velocity of the second glider after the collision? *(a) v1,i (b) −v1,i (c) 0 (d) −v1,i /2 (e) None of these Solution: This problem gives you extra information. It follows immediately from conservation of momentum that v2,f = v1,i . We could have come to the same conclusion without knowing the final velocity of the first glider. In general, in an elastic collision between two objects of the same mass with one of the objects initially at rest, the two objects will switch velocities after the collision. To see this, apply conservation of momentum and K.E.: ( ÷m m v1,i + 0 = m v1,f + m v2,f −−→ v1,i + 0 = v1,f + v2,f 1 2 mv1,i 2
÷(m/2)
2 2 2 2 2 + 0 = 12 mv1,f + 12 mv2,f −−−−→ v1,i + 0 = v1,f + v2,f
Solving the first equation for v2,f = v1,i − v1,f and substituting this expression into the second equation leads to 2 2 v1,i − v1,f = (v1,i − v1,f )2
factor the LHS and ÷ (v1,i −v1,f )
−−−−−−−−−−−−−−−−−−→ v1,i + v1,f = v1,i − v1,f −v1,f +v1,f
−−−−−−→ 2v1,f = 0 ⇒ v2,f = v1,i .
562 Problem 1550. (Derive Problems) At either end of a horizontal frictionless air track is a spring. The spring on the left has spring constant kL . The spring on the right has spring constant kR . A glider with a mass of mL is pressed against the left-hand spring, compressing it a distance xL from equilibrium. The glider is initially at rest and held in place by a trigger mechanism. The glider is then released, so that the spring propels it rightward. In the middle of the track is a second glider with a mass of mR . The two gliders have small magnets attached, so when the first glider strikes the second, they stick together and slide into the right-hand spring. Let xR be the maximum compression of the right-hand spring. Derive a formula for xR . (The diagram shows the situation before the left-hand glider is released, when the left-hand spring is compressed and neither glider is moving.)
mL E E E E
Solution:
mR
C C C C C C
This problem requires several steps.
Step 1: Apply the conservation of mechanical energy to the left-hand-glider-spring system. We will use the compression of the left-hand spring to calculate its potential energy. When it is released, the potential energy stored in the spring will be converted to kinetic energy of the left-hand glider; we will use that to determine the initial velocity of the glider, and hence the glider’s initial momentum. Step 2: Apply the conservation of momentum to the collision of the two gliders. When the gliders collide, they will not be in contact with either spring, so we can treat the pair of gliders as an isolated system. The collision will be perfectly inelastic, so applying the conservation of momentum will give us the speed of the combined gliders after they’ve stuck together. Step 3: Apply the conservation of mechanical energy to the right-hand-glider-pair-spring system. If we know the speed of the pair of gliders after the collision, then we know their kinetic energy. When the right-hand spring is fully compressed, all of that kinetic energy will
563 be stored in the spring; we’ll use that to determine the compression. We begin with step 1. Let xL be the compression of the left-hand spring. Then the initial potential energy in the spring, and hence the left-hand glider, is UL,i = 12 kL x2L . The initial kinetic energy of the left-hand glider is KL,i = 0, since the glider is at rest. After the left-hand spring is released, this energy is completely transferred to the lefthand glider. The final kinetic energy of the glider will be KL,f = 21 mL vL2 , and its potential is UL,f = 0, since the glider is no longer experiencing any forcing by the spring. Equating the initial and final mechanical energy of the left-hand-glider-spring system gives 1 1 kL x2L + 0 = UL,i + KL,i = UL,f + KL,f = 0 + mL vL2 2 2 s kL x2L mL
⇒
vL =
⇒
pL,i = mL vL =
⇒
(initial momentum of left-hand-glider before collision) q psys,i = pL,i + pR,i = mL kL x2L
q mL kL x2L
(initial momentum of two-glider system before collision) Notice that the initial momentum of the right-hand glider is zero since it is initially at rest. This completes step 1 and gives us the initial momentum for step 2. After the perfectly inelastic collision, the combined gliders will move at a speed vLR , conserving momentum. We’ll use mLR = mL + mR for the combined mass of the two gliders. The subscript LR on the mass and velocity is used to remind us that the masses are stuck together. Applying conservation of momentum just before and after the collision yields: p q mL kL x2L mLR vLR = psys,f = psys,i = mL kL x2L ⇒ vLR = mLR This completes step 2 and begins step 3. In the last step of the analysis, we again apply the conservation of mechanical energy. The kinetic energy of the combined gliders will be mL kL x2L 1 2 KR,i = mLR vLR = . 2 2mLR Since the right-hand spring is not yet compressed, the initial potential of the mass-spring system is zero. Thus the initial mechanical energy is MER,i = KR,i + UR,i =
mL kL x2L +0 2mLR
564 When the right-hand spring is fully compressed, all of the kinetic energy in the gliders will be stored in the right-hand spring. The final kinetic energy of the mass (gliders)spring system is KR,f = 0. The final potential energy of the mass-spring system is UR,f = Uspring = kR x2R /2, where xR is to be determined. Thus the final mechanical energy of the system is MER,f = KR,f + UR,f = kR x2R /2 + 0 We now apply the conservation of mechanical energy to arrive at the expression for xR : mL kL x2L kR x2R = UR,f + KR,f = UR,i + KR,i = 2 2mLR r r r r kL mL kL mL ⇒ xR = xL = xL kR mLR kR mL + mR Notice that if kL = kR and mR = 0 (no second glider), then xR = xL . Comment: Just remember, if you see this problem on an exam, say to yourself: I want me-mom-me! Use conservation of M.E., then momentum (mom), then one more application of M.E.
565 Problem 1551. (Dervive Problem [The Ballistic Pendulum]) Below is a sketch of the ballistic pendulum from Pima community college’s physics lab. The apparatus is designed to measure the speed of the steel ball as it leaves the mouth of the spring cannon. The ball is fired horizontally into a heavy bob attached to a light rod that is free to pivot at one end like a pendulum. The pendulum catches the ball as it leaves the muzzle of the cannon. We treat this as a perfectly-inelastic collision between the ball and the pendulum. The pendulum arm then swings away from the cannon and upward; a device on the apparatus allows one to measure the angle θ = Θmax that the rod swings through. When the arm swings, it lifts the ball to a height of hmax above the ball’s initial height when it was in the spring cannon. Although it would be very hard to measure this height directly, we can use trigonometry to compute the maximum height of the ball in terms of the measured angle Θmax . In the lab we can measure the values of the mass of the ball Mball ; the mass of the pendulum Mpend ; the distance between the pivot point of the pendulum and the center of mass of the ball-pendulum system, which we shall refer to as the length of the rod Lrod ; and the angle Θmax of the swing after the cannon is fired. (a) Find the initial velocity of the ball as it leaves the muzzle v0 as a function of the given data: Mball , Mpend , Lrod , Θmax . Your answer should not explicitly contain hmax , since it can be expressed as a function of these data variables. Note: I have used capital letters to denote known parameters that can be measured in the lab prior to the experiment. (b) If the spring is compressed from its equilibrium position by a distance X, what is the spring constant of the cannon as a function of the measured quantities? Hint: This is a 3-step problem. In step 1, use conservation of mechanical energy to find the relationship between the spring constant and the velocity of the ball as it leaves the cannon. In step 2, use conservation of momentum over the inelastic collision. Approximate the collision as a one-dimension collision in the horizontal direction. In step 3, apply conservation of mechanical energy to the ball-pendulum system. Take the initial state of the system just after the collision, and the final state of the system at the point where the pendulum is momentarily at rest at the highest point of its upswing.
