Instructor.pdf

Interactive Physics Curriculum Workbook Instructor Edition

Paul Mitiguy, Ph.D. Michael Woo Contributing Author Benjamin Reineman Artwork by Ellen Chong & Victor Orozco

ISBN 1-58524-999-8 (SI Units)

May 2010

c 2006-2010 Design Simulation Technologies, Inc. Copyright  c 2005 MSC.Software Corporation Portions Copyright 

Contents General instructions for using Interactive Physics curriculum . . . . . . . . . . . . . . 1 Speed, distance, and time 1.1 Distance, speed, and time (A) . . 1.2 Distance, speed, and time (B) . . 1.3 Speed, distance, and time (A) . . 1.4 Speed, distance, and time (B) . . 1.5 Time, distance, and speed . . . . 1.6 Constant speed and varying speed

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2 Acceleration and change in speed 2.1 Change in speed, acceleration, and time . . . 2.2 Acceleration, change in speed, and time (A) 2.3 Acceleration, change in speed, and time (B) 2.4 Constant and changing acceleration . . . . . 2.5 Free fall - how fast? . . . . . . . . . . . . . . 2.6 Upward projectile motion - how fast? . . . .

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3 Acceleration, speed, and distance 3.1 Free fall - how far? . . . . . . . . . . . . . . . . . 3.2 Upward projectile motion - how far? . . . . . . . 3.3 Upward projectile motion - maximum height . . . 3.4 Upward projectile motion - initial speed . . . . . 3.5 Time, acceleration, and distance . . . . . . . . . . 3.6 Speed, acceleration, and distance . . . . . . . . . 3.7 Distance, speed, acceleration, and highway safety 3.8 Speed, distance, acceleration, and highway safety

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iii

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4 Weight, mass, and gravity 44 4.1 Weight and mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.2 Weight, mass, and other planets . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.3 Mass and weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 5 Air 5.1 5.2 5.3 5.4

resistance Air resistance Air resistance Air resistance Air resistance

and falling objects - concepts & terminal velocity - concepts & gravity forces - concepts . . and projectile motion . . . . .

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6 Newton’s Law for motion along a line 6.1 Force, mass, and acceleration - Newton’s apple . 6.2 Force, mass, and acceleration - rocket . . . . . . 6.3 Acceleration, force, and mass (A) . . . . . . . . 6.4 Acceleration, force, and mass (B) . . . . . . . . 6.5 Acceleration, force, mass, and friction . . . . . . 6.6 Force, mass, acceleration, and air resistance . . 6.7 Force, mass, acceleration, and horizontal springs 6.8 Force, mass, acceleration, and vertical springs .

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7 Potential and kinetic energy 7.1 Gravitational potential energy 7.2 Potential energy in a spring . 7.3 Kinetic energy and mass . . . 7.4 Kinetic energy and speed . . .

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8 Conservation of energy 8.1 Potential and kinetic energy in a pendulum . . . . . . . . 8.2 Conservation of energy in a roller coaster (Windows only) 8.3 Conservation of energy for a snowboarder . . . . . . . . . 8.4 Conservation of energy in a spring . . . . . . . . . . . . . 8.5 Conservation of energy for a bungee jumper . . . . . . . 8.6 Energy loss due to friction . . . . . . . . . . . . . . . . .

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81 82 83 85 87 89 90

9 Temperature and heat 9.1 Temperature . . . . . 9.2 Heat capacity . . . . 9.3 Phase change . . . . 9.4 Thermal Expansion .

92 93 94 96 97

10 Heat transfer 10.1 Heat transfer . . . 10.2 Conduction . . . . 10.3 Convection . . . . . 10.4 Radiation (Windows

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11 Wave and sound 11.1 Wave motion . . . . . . . . . 11.2 Wave speed . . . . . . . . . . 11.3 Speed of sound . . . . . . . . 11.4 Mach number . . . . . . . . . 11.5 Doppler effect (Windows only)

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12 Intermolecular attractions 12.1 Hydrogen bonding between water molecules . . . . . . 12.2 Interaction of water molecules and a sodium ion . . . . 12.3 Interaction of water molecules and a chloride ion . . . . 12.4 Interaction of water with non-polar and polar molecules 12.5 Investigating the strength of hydrogen bonds . . . . . . 13 Discovery Learning

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124 ii

Why Interactive Physics? Physics is a challenging and rewarding subject to teach and learn. The laws of physics shed light on mysteries in the universe. Interactive Physics was specifically created to promote inquiry learning and discovery learning for younger scientists while introducing tools used by university researchers and by professionals in industries such as automotive, aerospace, machines, cameras, printers, and medical devices. A successful physical science classroom motivates students to explore their physical world and ask “why” and “what if ” questions while introducing calculator and graphing skills. An ideal high school physics program continues the conceptual exploration of physics and connects mathematics with the laws of physics. College physics and engineering serve a dual purpose. In addition to coupling advanced conceptual learning with higher mathematics, good university programs prepare students for research and professional careers with school-to-work skills. Level Middle School High School University Research Professional

Value Introductory principles Conceptual learning & basic math Advanced concepts & mathematics Validate motion theories Virtual prototyping

Interaction Interact with physics experiments Explore “what if” scenarios Build physics experiments Simulate reality Design real equipment

Why Simulation? Simulation provides insights through interaction. Textbooks are useful for providing structured curriculum, line drawings, and pictures. Classroom demonstrations together with computer and laboratory experiments provide an interactive, effective way to teach physical concepts and motivate students. Instructors appreciate Interactive Physics because it provides cost-efficient experiments that are safe, dependable, and time-effective. Students who learn with Interactive Physics appreciate the visually appealing, fast-paced, interactive experiments that provide a comprehensive understanding of real physical systems and allows them to move past a concept to test new theories.

Why Interactive Physics Curriculum? The Interactive Physics staff was recruited from the scientific and academic community. We understand the demands on an instructor’s time and budget and the need for experiments and curriculum that motivate, challenge, and test students. We understand the need to meet and exceed national and state content standards and objectives. We also understand that instructors need technical support from a reputable company with a firm commitment to education. If you have ideas on improving our software, experiments, or curriculum, please let us know. We have been listening to your suggestions and we will continue to support your excellence in education with excellence in software and physics experiments.

iii

Paul Mitiguy, Ph.D. From Milton, Massachusetts, Paul completed his bachelor’s degree in 1986 at Tufts University and later received his M.S. and Ph.D. in Mechanical Engineering at Stanford University. As a consulting associate professor at Stanford, Paul uses physics to teach dynamics and dynamic systems, researches and publishes in physics simulation, and reviews journal articles. Since 1990, Paul has been developing motion simulation software, including the physics program Autolev. After working at M.I.T. Lincoln Laboratory and NASA Ames, Paul developed educational and professional motion software at Knowledge Revolution and MSC.Software, including Interactive Physics, Working Model 2D/3D/4D, and MSC.visualNastran 4D.

Michael Woo, M.S. In 1999 and 2000, Michael received his Bachelor and Master of Science degrees in Mechanical Engineering from Stanford University. Michael was president of Tau Beta Pi (the engineering honor society) and was both an engineering classroom teaching assistant and science and math tutor at Stanford University. After graduate school, Michael designed database software at Tenfold Corporation. In 2001, he re-joined the Interactive Physics team to develop interactive educational software & curriculum. Paul and Michael have integrated software and curriculum into a wide number of physics and engineering textbooks, including: Textbook Conceptual Physics College Physics Physics for Scientists and Engineers College Physics Physics for Scientists and Engineers Principles of Physics Vector Mechanics 6th edition Foundations of Engineering Design of Machinery 3/e Mechanics of Machines Engineering Mechanics, Statics, & Dynamics Engineering Mechanics, Statics Engineering Mechanics, Dynamics

Author Hewitt Wilson Fishbane, Gasiorowicz, . . . Serway & Faughn Serway Serway Beer and Johnston Holtzapple and Reece Norton Cleghorn Bedford and Fowler Hibbeler Hibbeler

Publisher Addison-Wesley Prentice Hall Prentice Hall Saunders Saunders Saunders McGraw-Hill McGraw-Hill McGraw-Hill Oxford Prentice-Hall Prentice-Hall Prentice-Hall

Experiments on energy, temperature, and heat were created by Benjamin Reineman, a Mechanical Engineer in energy research in the Thermosciences Division at Stanford University. The graphics in some experiments were created by Ellen Chong and Victor Orozco. Ellen is an artist, educator, and a member of the Interactive Physics team. Victor is an artist, educator, and mechanical engineer from Stanford University..

iv

General Instructions for Using Interactive Physics (IP) Curriculum Requirements You must have installed a recent full or homework edition of Interactive Physics with curriculum.1 If you do not own Interactive Physics, visit http://www.interactivephysics.com. Running an IP Curriculum experiment (simulation) 1. On a Windows PC, simply click on StartCurriculum.html and follow the on-screen directions. Alternately, to open an experiment, browse to the IPCurriculum folder, then browse to the appropriate topic folder, and double-click on a file.2 2. Click to run the experiment and click to reset the experiment to the beginning. 3. Experiments contain sliders, check-boxes, or buttons. To select a value with a slider, click and drag the slider up and down (or left and right). The value is displayed in the box adjoining the slider. 4. After moving the slider, click to re-run the experiment with the new value. 5. Observe the experiment and answer the related question(s). 6. Repeat steps 2-6 and answer all the questions pertaining to that experiment. 7. If necessary, click the

or

button to rewind or advance one or more frames.

Closing an experiment or Exiting Interactive Physics 1. To close an experiment, choose the File menu and click Close (or click the lower 2. To exit Interactive Physics, choose the File menu and click Exit (or click the upper

1

). ).

Windows computers will automatically install the curriculum with the Interactive Physics program. To install curriculum on a Macintosh, insert the Interactive Physics CD into your CD drive, browse to your CD drive, and copy (drag) the IPCurriculum folder to your computer’s desktop or to a convenient location (e.g., to the folder where Interactive Physics was installed). 2 If you have difficulty opening an Interactive Physics file on Windows, click to the Windows Start Menu, click on Programs, then InteractivePhysics, and then InteractivePhysics again. This opens a blank Interactive Physics document. With Interactive Physics open, click on the File menu and then click Open . . . , then browse to the IPCurriculum folder that was copied to your computer, and double-click on the appropriate folder and experiment to open it.

v

Chapter 1 Speed, distance, and time Speed is a measure of the distance an object travels in a certain amount of time. Average speed is calculated by dividing the distance an object travels by the time it takes to travel that distance. The equation below relates average speed to distance traveled and elapsed time1 . AverageSpeed = DistanceTraveled ElapsedTime

Courtesy of MSC.Software

Each symbol in the equation is explicitly spelled out. Many textbooks use shorter symbols, e.g., using t in place of ElapsedTime and v in place of AverageSpeed. 1

1

1.1

Distance, speed, and time (A)

The distance that a car travels depends on the car’s average speed and travel time (elapsed time). In mathematical terms, DistanceTraveled = AverageSpeed ∗ ElapsedTime

1. Suppose the car has an average speed of 100 km/h (the abbreviation km/h stands for kilometers per hour) and the car travels for 2 hours. How many kilometers has the car traveled? Result: 100

km hour

∗ 2 hours =

200

km

To answer this question with Interactive Physics, click the button. Look at the meter that measures the distance traveled by the car and verify that the value reported by Interactive Physics matches the value you calculated with the formula above. When you are finished, click the

button.

2. Suppose the car has an average speed of 100 km/h and the car travels for 3 hours. How many kilometers has the car traveled? Result: 100

km hour

∗

3

hours =

To answer this question with Interactive Physics, click the

300

km

button (if necessary) and click and drag

the slider that controls the car’s travel time so that its value is 3 hours. Then, click the button. After the simulation stops, verify that the value reported by Interactive Physics matches the value you calculated with the formula above. Click the

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

button when you are finished.

2

Chapter 1: Speed, distance, and time

3. Suppose the car has an average speed of 100 km/h and the car travels for 1 hour. How many kilometers has the car traveled? Result: 100 Click the

km hour

∗ 1 hour =

100

button, change the slider to 1 hour, and click the

km button. Verify that your calcu-

lations match those reported by Interactive Physics. When you are finished, click the

button.

4. Suppose the car has an average speed of 100 km/h and the car travels for 2.5 hours. How many kilometers has the car traveled? Result: 100

km hour

∗ 2.5 hours =

250

km

Courtesy of MSC.Software c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

3

Chapter 1: Speed, distance, and time

1.2

Distance, speed, and time (B)

The distance that a car travels depends on the car’s average speed and travel time (elapsed time). In mathematical terms, DistanceTraveled = AverageSpeed ∗ ElapsedTime

1. Suppose the car has an average speed of 100 km/h (the abbreviation km/h stands for kilometers per hour) and the car travels for 3 hours. How many kilometers has the car traveled? Result: 100

km hour

∗ 3 hours =

300

km

To answer this question with Interactive Physics, click the button. Look at the meter that measures the distance traveled by the car and verify that the value reported by Interactive Physics matches the value you calculated with the formula above. When you are finished, click the

button.

2. Suppose the car is going more slowly, at an average speed of 50 km/h. How many kilometers does the car travel in 3 hours? Result: 50

km hour

∗

3

hours =

To answer this question with Interactive Physics, click the

150

km

button (if necessary) and click and drag

the slider that controls the car’s average speed so that its value is 50 km/h. Then, click the button. After the simulation stops, verify that the value reported by Interactive Physics matches the value you calculated with the formula above. Click the

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

button when you are finished.

4

Chapter 1: Speed, distance, and time

3. Suppose the car’s average speed is 75 km. How many kilometers does the car travel in 3 hours? Result: 75 Click the

km hour

∗ 3 hours =

225

km

button, change the slider to 75 km/h, and click the

button. Verify that your cal-

culations match those reported by Interactive Physics. When you are finished, click the

button.

4. Suppose the car’s average speed is 25 km/h. How many kilometers does the car travel in 3 hours? Result: 25

km hour

∗ 3 hours =

km

75

5. A car whose average speed is 75 km/h travels a shorter/ longer (circle one) distance in 3 hours than a car whose average speed is 25 km/h. 6. Bonus Question - answer without Interactive Physics. Suppose the car has an average speed of 100 km/h and the car travels for 9 hours. How many kilometers does the car travel? Result: 100

km hour

∗ 9 hours =

900

km

Courtesy of MSC.Software c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

5

Chapter 1: Speed, distance, and time

1.3

Speed, distance, and time (A)

When driving in traffic, the most important measurement of how fast you get to your destination is average speed. For example, there may be stretches of highway where a car can go 100 km/h and other parts of the highway where traffic slows the car to 30 km/h. If highway traffic is congested, it may be better to take a side-street where the car’s average speed is 60 km/h. The equation that helps answer the questions below is AverageSpeed = DistanceTraveled ElapsedTime

1. What is the average speed of a car that travels 100 km in 2 hours? Result: 100 km km = 50 hour 2 hours To answer this question with Interactive Physics, click the button. Look at the meter that measures the speed of the car and verify that the value reported by Interactive Physics matches the value you calculated with the formula above. When you are finished, click the

button.

2. What is the average speed of a car that travels 200 km in 2 hours? Result: 200 km = 2 hours To answer this question with Interactive Physics, click the so its value is 200 km. Next, click the

100

km hour

button and then click and drag the slider

button. After the simulation stops, verify that your calcu-

lations match those of Interactive Physics. When you are finished, click the

button.

3. The car that travels 100 km in 2 hours has a smaller /larger (circle one) average speed than a car that travels 200 km in 2 hours. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

6

Chapter 1: Speed, distance, and time

4. What is the average speed of a car that travels 300 km in 2 hours? Result: 300 km = 2 hours

150

To answer this question with Interactive Physics, click the slider to 300 km. Next, click the

km hour

button and then set the distance traveled

button and verify that the average speed displayed by Interactive

Physics matches your calculations. When you are finished, click the

button.

5. Bonus Question - answer without Interactive Physics A police officer figures out that a car travels 600 km in 4 hours. Does the officer know that the driver was speeding and should get a speeding ticket? Yes /No (circle one).

6. The distance between two toll-booths on a highway is 200 km. It was 2:00 p.m. when the driver took the ticket at the first toll-booth. It was 4:00 p.m. when the driver reached the second toll-booth. What was the average speed of the car between the two toll-booths?

Result: 200 km = 2 hours

100

km hour

7. A truck driver is driving on the highway and looks at her odometer when she enters a toll-booth. (The odometer measures how far the car has been driven since it was new.) The odometer displays 7,800 km. Two hours later, she reaches a second toll-booth and the odometer displays 7,950 km. What was the average speed of the car between the two toll-booths? Result: 150 km = 2 hours

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

7

75

km hour

Chapter 1: Speed, distance, and time

8. A bus driver enters a toll-booth and looks at her odometer and her watch. The odometer displays 12,300 km and the watch displays 1:15 p.m. At 3:15 p.m., she reaches a second toll-booth and the odometer displays 12,400 km. What was the average speed of the car between the two toll-booths? Result: 100 km = 2 hours

50

km hour

9. Bonus Question - answer without Interactive Physics. Suppose you are a toll-booth operator and a driver hands you a ticket from a toll booth located 240 km away. The time stamped on the ticket from the other toll-booth is 1:45 p.m. You look at your watch and see that the current time is 3:45 p.m. What was the average speed of the car between the two toll-booths?

Result: 240 km = 2 hours

120

km hour

10. Bonus Question - answer without Interactive Physics. Suppose you are a toll-booth operator and a driver hands you a ticket from a toll booth located 320 km away. The time stamped on the ticket from the other toll-booth is 9:45 a.m. The current time on your watch is 11:45 a.m. Was the driver speeding? Yes /No (circle one). What was the driver’s average speed? Result: 320 km = 2 hours

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

8

160

km hour

Chapter 1: Speed, distance, and time

1.4

Speed, distance, and time (B)

A police officer in an aircraft can determine a car’s average speed by measuring the time it takes for the car to travel a certain distance and then calculating the car’s average speed by AverageSpeed = DistanceTraveled ElapsedTime

1. What is the average speed of a car that travels 300 km in 3 hours? Result: 300 km = 3 hours Click the

100

km hour

button and verify that the value you calculated with the formula above matches the one

in the Interactive Physics speed meter. When you are finished, click the

button.

2. What is the average speed of a car that travels 300 km in 6 hours? Result: 300 km = 6 hours Click

50

km hour

and then click and drag the slider to 6 hours. Next, click

and verify that the speed

displayed by Interactive Physics matches your calculations. When you have finished, click

.

3. A car that travels 300 km in 3 hours has a smaller/ larger (circle one) average speed than a car that travels 300 km in 4 hours. 4. What is the average speed of a car that travels 300 km in 2 hours? Result: 300 km = 2 hours c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

9

150

km hour

Chapter 1: Speed, distance, and time

5. The distance between two toll-booths on a highway is 300 km. It was 2:00 p.m. when the driver took the ticket at the first toll-booth. It was 6:00 p.m. when the driver reached the second toll-booth. What was the average speed of the car between the two toll-booths?

