Physics%–%1st%Quarter% - Lorain City School District · PDF file11 L12text’ ......

Lorain City School District Scope, Sequence and Pacing Guides

Physics Grade 12

1 Common Formative Assessments will be implemented daily.

Physics – 1st Quarter

Strand Topic Content Statement Days Clear Learning Target Vocabulary Core

Resource Additional Resources Assessment

Scientific Inquiry “Motion”

Chapter 1: A Physics Toolkit 1. Mathematics and Physics

2. Measurement 3. Graphing Data

Chapter 2: Representing Motion.

1. Picturing Motion

2. Where & When?

3. Position-‐Time Graphs?

4. How fast?

Chapter 3: Accelerated Motion

1. Acceleration 2. Motion with Constant

Acceleration 3. Free Fall

RST.11-‐12.2. Determine the central ideas or

conclusions of a text; summarizes a

complex concepts, possesses, or information

presented in a text by paraphrasing them in simpler but still accurate terms.

RST.11-‐12.3.

Follow precisely a complex multi-‐step procedure when carrying out

experiments, taking measurements, or attempting technical tasks; analyze the

specific results based on explanations in

the text.

RST.11-‐12.9. Synthesize

information from a range of sources (e.g.,

55 I can demonstrate scientific methods understanding through various scientific phenomenon.

I can evaluate answers using dimensional analysis; especially, when dealing with graphs.

I can perform arithmetic operations using scientific notation.

I can apply data-‐collected using techniques such as CBL.

Chapter 1: Physics Dimensional analysis Significant digits Scientific method Hypothesis Scientific law Scientific theory Measurement Precision Accuracy Independent variable Dependent variable Line of best fit Linear relationship Quadratic relationship Inverse relationship

Labs #1, 2 and 3 on p.p. (20-‐21), (48-‐49), and visit

Physics Online for further accelerate Motion.

Error!

Hyperlink reference not valid. Is a series of videos-‐on-‐demand produced

by Annenberg that show classroom strategies

for implementing inquiry into the

high school

This simple animation above depicts some

additional information about the car's motion.

The velocity and acceleration of the car are depicted by vector arrows. The direction of these arrows is

representative of the direction of the velocity and

acceleration vectors. Note that the velocity vector is always

directed in the same direction which the car

is moving. A car moving eastward

would be described as having an eastward velocity. And a car moving westward

would be described as

Standardized Test Practices (1-‐>7) p.p. 29, 55 and 85 Exam 1, 2, and 3


Physics Grade 12


texts, experiments, simulations) into

coherent understanding of a

process, phenomenon, or concept, resolving

conflicting information when

possible.

RST.11-‐12.10. By the end of grade

12, read and comprehends

science/technical texts in the grades

11-‐12 text Complexity band independently and

proficiently.

WHST.11-‐12.2. Write

informative/explanatory texts, including the narration of historical events,

scientific procedures/experim

ent

I am able to use graphing techniques to plot data, and graph the relationship between independent vs. dependent variables as well as linear vs. nonlinear relationships. I can distinguish between accuracy and precision through different examples proposed by teacher. I can determine the precision of measured quantities.

I can interpret graphs, and recognize common relationships in graphs.

Chapter2: Motion diagram Practical Model Coordinate System Origin Position Distance Magnitude Vectors Scalars Resultant Time interval Displacement Position-‐ time graph Instantan-‐eous position Average velocity Average Speed Instantan-‐eous velocity Chapter 3: Velocity-‐ time graph Acceleration Average

classroom? While not all of the content is aligned to physical

science, the strategies can be

applied to any

content.

www.Physics

Online.com

having a westward velocity.

For more information on physical

descriptions of motion, visit The Physics

Classroom. Detailed information is

available there on the following topics:

Vectors and Scalars

Speed and Velocity

Acceleration

Describing Motion with Vector Diagrams

Motion Along Inclined Planes

Other animations can be seen at the

Multimedia Physics Studios.

Use the “Interactive Chalkboard” using the animations, video, and

power point presentation to bring


Physics Grade 12


acceleration Instantan-‐ eous acceleration Free fall Acceleration due to gravity

gravitational theories and laws picture’s closer to students. Spend at most 20

minutes. Use School Adaptation, video tape to illustrate

“Kepler’s Laws, Navigating in Space, and The Apple and The

Moon.”