566
Figure 12: Ballistic Pendulum found in lab
Figure 13: Ballistic Pendulum: Stage 1
Figure 14: Ballistic Pendulum: Stage 2
567
Figure 15: Ballistic Pendulum: Stage 3
568
16
Rotational kinetic energy and angular momentum
Problem 1552. Two physics students are on a merry-go-round that is rotating frictionlessly. Each student has a mass of 80 kg. Initially, one student is standing at the center, while the other is standing at the outer edge. The two change places, walking past one another at the same speed. How does the angular speed of the merry-go-round change as the students move? (a) The speed of the merry-go-round doesn’t change. (b) The speed decreases until the students meet halfway between the edge and the center; then it increases. *(c) The speed increases until the students meet halfway between the edge and the center; then it decreases. (d) The speed increases during the whole time that the students move, because they are adding energy to the system. Solution: There are no external torques, so angular momentum is conserved: L = Iω. Hence if I increases, ω decreases, and vice versa. We want to see if I changes as the students move. Let R be the merry-go-round’s radius, and m the mass of each student. When one is at the center and the other is at the edge, I = mR2 . When both are halfway from the center to the edge, I = (2m)(R/2)2 = mR2 /2. Since I gets smaller as the students walk toward the halfway point, then gets larger again, the merry-go-round first speeds up, then slows down.
569 Problem 1553. A physics instructor has a mass of M ; his wife has a mass of M/2. The two are on a merry-go-round that is rotating frictionlessly. Initially, the instructor is at the center and his wife is at the outer edge. The two trade places, passing one another halfway from the center to the edge. At which of the following three points is the merry-go-round’s angular speed the greatest? (a) When the instructor is at the center and the wife is at the edge. (b) When the wife is at the center and the instructor is at the edge. *(c) When the two are both halfway from the center to the edge. (d) The merry-go-round’s speed doesn’t change. Solution: There are no external torques on the system, so angular momentum is conserved: L = Iω ⇒ ω = L/I. The angular speed ω is the greatest when the moment of inertia I is the smallest. The contribution of each person to the total moment of inertia is mr2 , where m is their mass and r is their distance from the center. Let R be the radius of the merry-go-round, so that someone at the edge is at r = R. Look at the three situations: Instructor at center, wife at edge
I = M (0)2 +
Wife at center, instructor at edge
I=
Both halfway from center to edge
M 2 M R2 R = 2 2
M 2 (0) + M R2 = M R2 2 2 2 R M R 3M R2 I=M + = 2 2 2 8
Since I changes, the speed of the merry-go-round changes; so we can rule out (d). Since I is the smallest in situation (c), the merry-go-round’s speed is the greatest in that situation.
570 Problem 1554. A physics instructor has a mass of M ; his wife has a mass of M/2. The two are on a merry-go-round that is rotating frictionlessly. Initially, the instructor is at the center and his wife is at the outer edge. The two trade places, passing one another halfway from the center to the edge. At which of the following three points is the merry-go-round’s angular speed the smallest? (a) When the instructor is at the center and the wife is at the edge. *(b) When the wife is at the center and the instructor is at the edge. (c) When the two are both halfway from the center to the edge. (d) The merry-go-round’s speed doesn’t change. Solution: There are no external torques on the system, so angular momentum is conserved: L = Iω ⇒ ω = L/I. The angular speed ω is the smallest when the moment of inertia I is the largest. The contribution of each person to the total moment of inertia is mr2 , where m is their mass and r is their distance from the center. Let R be the radius of the merry-go-round, so that someone at the edge is at r = R. Look at the three situations: Instructor at center, wife at edge
I = M (0)2 +
Wife at center, instructor at edge
I=
Both halfway from center to edge
M 2 M R2 R = 2 2
M 2 (0) + M R2 = M R2 2 2 2 R M R 3M R2 I=M + = 2 2 2 8
Since I changes, the speed of the merry-go-round changes; so we can rule out (d). Since I is the largest in situation (b), the merry-go-round’s speed is the smallest in that situation. Problem 1555. A flywheel’s moment of inertia is 80 kg·m2 . Starting at rest, it is subjected to a torque of 15 N·m for 6 s. At the end of this time, what is its kinetic energy? Round your answer to two significant figures. (a) 41 J *(c) 51 J (e) None of these
(b) 46 J (d) 56 J
Solution: We know I, τ , and t. We want to know K. If we knew ω, we could use 1 2 K = 2 Iω . If we knew α, we could calculate ω = αt. We know I and τ , so we can calculate α using τ = Iα. 1 1 1 τ 2 τ 2 t 2 (15 N·m)2 (6 s)2 K = Iω 2 = I(αt)2 = I t = = = 51 J 2 2 2 I 2I 2(80 kg·m2 )
571 Problem 1556. A wind turbine has three blades. Each blade has a mass of 1000 kg and a length of 20 m. Each blade can be regarded as a thin rod connected at one end to the axis. If the turbine is turning at a rate of one revolution every 3.0 s, what is its kinetic energy? Round your answer to two significant figures. *(a) (c) (e)
8.8 × 105 J 1.1 × 106 J None of these
(b) (d)
9.7 × 105 J 1.2 × 106 J
Solution: We can calculate the moment of inertia of an individual blade, then multiply by 3 to get I for the whole turbine. We know the period T , so we can use ω = 2π/T to calculate ω. Then we can use K = 21 Iω 2 . mb R2 ⇒ I = 3Iblade = mb R2 3 2 1 2 1 2π 2 mb R2 2π 2π 2 (1000 kg)(20 m)2 2 K = Iω = (mb R ) = = = 8.8 × 105 J 2 2 T T2 (3.0 s)2 Iblade =
Problem 1557. A shaft can be regarded as a solid cylinder with radius 4 cm and mass 120 kg. If the shaft is turning at 20 revolutions per second, what is its kinetic energy? Round your answer to two significant figures. *(a) (c) (e)
760 J 890 J None of these
(b) (d)
830 J 960 J
Solution: We know the mass m and radius R of the shaft, and its frequency f . We want to know its kinetic energy K. If we knew I and ω, we could use K = 12 Iω 2 . We can use I = mR2 /2 and ω = 2πf . 1 2 1 mR2 K = Iω = (2πf )2 = π 2 mf 2 R2 2 2 2 = π 2 (120 kg)(20 s−1 )2 (0.04 m)2 = 760 J
572 Problem 1558. A windmill whose moment of inertia is I is initially at rest. The wind begins to blow, producing a torque of τ about the windmill’s axis. After a time t, what is the windmill’s kinetic energy? *(a) (c) (e)
τ 2 t2 2I τ t2 2I
(b) (d)
Iτ 2 t2 2 Iτ t2 2
None of these
Solution: We know I, τ , and t. We want to know K. If we knew ω, we could use 1 2 K = 2 Iω . If we knew L, we could calculate ω from L = Iω. We know τ and t, and ω0 = 0, so we can calculate L = τ t. 2 2 L τt 1 2 1 1 τ 2 t2 K = Iω = I = I = 2 2 I 2 I 2I Problem 1559. Two flywheels, one with moment of inertia 240 kg·m2 and one with moment of inertia 150 kg·m2 , turn on the same shaft. The first flywheel is fixed on the shaft; the second one is connected by a clutch, so that it can be made to turn either with the first or independently of it. Initially, the first wheel is turning at 30 rad/s; the second is at rest. The clutch is then operated, so that both wheels turn together. What is their common angular speed? Round your answer to two significant figures. (a) 17 rad/s (c) 20 rad/s (e) None of these