Result: 300 km = 4 hours

75

km hour

6. A driver enters a toll-booth and looks at her odometer and her watch. The odometer displays 3,600 km and the watch displays 10:30 a.m. At 1:30 p.m. she reaches a second toll-booth and the odometer displays 3,900 km. What was the average speed of the car between the two toll-booths? Result: 300 km = 3 hours

100

km hour

7. Suppose you are a toll-booth operator and a driver hands you a ticket from a toll booth located 300 km away. The time stamped on the ticket from the other toll-booth is 12:45 p.m. You look at your watch and see that the current time is 2:45 p.m. Was the driver exceeding 100 km/h? Yes /No (circle one). What was the driver’s average speed? Result: 300 km = 2 hours

150

km hour

8. Bonus Question - answer without Interactive Physics. Suppose you are a toll-booth operator and a driver hands you a ticket from a toll booth located 25 km away. The time stamped on the ticket from the other toll-booth is 8:45 a.m. You look at your watch and see that the current time is 9:15 a.m. What was the average speed of the car between the two toll-booths? Result: 25 km = 0.5 hours

50

km hour

9. Bonus Question - answer without Interactive Physics. A police officer in an aircraft observes a car that travels 30 km in 15 minutes. Should the officer issue the driver a speeding ticket? Yes /No. What was km the driver’s average speed? 120 hour

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

10

Chapter 1: Speed, distance, and time

1.5

Time, distance, and speed

It is time to take a trip to your favorite vacation spot located 300 km away. Before leaving, you want to to answer some basic questions such as “How long will it to drive there?”. The equation that helps answer the questions below is ElapsedTime = DistanceTraveled AverageSpeed

1. How long does it take a car whose average speed is 25 km/h to travel 300 km? Result: 300 km = 25 km/h

12

hours

2. How long does it take a car whose average speed is 100 km/h to travel 300 km? Result: 300 km = 100 km/h

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

11

3

hours

Chapter 1: Speed, distance, and time

3. Complete the table on the right which shows the amount of time it takes a car to travel 300 km for various speeds.

Speed (km/h) 25

Time to travel 300 km (hours) 12

50

6

75

4

100

3

125

2.4

150

2

4. Notice that the numbers on the first line of the previous table are 25 and 12, and that there is an X on the graph below at the location (25,12), i.e., 25 in the horizontal direction and 12 in the vertical direction. Similarly, there is a second X on the graph below whose location corresponds to the numbers on the second line of the previous table, i.e., 50 in the horizontal direction and 6 in the vertical direction. Continue until there are six Xs on the graph (one for each line of the table). Then draw a line connecting the first X to the second X and so on, until each X is connected to the next X.

12

Time to travel 300 km (hours)

11 10 9 8 7 6 5 4 3 2 1 0 25

50

75 100 Speed (km/h)

125

150

5. Bonus Question - answer with the previous graph. Estimate the time to travel 300 km at 30 km/h: 10 hours. Estimate the time to travel 300 km at 200 km/h: 1.5 hours.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

12

Chapter 1: Speed, distance, and time

6. You are driving in a car with cruise control set at 100 km/h. You pass a distance marker on the side of the road that reads 110 km. After driving for awhile, you see another distance marker that reads 410 km. How much time has elapsed since passing the first distance marker? Result: 300 km = 100 km/h

3

hours

7. Bonus Question - answer without Interactive Physics. You are driving on the speedway at an average speed of 150 km/h. Each lap is 5 km. How long does it take to go 60 laps? Result:

5

km lap

∗ 60 laps = 300 km ⇒

300 km = 150 km/h

2

hours

8. Bonus Question - answer without Interactive Physics. An accurate clock is one of the most important human inventions in the last thousand years. Clocks that accurately recorded the hour, minute, and second did not exist until 1700 A.D. Suppose you left your home at 1:15 p.m. but forgot to take your watch. Upon leaving your home, you pass a distance marker on the side of the road that reads 130 km. After driving for awhile at 80 km/h, you see another distance marker on the side of the road that reads 450 km. What time must it be? Result: 320 km = 4 hours ⇒ 80 km/h

5:15 p.m.

9. Bonus Question - answer without Interactive Physics. Suppose a fast car has an average speed of 100 km/h and travels for 3 hours. How much time does it take a slow car traveling at 80 km/h to go the same distance as the fast car? Result: 300 km = 80 km/h

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

3 hours

13

and

45 minutes

Chapter 1: Speed, distance, and time

1.6

Constant speed and varying speed

A car’s speed is constant if it does not change. Alternately, a car’s speed varies if it changes. AverageSpeed = DistanceTraveled ElapsedTime

1. Click and watch the number inside the box on the right-hand side of the gray car. Since the number does not change, the gray car has a constant /variable speed. 2. The green car has a constant/ variable (circle one) speed. 3. The red car has a constant/ variable speed. 4. The blue car has a

constant /variable speed.

5. The gray car’s speed is constant, so it has a minimum speed of 30 km and a maximum h km speed of 30 h . Since it travels a distance of 60 km in two hours, what is its average speed? Result: 60 km = 2 hours

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

14

30

km hour

Chapter 1: Speed, distance, and time

6. Determine the minimum, maximum, and average speed of the green car. To answer this question with Interactive Physics, click the button and then click on the arrow that is on the furthest right and bottom part of the screen. Each click will step the simulation forward. Observe the speed reported by the green car and record the smallest and largest values as you step the simulation until the end. When you are finished, click the

Result: Minimum speed =

0

km h

button.

Maximum speed =

Average speed = 100 km = 2 hours

50

100

km h

km hour

7. Complete the following table. Minimum Speed

Maximum Speed

Average Speed

Gray car

30

km h

30

km h

30

km h

50

km h

Green car

0

km h

100

km h

Red car

0

km h

200

km h

50

km h

50

km h

Blue car

50

km h

50

km h

8. When a car’s speed is constant, its average speed is different than its maximum or minimum speed. True/ False (circle one) 9. When a car’s speed varies, its average speed is different than its maximum or minimum speed. True /False (circle one)

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

15

Chapter 1: Speed, distance, and time

Chapter 2 Acceleration and change in speed Acceleration is a measure of the change in speed in a certain amount of time. Average acceleration is calculated by dividing the change in speed by the elapsed time as1 . ChangeInSpeed = CurrentSpeed − InitialSpeed

AverageAcceleration =

ChangeInSpeed ElapsedTime

Courtesy Department of Defense Each symbol in the equations are explicitly spelled out. Many textbooks use shorter symbols, e.g., using t in place of ElapsedTime and a in place of AverageAcceleration. 1

16

2.1

Change in speed, acceleration, and time

The change in the speed of a boat depends on the boat’s average acceleration and the amount of time the boat is accelerating (elapsed time). In mathematical terms, ChangeInSpeed = AverageAcceleration ∗ ElapsedTime

1. Suppose the boat has an average acceleration of 15 km/h/sec (the abbreviation km/h/sec stands for kilometers per hour per second) and the boat accelerates for 5 seconds. What is the change in the boat’s speed? Result: 15

km/h sec

∗ 5 sec =

75

km/h

2. Suppose the boat has an average acceleration of 10 km/h/sec and the boat accelerates for 5 seconds. What is the change in the boat’s speed? Result: 10

km/h sec

∗ 5 sec =

50

km/h

3. Complete the table for a boat that accelerates for 5 seconds. Average acceleration Change in speed km/h/sec km/h 15 75 50 10 5 25 0 0

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

17

Chapter 2: Acceleration and change in speed

Bonus Questions - answer without Interactive Physics. When a boat’s speed changes, its current speed (its speed after accelerating) depends on the boat’s average acceleration, the amount of time the boat is accelerating (elapsed time), and its initial speed. In mathematical terms, ChangeInSpeed = AverageAcceleration ∗ ElapsedTime CurrentSpeed = ChangeInSpeed + InitialSpeed A. Suppose the boat has an initial speed of 20 km/h and an average acceleration of 10 km/h/sec. What is the boat’s speed at 5 seconds? Result: ChangeInSpeed = 10 CurrentSpeed =

km/h sec

50

∗ 5 sec = km/h +

50

km/h

20

km/h =

70

km/h

B. Suppose the boat’s initial speed is 20 km/h and it has an average acceleration of 10 km/h/sec. What is the boat’s speed at 5 seconds? Result: 10

km/h sec

∗ 5 sec + 20 km/h =

70

km/h

C. Suppose the boat’s initial speed is 60 km/h and it slows down with an average acceleration of -10 km/h/sec. What is the boat’s speed at 2 seconds? Result: -10

km/h sec

∗ 2 sec + 60 km/h =

40

km/h

Courtesy Department of Defense c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

18

Chapter 2: Acceleration and change in speed

2.2

Acceleration, change in speed, and time (A)

To safely merge with highway traffic, it is important to change speed to match the speed of the other cars on the highway before entering the highway. The equation that helps answer the questions below is ChangeInSpeed AverageAcceleration = ElapsedTime

1. What is a car’s average acceleration when it increases its speed by 100 km/h in 5 seconds? Result: 100 km/h = 5 second

20

km/h sec

2. What is a car’s average acceleration when it increases its speed by 100 km/h in 4 seconds? Result: 100 km/h = 4 seconds

25

km/h sec

3. What is a car’s average acceleration when it increases its speed by 100 km/h in 2 seconds? Result: 100 km/h = 2 seconds

50

km/h sec

4. What is a car’s average acceleration when its speed changes from 0 km/h to 100 km/h in 5 seconds? Result: 100 km/h = 5 seconds c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

19

20

km/h sec

Chapter 2: Acceleration and change in speed

5. Bonus Question - without Interactive Physics. What is a car’s average acceleration when its speed changes from 20 km/h to 60 km/h in 2 seconds? Result: 40 km/h = 2 seconds

20

km/h sec

6. Bonus Question - without Interactive Physics. What is a car’s average acceleration when its speed changes from 30 km/h to 60 km/h in 6 seconds? Result: 30 km/h = 6 seconds

5

km/h sec

7. Bonus Question - without Interactive Physics. What is a car’s average acceleration when its speed changes from 60 km/h to 20 km/h in 4 seconds?

Result: -40 km/h = 4 seconds

-10

km/h sec

Courtesy of MSC.Software c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

20

Chapter 2: Acceleration and change in speed

2.3

Acceleration, change in speed, and time (B)

An measurement of a good sports car is its ability to accelerate. For example, a minivan can accelerate from 0 to 60 mph in 10 seconds whereas a high-priced sports car can accelerate from 0 to 60 mph in 5 seconds. The equation that helps answer the questions below is AverageAcceleration =

ChangeInSpeed ElapsedTime

1. Suppose the car accelerates from 0 to 50 km/h in 5 seconds. What is the average acceleration of the car? Result: 50 km/h = 5 seconds

10

km/h sec

2. What is the average acceleration of a car that accelerates from 0 to 75 km/h in 5 seconds? Result: 75 km/h = 5 seconds

15

km/h sec

3. What is the average acceleration of a car that accelerates from 0 to 100 km/h in 5 seconds? Result: 100 km/h = 5 seconds

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

21

20

km/h sec

Chapter 2: Acceleration and change in speed

4. The car that accelerates at 10 km/h/sec is probably a minivan /sports car (circle one) whereas the car that accelerates at 20 km/h/sec is probably a minivan/ sports car . Explain: A sports car usually has a larger engine than a minivan and can accelerate more quickly. 5. Bonus Question - answer without Interactive Physics. A driver enters a highway on-ramp at 20 km/h. To safely merge with traffic, the driver knows that she must increase her speed to 100 km/h in 4 seconds. What is the average acceleration of the car? Result: 80 km/h = 4 seconds

20

km/h sec

6. Bonus Question - answer without Interactive Physics. To accelerate a car, the driver should press on the gas /brake pedal (circle one). 7. Bonus Question - answer without Interactive Physics. A driver leaves a highway exit-ramp with an initial speed of 100 km/h. To safely merge with city traffic, the driver knows that she must reduce her speed to 30 km/h in 7 seconds. What is the average acceleration of the car? Result: -70 km/h = 7 seconds

-10

km/h sec

8. Bonus Question - answer without Interactive Physics. To decelerate (deceleration is a negative acceleration) the car, the driver should press on the gas/ brake pedal (circle one).

Courtesy of MSC.Software c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

22

Chapter 2: Acceleration and change in speed

2.4

Constant and changing acceleration

A boat’s acceleration is constant if it does not change. Alternately, a boat’s acceleration varies if it changes. The equation that helps answer the questions below is AverageAcceleration =

ChangeInSpeed ElapsedTime

1. Click the button and watch the number inside the box on the right-hand side of the gray boat. Since the number does not change, the gray boat has a constant /variable (circle one) acceleration. 2. The green boat has a 3. The red boat has a

constant /variable (circle one) acceleration. constant /variable acceleration.

4. The blue boat has a constant/ variable acceleration. 5. Because the gray boat’s acceleration is constant, it is easy to see that it has a minimum acceleration of 20 km/h/sec and a maximum acceleration of 20 km/h/sec. Since it changes its speed by 100 km/h in five seconds, what is average acceleration? Result: 100 km/h = 5 seconds

20

km/h sec

6. Determine the average acceleration of the green boat. Result: 0 km/h = 0 km/h sec 5 seconds

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

23

Chapter 2: Acceleration and change in speed

7. Determine the minimum, maximum, and average acceleration of the red boat. Result: Minimum acceleration = -20 km/h Maximum acceleration = -20 sec -100 km/h = 5 seconds

Average acceleration =

-20

km/h sec

km/h sec

8. Determine the minimum, maximum, and average acceleration of the blue boat. To answer this question with Interactive Physics, click the button and then click on the arrow that is on the furthest right and bottom part of the screen. Each click will step the simulation forward. Observe the acceleration reported by the blue boat and record the smallest and largest values as you step the simulation until the end. When you are finished, click the

Result: Minimum acceleration =

-62.7

km/h

button.

Maximum acceleration =

sec

Average acceleration =

0 km/h = 5 seconds

0

+62.7

km/h sec

km/h sec

9. When a boat’s acceleration is constant, its average acceleration is different than its maximum or minimum acceleration. True/ False (circle one) 10. When a boat’s acceleration varies, its average acceleration is different than its maximum or minimum acceleration. True /False (circle one)

Courtesy of United States Navy c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

24

Chapter 2: Acceleration and change in speed

2.5

Free fall - how fast?

When an object falls (without air resistance), it has a constant acceleration and its speed changes. The object’s speed depends on gravity, the amount of time the object is accelerating (elapsed time), and the object’s initial speed. In mathematical terms, ChangeInSpeed = AverageAcceleration ∗ ElapsedTime CurrentSpeed = ChangeInSpeed + InitialSpeed

1. Suppose the average acceleration due to gravity is 20 m/sec2 (the abbreviation m/sec2 stands for meters per second squared). Assuming the sky diver jumps from a plane with an initial speed of 0, how fast is the sky diver falling at 5 seconds? Result: 20

m sec2

∗ 5 sec =

100

m sec

2. Suppose the average acceleration due to gravity is 15 m/sec2 . Assuming the sky diver jumps from a plane with an initial speed of 0, how fast is the sky diver falling at 5 seconds? Result: 15

m sec2

∗ 5 sec =

75

m sec

3. For each value of gravity in the following table, determine how fast a sky diver falls at 5 seconds (assuming an initial speed of 0). After completing the table, mark the location on the graph of each pair of numbers with an X and then complete the graph by connecting the Xs. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

25

Chapter 2: Acceleration and change in speed

Gravity (m/sec2 ) 0

Speed (m/sec) 0

5

25

10

50

15

75

20

100

Free fall speed after 5 seconds

100

75

50

25

0 0

5

10

15

20

2

Gravity (m/sec )

4. Bonus Question - answer without Interactive Physics. Suppose the sky diver is falling at 20 m/sec and the average acceleration due to gravity is 10 m/sec2 . How fast is the sky diver falling 5 seconds later? Result: 20

m sec

+

10

m sec2

∗ 5 sec =

70

m sec

5. Bonus Question - answer without Interactive Physics. Suppose the sky diver is falling at 20 m/sec and the average acceleration due to gravity is 9.8 m/sec2 . How fast is the sky diver falling 10 seconds later? Result: 20

m sec

+

9.8

m sec2

∗ 10 sec =

118

m sec

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

26

Chapter 2: Acceleration and change in speed

2.6

Upward projectile motion - how fast?

When a baseball is thrown straight upward with a certain initial speed, it slows down, stops, then reverses direction and starts falling down. The baseball’s speed depends on its initial speed, gravity, and the time in flight. In mathematical terms, ChangeInSpeed = AverageAcceleration ∗ ElapsedTime CurrentSpeed = ChangeInSpeed + InitialSpeed

1. Suppose the average acceleration due to gravity is -10 m/sec2 and the baseball’s initial speed is 30 m/sec. What is the baseball’s speed at 5 seconds? Result: m -50 sec ChangeInSpeed = -10 secm2 ∗ 5 sec = CurrentSpeed = 30

m sec

+

-50

m sec

-20

=

m sec

2. Suppose the average acceleration due to gravity is -10 m/sec2 and the baseball’s initial speed is 20 m/sec. What is the baseball’s speed at 5 seconds? Result: ChangeInSpeed

=

-10

CurrentSpeed

=

20

m sec2 m sec

∗ 5 sec +

-50

= m sec

-50 =

m sec

-30

m sec

3. For each value of the baseball’s initial speed in the following table, determine the baseball speed at 5 seconds (assume the average acceleration due to gravity is -10 m/sec2 ). After completing the table, mark the location on the graph of each pair of numbers with an X and then complete the graph by connecting the Xs. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

27

Chapter 2: Acceleration and change in speed

Initial speed (m/sec) 0

Speed at 5 sec (m/sec) -50

10

-40

20

-30

30

-20

Baseball speed after 5 seconds (m/sec)

-10

-20

-30

-40

-50 0

10

20

30

40

Initial baseball speed (m/sec)

4. Bonus Question - answer with the previous graph. Estimate the baseball’s speed at 5 seconds if its initial speed is 15 m/sec? Result: -35 m sec 5. Bonus Question - answer with the previous graph. Estimate the baseball’s speed at 5 seconds if its initial speed is 35 m/sec? Result: m -15 sec 6. A ball is thrown upward with an initial speed of 30 m/sec. Determine the ball’s speed at each value of time in the following table (assume the average acceleration due to gravity is -10 m/sec2 ). After completing the table, complete the graph. 30

Baseball speed (m/sec) 30

1

20

2

10

3

0

4

-10

20 Baseball speed (m/sec)

Elapsed time (sec) 0

10

0

-10

-20 0

1

2

3

4

5

Time (sec)

7. Bonus Question - answer with the previous graph. A baseball is thrown upward at 30 m/sec. m Estimate the baseball’s speed at 2.5 seconds: 5 sec . m Estimate the ball’s speed at 4.5 seconds: -15 sec .

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

28

Chapter 2: Acceleration and change in speed

Chapter 3 Acceleration, speed, and distance Chapter 1 showed that speed is related to distance and time. Chapter 2 showed that acceleration is related to speed and time. By combining these relationships, it is possible to relate acceleration to speed and distance as shown in the following equations. DistanceTraveled = InitialSpeed ∗ ElapsedTime +

1 2

∗ AccelerationConstant ∗ ElapsedTime2

CurrentSpeed 2 = SpeedInitial2 + 2 ∗ AccelerationConstant ∗ DistanceTraveled

Courtesy NASA/JPL-Caltech

29

3.1

Free fall - how far?

When an object falls (without air resistance), it has a constant acceleration, its speed changes, and the distance it falls depends on the amount of time it is falling. In mathematical terms, DistanceTraveled = InitialSpeed ∗ ElapsedTime +

1 2

∗ AccelerationConstant ∗ ElapsedTime2

1. Suppose the acceleration (due to gravity) is 20 m/sec2 and the initial speed of the sky diver is 0 m/sec. How far does the sky diver fall in 5 seconds? Result: 0

m sec

∗ 5 sec

1 2

+

∗ 20

m sec2

2. Suppose a sky-diver jumps from a plane with an initial speed of 0 and falls for 5 seconds. Determine how far the sky-diver falls for each value of gravity in the table.

∗ 52 sec2 =

250

meters

Gravity Distance (m/sec2 ) (meters) 5 62.5 10

125

15

187.5

3. Bonus Question - answer without Interactive Physics. Suppose the sky diver’s initial speed is 20 m/sec and the acceleration (due to gravity) is 10 m/sec2 . How far does the sky diver fall in 5 seconds? Result: 20

m sec

∗ 5 sec

+

1 2

∗ 10

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

m sec2

30

∗ 52 sec2 =

225

meters

Chapter 3: Acceleration, speed, and distance

3.2

Upward projectile motion - how far?

When a baseball is thrown straight upward with a certain initial speed, it slows down, stops, and then starts falling down. The distance the baseball travels depends on its initial speed, gravity, and the time in flight. When the acceleration is constant, DistanceTraveled = InitialSpeed ∗ ElapsedTime +

1 2

∗AccelerationConstant ∗ ElapsedTime2

1. Suppose the acceleration (due to gravity) is -10 m/sec2 and the initial upward speed of the baseball is 30 m/sec. How far does the ball travel in 5 seconds? Result: 30

m sec

∗ 5 sec

+

1 2

∗ -10

m sec2

∗ 52 sec2 =

25

meters

Note: To answer this question mathematically, substitute 30 for InitialSpeed, substitute 5 for ElapsedTime, and substitute -10 for AccelerationConstant in the formula for DistanceTraveled above.

2. Suppose the acceleration (due to gravity) is -10 m/sec2 and the initial upward speed of the baseball is 20 m/sec. How far does the ball travel in 5 seconds? Result: 20

m sec

∗ 5 sec

+

1 2

∗ -10

m sec2

∗ 52 sec2 =

-25

meters

Note: The distance traveled by the ball is a negative number. This number represents the distance below (rather than above) the initial position of the ball.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

31

Chapter 3: Acceleration, speed, and distance

3. Suppose the acceleration (due to gravity) is -10 m/sec2 and the initial upward speed of the baseball is 10 m/sec. How far does the ball travel in 5 seconds? Result: 10

m sec

∗ 5 sec

1 2

+

∗ -10

m sec2

∗ 52 sec2 =

-75

meters

4. If the ball is upward launched with an initial speed of 30 m/sec, it reaches its maximum height when its speed is zero. True /False (circle one) 5. If the ball is upward launched with an initial speed of 20 m/sec or 10 m/sec, it reaches its maximum height when its speed is zero. True /False (circle one) 6. Bonus Question - answer without Interactive Physics. Suppose the baseball’s initial upward speed is 50 m/sec and the acceleration (due to gravity) is -10 m/sec2 . How far does the ball travel in 5 seconds? Result: 50

m sec

∗ 5 sec

1 2

+

∗ -10

m sec2

∗ 52 sec2 =

125

meters

7. Bonus Question - answer with and/or without Interactive Physics. Suppose that acceleration (due to gravity) is -10 m/sec2 and the initial upward speed of the baseball is 30 m/sec. Determine how far the ball travels for each value of time in the table and then complete the graph. 50

Time (sec) 0

Distance (meters) 0

1

25

2

40

3

45

4

40

5

25

6

0

-30

7

-35

-40

Distance upward (meters)

40 30 20 10 0 -10 -20

0

1

2

3 4 Time (seconds)

5

6

7

8. Bonus Question - answer with the previous graph. Estimate how far the ball travels in 5.5 seconds: 13.75 meters Estimate how far the ball travels in 8 seconds: -80 meters

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

32

Chapter 3: Acceleration, speed, and distance

3.3

Upward projectile motion - maximum height

When a basketball player jumps straight up, the player reaches a maximum height before falling downward. The player’s maximum jumping height depends on his/her initial vertical speed and gravity. When the acceleration is constant, MaximumHeight =

-(InitialSpeed2 ) 2 ∗ AccelerationConstant

1. The acceleration due to gravity near sea level on Earth is -9.8 m/sec2 . When a basketball player jumps upward with an initial speed of 4 m/sec, what is the maximum height between the player’s feet and the floor? Result: -(42 ) m2 /sec2 = 2 ∗ -9.8 m/sec2

0.82

meters

2. What is the maximum height a basketball player can jump when the player leaps upward m ? Assume the acceleration due to gravity is -9.8 secm2 . with an initial speed of 5 sec Result: -(52 ) m2 /sec2 = 2 ∗ -9.8 m/sec2

1.28

meters

3. The basketball player’s upward speed is zero when the player reaches maximum height. True /False (circle one)

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

33

Chapter 3: Acceleration, speed, and distance

4. Suppose that acceleration (due to gravity) is -9.8 m/sec2 . Determine the maximum height a basketball player jumps for each of the initial speeds in the table below, and then complete the graph. 1.4

Maximum height (meters) 0

1

0.05

2

0.20

3

0.46

4

0.82

5

1.28

1.2 Maximum height (meters)

Initial speed (m/sec) 0

1 0.8 0.6 0.4 0.2 0 0

1

2 3 Initial speed (seconds)

4

5

5. Bonus Question - answer with the previous graph. m : 0.62 meters Estimate the maximum height with an initial speed of 3.5 sec m Estimate the maximum height with an initial speed of 6 sec : 1.8 meters

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

34

Chapter 3: Acceleration, speed, and distance

3.4

Upward projectile motion - initial speed

To reach a certain maximum height, a basketball player needs to leap with a sufficiently high initial speed. In mathematical terms (when acceleration is constant, i.e., no air resistance), InitialSpeed =

√

-2 ∗ AccelerationConstant ∗ MaximumHeight

1. The acceleration due to gravity near sea level on Earth is -9.8 m/sec2 . What must be a basketball player’s initial vertical speed to jump a maximum height of 1.5 meters? Result:  -2 ∗ -9.8 m/sec2 ∗ 1.5 m =

2. Suppose that acceleration (due to gravity) is -9.8 m/sec2 . Determine the basketball player’s initial vertical speed to reach each of the maximum heights in the table to the right.

5.42

m sec

Maximum height (meters) 0.5

Initial speed (m/sec) 3.13

1.0

4.43

1.5

5.42

3. When a basketball player needs to jump very high, he/she should jump with a low vertical speed. True/ False (circle one)

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

35

Chapter 3: Acceleration, speed, and distance

3.5

Time, acceleration, and distance

When a rocket sled travels along a straight track, it is helpful to know how much time it takes to go from start to finish. When the rocket sled accelerates at a constant rate from an initial speed of 0, the answer to the question can be calculated with the formula  ElapsedTime =

2 ∗ DistanceTraveled AccelerationConstant

1. Assuming the rocket sled starts from rest (0 m/sec), how much time does the rocket sled take to travel 200 meters when it accelerates at a constant rate of 5 m/sec2 ? Result: 

2 ∗ 200 m = 5 m/sec2

8.94

sec

2. How much time does it take for the rocket sled to travel 200 meters when it accelerates from rest at a constant rate of 10 m/sec2 ? Result: 

2 ∗ 200 m = 10 m/sec2

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

36

6.32

sec

Chapter 3: Acceleration, speed, and distance

3. Determine how much time it takes for the rocket sled to accelerate from rest and travel 200 meters for each acceleration listed in the following table, and then complete the graph. 9 8.5

Time (seconds) 8.94

10

6.32

15

5.16

20

4.47

25

4.0

8 Time to reach 200 meters

Acceleration (m/sec2 ) 5

7.5 7 6.5 6 5.5 5 4.5 4 5

4. The higher the acceleration, the sled to travel 200 meters.

10

15 Acceleration (m/sec2)

20

25

shorter /longer (circle one) it takes for the rocket

5. Bonus Question - answer with the previous graph. Estimate the travel time for an acceleration of 7 secm2 : 7.6 sec Estimate the travel time for an acceleration of 30 secm2 : 3.6 sec

Courtesy of MSC.Software c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

37

Chapter 3: Acceleration, speed, and distance

3.6

Speed, acceleration, and distance

When a rocket sled travels along a straight track, it is helpful to know how fast it is going when it hits the finish line. When the rocket accelerates at a constant rate, the answer to the question can be calculated with the formula  CurrentSpeed = SpeedInitial2 + 2 ∗ AccelerationConstant ∗ DistanceTraveled

1. What is the speed of a rocket sled that accelerates from rest (0 m/sec) at a constant rate of 5 m/sec2 for a distance of 200 meters? Result:  02

m2 sec2

+

2∗5

m sec2

∗ 200 m =

44.72

m sec

2. What is the speed of a rocket sled that accelerates from rest at a constant rate of 10 m/sec2 for a distance of 200 meters? Result: 

02

m2 sec2

+

2 ∗ 10

m sec2

∗ 200 m =

63.25

m sec

3. What is the speed of a rocket sled that accelerates from rest at a constant rate of 15 m/sec2 for a distance of 200 meters? Result: 

02

m2 sec2

+

2 ∗ 15

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

38

m sec2

∗ 200 m =

77.46

m sec

Chapter 3: Acceleration, speed, and distance

4. The higher the acceleration the slower/ faster (circle one) a rocket sled is moving when it accelerates from rest for 200 meters. 5. Bonus Question - answer without Interactive Physics. What is the speed of a rocket sled that accelerates from an initial speed m2 m2 of 10 sec 2 at a constant rate of 20 sec2 for a distance of 200 meters? Result:  102

m2 sec2

+

2 ∗ 20

m sec2

∗ 200 m =

m sec

90

6. Bonus Question - answer without Interactive Physics. What is the speed of a rocket sled that accelerates from an initial speed m2 m2 of 10 sec 2 at a constant rate of 25 sec2 for a distance of 200 meters? Result: 

102

m2 sec2

+

2 ∗ 25

m sec2

∗ 200 m =

m sec

100.50

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

39

Chapter 3: Acceleration, speed, and distance

3.7

Distance, speed, acceleration, and highway safety

The distance it takes for a car to stop depends on its initial speed and its ability to decelerate (deceleration means slowing down). A car’s ability to decelerate depends on many factors, including road conditions (dry, wet, icy, oily, hill/flat, etc.), the car’s tires, its braking system, suspension system, and mass. When a car decelerates at a constant rate, the distance it takes to stop can be calculated with the formula SpeedInitial2 DistanceTraveled = 2 ∗ AccelerationConstant For the questions that follow, use AccelerationConstant= -5 m/sec2 (this valued is based on an Insurance Institute for Highway Safety study of passenger cars).

1. How far does it take to stop a car traveling at 10 Result:

m sec

102 m2 /sec2 -2 ∗ -5 m/sec2 = 2. How far does it take to stop a car traveling at 20 Result: 202 m2 /sec2 -2 ∗ -5 m/sec2 =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

40

? (10

10

m sec

is approximately 22

miles hour

.)

meters

? (20

40

m sec

m sec

≈ 45

miles hour

)

meters

Chapter 3: Acceleration, speed, and distance

3. Using your previous answers and additional calculations, fill in the following table. Initial speed (m/sec) (mph) 0 0 10 22 20 45 30 67 40 90

Stopping distance (meters) 0 10 40 90 160

Courtesy of MSC.Software

4. Using the previous table, plot the car’s stopping distance versus its initial speed. 160

Stopping distance (meters)

140 120 100 80 60 40 20 0 0

5

10

15 20 25 Initial car speed (m/sec)

30

35

40

5. Bonus Question - answer with the previous graph. Estimate the stopping distance of a car speeding at 35 Estimate the stopping distance of a car speeding at 50

m sec m sec

(78 mph): (111 mph):

122.5 250

meters meters

6. Bonus Question - answer without Interactive Physics. A milkman notices an accident 100 meters ahead and steps on the brake (100 meters is about m the size of a football field, including end-zones ). If the milk-truck is speeding at 40 sec (90 miles ), hour will the milkman be able to stop before the accident? Yes/ No (circle one) m (78 7. Referring to the last problem, if the milk-truck is traveling at 35 sec milkman be able to stop before the accident? Yes/ No (circle one)

miles hour

),

will the

8. Using the graph, estimate the maximum speed the milk-truck can have if it is to stop before the accident 100 meters ahead: 32 m/sec

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

41

Chapter 3: Acceleration, speed, and distance

3.8

Speed, distance, acceleration, and highway safety

Sometimes it is not possible to fully stop before hitting a tree, pedestrian, or another car. The injuries sustained to the people involved in an car accident are highly dependent on the car’s impact speed, which depends on its initial speed, its ability to decelerate, and the stopping distance. When a car decelerates at a constant rate, the impact speed can be calculated by CurrentSpeed =



SpeedInitial2 + 2 ∗ AccelerationConstant ∗ DistanceTraveled

For the first three questions, use AccelerationConstant= -5 m/sec2 (this valued is based on an Insurance Institute for Highway Safety study of passenger cars).

1. Suppose a sports car traveling at 20 m/sec (approximately 45 miles/hour) brakes to avoid hitting another car 40 meters away. How fast is the car moving when it hits the other car? Result: 

202

m2 sec2

+

2 ∗ -5

m sec

∗ 40 m =

0

m sec

2. Suppose a sports car traveling at 30 m/sec (approximately 67 miles/hour) brakes to avoid hitting another car 40 meters away. How fast is the car moving when it hits the other car? Result:  302

m2 sec2

+

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

2 ∗ -5

42

m sec

∗ 40 m =

22.4

m sec

Chapter 3: Acceleration, speed, and distance

3. Referring to the previous problem, suppose the sports car is traveling at 40 m/sec (144 km/hour). How fast is the car moving when it hits the other car? Result: 

402

m2 sec2

+

2 ∗ -5

m sec

∗ 40 m =

34.6

m sec

4. Bonus Question - answer without Interactive Physics. On hills or wet roads, the ability to decelerate drops substantially. Using 2 AccelerationConstant= 2 m/sec , calculate the speed that a car will strike another car located 40 meters away for each of the initial speeds in the table on the right.

Initial speed (m/sec) (km/h) 0 0 10 36 20 72 30 108 40 144

Impact speed (m/sec) (km/h) 0 0 0 0 15.5 56 27.2 98 37.9 136

5. Using the previous table, plot the car’s impact speed versus its initial speed. 40

Impact speed (m/sec)

35 30 25 20 15 10 5 0 0

5

10

15 20 25 Initial speed (m/sec)

30

35

40

6. Bonus Question - answer without Interactive Physics. Using your graph, estimate the initial speed that corresponds m m to an impact speed of 20 sec (72 km/h): 23.66 sec Courtesy of MSC.Software Note: Life-threatening injuries occur when a car’s impact speed is greater than 15

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

43

m sec

(50

km h

).

Chapter 3: Acceleration, speed, and distance

Chapter 4 Weight, mass, and gravity In everyday language, the words weight and mass are frequently used interchangeably. In physics, weight and mass are different. The mass of an object such as a book, spaceship, or person does not depend on the object’s location. Alternately, weight depends on an object’s location. For example, your weight on Earth is different from your weight on the Moon or on Mars, but your mass is the same. The equation that relates weight to mass is Weight = Mass ∗ Gravity where Gravity1 is related to the force caused by a very large object such as a planet2 on a much smaller object such as a human, animal, or aircraft carrier.

1

Experiments were done in September 2002 by the National Radio Astronomy Observatory to determine the speed of gravity. The results agree within 20% of an assumption made by Albert Einstein that gravity moves at the speed of light (approximately 300,000 km per second ). Edward Formalout, one of the researchers said that “Gravity is not well understood” and more accurate measurements need to be made. 2 Recent discoveries of planet-like objects in the outer fringes of the solar system have generated controversy about the definition of a planet. Many astronomers regard the solar system as having nine planets, but Pluto, discovered in 1930, may be one of the larger objects in the Kuiper Belt. The Kuiper Belt, first detected in 1992, may have as many as 70,000 objects measuring more than 100 kilometers in diameter. By comparison, the Earth’s diameter is approximately 12,800 km, the moon’s diameter is 3,500 km, Pluto’s is 2,300 km, and the recently discovered Quaoar’s is 1,200 km.

44

4.1

Weight and mass

Mass and weight are different. Mass is the measure of how much matter an object has. Weight depends on mass and gravity. The units of mass and weight are also different. In SI units, mass is measured in kilograms (kilograms is abbreviated kg) and weight is measured in newtons.3 In mathematical terms, Weight = Mass ∗ Gravity

1. Gravity near sea level on Earth is 9.8 secm2 . How much does 4 kg of fruit weigh on Earth? Result: 4 kg ∗ 9.8 secm2 =

39.2

newtons

49

newtons

2. How much does 5 kg of fruit weigh on Earth? Result: 5 kg ∗ 9.8

m sec2

=

3. How much does 3 kg of fruit weigh on Earth? Result: 3 kg ∗ 9.8

m sec2

=

29.4

newtons

3

In all countries other than the United States, Liberia, and Myanmar, SI units (International System of Units) are used to measure mass, weight, length, area, temperature, etc. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

45

Chapter 4: Weight, mass, and gravity

4. How much does 2 kg of fruit weigh on Earth? Result: 2 kg ∗ 9.8

m sec2

=

19.6

newtons Courtesy NASA/JPL-Caltech

5. How much does 1 kg of fruit weigh on Earth? Result: 1 kg ∗ 9.8

m sec2

=

9.8

newtons

6. Bonus Question - answer without Interactive Physics. Gravity on the Moon is 1.67 secm2 . How much does 4 kg of fruit weigh on the Moon? Result: 4 kg ∗ 1.67

m sec2

=

6.7

newtons

Courtesy NASA/JPL-Caltech

7. Bonus Question - answer without Interactive Physics. How much does 5 kg of fruit weigh on the Moon? Result: 5 kg ∗ 1.67

m sec2

=

8.4

newtons

8. Bonus Question - answer without Interactive Physics. Determine the ratio of the weight of 5 kg of fruit on Earth to the weight of 5 kg of fruit on the Moon. Result: 5 kg ∗ 9.8 secm2 5 kg ∗ 1.67 secm2

=

5.87

9. Bonus Question - answer without Interactive Physics. Objects on the Earth weigh 5.87 times heavier than objects on the Moon.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

46

Chapter 4: Weight, mass, and gravity

4.2

Weight, mass, and other planets

As far as scientists can tell, the mass of an object does not depend on where the object is in the universe. On the other hand, the weight of an object depends on gravity, and gravity depends on the proximity to a large object such as a planet or moon. In mathematical terms weight is related to mass and gravity by Weight = Mass ∗ Gravity

1. Gravity near sea level on Earth is 9.8 m/sec2 . How much does 2 kg of fruit weigh on Earth? Result: 2 kg ∗ 9.8

m sec2

=

19.6

newtons

Courtesy NASA/JPL-Caltech

2. Gravity on the surface of the Moon is 1.67 m/sec2 . How much does 2 kg of fruit weigh on the Moon? Result: 2 kg ∗ 1.67

m sec2

=

3.34

newtons Courtesy NASA/JPL-Caltech

3. Gravity on the surface of Jupiter is 24.81 m/sec2 ? How much does 2 kg of fruit weigh on Jupiter? Result: 2 kg ∗ 24.81

m sec2

=

49.62

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

newtons Courtesy NASA/JPL-Caltech

47

Chapter 4: Weight, mass, and gravity

4. How much does 2 kg of fruit weigh on each of the following celestial bodies. Planet Jupiter

Gravity Weight of 2 kg of fruit (m/sec2 ) (newtons) 24.81 49.62

Neptune

10.75

21.5

Saturn

10.46

20.92

Earth

9.8

19.6

Venus

8.88

17.76

Uranus

8.86

17.72

Mars

3.72

7.44

Mercury

3.71

7.42

Moon

1.67

3.34

Pluto

0.68

1.36

5. Use the previous table to create a bar chart of the fruit’s weight versus planet. 50

45

40

35

30

25

20

15

10

5

0 Jupiter

Neptune Saturn

Earth

Venus

Uranus

Mars

Mercury Moon

Pluto

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

48

Chapter 4: Weight, mass, and gravity

4.3

Mass and weight

Mass and weight are different. Usually, mass is determined by weighing an object on Earth and then calculating its mass using the formula Weight Mass = Gravity