*(b) (d)
18 rad/s 22 rad/s
Solution: Use conservation of angular momentum. The initial angular momentum is that of the first flywheel, since the second is at rest. After the two are connected, they turn at the same angular speed. We’ll use ω1 for the initial angular speed of the first flywheel, and ωc (for “combined”) for the final angular speed. I1 ω1 = (I1 + I2 )ωc
⇒
ωc =
(240 kg·m2 )(30 rad/s) I1 ω1 = = 18 rad/s I1 + I2 240 kg·m2 + 150 kg·m2
573 Problem 1560. A door’s moment of inertia is 8 kg·m2 . A bullet with a mass of 25 g moving at 300 m/s strikes the door perpendicularly at a distance of 75 cm from the hinges. The bullet remains embedded in the door. What is the door’s angular speed after the collision? Assume that the embedded bullet does not change the door’s moment of inertia. Round your answer to two significant figures. (a) 0.63 rad/s (c) 0.77 rad/s (e) None of these
*(b) (d)
0.70 rad/s 0.85 rad/s
Solution: Use conservation of angular momentum. The initial angular momentum is that of the bullet. The final angular momentum is that of the swinging door. mb vb R = Id ωd
⇒
ωd =
(0.025 kg)(300 m/s)(0.75 m) mb vb R = = 0.70 rad/s Id 8 kg·m2
(In case you’re wondering about the validity of the assumption, the correct moment of inertia for the door with the embedded bullet is I = Id + Ib = Id + mb R2 = 8 kg·m2 + (0.025 kg)(0.75 m)2 = 8.014 kg·m2 The difference is less than 1%; it shouldn’t affect the results rounded to two significant figures.) Problem 1561. A door’s moment of inertia is 8 kg·m2 . A bullet with a mass of 25 g, moving horizontally at 300 m/s, strikes the door at an angle of 60◦ to the door’s surface, at a distance of 75 cm from the hinges. The bullet remains embedded in the door. What is the door’s angular speed after the collision? Assume that the embedded bullet does not change the door’s moment of inertia. Round your answer to two significant figures. (a) 0.55 rad/s (c) 0.67 rad/s (e) None of these
*(b) (d)
0.61 rad/s 0.74 rad/s
Solution: Use conservation of angular momentum. The initial angular momentum is that of the bullet. The final angular momentum is that of the swinging door. Since the bullet doesn’t strike the door at a right angle, we need to calculate the angular momentum using only the component of its velocity perpendicular to the door. mb vb R cos θ = Id ωd (0.025 kg)(300 m/s)(0.75 m) cos 30◦ mb vb R cos θ = = 0.61 rad/s ⇒ ωd = Id 8 kg·m2
574 Problem 1562. A door’s moment of inertia is 10 kg·m2 . A ball with a mass of 40 g moving at 30 m/s strikes the door perpendicularly at a distance of 70 cm from the hinges; the ball bounces backward at 25 m/s. What is the door’s angular speed after the collision? Round your answer to two significant figures. *(a) (c) (e)
0.15 rad/s 0.19 rad/s None of these
(b) (d)
0.17 rad/s 0.21 rad/s
Solution: Use conservation of angular momentum. Before the collision, the total angular momentum of the system is that of the ball. After the collision, it consists of the angular momentum of the swinging door plus that of the rebounding ball. We’ll use m for the mass of the ball, vi for its initial velocity, and vf for its final velocity. We’ll take vi to be positive, so vf will be negative. Li = Lf = mvi R = mvf R + Iωf ⇒
ωf =
(0.040 kg)(0.70 m)[30 m/s − (−25 m/s)] mR(vi − vf ) = = 0.15 rad/s I 10 kg·m2
Problem 1563. A merry-go-round’s moment of inertia is 900 kg·m2 ; its radius is 2 m. It is initially at rest. A physics student with a mass of 60 kg is standing on the outer edge. The student jumps off at 2.5 m/s, in a direction tangent to the outer edge. After the student has jumped off, how long does it take for the merry-go-round to make one complete revolution? Round your answer to two significant figures. *(a) (c) (e)
19 s 23 s None of these
(b) (d)
21 s 25 s
Solution: Use conservation of angular momentum. Since the system is initially at rest, the final angular momentum of the merry-go-round must match the final angular momentum of the student. We want to know the period T that it takes to make one complete revolution; that’s related to the angular velocity ω by ω = 2π/T . 2πI T 2 2πI 2π(900 kg·m ) = = 19 s T = mvR (60 kg)(2.5 m/s)(2 m) Ls = Lm = mvR = Iω =
⇒
575 Problem 1564. A merry-go-round’s moment of inertia is 6000 kg·m2 . It is initially turning frictionlessly at a rate of one revolution per 15 s. A physics instructor with a mass of 100 kg is initially standing at the center of the merry-go-round. The instructor then moves to the outer edge, 4 m from the center. When he has reached the outer edge, how long does it take for the merry-go-round to make one revolution? Round your answer to two significant figures. (a) 15 s *(c) 19 s (e) None of these
(b) 17 s (d) 21 s
Solution: Since no external torques are applied to the system, angular momentum is conserved: L = Iω. I increases as the instructor moves outward, so ω must decrease to compensate. We’ll use ωi and ωf to indicate the initial and final angular velocities; Im for the moment of inertia of the merry-go-round by itself; and Ip for the physics instructor’s contribution to I. We want to know the period T , which is related to ω by ω = 2π/T . 2π(Im + Ip ) 2πIm = Ti Tf 2 2 (Im + Ip )Ti (Im + mR )Ti [6000 kg·m + (100 kg)(4 m)2 ](15 s) Tf = = 19 s = = Im Im 6000 kg·m2 Li = Lf = Im ωi = (Im + Ip )ωf =
⇒
Problem 1565. A small wind turbine has a moment of inertia of 400 kg·m2 , and blades 2 m long. The turbine is at rest, with one blade pointing upward at an angle of 30◦ to the vertical. A goose with a mass of 8 kg, flying horizontally in the plane of the turbine at 20 m/s, strikes the end of that blade. Upon colliding with the blade, the goose momentarily comes to a stop before falling to the ground. What is the turbine’s angular speed after the collision? Round your answer to two significant figures. *(a) (c) (e)
0.69 rad/s 0.84 rad/s None of these
(b) (d)
0.76 rad/s 0.92 rad/s
Solution: Use conservation of angular momentum. Initially, the total angular momentum is that of the goose. After the collision, the goose is at rest, so the total angular momentum is that of the turbine. The goose doesn’t hit the blade at a right angle, so in calculating its initial angular momentum, we need to use the component of its velocity perpendicular to the blade. Since the goose is flying horizontally and the blade is inclined at 30◦ to the vertical, the goose hits the blade at θ = 60◦ . Li = Lf = mvR sin θ = Iω ⇒
ω=
(8 kg)(20 m/s)(2 m) sin 60◦ mvR sin θ = = 0.69 rad/s I 400 kg·m2
576 Problem 1566. A hollow pipe has radius R. It is initially at rest on top of a hill with height y. The pipe is given a nudge so that it rolls down the hill. How fast is it going when it reaches the bottom? p p (b) v = g(y + R) (a) v = 2g(y + R)