1. Gravity near sea level on Earth is approximately 10 m/sec2 . What is the mass of fruit that weighs 40 newtons on Earth? Result: 40 newtons = 4 kg 10 m/sec2 Courtesy NASA/JPL-Caltech

2. What is the mass of fruit that weighs 50 newtons on Earth? Assume gravity is 10 m/sec2 . Result: 50 newtons = 10 m/sec2

5

kg

3. What is the mass of fruit that weighs 30 newtons on Earth? Result: 30 newtons = 10 m/sec2

3

kg

4. What is the mass of fruit that weighs 20 newtons on Earth? Result: 20 newtons = 10 m/sec2

2

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

kg 49

Chapter 4: Weight, mass, and gravity

Bonus Questions - answer without Interactive Physics. A. Gravity on the Moon is 1.67 m/sec2 . What is the mass of fruit that weighs 40 newtons on the Moon? Result: 40 newtons = 1.67 m/sec2

23.95

kg Courtesy NASA/JPL-Caltech

B. What is the mass of fruit that weighs 50 newtons on the Moon? Result: 50 newtons = 1.67 m/sec2

29.94

kg

C. What is the mass of fruit on Earth that weighs 30 newtons on the Moon? Assume Moon’s gravity is 1.67 m/sec2 and Earth’s gravity is 10 m/sec2 . Result: 30 newtons = 1.67 m/sec2

17.96

kg

D. What is the weight of fruit on Earth that weighs 30 newtons on the Moon? Assume Moon’s gravity is 1.67 m/sec2 and Earth’s gravity is 10 m/sec2 . Result: 30 newtons ∗ 10 m/sec2 = 1.67 m/sec2

179.6

newtons

E. What is the weight of fruit on the Moon that weighs 30 newtons on the Earth? Result: 30 newtons ∗ 1.67 m/sec2 = 10 m/sec2

5.01

newtons

Double Bonus Questions - answer without Interactive Physics. A. A student’s mass on Earth is 50 kg. Determine his weight on Earth: 490 newtons Determine his weight on the Moon: 83.5 newtons B. My weight on Earth is My weight on the Moon is

newtons. newtons.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

50

Chapter 4: Weight, mass, and gravity

Chapter 5 Air resistance In a vacuum or on a planet without an atmosphere, all objects fall at the same rate. In other words, a feather falls as quickly as a cannon ball. On Earth, the way an object falls depends on air resistance. For example, on Earth a feather falls more slowly than a cannon ball. When an object moves through air, it collides with a huge number of air molecules. The net effect of all the collisions is a force on the object. Depending on how the air and object are moving, the force usually slows the object down, resisting the object’s motion. Since the force of air usually resists an object’s motion, it is called air resistance. Scientists and engineers have found uses for these air forces, e.g., keeping a plane up in the air. Sometimes, the force of air resistance on a falling object can be ignored - particularly when the object is heavy and falls for a short distance. In other cases, the air resistance force has a major effect on how an object falls, particularly when the object is light or falls for a long distance. It is very difficult to write an exact equation for the air-resistance force on an object because it depends on many factors, including • the object’s motion relative to the air around it • the object’s shape, material, surface finish, and temperature • air pressure, temperature, density, humidity, and quality (the type of molecules) Two equations that approximate Fair , the air resistance force on an object moving in a straight line in still air, are given below. Fair = CoefficientOfLowSpeedAirResistance ∗ CrossSectionalAreaOfObject ∗ VelocityOfObject Fair = CoefficientOfHighSpeedAirResistance ∗ CrossSectionalAreaOfObject ∗ VelocityOfObject2

51

5.1

Air resistance and falling objects - concepts

It is very difficult to accurately predict how an object falls with air resistance. It is significantly easier to predict how objects move in a vacuum (there is no air or air resistance in a vacuum).

1. A rock and a feather are released from rest from the same height with air resistance on. Do they hit at the same time? Yes/ No (circle one) 2. A rock and a feather are released from rest from the same height with air resistance off. Do they hit at the same time? Yes /No (circle one) 3. A rock and a feather are released from rest from the same height with air resistance on. Determine the time and the speed that each object hits the ground. Repeat this exercise with air resistance off. Air resistance on Object Rock Feather

Time to impact

2.5 sec 25

sec

Air resistance off

Speed at impact

7 0.6

Time to impact

m sec m sec

4. A rock falls fastest when there is no air resistance. 5. A feather falls fastest when there is no air resistance.

15.7

m sec

1.6 sec

15.7

m sec

True /False (circle one) True /False (circle one)

6. Without air resistance, a rock and a feather fall at the same speed. 7. With air resistance, a rock and a feather fall at the same speed.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

52

Speed at impact

1.6 sec

True /False

True/ False

Chapter 5: Air resistance

5.2

Air resistance & terminal velocity - concepts

When an object falls in air, it eventually reaches a maximum speed. This maximum speed is called terminal velocity and depends on air resistance and an object’s weight.

Parachutist weight (N) 400

Terminal velocity (km/h) 17.5

800

24.8

1200

30.3

Terminal velocity (km/h)

1. Complete the table and graph the terminal velocity for each of the following parachutists. 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 400

2. A heavier object has a higher terminal velocity.

600

800 Parachutist weight (N)

1000

1200

True /False (circle one)

3. Without air resistance, the parachutist reaches a terminal velocity. True/ False With air resistance, the parachutist reaches a terminal velocity. True /False 4. With air resistance, a 400 N and 1200 N parachutist fall at the same speed. True/ False 5. In a vacuum, a 400 N and a 1200 N parachutist fall at the same speed. True /False 6. Estimate the terminal velocity of a 600 N parachutist: 21.0 km/h. Estimate the terminal velocity of a 1600 N parachutist: 34.3 km/h.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

53

Chapter 5: Air resistance

5.3

Air resistance & gravity forces - concepts

When an object falls in air, it eventually reaches a maximum speed. It does not accelerate further because the upward force due to air resistance exactly balances the downward force due to gravity.

1. At the instant a 800 N (N is an abbreviation for newtons) parachutist starts to fall, the gravity force on the parachutist is 800 N and the air-resistance force on the parachutist is 0 N. After the parachutist reaches terminal velocity, the gravity force on the parachutist is 800 N and the air-resistance force on the parachutist is 800 N. 2. At the instant a 1200 N parachutist starts to fall, the gravity force on the parachutist is 1200 N and the air-resistance force on the parachutist is 0 N. After the parachutist reaches terminal velocity, the gravity force on the parachutist is 1200 N and the airresistance force on the parachutist is 1200 N. 3. When a 400 N parachutist starts to fall, the gravity force on the parachutist is 400 N and the air-resistance force on the parachutist is 0 N. After the parachutist reaches terminal velocity, the gravity force on the parachutist is 400 N and the air-resistance force on the parachutist is 400 N. 4. The parachutist reaches terminal velocity when the downward gravitational force on the parachutist is less than/ equal to /greater than (circle one) the upward air resistance force on the parachute. 5. There is a larger gravitational force on a heavier parachutist.

True /False (circle one)

6. It takes less/the same/ more (circle one) air resistance force to balance the gravitational force of a heavier parachutist. 7. With the same parachute, a lighter/ heavier parachutist will hit the ground harder.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

54

Chapter 5: Air resistance

5.4

Air resistance and projectile motion

Air resistance plays an important role in sports. It affects a runner’s speed, a jumper’s leap, and the distance a baseball player can hit a ball.

With the baseball’s Speed at 50 m/sec, Angle at 45 deg, and Air resistance OFF, click

1. What is the horizontal distance traveled by the baseball?

256

.

m

2. What is the maximum height reached by the baseball in its flight? Click

. Check the Air resistance box and click

65

m

36

m

.

3. What is the horizontal distance traveled by the baseball?

102

4. What is the maximum height reached by the baseball in its flight?

m

5. With air resistance, the ball travels a shorter /the same/a longer horizontal distance. 6. Complete the table below which shows the horizontal distance traveled by a baseball at various initial speeds, with and without air resistance. Use a launch angle of 45 degrees. Speed m/sec mph 35 78

Distance without air resistance meters feet 126 413

Distance with air resistance meters feet 69 226

40

89

164

539

81

266

45

101

208

681

92

302

50

112

256

840

102

335

55

123

310

1016

111

364

60

134

368

1208

121

397

7. Air resistance decreases /increases the distance traveled by a baseball. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

55

Chapter 5: Air resistance

8. Complete the graph of horizontal distance versus angle. Use an initial speed of 55 m/sec. Angle deg 25

Distance without air resistance meters feet 239 783

Distance with air resistance meters feet 102 335

30

269

883

109

358

35

291

956

112

367

40

305

1001

113

371

45

310

1016

111

364

50

305

1000

108

354

9. Use the previous table to complete the graph below. Use Xs for the curve that represents distance traveled by a baseball without air resistance. Use Os for the curve that represents distance traveled by a baseball with air resistance.

Horizontal distance traveled (m)

350 300 250 200

without air resistance with air resistance

150 100 50 25

30

35 40 Launch angle (degree)

45

50

10. Without air resistance, the ball travels farthest when the launch angle is about 45 deg. 11. With air resistance, the ball travels farthest when the launch angle is about

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

56

40

deg.

Chapter 5: Air resistance

Chapter 6 Newton’s Law for motion along a line Force = Mass ∗ Acceleration ... the whole burden of philosophy seems to consist in this, from the phenomena of motions to investigate the forces of nature, and from the forces to demonstrate the other phenomena. -Isaac Newton, Principia Philosophiae (1686) Dynamics is a scientific study whose purpose is to describe how mechanical systems move when acted upon by given forces, or to determine what forces must be applied to a system in order to cause it to move in a specified manner. Dynamics is useful in the design of automobiles, aircraft, spacecraft, watercraft, farm equipment, cranes, trucks, plows, construction tools, unicycles, bicycles, tricycles, motorcycles, skateboards, baggage handling machines, disk drives, printing equipment, laundry and kitchen equipment, packaging equipment, robotic devices, artificial limbs, exercise equipment, toys, position and rate sensors, military equipment, gears, bearings, vibration isolators, and nearly every other mechanism. Additionally, dynamics is useful in motion prediction for accident reconstruction, earthquakes, artificial gravity, human blood circulation, planetary exploration, crash-testing, medical research, and human and animal locomotion. Prior to the advent of computers, extensive knowledge of mathematics was required to analyze even simple systems such as a spinning top. When personal computers became available in the 1980s, it became much easier to analyze complicated systems with sophisticated computer programs such as Interactive Physics.

Courtesy NASA/JPL-Caltech

57

6.1

Force, mass, and acceleration - Newton’s apple

The legend is that Newton discovered the important law of physics Force = Mass ∗ Acceleration when he was struck by an apple falling from an apple tree. By studying the motion of the apple, Newton determined its acceleration. By knowing the mass of the apple and its acceleration, Newton could determine the resultant vertical force on the apple.

1. Using time-motion studies, the acceleration of a 0.1 kg apple is determined to be 9.8 m/sec2 . What is the resultant vertical force on the apple? Result: 0.1 kg ∗ 9.8

m sec2

=

0.98

newtons

2. A 0.2 kg apple falls with an acceleration of 9.8 m/sec2 . What is the resultant vertical force on the apple? Result: 0.2 kg ∗ 9.8

m sec2

=

1.96

newtons

3. What is the resultant force on a 0.3 kg apple that accelerates at 9.8 m/sec2 ? Result: 0.3 kg ∗ 9.8 c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

m sec2

58

=

2.94

newtons

Chapter 6: Newton’s Law for motion along a line

4. What is the resultant force on a 0.4 kg apple that accelerates at 9.8 m/sec2 ? Result: 0.4 kg ∗ 9.8

m sec2

=

3.92

newtons

5. Bonus Question - answer without Interactive Physics. The downward gravitational force on any object near Earth is equal to the object’s mass multiplied by Earth’s gravitational constant, 9.8 m/sec2 , i.e., ForceFromGravity = Mass ∗ Gravity What is the gravitational force on a woman whose mass is 50 kg? Result: 50 kg ∗ 9.8

m sec2

=

490

Courtesy NASA/JPLCaltech

newtons

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

59

Chapter 6: Newton’s Law for motion along a line

6.2

Force, mass, and acceleration - rocket

In 1687, Sir Isaac Newton postulated that force, mass, and acceleration were related by Force = Mass ∗ Acceleration By videotaping the motion of a 400 kg rocket sled, its acceleration can be determined. Calculate the resultant horizontal force that caused the motion.

1. Suppose the 400 kg rocket has an acceleration of 5 m/sec2 (the abbreviation m/sec2 stands for meters per second squared). What is the resultant horizontal force on the rocket? Result: 400 kg ∗ 5 secm2 = 2000 newtons To answer this question with Interactive Physics, click the button and verify that the meter that measures the resultant horizontal force matches the value you calculated with the formula above.

2. What is the resultant horizontal force on a 400 kg rocket whose acceleration is 10 Result: 400 kg ∗ 10

m sec2

=

To answer this question with Interactive Physics, click the

4000

m sec2

?

newtons

button (if necessary) and click and drag

2

the slider so that its value is 10 m/sec . Then, click the button, and verify that the meter that measures the resultant horizontal force matches the value you calculated with the formula above.

3. What is the resultant horizontal force on the rocket if it has an acceleration of 15 Result: 400 kg ∗ 15 c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

m sec2

60

=

6000

m sec2

?

newtons

Chapter 6: Newton’s Law for motion along a line

4. What is the resultant horizontal force on the rocket if it has an acceleration of 20 Result: 400 kg ∗ 20

m sec2

=

8000

m sec2

=

10,000

?

m sec2

?

newtons

5. What is the resultant horizontal force on the rocket if it has an acceleration of 25 Result: 400 kg ∗ 25

m sec2

newtons

6. A rocket that has a larger acceleration requires a larger force.

True /False (circle one)

7. Bonus Questions - answer without Interactive Physics. A. A rocket accelerating at 1 g is accelerating at a rate equal to the acceleration due to gravity on Earth (approximately 10 m/sec2 ). How many g s does a rocket have when its acceleration is 5 m/sec2 ? Result: 5 m/sec2 = 0.5 g s 10 m/sec2 B. How many g s does a rocket have when its acceleration is 15 m/sec2 ? Result: 15 m/sec2 = 10 m/sec2

1.5

g s

C. How many g s does a rocket have when its acceleration is 25 m/sec2 ? Result: 25 m/sec2 = 10 m/sec2

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

61

2.5

g s

Chapter 6: Newton’s Law for motion along a line

6.3

Acceleration, force, and mass (A)

One way an astronaut or computer can control a spaceship is by controlling the force on its rockets. When the mass of the spaceship is known, its acceleration can by calculated by Acceleration = Force Mass

1. A rocket applies a horizontal force of 1000 newtons to a 400 kg rocket sled that slides on smooth tracks. What is the acceleration of the rocket sled? Result: 1000 newtons = 400 kg

2.5

m sec2

2. A rocket applies a 6000 newton horizontal force to a 400 kg rocket sled that slides on smooth tracks. What is the rocket sled’s acceleration? Result: 6000 newtons = 400 kg

15

m sec2

3. A larger force produces a smaller/ larger (circle one) acceleration.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

62

Chapter 6: Newton’s Law for motion along a line

4. A rocket applies a force to a 400 kg rocket sled. For each of the forces in the following table, calculate the resulting acceleration and then complete the graph. 20

Acceleration (m/sec2 ) 0

1000

2.5

2000

5.0

4000

10.0

5000

12.5

8000

20.0

15 Acceleration (m/sec2)

Force (newtons) 0

10

5

0 0

1000

2000

3000 4000 5000 Force (newtons)

5. Bonus Question - answer with the previous graph. Estimate the acceleration with a 6000 newton force: 15 Estimate the acceleration with a 9000 newton force:

22.5

6000

7000

8000

m sec2 m sec2

Courtesy NASA/JPL-Caltech

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

63

Chapter 6: Newton’s Law for motion along a line

6.4

Acceleration, force, and mass (B)

To make it easier to control the motion of spaceships, they are made with as little mass as possible. In mathematical terms, acceleration and mass are related by Acceleration = Force Mass

1. A rocket applies a force of 1200 newtons to a 100 kg rocket sled. What is the acceleration of the rocket sled? Result: 1200 newtons = 100 kg

12

m sec2

2. A rocket applies a 1200 newton force to a 400 kg rocket sled. What is the rocket sled’s acceleration? Result: 1200 newtons = 400 kg

3

m sec2

3. A 1200 newton force produces a smaller /larger (circle one) acceleration on a 400 kg rocket sled than on a 100 kg rocket sled.

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Chapter 6: Newton’s Law for motion along a line

4. Bonus Question - answer without Interactive Physics. A rocket applies a 1200 newton force to various rocket sleds. For each rocket sled in the following table, calculate the acceleration and then complete the graph. 12

Acceleration (m/sec2 ) 12.0 6.0 4.0 3.0 2.4 2.0

10 Acceleration (m/sec2)

Mass (kg) 100 200 300 400 500 600

8

6

4

2 100

150

200

250

300

350

400

450

500

550

600

Mass (kg)

5. Bonus Question - answer without Interactive Physics. Estimate the acceleration of a 350 kg sled: 3.4 secm2 Estimate the acceleration of a 50 kg sled:

24

m sec2

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65

Chapter 6: Newton’s Law for motion along a line

6.5

Acceleration, force, mass, and friction

Because a rocket sled rides on train tracks that are not perfectly smooth, there is a frictional force that resists the motion of the rocket. This frictional force depends on the weight of the rocket and a quantity called the coefficient of friction. The equation that relates the rocket sled’s acceleration with the rocket sled’s mass and the resultant horizontal force (the rocket’s horizontal force minus the friction force), is Acceleration = Force Mass

1. A rocket applies a horizontal force of 1000 newtons to a 400 kg rocket sled that slides on rough tracks. The friction force between the sled and the tracks is 400 newtons. Determine the resultant horizontal force on the rocket sled and its acceleration. Result: ResultantForce =

600

newtons

600 newtons = 1.5 400 kg

Acceleration =

m sec2

2. A rocket applies a horizontal force of 2000 newtons to a 400 kg rocket sled that slides on rough tracks. The friction force between the sled and the tracks is 400 newtons. Determine the resultant horizontal force on the rocket sled and its acceleration. Result: ResultantForce = Acceleration =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

1600

newtons

1600 newtons = 4 400 kg

66

m sec2

Chapter 6: Newton’s Law for motion along a line

3. A rocket applies a horizontal force of 4000 newtons to a 400 kg rocket sled that slides on rough tracks. The friction force between the sled and the tracks is 400 newtons. Determine the resultant horizontal force on the rocket sled and its acceleration. Result: ResultantForce = Acceleration =

3600

newtons

3600 newtons = 9 400 kg

m sec2

4. A rocket applies a horizontal force of 6000 newtons to a 400 kg rocket sled that slides on rough tracks. The friction force between the sled and the tracks is 400 newtons, Determine the resultant horizontal force on the rocket sled and its acceleration. Result: ResultantForce = Acceleration =

5600

newtons

5600 newtons = 14 400 kg

m sec2

5. Bonus Question - answer without Interactive Physics. A rocket applies a horizontal force of 6000 newtons to a 400 kg rocket sled that slides on rough tracks. The friction force between the sled and the tracks is 400 newtons, and, at a certain speed, the air-resistance force opposing the rocket sled’s motion is 800 newtons. Determine the resultant horizontal force on the rocket sled and its acceleration. Result: ResultantForce = Acceleration =

4800

newtons

4800 newtons = 12 400 kg

m sec2

6. A rocket sled accelerates more slowly/ quickly (circle one) when there is no friction.

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67

Chapter 6: Newton’s Law for motion along a line

6.6

Force, mass, acceleration, and air resistance

The resultant force on a parachutist is the difference between two forces, namely the downward force due to Earth’s gravitational pull and an upward force due to air resistance. • The downward force on a parachutist due to Earth’s gravitational pull is equal to the mass of the parachutist multiplied by Earth’s gravitational constant, 9.8 m/sec2 , i.e., ForceFromGravity = Mass ∗ Gravity • Although it is known that the upward force on the parachutist from air resistance depends on the size of the parachute, the parachutist’s speed, air pressure, etc., it is difficult to write an equation for the forces due to air resistance. One way to exactly calculate the air resistance forces is to do time-motion studies on the parachutist and then use Newton’s law which relates the sum of all forces acting on the parachutists with the parachutist’s mass and acceleration, i.e., ForceFromAirResistance + -ForceFromGravity = Mass ∗ Acceleration • It is worth mentioning that there is a minus sign before ForceFromGravity because it points downward whereas the ForceFromAirResistance is treated as positive because it points upward . Similarly, a negative acceleration means the acceleration is downward . By rearranging the previous equation, one finds ForceFromAirResistance = ForceFromGravity + Mass ∗ Acceleration

1. What is the resultant upward force on a 100 kg parachutist due to air resistance if the parachutist’s acceleration is 0? Result: 980 newtons 100 kg ∗ 9.8 secm2 + 100 kg ∗ 0 secm2 = c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

68

Chapter 6: Newton’s Law for motion along a line

2. What is the resultant upward force on a 100 kg parachutist due to air resistance if the parachutist has an acceleration of -1.4 m/sec2 ? (For this problem, a negative acceleration means an acceleration downward.)

Result: 100 kg ∗ 9.8

m sec2

100 kg ∗ -1.4

+

m sec2

=

840

newtons

3. What is the resultant upward force on a 100 kg parachutist due to air resistance if the parachutist has an acceleration of -4.2 m/sec2 ? Result: 100 kg ∗ 9.8

m sec2

+

100 kg ∗ -4.2

m sec2

=

560

newtons

4. What is the resultant upward force on a 100 kg parachutist due to air resistance if the parachutist has an acceleration of -7.0 m/sec2 ? Result: 100 kg ∗ 9.8

m sec2

+

100 kg ∗ -7.0

m sec2

=

280

newtons

5. What is the resultant upward force on a 100 kg parachutist due to air resistance if the parachutist has an acceleration of -9.8 m/sec2 ? Result: 100 kg ∗ 9.8

m sec2

100 kg ∗ -9.8

+

m sec2

=

0

newtons

6. Determine the upward force on a 100 kg parachutist due to air resistance when the parachutist is falling at a constant speed. Result: 100 kg ∗ 9.8

m sec2

+

100 kg ∗ 0

m sec2

=

980

newtons

Courtesy NASA c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

69

Chapter 6: Newton’s Law for motion along a line

6.7

Force, mass, acceleration, and horizontal springs

The force in a linear spring depends on its spring constant and its stretch. ForceInSpring = SpringConstant ∗ SpringStretch If a block on a smooth horizontal table is attached to a spring and the spring is stretched, the block will slide on the table in a manner predicted by Newton’s law Force = Mass ∗ Acceleration

1. A linear spring has a spring constant of 50 N/m (abbreviation N/m means newtons per meter). Determine the magnitude and direction of the initial spring force on the block and the magnitude and direction of the initial acceleration of the 2 kg block for the following spring stretches and spring compressions.1 Spring Stretch or Compression (m)

1

Initial Spring Force (newtons)

0.5 (stretch)

50

N m

∗ 0.5 m = 25

1.0 (stretch)

50

N m

∗ 1.0 m = 50

0.5 (compression)

50

N m

∗ 0.5 m = 25

1.0 (compression)

50

N m

∗ 1.0 m = 50

Initial Acceleration ( secm2 ) 25 newtons = 12.5 2 kg 50 newtons = 25 2 kg 25 N = 12.5 2 kg 50 N = 25 2 kg

Direction of Initial Force or Acceleration (circle one) to the right/ left to the right/ left to the right /left to the right /left

A spring that has a negative stretch is said to be in compression.

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70

Chapter 6: Newton’s Law for motion along a line

6.8

Force, mass, acceleration, and vertical springs

The resultant force on a block that is suspended by a spring is the difference between two forces, namely the downward force due to Earth’s gravitational pull and the force due to the spring. The downward gravitational force on the block is equal to the mass of the block multiplied by Earth’s gravitational constant, 9.8 m/sec2 , i.e., ForceFromGravity = Mass ∗ Gravity The upward force on the block from the linear spring depends on the spring constant and the stretch in the spring, i.e., ForceInSpring = SpringConstant ∗ SpringStretch To calculate the initial acceleration of the block, use Newton’s law Force = Mass ∗ Acceleration

1. A linear spring with a spring constant of 50 Determine the following quantities:

N m

is stretched downward 0.5 meters.

• magnitude and direction of the initial spring force on the block • magnitude and direction of the gravity force on the 2 kg block • magnitude and direction of the initial resultant force on the block • magnitude and direction of the initial acceleration of the block

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71

Chapter 6: Newton’s Law for motion along a line

Result: InitialSpringForce = GravityForce = InitialResultantForce = InitialAcceleration =

50

N m

∗ 0.5 m = 25

2 kg ∗ 9.8

m sec2

N

= 19.6

25 N − 19.6 N = 5.4 5.4 N = 2.7 2 kg

up /down

N N

m sec2

up/ down up /down up /down

2. Answer the same questions when the spring is stretched 1.0 meter. Result: 50

InitialSpringForce =

∗ 1.0 m = 50 N

2 kg ∗ 9.8

GravityForce = InitialResultantForce =

N m

m sec2

= 19.6 N

50 N − 19.6 N = 30.4 N 30.4 N = 15.2 2 kg

InitialAcceleration =

m sec2

up /down up/ down up /down up /down

3. Answer the same questions when the spring is compressed 0.5 meters. Result: 50

InitialSpringForce =

∗ 0.5 m = 25 N

2 kg ∗ 9.8

GravityForce = InitialResultantForce =

N m

m sec2

= 19.6 N

25 N + 19.6 N = 44.6 N 44.6 N = 22.3 2 kg

InitialAcceleration =

m sec2

up/ down up/ down up/ down up/ down

4. Answer the same questions when the spring is compressed 1.0 meters. Result: 50

InitialSpringForce =

∗ 1.0 m = 50 N

2 kg ∗ 9.8

GravityForce = InitialResultantForce =

N m

m sec2

= 19.6 N

50 N + 19.6 N = 69.6 N

InitialAcceleration =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

69.6 N = 34.8 2 kg

72

m sec2

up/ down up/ down up/ down up/ down

Chapter 6: Newton’s Law for motion along a line

Chapter 7 Potential and kinetic energy Energy exists in many forms, including mechanical, chemical, electromagnetic, and nuclear, and can be converted from one form to another. For example, putting gasoline in a car adds chemical energy, which is converted during combustion into the mechanical energy required to move the engine’s pistons, and ultimately, the wheels and the car. At the same time, the engine runs the alternator, generating electrical energy to run the radio and the headlights. This chapter focuses on the mechanical energy associated with the position (potential energy) or movement (kinetic energy) of objects. Total mechanical energy of a system is the sum of potential energy and kinetic energy in the system. Two specific types of potential energy are gravitational potential energy and spring potential energy. GravitationalPotentialEnergy = Mass ∗ Gravity ∗ Height SpringPotentialEnergy = 12 ∗ SpringConstant ∗ SpringStretch2 KineticEnergy = 21 ∗ Mass ∗ Speed2

Courtesy NASA/JPL-Caltech

73

7.1

Gravitational potential energy

Gravitational potential energy is energy associated with an object’s height and is related to gravity and the object’s mass. In mathematical terms, GravitationalPotentialEnergy = Mass ∗ Gravity ∗ Height The SI units for energy are joules. One joule is defined as 1 joule = 1 newton ∗ m = 1

kg∗m sec2

∗m = 1

kg∗m2 sec2

1. Find the gravitational potential energy of the block when it is 1 m above the ground. Result: 10

kg ∗

9.8

m sec2

∗

1 m =

98 newtons ∗ m =

98 joules

To answer this question with Interactive Physics, click and verify that the value in the potential energy meter when the block’s height is 1 m matches the value you calculated with the formula above.

2. Find the gravitational potential energy of the block when it is 2 m above the ground. Result: 10 kg ∗ 9.8

m sec2

∗ 2m =

196

joules

3. The higher the block, the more potential energy it has. True /False 4. If the block is on the ground, its gravitational potential energy is zero. True /False 5. If the block is below the ground, its gravitational potential energy is negative. True /False 6. Bonus Question - answer without Interactive Physics. Find the gravitational potential energy of a 20 kg block 4 m above the ground. Result: 20 kg ∗ 9.8

m sec2

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

∗ 4m = 74

784

joules

Chapter 7: Potential and kinetic energy

7.2

Potential energy in a spring

Potential energy is a measure of stored energy. One way to store energy is in a spring. For example, when you wind a mechanical watch, energy is stored by stretching the spring inside the watch. This energy is then used to make the watch run. The potential energy stored in a linear spring is calculated by the equation SpringPotentialEnergy = 12 ∗ SpringConstant ∗ SpringStretch2

For the following questions, assume the spring has a spring constant of 50 mN . 1. Find the potential energy in a spring that is stretched 1.0 m. Result: 1 ∗ 50 2

N m

∗ ( 1.0

m )2 =

25

newtons ∗ m

=

25 joules

2. Find the potential energy in a spring that is stretched 2.0 m. Result: 1 ∗ 50 2

N m

∗ (2.0 m)2 =

100

joules

3. Find the potential energy in a spring that is compressed 1.0 m (i.e., stretched -1.0 m). Result: 1 ∗ 50 2

N m

∗ (-1.0 m)2 =

25

joules

4. Find the potential energy in a spring that is compressed 2.0 m. Result: 1 ∗ 50 2

N m

∗ (-2.0 m)2 =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

75

100

joules

Chapter 7: Potential and kinetic energy

5. Using your previous answers and additional calculations, fill in the table and complete the following graph. Assume that the spring constant is 50 N/m. Spring stretch (m) -2.0

Spring potential energy (joules) 100

-1.5

56.3

-1.0

25.0

-0.5

6.3

0.0

0

0.5

6.3

1.0

25.0

1.5

56.3

2.0

100

Potential energy in spring (joules)

100

80

60

40

20

0 -2

-1.5

-1

-0.5 0 0.5 Spring stretch (m)

1

6. The potential energy in a spring is never a negative number.

1.5

2

True /False

7. The shape of the curve you drew is a straight line/ parabola .

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76

Chapter 7: Potential and kinetic energy

7.3

Kinetic energy and mass

Kinetic energy measures the energy of an object in motion. It is related to the speed and mass of an object. The more massive a moving object is, the more kinetic energy it has. A moving train has more kinetic energy than a car travelling at the same speed. In mathematical terms, KineticEnergy = 21 ∗ Mass ∗ Speed2

1. Find the kinetic energy1 of a 5 kg toy car traveling at 5 m/sec. Result:

1 ∗ 5 kg ∗ ( 5 2

m 2 sec

) =

62.5

kg∗m2

62.5 newtons ∗ m =

=

sec2

62.5 joules

2. Find the kinetic energy of a 10 kg toy car traveling at 5 m/sec. Result: 1 ∗ 10 2

kg ∗ ( 5

m 2 sec ) = 125

kg ∗ m sec 2

2

= 125

newtons ∗ m

= 125

joules

3. Find the kinetic energy of a 15 kg toy car traveling at 5 m/sec. Result: 1 ∗ 15 kg ∗ ( 5 2

1

m sec

)2 =

187.5

joules

This problem neglects the rotational kinetic energy associated with the car’s spinning wheels.

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77

Chapter 7: Potential and kinetic energy

4. Using your previous answers and additional calculations, fill in the table and complete the following graph. Assume the car’s speed is 5 m/sec. 250

5

62.5

10

125.0

15

187.5

20

250.0

200 Kinetic energy (joules)

Mass Kinetic energy (kg) (joules) 0 0.0

150

100

50

0 0

5

10

15

20

Mass of car (kg)

5. Bonus Question - answer without Interactive Physics. The kinetic energy of a car moving at 5 m/sec is 220 joules. A. Use the previous graph to estimate its mass. ≈ 17.5 kg B. Use the equation for kinetic energy to exactly calculate its mass. Result: 220 joules = 1 2 m ∗ ( 5 sec ) 2

17.6

kg

6. The kinetic energy of a 0.0 kg car is zero. True /False It is possible to build a 0.0 kg car. True/ False

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78

Chapter 7: Potential and kinetic energy

7.4

Kinetic energy and speed

Kinetic energy measures the energy of an object in motion. It is related to the speed and mass of an object. The faster an object moves, the more kinetic energy it has. A car traveling at high speeds on the freeway has more kinetic energy than it does when it is slowly pulling into a driveway. In mathematical terms, KineticEnergy = 21 ∗ Mass ∗ Speed2

1. Find the kinetic energy2 of a 10 kg toy car traveling at 2 m/sec. Result: kg ∗ m 2 1 m 2 ∗ 10 kg ∗ ( 2 sec ) = 20 = 20 newtons ∗ m sec 2 2 2. Find the kinetic energy of a 10 kg toy car travelling at 4 m/sec. Result: kg ∗ m 2 1 m ∗ 10 kg ∗ ( 4 sec )2 = 80 = 80 newtons ∗ m sec 2 2

= 20 joules

= 80

joules

3. Find the kinetic energy of a 10 kg toy car travelling at 6 m/sec. Result: 1 ∗ 10 kg ∗ ( 6 m )2 = 180 joules sec 2

2

This problem neglects the rotational kinetic energy associated with the car’s spinning wheels.

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79

Chapter 7: Potential and kinetic energy

4. Using your previous answers and additional calculations, fill in the table and complete the following graph. Assume the car’s mass is 10 kg. 500 450 400 Kinetic energy (joules)

Speed Kinetic energy (m/sec) (joules) 0 0

350 300

2

20

4

80

6

180

8

320

100

10

500

50

250 200 150

0 0

2

4 6 Speed of car (m/s)

8

10

5. Bonus Question - answer without Interactive Physics. The kinetic energy of a 10 kg car is 400 joules. m A. Use the previous graph to estimate its speed. 9 sec B. Use the equation for kinetic energy to exactly calculate its speed. Result:  400 joules m 8.94 sec 1 ∗ 10 kg = 2 6. Bonus Question - answer without Interactive Physics. Is it possible for a 10 kg car to have -80 joules of kinetic energy? Yes/ No Try to calculate the speed at which the car must be travelling.

The car’s speed is 4 ∗

√

-1, which is not physically meaningful (it is an imaginary number).

7. Bonus Question - answer without Interactive Physics.

A 10 kg red car is moving at 5 m/sec and a 5 kg blue car is moving at 10 m/sec. The red car has less /the same/more kinetic energy than the blue car. Explain: m Red car: 12 ∗ 10 kg ∗ ( 5 sec )2 = 125 joules m Blue car: 21 ∗ 5 kg ∗ ( 10 sec )2 = 250 joules

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80

Chapter 7: Potential and kinetic energy

Chapter 8 Conservation of energy Potential energy can be converted to kinetic energy and vice versa. For example, a bungee jumper on the platform before a jump is at rest and has potential energy and no kinetic energy. Once he jumps, this potential energy is converted into kinetic energy as he falls towards the river. When he bounces back up, the kinetic energy is converted into potential energy. Sometimes, the mechanical energy in the system is conserved, i.e. the sum of the initial potential energy and initial kinetic energy is the same as the final sum. However, when nonconservative forces such as friction and air resistance act on the system, total mechanical energy is not conserved. For example, a book sliding on a flat frictionless floor will have a constant speed and constant mechanical energy. On the other hand, a book sliding on a real floor (with friction) loses speed and loses mechanical energy in the form of heat. With high-precision temperature sensors, you can measure an increase in temperature in both the book and the floor. Lubricants are used to decrease friction and energy loss in machines. It takes less effort (energy) to ride a bicycle with well-oiled gears and chain than one with rusty components. The same is true for a car, lawn-mower, scissors, or sliding books. InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy

GravitationalPotentialEnergy = Mass ∗ Gravity ∗ Height 1 ∗ SpringConstant ∗ SpringStretch2 2 1 ∗ Mass ∗ Speed2 KineticEnergy = 2

SpringPotentialEnergy =

81

8.1

Potential and kinetic energy in a pendulum

When a pendulum swings back and forth, there is an exchange between potential and kinetic energy. At the highest part of its swing, it moves most slowly, while at the lowest part of its swing, it moves most quickly.1

With air resistance set to 0, click

.

1. The potential energy is largest when the pendulum is at the top /bottom of its swing. When the potential energy is largest, the kinetic energy is smallest /largest . 2. The kinetic energy is largest when the pendulum is at the top/ bottom of its swing. When the kinetic energy is largest, the potential energy is smallest /largest . Click

, set the pendulum amplitude to Small, and click

.

3. When the pendulum swings down, potential/ kinetic energy increases and potential /kinetic energy decreases. 4. When the pendulum swings up, potential /kinetic energy increases and potential/ kinetic energy decreases. 5. The sum of potential energy and kinetic energy is constant (does not change). True /False 6. With air resistance set to 1.0 (set the value in the air resistance slider to 1.0) , the sum of potential energy and kinetic energy is constant. True/ False

1

Similarly, in an idealized setting with no friction, a pendulum clock does not require winding. However, in the real world, air resistance and friction cause the clock to slow down and lose energy. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

82

Chapter 8: Conservation of energy

8.2

Conservation of energy in a roller coaster

(Windows only)

As a free-moving roller coaster moves up and down, there is an exchange between potential and kinetic energy. The cart slows down as it goes higher and speeds up as it goes lower.2 GravitationalPotentialEnergy = Mass ∗ Gravity ∗ Height KineticEnergy =

Click

1 ∗ Mass ∗ Speed2 2

, and observe the cart’s speed and height as it moves along the track.

1. The higher the cart is, the more potential /kinetic energy it has. The faster the cart moves, the more potential/ kinetic energy it has. 2. As the cart moves uphill, potential /kinetic energy increases and potential/ kinetic energy decreases. Alternately, as the cart moves downhill, potential/ kinetic energy increases and potential /kinetic energy decreases. 3. Calculate the sum of potential energy and kinetic energy of the 150 kg cart at the bottom m of the track (where the height is 0 m), when its speed is 15 sec . Verify your calculations with the results from Interactive Physics. Result: PotentialEnergy = 150 kg ∗ 9.8

m sec2

∗ 0.0 m =

1 m 2 ∗ 150 kg ∗ (15 sec ) = 2 PotentialEnergy + KineticEnergy = 16, 875 joules KineticEnergy =

0

joules

16, 875 joules

2

This problem assumes no energy loss due to air resistance or friction along the track. In reality, these can be important factors when designing a roller coaster. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

83

Chapter 8: Conservation of energy

Use the meter information in Interactive Physics for cart height and speed at the top of the track.