(c)
R v = 2g y + 2
(e)
None of these
1/2 *(d) v =
√
gy
Solution: Use conservation of mechanical energy. Initially, the pipe is at rest, so the total mechanical energy is the gravitational potential energy. At the bottom of the hill, the potential energy is zero, and the total energy is kinetic: partly the kinetic energy of the pipe’s center of mass moving in a line, and partly the kinetic energy of the pipe’s rotation. To determine how to partition the kinetic energy among these, use the no-slip condition: if a point where the pipe touches the ground is at rest relative to the ground, then relative to the pipe’s center, a point on the outer surface is rotating with a speed equal to the center’s linear speed relative to the ground. Thus ω = v/R. v 2 1 1 1 1 = mv 2 U0 = Kf = mgy = mv 2 + Iω 2 = mv 2 + (mR2 ) 2 2 2 2 R √ ⇒ v = gy Problem 1567. A thin cylindrical pipe filled with concrete has radius R. It is initially at rest on top of a hill with height y. The pipe is given a nudge so that it rolls down the hill. How fast is it going when it reaches the bottom? 1/2 p R (b) v = 2g y + (a) v = 2g(y + R) 2 r r 4gy 3gy (d) v = *(c) v = 3 2 (e)
None of these
Solution: Use conservation of mechanical energy. Initially, the pipe is at rest, so the total mechanical energy is the gravitational potential energy. At the bottom of the hill, the potential energy is zero, and the total energy is kinetic: partly the kinetic energy of the pipe’s center of mass moving in a line, and partly the kinetic energy of the pipe’s rotation. To determine how to partition the kinetic energy among these, use the no-slip condition: if a point where the pipe touches the ground is at rest relative to the ground, then relative to the pipe’s center, a point on the outer surface is rotating with a speed equal to the center’s linear speed relative to the ground. Thus ω = v/R. 1 2 1 2 1 2 1 mR2 v 2 3mv 2 U0 = Kf = mgy = mv + Iω = mv + = 2 2 2 2 2 R 4 r 4gy ⇒ v= 3
577 Problem 1568. (Requires calculus) A physics instructor has mass mi ; his wife has mass mw . The two are standing on a merry-go-round that is rotating frictionlessly. Initially, the instructor is standing at the outer edge, at a distance R from the center; the wife is standing at the center. The two change places, walking past one another at the same speed in opposite directions. How far from the center will the instructor be when the merry-go-round is rotating at its maximum speed? mw R mi R *(b) (a) mi + mw mi + mw s s mi R2 mw R2 (c) (d) mi + mw mi + mw (e)
None of these
Solution: There are no external torques, so angular momentum is conserved: L = Iω. Hence ω will be at its maximum when I is at its minimum. If the instructor is at distance r from the center, his wife is at distance R − r; so their contribution to the moment of inertia is I(r) = mi r2 + mw (R − r)2 To find the value of r at which I(r) is at a minimum, we want dI = 0 = 2mi r − 2mw (R − r) = 2(mi + mw )r − 2mw R dr
⇒
r=
mw R mi + mw
To check that this is really a minimum, note that d2 I/dr2 = 2(mi + mw ) > 0; so I(r) is concave upward, and therefore has a minimum at our value of r.
578
Part VI
Applications of Mechanics
579
17 17.1
Applications of Mechanics Harmonic Motion
Problem 1569. A pendulum goes through 20 complete cycles in 60 seconds. What is its frequency? 2π 1 Hz (b) Hz *(a) 3 3 3 (c) 3 Hz (d) Hz 2π (e)
None of these
Solution:
1 Hz = 1 cycle/s; so f=
20 cycles 1 = Hz 60 s 3
Problem 1570. An oscillator is made using a 10 kg block sliding on a frictionless horizontal surface, attached to a horizontal spring with a force constant of 4000 N/m. The spring is initially stretched by 10 cm and held for a moment, then released. As the block oscillates, what is its maximum speed? (a) 1 m/s (c) 4 m/s (e) None of these
*(b) 2 m/s (d) 8 m/s
Solution: Use conservation of energy. Initially, the system is not moving, so its total energy is the potential energy of the spring. When the block is at its maximum speed, the potential energy of the spring is zero, so the total energy of the system is the kinetic energy of the block. Hence: 1 U0 + K0 = U0 = kx20 2 1 Uf + Kf = Kf = mvf2 2 2 2 ⇒ kx0 = mvf kx20 m r 1/2 k 4000 N/m ⇒ vf = x0 = (0.1 m) = 2 m/s m 10 kg
⇒ vf2 =
580 Problem 1571. An oscillator is made using a 5 kg mass sliding on a frictionless horizontal surface, attached to a horizontal spring with a force constant of 20 N/m. What is the period of the oscillator in seconds? (a) 1 s (c) 2 s (e) None of these
*(b) π s (d) 2π s
Solution: We know the mass m and the spring constant k. We want to know the period T . We don’t have a formula that gives us the period directly, but we have one for the angular frequency ω, and we know that ω = 2π/T . r r 1/2 k m 2π 5 kg ω= = ⇒ T = 2π = 2π =π s m T k 20 N/m Problem 1572. An astronaut moves his furniture to Planet X. On Earth, the pendulum in his grandfather clock has a period of 1.0 s. On Planet X, the pendulum’s period is 0.5 s. What is the gravitational acceleration gX of Planet X in terms of Earth’s gravity gE ? (a) gE /4 (c) 2gE (e) None of these
(b) gE /2 *(d) 4gE
Solution: We have a formula for the angular frequency ω of a pendulum given the mass m and the gravitational acceleration g. We want to know the period T . We know that ω = 2π/T . We don’t know the pendulum length L, but it is the same on both planets. r g 2π = ω= T L √ √ √ √ ⇒ 2π L = T g = TE gE = TX gX ⇒ TE2 gE = TX2 gX 2 TE2 gE 1.0 s ⇒ gX = = gE = 4gE TX2 0.5 s
581
17.2
Mechanical waves and sound
Problem 1573. A tsunami consists of a series of waves travelling at a speed of 800 km/hr, with a wavelength of 200 km. What is the period of the waves? (a) 15 s *(c) 15 min (e) None of these
(b) (d)
4 min 4 hr
Solution: Notice that the speed is given in km/hr, and that the wavelength is given in km. This means that we don’t need to convert units. We would have had to do so if, for instance, the speed had been given in m/s and the wavelength in km. v=
λ T
⇒
T =
λ 200 km 1 = = hr = 15 min v 800 km/hr 4
Problem 1574. A cable with a linear density of 3 kg/m is stretched between two poles 30 m apart. If the cable is plucked, the resulting transverse wave travels from one pole to the other in 1 second. What is the tension in the cable? (a) 100 N (b) 300 N (c) 900 N *(d) 2700 N (e) None of these Solution: We know the linear density µ of the cable, and we can calculate the speed v from the length l and the time t. We want the tension FT . s FT l2 µ (30 m)2 (3 kg/m) FT ⇒ v2 = ⇒ FT = v 2 µ = 2 = = 2700 N v= µ µ t (1 s)2 Problem 1575. A violin string is tuned so that its third harmonic has a frequency of 1200 Hz. What is the fundamental frequency of the string? (a) 150 Hz (c) 3600 Hz (e) None of these Solution:
fn = nf1
*(b) (d)
⇒
f1 =
400 Hz 9600 Hz fn n
⇒
f1 =
f3 1200 Hz = = 400 Hz 3 3
582 Problem 1576. An obscure stringed instrument has two strings of identical length and composition, called the P and Q strings. The P string has a tension of TP , so that its fundamental frequency is fP 1 . The Q string is tuned so that its fundamental frequency fQ1 is the same as the second harmonic of the P string. What is the tension of the Q string? √ 2 TP (a) TP /2 (b) (c) 2TP *(d) 4TP (e) None of these Solution:
We will start by finding the relationship between fQ1 and fP 1 . fQ1 = fP 2 = 2fP 1
Since f1 = v/2L, and since the strings have equal length (LP = LQ = L), fQ1 =
vQ 2vP = 2fP 1 = 2L 2L
⇒
vQ = 2vP
Since the strings have the same composition, they have the same length density: µP = µQ = µ. Hence s s TQ 4TP TQ TP = 2vP = 2 ⇒ = ⇒ TQ = 4TP vQ = µ µ µ µ Problem 1577. A violinist is playing a note with a frequency of 480 Hz. The violinist next to him is playing a note with a frequency of 500 Hz. What is the beat frequency produced by the two notes? *(a) (c) (e)
20 Hz 490 Hz None of these
Solution:
(b) 460 Hz (d) 520 Hz
fbeat = f1 − f2 = 500 Hz − 480 Hz = 20 Hz
583 Problem 1578. Two wires of equal length L are stretched between two supports. Both wires are subjected to the same tension FT . The first wire has a mass of MA ; the second, of 3MA . If the fundamental frequency of the first wire is fA and that of the second wire is fB , which of the following equations is true? fA fA *(b) fB = √ (a) fB = 3 3 √ (c) fB = 3 fA (d) fB = 3fA (e)
None of these
Solution: We have an equation for fundamental frequency, which requires the wave speed v. We have an equation for wave speed, which requires the mass per unit length µ = M/L. s √ √ √ v FT FT FT L F √ = √ T and v = ⇒ f1 = f1 = √ = 2L µ 2L µ 2L M 2 LM In this problem, L and FT are the same for both wires; M and f1 are different. We separate the constant part from the variable one: p p √ FT f1 M = √ = fA MA = fB 3MA 2 L √ fA MA fA ⇒ fB = √ =√ 3MA 3 Problem 1579. At a distance of 4 m from an omnidirectional speaker, the sound intensity is 9 W/m2 . At what distance is the intensity 1 W/m2 ? (a) 8 m (c) 18 m (e) None of these
*(b) 12 m (d) 36 m
Solution: The intensity of sound varies inversely with the square of the distance. Let r1 = 4 m, I1 = 9 W/m2 , and I2 = 1 W/m2 . We want the distance r2 . I0 r2 ⇒ I0 = Ir2 = I1 r12 = I2 r22 I1 r12 ⇒ r22 = I2 #1/2 " r I1 9 W/m2 (4 m) = 12 m ⇒ r2 = r2 = I2 1 W/m2 I=
584 Problem 1580. A physics professor is in a car stalled on a railroad track. An oncoming train sounds its horn, and the professor notes that he hears it at a frequency of fp . He knows that when the train is standing still, its horn has a frequency of ft ; and that the local speed of sound is v0 . How fast is the train approaching him? fp − ft fp − ft v0 *(b) v0 (a) ft fp ft fp (c) v0 (d) v0 fp − ft fp − ft (e)
None of these
Solution: We know the source frequency fS = ft and the listener frequency fL = fp , the listener speed vL = 0, and the local speed of sound v = v0 . We have a formula for Doppler shift. We want to know the train’s approach speed vt . We will modify the textbook’s formula slightly, since in it the source speed is positive when it’s going away from the listener; we want vt to be the speed toward the professor, so we’ll use vS = −vt . v + vL fS v + vS v0 substitute −−−−−→ fp = ft v0 − vt fL =
×(v0 −vt )
−−−−−→ v0 fp − vt fp = v0 ft +vt fp −v0 ft
−−−−−−→ vt fp = v0 fp − v0 ft = v0 (fp − ft ) fp − ft ÷fp −−→ vt = v0 fp
17.3
Elasticity
Problem 1581. A mop handle makes an angle of θ to the horizontal. A force of F is applied along the mop handle; the mop contacts the floor over an area of A. What is the pressure exerted by the mop on the floor? F F (b) (a) A cos θ A sin θ F cos θ F sin θ (c) *(d) A A (e)
None of these
Solution: The component of F perpendicular to the floor is Fy = F sin θ. This force is distributed over an area of A; so the pressure is Fy /A = F sin θ/A.
585 Problem 1582. Two wires hang from the ceiling. The wires are identical, except that one is twice as long as the other. If a 20 kg mass is attached to the shorter wire, it stretches by 1 mm. If the same mass is attached to the longer wire, how much will it stretch? √ 2 mm (a) 1 mm (b) *(c) 2 mm (d) 4 mm (e) None of these Solution: This is a situation of tensile stress. Since the two wires are identical in composition, they have the same Young’s modulus Y ; and they have the same crosssectional area A. Their lengths are different: l1 and l2 = 2l1 . Each is subjected to the same tensile force, F⊥ = 20 kg · g. The first wire stretches by ∆l1 = 1 mm. We want to know the length change ∆l2 of the second wire. In this situation, Hooke’s law is ∆l F⊥ =Y A l Since only l and ∆l differ from one wire to the other, we can write: F⊥ ∆l ∆l1 ∆l2 = = = YA l l1 l2
⇒
∆l2 =
l2 ∆l1 2l1 ∆l1 = = 2∆l1 = 2(1 mm) = 2 mm l1 l1
586 Problem 1583. Two balls made of different materials are dropped from the deck of a ship and sink to the bottom of an ocean trench. On the surface, the first ball has a volume of 60 cm3 and the second ball a volume of 120 cm3 . At the bottom of the trench, the first has a volume of 50 cm3 and the second a volume of 110 cm3 . If the bulk modulus of the first ball is B1 and that of the second is B2 , which of the following is true? *(a) B1 < B2 (c) B1 > B2 (e) None of these
(b) (d)
B1 = B2 Not enough information
Solution: We know the initial and final volumes for the two balls. We can calculate the volume changes: ∆V1 = 60 cm3 − 50 cm3 = 10 cm3 ∆V2 = 120 cm3 − 110 cm3 = 10 cm3 We don’t know the pressure at the bottom of the trench, but we know that it’s the same for both balls. We want to know about the bulk moduli B1 and B2 . The relationship between volume, pressure, and bulk modulus is P =B
∆V V
Since the pressure is the same for both balls, we can write B2 ∆V2 B1 ∆V1 = V1 V2 1 B1 V1 ∆V2 60 cm3 10 cm3 · = ⇒ = · = 3 3 B2 ∆V1 V2 10 cm 120 cm 2 ⇒ 2B1 = B2 Since B is positive, B1 < B2 . Problem 1584. A cable has a cross-sectional area of 4 cm2 . It will break if subjected to a tensile stress greater than 2 × 108 Pa. How much force must be applied to the cable in order to break it? *(a) 8 × 104 N (b) 5 × 107 N 8 (c) 8 × 10 N (d) 5 × 1012 N (e) None of these Solution: N/m2 . P =
Don’t forget to convert centimeters to meters. Remember that 1 Pa = 1 F A
⇒
F = P A = (2.0 × 108 N/m2 )(4 × 10−4 m2 ) = 8 × 104 N
587
17.4
Static Fluids
17.4.1
Pascal’s Law and Archimedes’s principle
Problem 1585. Which of the following is a statement of Pascal’s law? (a) If a gas is maintained at a constant temperature, then its volume will be inversely proportional to its pressure. *(b) Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. (c) A body immersed in a fluid is buoyed up with a force equal to the weight of the fluid displaced by the body. (d) One mole of any ideal gas at standard temperature and pressure will have a volume of 22.4 liters. Problem 1586. Which of the following is a statement of Archimedes’s principle? (a) If a gas is maintained at a constant pressure, then its volume will be proportional to its absolute temperature. (b) Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. *(c) A body immersed in a fluid is buoyed up with a force equal to the weight of the fluid displaced by the body. (d) Don’t be rude to armed Romans.