4. Calculate the sum of the potential energy and kinetic energy of the 150 kg cart at the top of the track, where the height is 9.0 m and the speed is 6.971 m/sec. Verify your calculations with Interactive Physics.3 Result: 150 kg ∗ 9.8

PotentialEnergy =

m sec2

∗ 9.0 m =

1 ∗ 150 kg ∗ (6.971 2

KineticEnergy = PotentialEnergy + KineticEnergy =

m 2 sec

) =

13, 230

joules

3, 645 joules

16, 875 joules

5. The sum of the cart’s potential energy and kinetic energy at the top and bottom of the track are equal. True /False 6. Using Interactive Physics, determine which speed(s) give the cart enough kinetic energy m to reach the top of the track: 10/12.5/ 15 sec . 7. Bonus Question - answer without Interactive Physics. Use conservation of energy to calculate the minimum kinetic energy the cart needs at the bottom of the track in order to reach the top of the track. Hint: One way to write conservation of energy is

InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy where the initial potential energy and initial kinetic energy are calculated when the cart is at the bottom of the track, and final potential energy and final kinetic energy are calculated when the cart is at the top. The first step to solve this problem is to calculate the following quantities:

InitialPotentialEnergy = FinalPotentialEnergy = FinalKineticEnergy =

0

joules

150 kg ∗ 9.8 0

m sec2

∗ 9.0 m =

13, 230

joules

joules

Now you are ready to plug in to the conservation of energy equation:

Result: InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy

0

joules + InitialKineticEnergy =

13, 230 joules

InitialKineticEnergy =

13, 230 joules

+

0

joules

8. Bonus Question - answer without Interactive Physics. Using results from the previous question, calculate the speed the cart needs at the bottom of the track in order to reach the top of the track. Result:    13, 230 joules m Speed =  = 13.28 sec 1 ∗ 150 kg 2 3

The numbers displayed in Interactive Physics are rounded. Use a height of 9 m and a speed of 6.971 m/s in your calculation. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

84

Chapter 8: Conservation of energy

8.3

Conservation of energy for a snowboarder

A snowboarder gains potential energy as he is lifted by the ski lift to the top of the mountain. As he goes down a slope, his potential energy decreases as his kinetic energy and speed increase. GravitationalPotentialEnergy = Mass ∗ Gravity ∗ Height KineticEnergy =

1 ∗ Mass ∗ Speed2 2

Set the initial height of the snowboarder to 10 m.

1. Calculate the sum of the potential energy and kinetic energy of the 50 kg snowboarder whose initial height is 10 m. Verify your calculations with Interactive Physics.4 Result: InitialPotentialEnergy = 50 kg ∗ 9.8

m sec2

∗ 10.0 m = 4900 joules

1 m 2 ∗ 50 kg ∗ (0 sec ) = 0 joules 2 InitialPotentialEnergy + InitialKineticEnergy = 4900 joules InitialKineticEnergy =

Click

, and note the snowboarder’s height and speed at the bottom.

2. Calculate the sum of the potential energy and kinetic energy of the 50 kg snowboarder when he reaches the bottom of the slope. Verify your calculations with Interactive Physics. Result: FinalPotentialEnergy = FinalKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy = 4

50 kg ∗ 9.8

m sec2

∗ 0.0 m =

1 ∗ 50 kg ∗ (14.0 2 4900

m 2 sec

) =

0 joules 4900 joules

joules

The symbol kJ in the Interactive Physics meter is an abbreviation for kilojoules (1000 joules).

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

85

Chapter 8: Conservation of energy

3. The sum of the potential energy and kinetic energy at the bottom of the slope is equal to the sum of the potential energy and kinetic energy at the top of the slope. True /False 4. Bonus Question - answer without Interactive Physics. A. Calculate the potential energy and kinetic energy of a 50 kg snowboarder who starts from a height of 15 m with an initial speed of 0 m/sec. Result: InitialPotentialEnergy = InitialKineticEnergy =

50 kg ∗ 9.8

∗ 15.0 m =

m sec2

1 ∗ 50 kg ∗ (0.0 2

m 2 sec

) =

7350

0

joules

joules

B. Calculate the potential energy of the 50 kg snowboarder when he reaches the bottom of the slope (height = 0 m). Result: FinalPotentialEnergy =

50 kg ∗ 9.8

m sec2

∗ 0.0 m =

0

joules

C. Use conservation of energy to calculate the snowboarder’s kinetic energy when he reaches the bottom of the slope. Hint: One way to write conservation of energy is

InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy where the initial potential energy and kinetic energy are calculated when the snowboarder is at the top, and final potential energy and kinetic energy are calculated when he is at the bottom.

Result: InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy

7350 joules

+

0 joules

FinalKineticEnergy

=

0

=

7350

joules

+ FinalKineticEnergy

joules

D. A 50 kg snowboarder starts at the top of a slope that is 15 m high. Calculate the snowboarder’s speed at the bottom of the slope. Result:    7350 joules m = Speed =  17.15 sec 1 ∗ 50 kg 2

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86

Chapter 8: Conservation of energy

8.4

Conservation of energy in a spring

When you pull back a pebble in a sling shot, you store energy in the springy elastic. When you release it, the stored energy is converted to kinetic energy as the pebble zooms away. 1 ∗ SpringConstant ∗ SpringStretch2 2 1 ∗ Mass ∗ Speed2 KineticEnergy = 2

SpringPotentialEnergy =

With the initial spring stretch set to 2.0 m, click

and watch the block move back and forth.

1. Calculate the sum of potential energy and kinetic energy when the spring is stretched 2.0 m. (Note: The spring has a spring constant of 50 N/m.) Result: 1 ∗ 50 mN ∗ (2.0 m)2 = 100.0 joules InitialPotentialEnergy = 2 1 m 2 ∗ 10 kg ∗ (0 sec InitialKineticEnergy = ) = 0 joules 2 InitialPotentialEnergy + InitialKineticEnergy = 100.0 joules 2. The sum of potential energy and kinetic energy (as shown in Interactive Physics) increases/decreases/ remains the same . 3. The spring potential energy is largest when the spring stretch is 2.0 The kinetic energy is largest when the spring stretch is 0.0 m.

m or

-2.0 m.

4. Use conservation of energy to calculate the kinetic energy of the 10 kg block when it is at the middle of its travel (spring stretch = 0.0 m). Result: InitialPotentialEnergy + InitialKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy 100.0 joules +

0

joules = FinalKineticEnergy =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

87

0 joules + FinalKineticEnergy 100.0 joules Chapter 8: Conservation of energy

5. Bonus Question. Using your results from the previous question, calculate the maximum speed of the 10 kg block that is initially stretched 2.0 m. Result:    100.0 joules m  = 4.47 sec 1 ∗ 10 kg 2 To verify this result with Interactive Physics, click the

button, then

click on the arrow that is on the furthest right and bottom part of the screen. Each click will step the simulation forward. Observe the block’s speed and record the largest value.

6. Bonus Question. Calculate the maximum speed of a 10 kg block that is initially stretched 1.0 m. Verify your results with Interactive Physics. Result: N InitialPotentialEnergy = 12 ∗ 50 m ∗ (1.0 m)2 = 25.0 joules InitialKineticEnergy =

1 2

∗ 10 kg ∗ (0

FinalPotentialEnergy =

1 2

∗ 50

InitialPotentialEnergy + InitialKineticEnergy

25.0

joules

+

0

joules

N m

m 2 sec

) =

0

∗ (0.0 m)2 =

=

joules

0 joules

FinalPotentialEnergy + FinalKineticEnergy

=

0

joules

+ FinalKineticEnergy

FinalKineticEnergy = 25.0 joules

   25.0 joules Speed =  1 ∗ 10 kg 2

=

m sec

2.24

7. Bonus Question - answer without Interactive Physics. A 10 kg block is attached to a spring with a spring constant of 50 N/m. The block is observed to have a speed of 9.0 m/sec when the spring is unstretched. Determine the maximum stretch of the spring. Result: N 2 InitialPotentialEnergy = 1 2 ∗ 50 m ∗ (0.0 m) = 0 joules InitialKineticEnergy =

1 ∗ 10 kg ∗ (9.0 2

m 2 sec

405 joules

FinalKineticEnergy =

1 ∗ 10 kg ∗ (0.0 2

m 2 sec

0

InitialPotentialEnergy + InitialKineticEnergy

0

joules

+

405 joules

joules

= FinalPotentialEnergy +

0

joules

405 joules

   405 joules SpringStretch =  1 ∗ 50 N m 2 88

) =

= FinalPotentialEnergy + FinalKineticEnergy

FinalPotentialEnergy =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

) =

=

4.02 m

Chapter 8: Conservation of energy

8.5

Conservation of energy for a bungee jumper

Energy can be exchanged between gravitational potential energy, spring potential energy, and kinetic energy. When a bungee jumper is at the bottom of her bounce, all of her gravitational potential energy and kinetic energy has been converted to spring potential energy.

Click

and watch the bungee jumper bounce up and down a few times.

1. The gravitational potential energy is largest when the jumper is at the top /middle/bottom, and smallest when she is at the top/middle/ bottom . 2. When the bungee jumper is at the top, there is no stretch in the bungee cord. Therefore, the spring potential energy is smallest /largest . At the bottom, the bungee cord is highly stretched, and the spring potential energy is smallest/ largest . 3. The kinetic energy seems to be highest when the jumper is at the top/ middle /bottom of a bounce. At this point, her speed is smallest/ largest . 4. The sum of potential energy and kinetic energy (as shown in Interactive Physics) increases/decreases/ remains the same . 5. Bonus Question - answer without Interactive Physics. Complete the following table. Each row represents a different bungee jumper height.

potential energy (joules) 1000

Spring potential energy (joules) 0

Kinetic energy (joules) 0

Total mechanical energy (joules) 1000

top /bottom/in between

356

415

229

1000

top/bottom/ in between

0

1000

0

1000

top/ bottom /in between

276

524

200

1000

top/bottom/ in between

Gravitational

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89

Bungee jumper height

Chapter 8: Conservation of energy

8.6

Energy loss due to friction

In the real world, the sum of potential energy and kinetic energy is not usually constant. Friction, air resistance, and other forces cause some energy to be converted to heat. 1 ∗ SpringConstant ∗ SpringStretch2 2 1 KineticEnergy = ∗ Mass ∗ Speed2 2

SpringPotentialEnergy =

Click

, and watch the block move back and forth.

1. Calculate the sum of the initial potential energy and initial kinetic energy when the spring is stretched 2.0 m. (Note: The spring has a spring constant of 50 N/m.) Result: 1 ∗ 50 mN ∗ (2.0 m)2 = 100.0 joules InitialPotentialEnergy = 2 1 m 2 ∗ 10 kg ∗ (0 sec InitialKineticEnergy = ) = 0 joules 2 InitialPotentialEnergy + InitialKineticEnergy = 100.0 joules 2. With no friction or air resistance, the sum of potential energy and kinetic energy (as shown in Interactive Physics) increases/decreases/ remains the same .

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90

Chapter 8: Conservation of energy

Click

, and check the Friction box. Click

may need to use the speed at 2.50 sec.

or

, then click

when the timer reads 2.50 sec. You

arrows to rewind or advance the simulation. Record the spring stretch and block

3. With friction, calculate the sum of potential energy and kinetic energy at 2.50 seconds. Result: 1 ∗ 50 N ∗ (1.01 m)2 = 25.5 joules FinalPotentialEnergy = m 2 FinalKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy =

1 ∗ 10 kg ∗ (2.17 2

m 2 sec

) =

23.5 joules

49.0 joules

This sum is less than /greater than/equal to the sum of the initial potential energy and kinetic energy. 4. Over time, friction causes the sum of potential energy and kinetic energy to increase/ decrease . Click

. Uncheck the Friction box and check the Air resistance box. Click

the timer reads 5.00 sec. You may need to use the the spring stretch and block speed at 5.00 sec.

or

, then click

when

arrows to rewind or advance the simulation. Record

5. With air resistance, calculate the sum of potential energy and kinetic energy at 5.00 seconds. Result: 1 ∗ 50 N ∗ (0.13 m)2 = 0.4 joules FinalPotentialEnergy = m 2 FinalKineticEnergy = FinalPotentialEnergy + FinalKineticEnergy =

1 ∗ 10 kg ∗ (2.42 2

m 2 sec

) =

29.3 joules

29.7 joules

This sum is less than /greater than/equal to the sum of the initial potential energy and kinetic energy. 6. Over time, air resistance causes the sum of potential energy and kinetic energy to increase/ decrease . 7. Bonus Question. With air resistance ON, the sum of potential energy and kinetic energy decrease most quickly in Interactive Physics when the block is moving slowly/ quickly .

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91

Chapter 8: Conservation of energy

Chapter 9 Temperature and heat Temperature is a measure of how hot or cold an object is. Common tools for measuring temperature are alcohol, mercury, electronic, liquid crystal, and infrared thermometers. Two common scales used in measuring temperature are the Fahrenheit temperature scale and the Celsius temperature scale. In the Fahrenheit scale, developed by German physicist Daniel Fahrenheit (1686-1736), water freezes at 32 ◦ F and boils at 212 ◦ F. The Celsius scale, in which water freezes at 0 ◦ C and boils at 100 ◦ C, is named after Swedish astronomer Anders Celsius (1701-1744). Until the 1970’s, the Fahrenheit scale was widely used in English-speaking countries. Today, most countries in the world use the Celsius scale. Another temperature scale widely used in science named after British physicist William Kelvin (1824-1907). Kelvin (K) is the official unit of temperature measurement in the metric system. The most famous reference point in the Kelvin scale is 0 K, also known as absolute zero. Absolute zero (approximately -459.69 ◦ F or -273.16 ◦ C) is the lowest temperature possible theoretically. At absolute zero, atoms have no kinetic energy and no molecular motion.

Courtesy U.S. Fish and Wildlife Service

92

9.1

Temperature

Temperature is a measure of how hot or cold an object is. For a gas, temperature is directly related to the average kinetic energy of the molecules.1

Set the gas temperature to Cold, click . After observing the movement of the molecules, slide the temperature to Hot and notice how the change in temperature affects the motion of the molecules. Slide the temperature to Cold and again notice the change in the motion of the molecules.

1. At higher temperatures, the average speed of the gas molecules increases /decreases/stays constant. 2. At lower temperatures, the average speed of the gas molecules increases/ decreases /stays constant. 3. The Fahrenheit scale (◦ F) and the Celsius scale (◦ C) are related as followed. 9 ◦ ◦ ∗ C + 32 F = 5 5 ◦ ∗ (◦ F − 32) C = 9 Complete the following table. ◦

Description

Water freezes at Water boils at Normal body temperature is about The surface of the sun is about Paper burns at Absolute zero2 1

F 32 212 98.6 10832 451 -459.4

◦

C 0 100 37 6000 232.8 -273

The temperature of an ideal gas is related to its average kinetic energy by T =

kB = 1.38 x 10 23

joules K

2 3 kB

1 2

m v2



where

1 2

m v is the average translational kinetic energy of the

93

Chapter 9: Temperature and heat

is the Boltzmann’s constant and

2

molecules. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

9.2

Heat capacity

Specific heat is a measure of how much heat is required to change the temperature of an object with a specified mass and material. In mathematical terms3 Heat = Mass ∗ SpecificHeat ∗ TemperatureChange

With the Mass set to 0.25 kg, click

. Observe the 80 ◦ C temperature change over 10 sec.

1. Calculate the heat added to 0.25 kg of water (which has a specific heat of 4180 when the candle changes the water’s temperature by 80 ◦ C. Result: 0.25 kg ∗ 4180 kgjoules ∗ 80 ◦ C = 83, 600 joules ∗ ◦C Click

, set the Mass to 0.50 kg, and click

joules kg ∗ ◦ C

)

.

2. Calculate the heat added to 0.50 kg of water (which has a specific heat of 4180 when the candle changes the water’s temperature by 40 ◦ C. Result: 0.50 kg ∗ 4180 kgjoules ∗ 40 ◦ C = 83, 600 joules ∗ ◦C

joules kg ∗ ◦ C

)

The temperature change in the 0.50 kg of water is less than /greater than/equal to the temperature change in the 0.25 kg of water. The heat added to the 0.50 kg of water is less than/greater than/ equal to the heat added to the 0.25 kg of water. Click

, set the Mass to 1.00 kg, and click

.

3. Calculate the heat added to 1.00 kg of water (which changes the temperature by 20 ◦ C). Result: 1.00 kg ∗ 4180 kgjoules ∗ 20 ◦ C = 83, 600 joules ∗ ◦C 3

A positive value of heat means that heat flows into the object and its temperature increases. A negative value of heat means that heat flows out of the object and its temperature decreases. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

94

Chapter 9: Temperature and heat

4. Bonus Question - answer without Interactive Physics. How much heat must be added to 1.0 kg of water to raise its temperature by 80 ◦ C? Result: 1.00 kg ∗ 4180 kgjoules ∗ 80 ◦ C = 334, 400 joules ∗ ◦C 5. Bonus Question - answer without Interactive Physics. 40, 000 joules of heat is added to 0.15 kg of water that is initially at 10 ◦ C. Calculate its final temperature. Result: TemperatureChange = FinalTemperature =

40, 000 joules = 0.15 kg ∗ 4180 kgjoules ∗ ◦C 10 ◦ C + 63.8 ◦ C =

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

95

◦

63.8 73.8

◦

C

C

Chapter 9: Temperature and heat

9.3

Phase change

Heating a substance does not always change its temperature. Sometimes, heating a substance changes the phase of a substance from solid to liquid, liquid to gas, or solid to gas.4

Click

. Press and hold the Add heat button until the ice begins to melt. Observe the temperature change.

1. Before the ice begins to melt, its temperature increases /decreases/stays constant. 2. The ice begins to melt at 0

◦

C.

3. While the ice is melting, the temperature increases/decreases/ stays constant. 4. While the ice is melting, the ice and liquid water are at the same temperature. True /False Press and hold the Add heat button until the water begins to boil. Observe the temperature change.

5. Before the water begins to boil, the temperature increases /decreases/stays constant. 6. The water begins to boil at 100

◦

C.

7. While the water is boiling, the temperature increases/decreases/ stays constant. Note: To re-run the experiment, close and re-open the file.

4

One reason that Earth is unique in our solar system is that water is naturally found in three phases - solid ice, liquid water, and gaseous water vapor. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

96

Chapter 9: Temperature and heat

9.4

Thermal Expansion

As temperature increases, the molecules in a material tend to move farther apart from each other. The resulting expansion is known as thermal expansion. With few exceptions, all solids, liquids, and gases expand as temperature increases.5 For a solid, the amount of thermal expansion is calculated with the following equation ChangeInLength = ExpansionCoefficient ∗ OriginalLength ∗ TemperatureChange

With the temperature set to Low, click

.

1. The brass strip and the iron strip start at the same length. True /False Slide the temperature to Medium and then Hot and observe how the brass and iron strips change length.

2. An increase in temperature causes the brass /iron to expand more. 3. When the brass and iron are glued together as a single strip with the brass on top, a temperature increase causes the strip to bend up/ down . 4. Based on your observations, brass /iron has the larger expansion coefficient. 5. Bonus Question - answer without Interactive Physics. The 110-story Sears Tower in Chicago is about 440 m high on a cold winter night when the temperature is -15 ◦ C (5 ◦ F). The structure of the building is mostly steel and concrete, both of which have an expansion coefficient of .000011 ◦1C . Calculate how much taller the building is during a warm summer day when the temperature is 40 ◦ C (104 ◦ F). Result: 40 ◦ C − -15 ◦ C = 55 ◦ C ChangeInTemperature = ChangeInLength =

.000011

1 ◦C

∗ 440 m ∗ 55 ◦ C = .266 m = 26.6 cm

5

Water has the uncommon property that it is most dense at 4 ◦ C, (39 ◦ F) and is actually less dense as a solid than as a liquid. This is why ice floats in a cup of water. If ice did not float on oceans and lakes, earth’s ecosystem would be dramatically different. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

97

Chapter 9: Temperature and heat

Chapter 10 Heat transfer Heat transfer is the flow of energy from a warm object to a colder one. Heat is transfered from the warm object to the cold one until both objects are the same temperature. Heat may be transferred by conduction, convection, or radiation. Conduction is heat transfer between or within objects in direct contact. For example, when you put a cold metal spoon in hot soup, the spoon becomes warmer because heat flows from the hot soup to the cold spoon. Materials that conduct heat well are called conductors; those that conduct heat poorly are called insulators. Diamond is the best conductor known,a far better than silver, copper, and aluminum, which are all considered good conductors. Diamond’s conducivity is so high that you can melt and cut through ice by simply holding a small piece of diamond with your warm fingers and pressing it against the ice. a

Diamond is also the hardest substance known and has the highest melting point.

Convection occurs by the movement of a substance. When the movement is caused by a pump or a fan, such as cooling of computer chips by a fan, it is called forced convection. When the movement occurs because of difference in density, such as water being boiled in a pot or air around a fire, it is called natural convection. When water is boiled, the water closest to the fire is warmed first. This heated water expands, becomes less dense, and rises to the top of the pot. This allows the cooler water to sink to the bottom of the pot, where it is heated. Radiation is the third means of heat transfer. Any object with a temperature above absolute zero emits radiation continuously. This radiation is absorbed and/or reflected by surrounding objects.a An object that absorbs more radiation than it emits becomes warmer while an object that emits more radiation than it absorbs becomes cooler. When you stand in the sun, radiation emitted by the sun is absorbed/reflected by you; at the same time, radiation emitted by you is absorbed/reflected by the sun. You become warmer because you absorb more radiation than you emit. a

A black body is a theoretical object that absorbs 100% of the radiation that hits it and emits the maximum amount of radiation possible at a given temperature.

98

Ouch!

10.1

Heat transfer

Heat flows from warmer objects to colder objects, and continues to flow until the objects are at the same temperature. When you heat a pot of water on the stove, heat from the burner flows into the pot and the water. When you place a cup of warm milk in the refrigerator, heat flows from the milk to the cold surroundings.1

Initialize the thermometer by moving the slider to Cold. Set the temperature to Medium. Click bring the right hand in contact with the left hand.

to

1. The temperature of the right hand is less than/ equal to /greater than the temperature of the left hand. Is there a net transfer of heat? Yes/ No 2. The temperature of the left hand increases/decreases/ stays constant . Slide the temperature to Hot.

3. The temperature of the cocoa mug is less than/equal to/ greater than that of the hand. As heat flows from the cocoa mug /hand to the cocoa mug/ hand , the temperature of the hand increases /decreases/stays constant . Slide the temperature to Cold.

4. The temperature of the ice is less than /equal to/greater than the temperature of the hand. As heat flows from the ice/ hand to the ice /hand, the temperature of the hand increases/ decreases /stays constant. 5. Heat always flows from a warmer /colder object to a warmer/ colder object.

1

For this problem, assume the object on the right has a fixed temperature.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

99

Chapter 10: Heat transfer

10.2

Conduction

Conduction is the means by which heat flows between (or within) objects that are in direct contact. Materials that conduct heat well are called heat conductors. Materials that conduct heat poorly are called heat insulators.

Good conductors and insulators

Click

Material

Conductivity ( m∗W◦ C )

Good conductors Diamond Silver Copper

2300 406 385

Good insulators Concrete Air (at 0 ◦ C) Styrofoam

0.8 0.02 0.01

so the hot iron is in contact with the steel rod. Observe how the temperature along the rod changes.

1. The steel rod starts at a uniform (all the same) temperature. The part of the rod in contact with the hot iron increases its temperature most quickly. True /False 2. Heat flows along the rod to the left/ right , from where it is warmer /colder to where it is warmer/ colder . Click

, bring the slider to Clay, and click

. Observe how the temperature along the rod changes.

3. The clay rod starts at a uniform temperature. The part of the rod in contact with the hot iron increases its temperature most quickly. True /False 4. Heat flows along the rod from where it is warmer /colder to where it is warmer/ colder. 5. Compared to the steel rod, the temperature of the clay changes more slowly /quickly . 6. Clay is a better heat conductor/ insulator . Steel is a better heat conductor /insulator. 7. Bonus Question - answer without Interactive Physics. Which material would make a better mug to hold your hot cocoa, so that the cocoa stays warm and your hand doesn’t burn? Clay /Steel Which material would make a better cooking pot, so that heat from the burner easily transfers to whatever is inside the pot? Clay/ Steel

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

100

Chapter 10: Heat transfer

10.3

Convection

Convection is another form of heat transfer. In liquids and gases, convection is what allows heat to be transferred when molecules move. When the movement is caused by a pump or a fan, such as cooling of computer chips by a fan, it is called forced convection. When the movement occurs because of difference in density, such as water being boiled in a pot or air around a fire, it is called natural convection.

Click

to begin the exercise.

1. The flame makes the bottom of the pot hottest near the center /sides. 2. Heat is transferred from the flame to the water at the bottom of the pot.2 The water becomes less /more dense and rises /sinks. 3. The water farthest from the flame is cooler /warmer than the water near the flame, hence the water at the top of the pot is less/ more dense, which causes it to rise/ sink. 4. Convection allows water in different parts of the pot to quickly reach the same temperature even though it is only being heated at the bottom. True /False 5. Research Questions - answer without Interactive Physics. A. Convection occurs in gases as well as liquids. True /False B. Due to convection, air above a candle flame is hotter /colder than air beside it. C. Large scale convection causes our Earth’s winds to blow. True /False

2

Assume the water’s temperature is above 4 ◦ C (39 ◦ F) and hence becomes less dense as it gets hotter.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

101

Chapter 10: Heat transfer

10.4

Radiation

(Windows only)

Radiation is the means by which heat is transferred from one object to another without direct contact or movement of molecules. All objects emit radiation, and warmer objects emit more than cooler ones. When radiation hits an object, it is partly reflected and partly absorbed. The part that is absorbed heats the object.

With the man inside the shelter during the day, click

.

1. If the air temperature inside the shelter is the same as it is outside, the man feels warmer/ cooler inside than he would outside because (circle one) a. the shelter blocks some heat radiated by the sun. b. the shelter blocks some of the heat radiated by the man. Move the man out from the shelter by changing the bottom slider to Outside.

2. When the man is outside and exposed to the sun, more heat radiates from the man/ sun to the man /sun, and the man feels hot /cold. Change the top slider to Night, and keep the man Outside the shelter.

3. At night, when the man is outside the shelter, more heat radiates from the man /night sky to the man/ night sky , and the man feels hot/ cold . Bring the man back inside the shelter by changing the bottom slider to Inside.

4. At night, if the air temperature inside the shelter is the same as it is outside, the man feels warmer inside /outside because a. the shelter blocks some heat radiated by the night sky. b. the shelter blocks some of the heat radiated by the man.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

102

Chapter 10: Heat transfer

5. Bonus Question - answer without Interactive Physics. Your mom is sitting in front of a closed, well insulated, triple-paned window on a cold clear night in a room at 21 ◦ C (70 ◦ F). She complains that she feels cold. You explain that she will feel warmer if she closes the drapes mostly because a. the drapes are good insulators. b. the drapes block some of her radiation.

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

103

Chapter 10: Heat transfer

Chapter 11 Wave and sound A wave is a disturbance that carries energy from one place to another with relatively small disturbance to the medium through which it travels. Waves are characterized by their frequency, wavelength, and amplitude. Water waves and sound waves are two common waves. A pebble thrown into a lake generates a water wave. The wave begins at the impact point where the pebble hits the water and spreads to other parts of the water, in increasingly larger circles centered on the impact point. The speed at which the wave moves outward is called wave speed. It is important to note that even though the wave is spread to a location far away from the impact point, the water molecules are not displaced - they simply oscillate up and down and forward and backward.1 WaveSpeed = WaveFrequency ∗ WaveLength Sound waves are important for human communication.2 Transmission of sound waves requires a medium, such as air, water, or a string. The speed of sound is related to the elasticity and density of the medium. For example, sound travels more quickly in warmer air than in cooler air because air molecules are closer together in warmer air. SpeedOfSound = DistanceTraveled ElapsedTime

1

The oscillation of water waves is eventually damped out by friction and air resistance. Sound waves that can be heard by humans have frequencies between 20 Hz and 20,000 Hz. Ultrasonic waves are above the audible range (above 20,000 Hz) and infrasonic waves are below the audible range (below 20 Hz). 2

104

11.1

Wave motion

The frequency of a wave is the number of cycles per unit time and is usually measured in cycles per second, or hertz (Hz). The period of a wave is the amount of time required per cycle and is equal to 1/frequency. The amplitude of an undamped wave is half the vertical distance from the wave’s crest (its high point) to its trough (its low point).

With Frequency at 2 Hz and Amplitude at 1 m, click

1. A wave at 2 Hz has Click

2 cycles in 1 sec.

, set Frequency to 4 Hz, and click

2. A wave at 4 Hz has

.

.

4 cycles in 1 sec.

3. A higher frequency means more cycles in a given amount of time. True /False Click

, set Frequency to 3 Hz, and click

4. What is the wave’s amplitude?

1

. Observe the amplitude of the wave.

m

5. What is the vertical distance between the crest and the trough?

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

105

2 m

Chapter 11: Wave and sound

11.2

Wave speed

The speed at which a wave travels through a medium such as air or water is called wave speed and is related to the wave’s frequency and wavelength by WaveSpeed = WaveFrequency ∗ WaveLength

With Frequency at 1 Hz, click

.

1. Find the wavelength of a wave traveling at 60 m/s with a frequency of 1 Hz. Result: m m 60 sec 60 sec m = 60 cycle = 1 Hz 1 cycle sec

Click

, set Frequency to 2 Hz, and click

.

2. Find the wavelength of a wave traveling at 60 m/s with a frequency of 2 Hz. Result: m m 60 sec 60 sec m = 30 cycle = 2 Hz 2 cycle sec

Click

, set Frequency to 3 Hz, and click

.

3. Find the wavelength of a wave traveling at 60 m/s with a frequency of 3 Hz. Result: m m 60 sec 60 sec m = 20 cycle = 3 Hz 3 cycle sec 4. At a given wave speed, an increase in frequency causes the wavelength to increase/ decrease. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

106

Chapter 11: Wave and sound

5. Complete the following chart about the various waves.

(A)

(B)

(C)

(D) Wave

Wavelength (m/cycle)

Wave Frequency (Hz or cycle ) sec m 100 sec = 20 cycle m sec 5 cycle

A

5

B

10

C

7

m 280 sec = 40 m 7 cycle

D

8

30

20

cycles sec

Wave Speed (m/sec) 100 20

cycle sec

∗ 10

m cycle

cycle

280

sec

cycles sec

= 200

30

cycle sec

∗ 8

m cycle

= 240

m sec

m sec

m sec

m sec

Courtesy Department of Defense c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

107

Chapter 11: Wave and sound

11.3

Speed of sound

Sound is a wave and travels through a medium. The speed of sound depends on the density and elasticity of the medium, which can be solid, liquid, or gas. Without a medium (i.e., in a vacuum), sound is not possible. In general, sound travels faster in solids than liquids, and faster in liquids than air. The speed of sound is calculated by the equation SpeedOfSound = DistanceTraveled ElapsedTime

Click

. Choose Air, Water, or Iron and click

.

1. Determine the time it takes for the sound of a hammer strike to travel 1000 m in the various substances listed below. Substance

Speed of sound m sec

Time for sound to travel 1000 m 1000 mm = 2.92 sec 343 sec

Air at 20 ◦ C

343

Water at 25 ◦ C

1493

m sec

1000 m m 1493 sec

=

0.67 sec

Iron

5130

m sec

1000 m m 5130 sec

=

0.19 sec

2. Sound travels more quickly in iron than in air. True /False 3. Sound travels more quickly in air than in water. True/ False 4. Sound travels more quickly in water than in iron. True/ False 5. Bonus Question - answer without Interactive Physics. A starting gun is fired to begin a 100 m race. At the finish line, one official started her timer the instant she saw the gun fired whereas a second official started his timer the instant he heard the gun fired. Assume the air temperature is 20 ◦ C (68 ◦ F). (a) What is the difference in the race-time recorded by the two officials? Result: 100 m = 0.29 sec m 343 sec (b) The correct race time is reported by the official who saw /heard the gun fired. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

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Chapter 11: Wave and sound

6. Bonus Question - answer without Interactive Physics.

Determine the time difference reported by the various observers who see the lightning and hear the thunder. Assume the air temperature is 20 ◦ C (68 ◦ F). Observer

Distance from lightning

Time difference 1000 mm = 2.92 sec 343 sec

A

1000 m

B

1609 m (1 mile)

1609 mm 343 sec

=

4.69 sec

C

2000 m

2000 mm 343 sec

=

5.83 sec

D

3000 m

3000 mm 343 sec

=

8.75 sec

7. Bonus Question - answer without Interactive Physics. An observer reports hearing thunder 1 second after seeing the lightning. How far away was the lightning? Assume the air temperature is 20 ◦ C (68 ◦ F). Result: 1 sec ∗ 343

m sec

=

343 m

8. Bonus Question - answer without Interactive Physics. A person sees lightning and hears its thunder 14 seconds later. How far away was the lightning? Assume the air temperature is 20 ◦ C (68 ◦ F). Result: 343

m sec

∗ 14 sec

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

=

109

4802 m

≈

3 miles

Chapter 11: Wave and sound

11.4

Mach number

The Austrian physicist Ernst Mach developed the Mach number (M) to describe high-speed flight. At Mach 1, an aircraft travels at the speed of sound whereas at Mach 2, the aircraft travels at twice the speed of sound. Mach number and the speed of sound depend on air temperature and density, which vary with altitude. At sea level, the speed of sound is 1223 km (340 ms ) whereas 12,000 m above sea h (295 ms ). Since the speed of sound depends on altitude, level, the speed of sound is 1061 km h Mach 1 at sea level is faster than Mach 1 at an altitude of 12,000 m. Mach number is calculated by the equation MachNumber =

Click

FlightSpeed SpeedOfSound

. Select the appropriate Altitude and Speed of the jet, and click

.

1. Determine the Mach number of a jet traveling at the altitudes and speeds listed below. Speed of jet (m/sec) Altitude of jet (m) Mach number 300 12000 1.02

900

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

110

9000

0.99

6000

0.95

3000

0.91

12000

3.05

9000

2.96

6000

2.85

3000

2.74 Chapter 11: Wave and sound

2. Bonus Question - answer without Interactive Physics. A jet travels at Mach 2.4 at an altitude of 12,000 m. What is its speed? Result: 2.4 ∗ 294.7

m sec

=

707.28

m sec

3. Bonus Question - answer without Interactive Physics. The SR-71 Blackbird, one of the fastest and highest-flying aircraft, can fly at a speed m of Mach 3.3 at an altitude of 23,000 m where the speed of sound is 295 sec . How many minutes does the SR-71 take to travel the 4600 km from Los Angeles to New York? Result: 3.3 ∗ 295

m sec

m ∗ 1 km ∗ 60 sec 973.5 sec 1000 m 1 min 4600 km ∗

= =

1 min = 58.41 km

m sec

973.5 58.41 78.8

km min

min

Courtesy NASA Dryden Flight Research Center Photo Collection

c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

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Chapter 11: Wave and sound

11.5

Doppler effect

(Windows only)

As a sound source moves towards an observer (like a fire truck approaching you), the sound frequency heard by the observer increases. As the sound source moves away from an observer, the sound frequency heard by the observer decreases. The change in frequency due to the motion of either the sound source (fire truck) or the observer (you) is an example of the Doppler Effect.

With Fire truck (source) velocity and Skateboarder (observer) velocity at 0.00 m/s, click

.

1. When both the fire truck and the skateboarder are stationary, the siren frequency observed by the skateboarder is the same as the source frequency. True /False Click

, set the Fire truck (source) velocity to -20.00 m/s, and click

.

2. As the fire truck approaches the stationary skateboarder, the siren frequency observed by the skateboarder is lower than/ higher than /equal to the source frequency. As the fire truck moves away from the stationary skateboarder, the siren frequency observed by the skateboarder is lower than /higher than/equal to the source frequency. Click

, set the Skateboarder (observer) velocity to 20.00 m/s, and click

.

3. As the fire truck and the skateboarder approach each other, the siren frequency observed by the skateboarder is lower than/ higher than /equal to the source frequency. As they move away from each other, the siren frequency observed by the skateboarder is lower than /higher than/equal to the source frequency. Click

, set the Fire truck (source) velocity to 0.00 m/s, and click

.

4. As the skateboarder approaches the stationary fire truck, the siren frequency observed by the skateboarder is lower than/ higher than /equal to the source frequency. As the skateboarder moves away from the stationary fire truck, the siren frequency observed by the skateboarder is lower than /higher than/equal to the source frequency. c 2006-2010 DST. Portions Copyright  c 2005 MSC.Software Copyright 

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Chapter 11: Wave and sound

Chapter 12 Intermolecular attractions A molecule is a collection of atoms held together by covalent bonds. Molecules make up the air you breathe, the water you drink, the food you eat, and the homes you live in. A molecule’s behavior is primarily determined by intermolecular attractions, i.e., a molecule’s ability to attract other molecules. Attractions between molecules create sugar crystals, snowflakes, and the double helix of DNA. Intermolecular attractions also explain why salt dissolves in water while oil does not. Most intermolecular attraction results from the polarity of bonds within each molecule. The term covalent bond describes a bond between atoms due to the sharing of electrons. When atoms share electrons equally, the bond is called non-polar. Alternately, when atoms share electrons unequally, the bond is called a polar bond. In a polar bond, one of the bonding atoms is slightly positive as its electrons are pulled away from it, while the other bonding atom is slightly negative as it pulls electrons towards itself. The symbol δ+ describes the charge on an atom that is slightly positive, while the symbol δ− describes the charge on an atom that is slightly negative. An ionic substance or salt is a collection of atoms held together by ionic bonds. An ionic bond occurs when one atom takes one or more electrons from another atom (no sharing). The atom with extra electrons is called a negative ion whereas the atom losing electrons is called a positive ion.

Summary of bonds Type of bond/force and description Non-polar (pure) covalent bond Electrons are equally shared Polar covalent bond Electrons are unequally shared (δ + and δ − ) Ionic bond Electrons are transferred (no sharing) Hydrogen bond Attraction between δ + and δ − atoms Van Der Waals (dispersion) force Attraction between instantaneous dipoles

Between atoms

Strength Strong

atoms

Strong

atoms

Strong

polar molecules

Weak (≈ 50x weaker than covalent bonds)

non-polar molecules

Weaker than a hydrogen bond

113

12.1

Hydrogen bonding between water molecules

Liquid water is the most common polar molecule on Earth. Its polar atoms enable water to attract many other types of molecules. (which is why it is called the ”universal solvent”). Water molecules (H2 O) are made from hydrogen and oxygen atoms. Although both hydrogen and oxygen are gases at room temperature, water is a liquid at room temperature. Water molecules are capable of forming strong intermolecular attractions known as hydrogen bonds. When hydrogen bonds exists H2 O is in a liquid or solid state. These hydrogen bonds are what make water a liquid instead of a gas at room temperature.