17.4.2
Density, specific volume, specific weight
Problem 1587. The density of air is 1.3 kg/m3 . A room is 5.6 m long by 4.9 m deep by 2.4 m high. What is the mass of the air in the room? Round your answer to the nearest km. (a) 62 kg (b) 69 kg (c) 77 kg *(d) 86 kg (e) None of these Solution: The mass equals the density times the volume. We know the density ρ, the length l, the width w, and the height h. Hence m = ρV = ρlwh = (1.3 kg/m3 )(5.6 m)(4.9 m)(2.4 m) = 86 kg
588 Problem 1588. The density of air at sea level is 1.20 kg/m3 . The living room in your seaside cottage is 5.0 m wide, 4.0 m deep, and 3.0 m high. What is the weight of the air in the room? Give your answer to the nearest pound. Recall: 1 lb = 4.448 N. (a) 18 lb (b) 39 lb *(c) 160 lb (d) 1752 lb (e) None of these Solution: The weight of a volume of gas is the mass times gravity = density times volume times gravity, and the the volume of a rectangular box is Volume = length × width × height: w = mg = (ρV )g = gρlwh = (1.20 kg/m3 )(5.0 m)(4.0 m)(3.0 m) = 710 N. Lastly, we must convert Newtons to pounds: 1 lb = 160 lb. 710 N = 710 N 4.448 N Problem 1589. Your new roommate has left the water running in the sink of your rectangular water-tight dorm room, and the room is now full of water. If the dimensions of the room are: length = 5.0 m, width = 4.0 m, and height = 3.0 m, and the density of water is 1.0 × 103 kg/m3 , then what is the weight of the water in the room? Round your answer to two significant figures. (a) 6.5 × 106 N *(b) 5.9 × 105 N 4 (c) 6.0 × 10 N (d) 5.4 × 103 N (e) None of these Solution: wwater = mwater g = ρwater V g, where V is the volume of the container. We have used the definition of density ρ = m/V to find the mass of the water. The volume of a rectangular box is length × width × height = LW H. Thus w = ρwater V g = ρwater LW Hg = (103
kg )(5m)(4m)(3m)gm/s2 = 64 (9.8)N ≈ 65 . m3
Problem 1590. Your new roommate has filled his room with nitrous oxide gas, which has a density of 1.96 kg/m3 . The room is 4.0 m long by 5.0 m wide by 3.0 m high. What is the mass of the N2 O in the room? Round your answer to the nearest kilogram. (a) 115 kg (b) 57 kg *(c) 118 kg (d) 164 kg (e) None of these Solution: From the definition of density ρ = m/V we have mN2 O = ρN2 O V , where the volume of the rectangular room is V = LW H. Thus mN2 O = ρwater LW H = (1.96
kg )(4m)(5m)(3m)m/s2 ≈ 120kg . 3 m
589 Problem 1591. Your new roommate has filled his room with helium, which has a density of 0.18 kg/m3 . The room is 4.0 m long by 5.0 m wide by 3.0 m high. What is the mass of the helium in the room? Round your answer to two significant figures. (a) 0.71 kg *(b) 11 kg (c) 1.0 kg (d) 28 kg (e) None of these Solution: From the definition of density and the volume of a rectangular room: mhelium = ρhelium LW H = (0.18
kg )(4m)(5m)(3m)m/s2 = 10.8 ≈ 11kg , m3
where L = length, W = width, and H = height. Problem 1592. The mass of an unknown gas mixture in a room that is 3 m × 4 m × )? 5 m is known to be 500 kg. What is the density of the gas (ρ = m V Solution:
From the definition of density and the volume of a rectangular room: ρ=
m 500 kg 25 kg kg = = = 8.33 , LW H (5 m)(4 m)(3 m) 3 m3 m3
where m = mass, L = length, W = width, and H = height. 17.4.3
Static Pressure
Problem 1593. A solar-water heating system uses solar panels on the roof of a large build, 40.0 m above the storage tank. The pressure at the panels is 1 atmosphere. What is the absolute pressure at the top of the tank? Solution:
Use the formula for static pressure: Pabs = Patm + ρgh.
Problem 1594. A solar-water heating system uses solar panels on the roof of a large build 10 m above the top of a large storage tank. The pressure at the panels is 1 atmosphere. What is the gage pressure at the top of the tank? Take the density of water to be ρ = 1000 mkg3 and g = 10 sm2 . (a) 1 kPa (b) 10 kPa 2 *(c) 10 kPa (d) 103 kPa (e) None of these Solution:
Use the equation
Pgage = ∆P = Pabs − Patm = ρgh = (103
kg m N )(10 2 )(10m) = 105 2 = 105 Pa = 102 kPa 3 m s m
590 Problem 1595. A town in the middle-of-nowhere is built up around a large 100 meter hill. Since the hill is in the center of a town, it is proposed at a town-hall meeting that a water tower be placed at the top of the hill. However, the town barber points out that when water flows through a pipe there is a great amount of resistance owing to viscosity and turbulent mixing of the fluid. He estimates that the gauge pressure in the pipe at the bottom of the hill must be at least 100 kPa in order for water to make it to the outskirts of town. Find what the gauge pressure would be at the bottom of the hill. Solution:
Use Pgage = Pabs − Patm = ρgh.
Problem 1596. A submarine has a circular top hatch with a 1 m diameter. What is the force exerted on the hatch from the pressure due to the water above the hatch when the sub dives down to a depth of 200 m? Take the approximate density of sea water to be ρ = 1000 mkg3 and g = 10 sm2 . Note: we are ignoring atmospheric pressure Patm in this problem and we are assuming the submarine is a rigid body (i.e., we can ignore the pressure from inside the sub). (a) 2π × 105 N (b) π × 106 N (c) 2π × 106 N *(d) π2 × 106 N (e) None of these Solution: First we’ll find the pressure on the hatch, then we’ll use the definition of pressure to find the force Fhatch = Phatch Ahatch , where Fhatch is the force on the hatch from the pressure of the sea water above the hatch, Phatch is the pressure on the hatch, and Ahatch is the area of the hatch. Phatch = ρgh = (103
m N kg )(10 2 )(2 × 102 m) = 2 × 106 Pa = 2 × 106 2 . 3 m s m
Using the definition of pressure we find: Fhatch = Phatch Ahatch =
2 (πrhatch ) Phatch
2 N 1 π =π m2 2 × 106 = × 106 N 2 2 m 2
Problem 1597. A vertical, frictionless piston-cylinder device contains a gas that is in equilibrium with the weight of the piston balanced by the internal pressure for the gas. Suppose the mass of the piston is mpiston , the area of the piston cylinder is Acyl , and the atmospheric pressure outside the cylinder is Patm . Derive a formula for the gas contained in the cylinder Pgas in terms of the given information. Solution:
Using a free-body diagram it can be seen that Fgas = Fatm + mpiston g
Acyl
−−−−−−→
Pgas = Patm +
mpiston g . Acyl
591 Problem 1598. The basic barometer can used to measure the height of a building. To do this a measurement is made at the bottom of the building at ground level and a second measurement is taken at the top of the building on the roof. One can then relate these readings to pressure using the equation for a mercury-filled barometer: Patm (hHg ) = ρHg ghHg , where ρHg = 13, 600 kg/m3 is the density of mercury, g is the acceleration due to gravity, and hHg is the height of the mercury column in the barometer measured in millimeters. The approximate height of the building can then be found by making a few simplifying assumptions. For moderate sized buildings a reasonable assumption for the average density of air under fair weather conditions is approximately ρair ≈ 1.18 Kg/m3 . If the barometric readings at the top and bottom of a certain building are 730 mmHg and 755 mmHg respectively, what is the height of the building? Solution:
See handwritten solutions
Problem 1599. A gas is contained in a vertical, frictionless piston-cylinder device. The piston has a mass of 4 kg and cross-sectional area of 35 cm2 . A compressed spring above the piston exerts a force of 60 N on the piston. If the atmospheric pressure is 95 kPa, determine the pressure inside the cylinder. Solution:
See handwritten solutions
Problem 1600. High-altitude balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Fb = ρair gVballoon , will push the balloon upward. If we approximate the shape of the balloon as a sphere with a radius of 5 m and a total payload of 140kg (including the ropes and cage), determine the acceleration of the balloon when it is first released. Assume the density of air is ρair = 1.16 kg/m3 . Solution:
See handwritten solutions
Problem 1601. A glass tube is attached to a water pipe as shown in the figure below. If the water pressure at the bottom of the tube is 115 kPa and the local atmospheric pressure is Patm = 92 kPa, determine how high the water will rise in the tube, in m. Assume g = 9.8 m/s2 at that location and take the density of water to be 103 kg/m3 .
592 Patm
h=?
direction of fluid flow
Figure 16: Vertical glass tube attached to horizontal water pipe. Solution:
See handwritten solutions
Problem 1602. A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside the cooker. The lid of a pressure cooker is well sealed, and the steam can escape only through an opening in the middle of the lid. A separate piece of metal, the petcock, sits on top of this opening and prevents steam from escaping until the pressure inside the cooker provides enough force on the petcock to over come the weight of the petcock allowing the hot gas to escape. This mechanism provides a safety device that limits the maximum pressure that can form in the pressure cooker by providing a periodic release of the high-pressure gas in the container. This in turn prevents extremely high pressures from building up in the cooker that could lead to a potentially deadly explosion of hot gas and metal shards. Assuming that once the cooker reaches the maximum pressure that the gas remains constant in the pressure cooker, determine the mass of the petcock needed for a pressure cooker that is designed to have an operation gage pressure of 100 kPa and has an opening cross-sectional area of 4 mm2 . Assume an atmospheric pressure of 101 kPa. Be sure and draw a free-body diagram of the petcock to accompany your solution. Solution:
See handwritten solutions
593
18
Introduction to Thermodynamics
18.1
Introductory Concepts and Definitions
18.1.1
What is the study of thermodynamics?
Problem 1603. Which of the following items does not describe the main focus of the study of thermodynamics. (a) energy storage (b) the transfer of energy through heat and work (c) how energy transforms from one form of energy, such as kinetic, into another form *(d) how energy is bought and sold on the open market Solution: While holding down cost and making equipment energy efficient is always a concern for engineerings, the particulars of how energy is bought and sold is typically not a concern. Answer (d). 18.1.2
Defining Systems
Problem 1604. Which statement best describes the concept of a system? *(a) the subject of the analysis (b) an object with fixed set of molecules (c) a fluid together with some sort of solid (d) an object that radiates heat into its environment Solution:
(a)
Problem 1605. Consider a refrigerator in a kitchen. Take the refrigerator and everything in it to be our system. Which best describes the system’s surroundings? (a) All of the air in the kitchen. (b) Any one standing in the kitchen. (c) The air inside the refrigerator. *(d) Everything in the universe external to the system. Solution:
(d)
Problem 1606. Consider a refrigerator in a kitchen. Take the refrigerator and everything in it to be our system. Which best describes the system’s boundary? (a) All of the air in the kitchen. (b) The thin region separating the system from everything else. (c) The air inside the refrigerator. (d) The walls of the refrigerator. Solution: Only (b) is correct. The walls are part of the system, but out could think of the thin metal layer as the boundary in practice.
594 18.1.3
Closed systems
Problem 1607. Identify all of the true statements: (a) A closed system is necessarily an isolated system *(b) An isolated system is necessarily a closed system (c) A system cannot be both closed and isolated. (d) The concepts of closed and isolated in regards to a system are independent concepts. Solution: Only (b) is true. In an isolated system no energy can escape. If mass were to escape, then by e = mc2 , energy would also escape. 18.1.4
Control volume
Problem 1608. Identify all of the true statements: *(a) closed system = control mass (b) closed system = control volume (c) open system = control mass *(d) open system = control volume Solution:
Only (a) and (d) are true.
Problem 1609. Which of the following situations would be well suited for using a control volume in the thermodynamic analysis of the system? Do not worry about the details of the analysis. (a) compression of air in a cylinder (b) expansion of gases in a cylinder after a combustion (c) the air in a balloon *(d) filling a bike tire with air from a compressor Solution: The answer is (d). The bike tire is not a closed system, but the volume is clearly defined as the inside of the tube in the tire. Notice that the control volume can change shape. Problem 1610. Which of the following situations would be well suited for using a control mass in the thermodynamic analysis of the system? Do not worry about the details of the analysis. *(a) compression of air in a sealed cylinder (b) a window unit air conditioner (c) a jet engine (d) a fire hose spurting out water Solution:
Only (a) the system in (a) is a closed system.
595 18.1.5
Property, state, and process
Problem 1611. Consider the following two statements: (i) A system is in steady state if its properties are independent of space. (ii) A system is in steady state if its properties are independent of time. Which of the following statements is true: (a) Statement (i) is true and statement (ii) is true (b) Statement (i) is true and statement (ii) is false *(c) Statement (i) is false and statement (ii) is true (d) Statement (i) is false and statement (ii) is false Solution: Steady-state ⇒ time-independent 18.1.6
Extensive and intensive properties
Problem 1612. Which of the following is not an extensive property? (a) Kinetic Energy (b) Momentum (c) Mass *(d) Density (e) None of these Problem 1613. Which of the following is not an intensive property? (a) Velocity *(b) Volume (c) Pressure (d) Temperature (e) None of these
18.1.7
Equilibrium, quasi-equilibrium, and processes
Problem 1614. Two metal blocks, one at 50◦ , and the other at 0◦ are set next to each other in a perfectly insulted box at time t = 0. At this instant are the blocks in thermal equilibrium? Now suppose you wait a long time. Are they in equilibrium now? Solution: At time t = 0 the blocks are not in thermal equilibrium. As time increases (i.e., t → ∞) the system approaches equilibrium. Problem 1615. In a quasi-equilibrium process, the pressure in a system (a) is held constant throughout the entire process *(b) is approximately spatially uniform throughout the system at each moment in time (c) increases if volume increases (d) always varies with temperature Problem 1616. Which of the following process is a quasi-equilibrium process? (a) the stirring and mixing of cold creamer in hot coffee (b) a balloon bursting (c) combustion *(d) the slow and steady compression of air in a cylinder
596 18.1.8
Base SI units for mass, length, time, Force, energy, and pressure
Problem 1617. Express kinetic energy in terms of the base SI units: the kilogram, meter, and second. Write your answer in terms of the abbreviations ({kg, m, s}), and then identify the resulting derived unit. Solution:
Apply the uninator: m 2 m 1 2 2 [KE] = [ mv ] = [m] · [v] = kg = kg 2 · m = N · m = J 2 s s
Problem 1618. Express work (force × distance) in terms of the base SI units: the kilogram, meter, and second. Express your answer in terms of the abbreviations (kg, m, s). Solution:
Apply the uninator: [W ] = [F · d] = [F ] · [d] = N · m = J
Problem 1619. Express P V , where P is pressure and V is volume, in terms of the base SI units: the kilogram, meter, and second. Express your answer in terms of the abbreviations (kg, m, s). Compare this answer to the previous one. What do you notice? Solution:
Apply the uninator: N F ·− V = · m3 = J [P − V] = A m2
Thus, preesure times volume has the units of work. In fact, it will be the most common form of work that we will encounter. It is the work done by an expanding gas in moving a boundary. Problem 1620. Express power in terms of the base SI units: the kilogram, meter, and second. Write your answer in terms of the abbreviations ({kg, m, s}), and then identify the resulting derived unit. Solution:
Apply the uninator:
W [P] = t
=
J = watt s
Problem 1621. Express specific weight in terms of the base SI units: the kilogram, meter, and second. Express your answer in terms of the abbreviations ({kg, m, s}), and then identify the resulting derived unit. Solution:
Apply the uninator: m kg N Pa [ρg] = g = 3· = − V m kg m
⇒
[ρgh] = Pa,
as expected, since ρgh is known to be the pressure owing to the weight the fluid h units above the point where the pressure is measured.