Open the Interactive Physics experiment HydrogenBondingBetweenWaterMolecules.ip.

1. An H2 O molecule is shown to the right. Circle the oxygen atom and put a square around each hydrogen atom. Oxygen is bonded to one/ two hydrogens. Hydrogen is bonded to one /two oxygen. With Hydrogen Bonding ON

, click

.

2. The water molecules attract /repel each other. 3. The oxygen atoms in one water molecule attract the oxygen/ hydrogen atoms in a neighboring water molecule. 4. The term “hydrogen bond” describe the attraction between a polarized hydrogen (δ+ ) in one molecule and a negatively polarized atom (δ− ) in a neighboring molecule. In this simulation, which atom is negatively polarized? Oxygen /Hydrogen . 5. Circle the picture that best describes the hydrogen bonding between two water molecules. Note: the three dots (. . . ) symbolize a hydrogen bond.

c 2006 by Steve Sogo Portions Copyright 

(with permission)

114

Chapter 12: Intermolecular attractions

Turn Hydrogen Bonding OFF and click

.

6. With hydrogen bonding OFF, the molecules show no attraction to each other and move off in random directions. True /False . 7. Bonus Question - answer without Interactive Physics. On Earth, H2 O molecules exist in three states: solid (ice), liquid (water), and gas (steam/water vapor). Knowing that increasing temperature breaks hydrogen bonds, guess which states have hydrogen bonds between the H2 O molecules? (circle all that apply). Solid H2 O (ice)

Liquid H2 O (water)

Gas H2 O (steam/water vapor)

Courtesy USAF

Courtesy NOAA

Courtesy USAF

8. Bonus Question - answer without Interactive Physics. Hydrogen molecules (H2 ) and oxygen molecules (O2 ) do not form hydrogen bonds (they do not attract each other and cluster together like water molecules). Why are H2 and O2 incapable of forming hydrogen bonds with each other? (circle all that apply) • Both H2 and O2 are non-polar molecules • Both H2 and O2 lack the δ+ and δ− charges required for forming hydrogen bonds • Gases do not have hydrogen bonds • Both H2 and O2 are unstable

c 2006 by Steve Sogo Portions Copyright 

(with permission)

115

Chapter 12: Intermolecular attractions

12.2

Interaction of water molecules and a sodium ion

An atom with extra electrons is called a negative ion (anion) whereas an atom which lost electrons is called a positive ion (cation). This experiment investigates how water molecules interact with a positively charged sodium ion.

Open the Interactive Physics experiment InteractionOfWaterMoleculesAndASodiumIon.ip

√ 1. Before running the experiment, choose ( ) one of the following hypotheses. • The water molecules will repel the sodium ion • The water molecules will attract the sodium ion • The water molecules will attract each other and leave the sodium ion alone After making a hypothesis, click

and then circle the correct statement above.

2. The oxygen in the water molecule is attracted to the sodium ion. True /False The hydrogen in the water molecule is attracted to the sodium ion. True/ False Circle the best reason for your answer. • The positively charged sodium ion attracts the δ− polarized oxygen atom. • The positively charged sodium ion attracts the δ+ polarized hydrogen atom. • The sodium ion has more charge than water. • Sodium and water are different substances.

c 2006 by Steve Sogo Portions Copyright 

(with permission)

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Chapter 12: Intermolecular attractions

The lower-left water molecule does not attach to the sodium ion. Click left water molecule, and click

, use the slider to rotate the lower-

. If necessary, repeat this process until the lower-left water molecule attaches

to the sodium ion.

3. The lower-left water molecule attaches to the sodium ion when: • The hydrogens are closest to the sodium ion. • The oxygen is closest to the sodium ion. 4. Bonus Question - answer without Interactive Physics. Knowing that solubility depends on intermolecular attraction, circle one of the following: • Sodium ions do not dissolve in water. • Sodium ions dissolve easily in water. 5. Bonus Question - answer without Interactive Physics. (A). Circle the following ions which attract water molecules like a sodium ion. √ (B). Check ( ) the ion that you think attracts water molecules most. •

Potassium ion (K + )

•

Calcium ion (Ca2+ )

•

Aluminum ion (Al3+ )

√

Courtesy of Dr. Alain Laderach

Courtesy of Dr. Chris Bruns

NIH Center for Biomedical Computation at Stanford

NIH Center for Biomedical Computation at Stanford

c 2006 by Steve Sogo Portions Copyright 

(with permission)

117

Chapter 12: Intermolecular attractions

12.3

Interaction of water molecules and a chloride ion

An atom with extra electrons is called a negative ion (anion) whereas an atom which lost electrons is called a positive ion (cation). This experiment investigates how water molecules interact with a negatively charged chloride ion.

Open the Interactive Physics experiment InteractionOfWaterMoleculesAndAChlorideIon.ip

√ 1. Before running the experiment, choose ( ) one of the following hypotheses. • The water molecules repel the chloride ion • The water molecules attract the chloride ion • The water molecules attract each other and leave the chloride ion alone After making a hypothesis, click

and then circle the correct statement above.

2. The oxygen in the water molecule is attracted to the chloride ion. True /False The hydrogen in the water molecule is attracted to the chloride ion. True/ False Circle the best reason for your answer. • The negatively charged chloride ion attracts the δ− polarized oxygen atom. • The negatively charged chloride ion attracts the δ+ polarized hydrogen atom. • The chloride ion has more charge than water. • Chloride and water are different substances.

c 2006 by Steve Sogo Portions Copyright 

(with permission)

118

Chapter 12: Intermolecular attractions

The upper-middle water molecule does not attach to the chloride ion. Click upper-middle water molecule, and click

, use the slider to rotate the

. If necessary, repeat this process until the upper-middle water

molecule attaches to the chloride ion.

3. The upper-middle water molecule attaches to the chloride ion when: • The hydrogens are closest to the chloride ion. • The oxygen is closest to the chloride ion.

4. Bonus Question - answer without Interactive Physics. Knowing that solubility depends on intermolecular attraction, circle one of the following: • Chloride ions do not dissolve in water. • Chloride ions dissolve easily in water.

5. Bonus Question - answer without Interactive Physics. Knowing that noble gases such as helium, neon, and argon are reluctant to gain or lose electrons, explain why noble gases do not dissolve well in water? (circle all that apply). • Noble gas atoms have positive charges • Noble gas atoms have negative charges • Noble gas atoms have no charge • Noble gas atoms do not attract water molecules

6. Bonus Question - answer without Interactive Physics. Aluminum atoms exist in two forms, namely neutral (uncharged) aluminum (Al) atoms which forms pure, elemental aluminum or positively charged Al3+ which forms compounds with negatively charged ions such as chloride (Cl− ). Based on how water interacts with positive and negative ions, guess which of the following would be soluble in water: • Neutral Al atoms • Positively charged Al3+ ions • Aluminum chloride AlCl3 • Pure aluminum metal, e.g., aluminum foil

c 2006 by Steve Sogo Portions Copyright 

(with permission)

119

Chapter 12: Intermolecular attractions

12.4

Interaction of water with non-polar and polar molecules

Experiments 12.1, 12.2, and 12.3 show that water molecules are attracted to the positive and negative ions inherent to ionic substances (salts). There are many substances in the world that contain molecules rather than charged ions. The interaction of water with these molecular substances is greatly influenced by the polarity of the molecular substance. This experiment investigates how water molecules interact with non-polar molecules and polar molecules.

Open the Interactive Physics experiment InteractionOfWaterMoleculesWithNonPolarAndPolarMolecules.ip

√ 1. Before running the experiment choose ( ) one or more of the following hypotheses. • The non-polar molecule will repel water • The non-polar molecule will attract water • The non-polar molecule will not attract or repel water • The water molecules will attract each other After making a hypothesis about non-polar molecules, click

and then circle the correct statement above.

2. Based on the behavior seen in this experiment, circle one of the following: • The non-polar molecule dissolves in water • The non-polar molecule will not dissolve in water 3. Circle the bonds that were formed during this simulation. • Covalent bonds • Hydrogen bonds • Ionic bonds

c 2006 by Steve Sogo Portions Copyright 

(with permission)

120

Chapter 12: Intermolecular attractions

Set the slider to Polar Molecule and click

.

4. Circle the best description of the interaction of water molecules with a polar molecule. • The polar molecule repels the water molecules • The polar molecule attracts the water molecules • The water molecules attract each other and leave the polar molecule alone 5. In observing the interaction of the polar molecule, it seems that the pink atom is δ+ /δ− (circle one) and the green atom is δ+ / δ− (circle one). 6. The behavior of this simulation suggests that the polar molecule is soluble /insoluble in water. (Soluble means it dissolves, insoluble means it does not dissolve.) 7. Based on your observation of these simulations, it seems that • Polar molecules can form hydrogen bonds. True /False • Non-polar molecules can form hydrogen bonds. True/ False 8. Bonus Question - answer without Interactive Physics. The word ”hydrophilic” comes from the Greek words hydro (water) and philic (love). Hence, a hydrophilic molecules ”loves water” (is attracted to water). Circle the hydrophilic molecules in the following list. • Alcohol (a polar molecule) • Sugar (a polar molecule) • Oil (a non-polar molecule) • Propane (a non-polar molecule) • Ammonia (a polar molecule) 9. Bonus Question - answer without Interactive Physics. Circle the word that describe molecules that are ”afraid of water”. (Hint: Use a dictionary, encyclopedia, or the web.) • Hydrometer • Hydraulic • Hydrophobic • Hydrolysis 10. Bonus Question - answer without Interactive Physics. Molecules that readily dissolve in water are hydrophilic /hydrophobic .

c 2006 by Steve Sogo Portions Copyright 

(with permission)

121

Chapter 12: Intermolecular attractions

12.5

Investigating the strength of hydrogen bonds

As summarized in the table at the beginning of this chapter, there are many types of bonds between atoms and ions. By using a “bowling ball” to test the strength of various bonds, this simulation helps you compare the strength of hydrogen bonds relative to covalent bonds. In general, strong bonds are called stable (the bond is not easily broken). This simulation demonstrates how the stability of a hydrogen bond depends on its surrounding temperature.

Open the Interactive Physics experiment InvestigatingStrengthOfHydrogenBonds.ip and click

.

1. Circle the best description of the changes in bonding that occur when the bowling ball strikes the water molecules. • Both hydrogen bonds and covalent bonds remain intact • Hydrogen bonds are broken, while covalent bonds remain intact • Covalent bonds are broken, while hydrogen bonds remain intact • Both hydrogen bonds and covalent bonds are broken . Set the bowling ball speed slider to Low and click

Click Then, click

, set the bowling ball speed slider to High and click

. .

2. Circle the statements that are true at low temperature (represented by a slow bowling ball). • Hydrogen bonds are more likely to hold together (hydrogen bonds are more stable) • Hydrogen bonds are less likely to hold together (hydrogen bonds are less stable) • Temperature has no effect on the stability of hydrogen bonds

c 2006 by Steve Sogo Portions Copyright 

(with permission)

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Chapter 12: Intermolecular attractions

3. Bonus Question - answer without Interactive Physics. On Earth, H2 O molecules exist in three states: solid (ice), liquid (water), and gas (steam/water vapor). Circle the state of H2 O with the most stable hydrogen bonds. Solid H2 O (ice)

Liquid H2 O (water)

Gas H2 O (steam/water vapor)

Courtesy USAF

Courtesy NOAA

Courtesy USAF

4. Bonus Question - answer without Interactive Physics. Circle the types of bonds you must break to swim through water. • Hydrogen bonds • Covalent bonds • Both hydrogen bonds and covalent bonds

c 2006 by Steve Sogo Portions Copyright 

(with permission)

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Chapter 12: Intermolecular attractions

Chapter 13 Discovery Learning In the previous chapters, you used Interactive Physics for inquiry learning. You answered questions using formulae, calculations, and structured Interactive Physics experiments. Although inquiry learning is excellent for a conceptual understanding of physics, scientific progress is made through new discoveries. The most important feature in Interactive Physics is discovery learning, where you create your own question and build experiments to answer them. With Interactive Physics, the possibilities in building cars, airplanes, robots, pinball machines, humans, animals, etc., are limited only by your imagination. This chapter shows you how to build an experiment. When you have finished, you can start the exciting exploration of physics with the same tools used by professional scientists! Who knew learning physics could be so much fun!

Step

Related Physics Concepts

Creating a falling block

Mass; freely falling objects; laws of motion; linear kinematics

Adding a velocity vector

Vector and scalar quantities; vector components; unit vector

Making a pendulum

Oscillatory motion; frequency and amplitude; rotational kinematics

Graphing pendulum motion

Graphs and measurements; motion diagrams

Changing gravity

Law of gravity; Newtons second law

Adding air resistance

Air resistance; non-conservative forces

Adding a spring

Oscillations; conservative forces; conservation of kinetic/potential energy

Controlling spring constant

Spring constant; natural spring length; equilibrium spring length

Collisions with a circle

Collision; elasticity; frictional forces; impulse and momentum

Attaching pictures to objects

Attaching pictures makes physics experiments realistic and fun

Adding sound

Sound waves; speed of sound; Doppler effect; sound frequency and intensity

Adding a curved slot joint

Roller coaster physics; motion in two dimensions; conservation of energy and momentum

Adding a force

Concept of force; Newtons first law; work and energy

Running demo files

Interactive Physics allows you to explore other essential physics topics, including: electrostatics, evaporation and condensation, gears, kinetic theory of gas, machines, magnetism, particle dynamics, projectiles and rockets, pulleys, rotational dynamics, static equilibrium, superposition of waves, and many more.

124

Starting Interactive Physics

4 Graphing the Pendulum’s Motion

1. Ensure that Interactive Physics is installed on your computer.

1. To graph the pendulum’s motion, click on the rectangle. Under the Measure menu, select Position, then select Rotation Graph. . Note: Data can be displayed as a graph, bar chart, 2. To collect data, click or number, and can be changed while running the experiment. Click . 3. The graph shows the pendulum’s amplitude and frequency. To make the graph larger, click on the graph and drag its lower right-hand corner to the right.

menu, click on Programs and then Interactive Physics 2. From the Windows and then Interactive Physics. This opens a new experiment.

1 Creating a Falling Block 1. The first simulation is Newton’s first experiment, dropping a block. 2. To draw a rectangle, click on the Rectangle tool, then click in the workspace and draw a long thin rectangular block. 3. To run the experiment and see the block fall due to gravity, click . 4. Click to reset the experiment.

5 Changing Gravity 1. To change gravity, click on the World menu, select Gravity, slide the slider to the top for the value 20 m/sec2, and click [OK]. 2. Click and observe that, in agreement with theoretical and experimental . predictions, the pendulum has a higher natural frequency. Click

2 Adding a Velocity Vector 1. To add a velocity vector, click on the rectangle. 2. From the Define menu, click on Vectors and then Velocity. 3. Click and observe that the vector changes magnitude as the block falls. . 4. Click Optional: To add a numerical value to the velocity vector (or its components), click on the Define menu, select Vector Display, and check the Value box.

3 Making a Pendulum 1. To make a pendulum, click on the Pin joint tool and then click on the upper lefthand corner of the rectangle. 2. Click and observe that the vector changes magnitude and direction as the pendulum moves. Click .

6 Adding Air Resistance 1. Under the World menu, select Air Resistance, click on Low Speed, and accept the default air resistance value of 0.3 kg/(m*s) by clicking [OK]. 2. Click and observe the exponentially decaying oscillations. Click .

125

7 Adding a Spring

9 Collisions with a Circle

1. To add a spring, click on the Spring tool. Click on the upper right-hand corner of the block and stretch the spring up and to the right. and observe the pendulum's higher natural frequency and new 2. Click equilibrium position. Click .

1. To create a circle, click on the Circle tool, then click in the workspace and draw a to zoom to extents.) circle. (If your rectangle is high on the screen, click to start the experiment and observe that the circle bounces and rolls 2. Click on top of the rectangle. Automatic collision and contact is a very useful feature in Interactive Physics (even the elastic and frictional properties of objects may be . varied). Click

8 Controlling the Spring Constant 1. To control the spring constant, click on the spring. Under the Define menu, select New Control, then select Spring Constant. 2. The slider that controls the spring will appear in the left-hand side of the workspace. To move the slider's location closer to the spring, click on its title and drag it next to the spring. 3. To see the effect of varying the spring constant, click and observe that the pendulum angle is a function of the spring-constant (move the slider up and down . while the experiment is running). Click

10 Attaching a Picture to an Object 1. To find the spaceman picture on Windows, select menu, then Programs, then Interactive Physics, then Interactive Physics Files, then the IPIntroduction folder. 2. Double-click on the bitmap file “Spaceman.bmp.” This should open the bitmap in a program such as Paint. 3. In Paint, select the Edit menu and then choose Select All. Next, select the Edit menu again and then choose Copy. 4. Go back to Interactive Physics and select its Edit menu and then choose Paste. 5. To attach the spaceman bitmap to the circle, click on the spaceman, then hold down [Shift] while you click on the circle. 6. Select the Object menu and then Attach Picture. Notice that the circle object has disappeared and has been replaced by the spaceman image. 7. Click to run the experiment. Click . Note: Interactive Physics was designed to be easy-to-use. In this exercise, the only time you need to touch the keyboard is to hold down the [Shift] key.

126

11 Adding Sound

14 Running Demo Files

1. Click on the spaceman, select the Measure menu and choose Hear the Collision. 2. Click to start the experiment and hear the sound when the spaceman . contacts the block. Click

Windows users: 1. Under the Script menu, click on "Run All Demo Files." 2. Sit back and enjoy a series of demos on a variety of physics topics. 3. To quit, select the File menu and choose Exit. Mac Users: 1. Browse to the For Demo Users folder installed with Interactive Physics. 2. Double-click on each of the files, then click Run. 3. To quit, select the File menu and choose Quit.

12 Adding a Curved Slot Joint 1. To add a Curved Slot Joint, click on the Curved Slot joint tool. 2. Click on the spaceman and then click on a couple of other places to the right of the spaceman, and then double-click to complete the slot (see figure below). to start the experiment and observe that the spaceman slides down 3. Click the curved slot. Click .

15 Curriculum Workbook Supplementary workbooks with Interactive Physics exercises of varying difficulty are available with purchase. To try the instructional curriculum: Windows users: Go to the Windows , then Programs, then Interactive Physics, then click on StartCurriculum.html, then choose Demo Users. Note: The Demo Edition can open only Demo files. The Full Edition must be purchased to open the curriculum and additional 150+ physics experiments.

13 Adding a Force 1. To add thrust to the spaceman to overcome air resistance, click on the Force tool, then click on the spaceman, then move the mouse to the left and click again. 2. Click to start the experiment and observe that the spaceman overcomes . air resistance and moves more quickly along the curved slot. Click

Interactive experiments explore concepts in Energy, Temperature, Heat Transfer, Waves, and Sound Plus experiments in speed, distance, time, acceleration, force, weight, mass, gravity, and air resistance New curriculum workbook correlated with National and State Educational Standards and Objectives Full-color teacher edition and black-line master student edition

http://www.interactivephysics.com

127