597 Problem 1622. Express specific volume in terms of the base SI units: the kilogram, meter, and second. Express your answer in terms of the abbreviations ({kg, m, s}). Solution: The specific volume is just the reciprocal of density. Work it out! Problem 1623. Which one of the following expressions can be converted to the unit of a Joule? (a) Pa · m2 *(b) Pa · m3 2 (c) Pa/m (d) N/kg (e) None of these Solution:
Since Pa = N/m2 and J = N·m, it follows that (b) is the answer.
Problem 1624. Which of the following is not an acceptable “extended” SI unit? Recall: The SI system is the MKS system (Meter, Kilogram, Second), but we’ll allow the extended SI system to include the cgs system (centimeter, gram, second), but you can’t mix these two systems. (a) distance measured in centimeters (b) pressure measured in newtons per square meter (c) volume measured in cubic meters *(d) density measured in grams per cubic meter Solution:
Can’t mix grams and meters.
598
18.2
Temperature Scales
Use the equations TC = TK − 273 and TF = 95 TC + 32 to answer the following questions. Round your answers to the nearest integer. Problem 1625. If heat is added to a system and the temperature of a system increases, without knowing anything else, which form of energy will be definitely increase? (a) The kinetic energy of the system (b) The potential energy of the system (c) The work done by the system *(d) The internal energy (i.e., the molecular energy) of the system Solution: If the temperature increases, then the internal energy, which depends on temperature, will certainly rise. But adding heat to a system need not result in an increase in bulk energy of the system. For example, consider a pot of water sitting on a hot-plate. If we take the water as our system, then clearly heat is being added to the system, but since the pot is not moving, its potential and kinetic energy are not changing. Since the boundary of the system is not moving, there is no work being done to, or by the system. Problem 1626. Convert 98◦ F to ◦ C. (a) 32◦ C (b) 208◦ C (c) 20◦ C *(d) 37◦ C (e) None of these 5 5 5 5 5 TC = (TF −32) = (98−32)◦ C = ·66◦ C = ·22◦ C = (35+ )◦ C ≈ 37◦ C 9 9 9 3 3 Problem 1627. Convert 68◦ F to ◦ C. (a) 12◦ C (b) 17◦ C *(c) 20◦ C (d) 37◦ C (e) None of these Solution:
5 TC = (TF − 32) = 20◦ C 9 Problem 1628. Convert 110◦ F to ◦ C. (a) 50◦ C *(b) 43◦ C ◦ (c) 20 C (d) 37◦ C (e) None of these
Solution:
5 TC = (TF − 32) = 43.3◦ C ≈ 43◦ C 9 Problem 1629. Convert 20◦ C to ◦ F. *(a) 68◦ F (b) 98◦ F ◦ (c) 120 F (d) 212◦ F (e) None of these Solution:
Solution:
9 TF = TC + 32 = 68◦ F 5
599 Problem 1630. Convert 100◦ C to ◦ F. (a) 68◦ F (b) 98◦ F (c) 120◦ F *(d) 212◦ F (e) None of these Solution:
9 TF = TC + 32 = 212◦ F 5
Problem 1631. Convert 20◦ C to K. (a) −253 K *(b) 293 K (c) 68 K (d) 0 K (e) None of these Solution:
TK = TC + 273 K = (20 + 273) K = 293 K.
Problem 1632. Convert 10 K to ◦ C. *(a) −263 K (b) 293 K (c) 68 K (d) 0 K (e) None of these Solution:
TC = TK − 273◦ C = (10 − 273)◦ C = −263◦ C.
Problem 1633. Convert 98◦ F to ◦ R (degrees rankine). (a) 460◦ R *(c) 558◦ R (e) None of these
(b) (d)
−558◦ R 200◦ R
Solution: TR = TF + 459.67 ≈ TF + 460 = 558◦ R. Note: Some books say do not use the degree symbol with the Rankine scale, since it’s an absolute scale like the Kelvin scale.
600 Problem 1634. (Measuring wind chill) It is well-known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the “chilling effect” of the wind. This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities. The equivalent wind chill temperature in ◦ F is given by (all variables with an asterisk superscript (∗ ) are dimensional quantities) ∗ ∗ ∗ ∗ Tequiv,F (◦ F) = Tadj,F (◦ F) + (Tambient,F (◦ F) − Tadj,F (◦ F)) f (vb )
where f is the dimensionless function (which must have dimensionless arguments in order to be dimensionally consistent! - for any smooth function this can be proved via a Taylor series expansion) √ f (vb ) = 0.475 − 0.0203vb + 0.304 vb , ∗ and Tadj,F = 91.4◦ F is a temperature adjustment term, vb is a dimensionless wind speed that has been non-dimensionalized by miles per hour (i.e. vb∗ = vb , (miles/hour and ∗ Tambient is the ambient air temperature in ◦ F in calm air, which is taken to be light winds at speeds of no more than 4 mph. The constant 91.4◦ F in the above equation is the mean temperature of a resting person in a comfortable environment. Windy air at temperature Tambient and velocity v will feel as cold as the calm air at temperature Tequiv . Notice that the units of the coefficients in the second factor of the second term are, in order: dimensionless, hours per mile, and the square root of hours per mile so that the resulting second factor is dimensionless after the dimensional variables are substituted into the equation.
Using proper conversion factors, obtain an equivalent relation in SI units where vSI is the dimensionless wind speed based on m/s and Tambient is ambient air temperature in ◦ C. Hint: When working with physical equations all terms must be dimensionally consistent. To preserve the dimensional consistency whatever we do to one side of the equation, we must do to the other, just as you learned in algebra. Solution: Starting from the original equation we first transform the dimensional component (i.e., the temperature component has physical dimensions, but the factor f (vb ) is dimensionless). ∗ ∗ ∗ ∗ (◦ F)) f (vb ) (◦ F) = Tadj,F (◦ F) + (Tambient,F (◦ F) − Tadj,F Tequiv,F −32◦ F
∗ −−−−−−−→ Tequiv,F (◦ F) − 32◦ F
5◦ C 9◦ F
−−−−−−→
∗ ∗ ∗ = (Tadj,F (◦ F) − 32◦ F) + (Tambient,F (◦ F) − Tadj,F (◦ F)) f (vb ) ◦ 5C ∗ Tequiv,F (◦ F) − 32◦ F ◦ 9F ◦ 5C ∗ = (Tadj,F (◦ F) − 32◦ F) 9◦ F ◦ 5C ∗ ◦ ∗ ◦ + ( F) f (vb ) . T ( F) − T ambient,F adj,F 9◦ F
601
Using the formula
TC∗ (◦ C)
∗ (◦ C) Tequiv,C
=
5◦ C 9◦ F
(TF∗ (◦ F) − 32◦ F) we can rewrite the equation as
◦
= 33 C +
5◦ C 9◦ F
∗ Tambient,F
◦
( F) −
∗ Tadj,F
◦
( F) f (vb )
rewrite
∗ −−−−−−−→ Tequiv,C (◦ C) ◦
= 33 C +
5◦ C 9◦ F
∗ ∗ (Tambient,F (◦ F) − 32◦ F) − (Tadj,F (◦ F) − 32◦ F) f (vb )
= 33◦ C ◦ 5◦ C 5C ∗ ◦ ◦ ∗ ◦ ◦ + (T ( F) − 32 F) − (T ( F) − 32 F) f (vb ) ambient,F adj,F 9◦ F 9◦ F ∗ = 33◦ C + Tambient,C (◦ C) − 33◦ C f (vb )
The last step is to convert the correction factor f (vb ).
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