Movie segments can also be selected from: “Annehttp://www.learner.org/resources/series42.html?pop=yes&pi

d=582”

https://ieeetv.ieee.org/channels/pels?utm_source=google&utm_medium=cpc&utm_content=Electromagnetism&utm_campaign=IEEEtv+-‐

+Channels+-‐+Power+Electronics


Physics Grade 12


Content Elaborations: Motion In physical science, the concepts of position, displacement, velocity and acceleration were introduced and straight-‐line motion involving either uniform velocity or uniform acceleration was investigated and represented in position vs. time graphs, velocity vs. time graphs, motion diagrams and data tables. Forces, Momentum and Motion: In earlier grades, Newton’s laws of motion were introduced; gravitational forces and fields were described conceptually; the gravitational force (weight) acting on objects near Earth’s surface was calculated; friction forces and drag were addressed conceptually and quantified from force diagrams; and forces required for circular motion were introduced conceptually. In this course, Newton’s laws of motion are applied to mathematically describe and predict the effects of forces on more complex systems of objects and to analyze objects in free fall that experience significant air resistance.

Strand Topic Content Statement Days Clear

Learning Target

Vocabulary Core Resource

Additional Resources Assessment

Scientific Inquiry and Physical Sciences

And Forces, Momentum and Motion

Chapter 4: Forces in One Dimension 1. Forces and

Motion 2. Using Newton’s Laws

3. Interaction Forces

Chapter 5: Forces in Two

Dimensions 1. Vectors 2. Friction 3. Forces and Motion in Two

Dimensions

RST.11-‐12.5. Analyze how the text

structures information or ideas into categories or

hierarchies, demonstrating understanding of the information or ideas.

RST.11-‐12.7.

Integrate and evaluate multiple sources of

information presented in diverse formats and media (e.g., quantitative data, video, multimedia) in order to

address a question or solve a problem.

RST.11-‐12.7.

Integrate and evaluate multiple sources of

information presented in

55 I can restate and Explain the movement of object by applying Newton’s Three Laws of motion. I can apply principles of forces and motion to mathematically analyze, describe and predict the net forces on objects or systems. I can relate

Chapter 4: Force Free-‐body diagram Net force Newton’s Second Law Newton’s First Law Inertia Equilibrium Apparent weight Weightlessness Drag force Terminal velocity Interaction Pair Newton’s Third Law Tension Normal force

1. “How it Works-‐bathroom scale” p.p. 110 2. Lab

Experiment # 4:

“Forces in elevators” (Text Book p.p. 108-‐109)

Problem Solving

Many problems can be

solved from interpreting graphs and charts as detailed in the motion graphs

section. In addition, when

acceleration is constant, average

velocity can be

calculated

Review packets (chapters 1, 2, and 3) Open Book Quizzes (1, 2, and 3) Exams 1, 2, and 3


Physics Grade 12


Chapter 6: Motion in Two

Dimensions

1.Projectile Motion

2. Circular Motion

3. Relative Velocity

diverse formats and media (e.g., quantitative data, video, multimedia) in order to

address a question or solve a problem.

and explain how societal issues and considerations affect the progress of science and technology.

Chapter 5: Components Vector resolution Kinetic friction Static friction Coefficient of kinetic friction Coefficient of static friction Equilibrant Chapter 6: Projectile Trajectory Uniform circular motion Centripetal acceleration Centripetal force

by taking the

average of the initial and final

instantaneous velocities (??𝒂𝒗𝒈 = (𝒗𝒇 + 𝒗𝒊 )/2). This relationship does not hold true when the acceleration changes. The

equation can be used

in conjunction with other Kinematics equations to solve

increasingly complex problems, including those

involving free fall with

negligible air


Physics Grade 12


resistance in which objects fall

with uniform

acceleration. Near the surface of Earth, in the absence of other

forces, the acceleration of freely falling

objects is 9.81 m/s2. Assessments of motion problems, including projectile motion, will not include problems that require

the quadratic equation to solve.

Projectile Motion When an object has


Physics Grade 12


both horizontal and vertical components of motion, as in a

projectile, the

components act

independently of each other. For a projectile in the absence

of air resistance, this means

that horizontally

, the projectile will

continue to travel at constant speed just like it would if there were no vertical motion. Likewise, vertically the object


Physics Grade 12


will accelerate just as it would

without any horizontal motion. Problem solving will be limited to solving for the

range, time, initial height, initial

velocity or final

velocity of horizontally launched projectiles with

negligible air

resistance. While it is

not inappropria

te to explore more

complex projectile problems, it


Physics Grade 12


must not be done at the expense of other parts of the

curriculum.

Error! Hyperlink reference not valid. is a series of videos-‐on-‐demand produced

by Annenberg that show classroom strategies

for implementing inquiry into the

high school classroom. While not all of the content is aligned to physical

science, the strategies can be

applied to


Physics Grade 12


any content. “Annehttp://www.learner.org/resources/series42.html?pop=yes&pid=

582”

https://ieeetv.ieee.org/channels/pels?utm_source=google&utm_medium=cpc&utm_content=Electromagnetism&utm_campaign=IEEEtv+-‐+Channels+

-‐+Power+Electronics

Content Elaborations: Forces, Momentum and Motion: In earlier grades, Newton’s laws of motion were introduced; gravitational forces and fields were described conceptually; the gravitational force (weight) acting on objects near Earth’s surface was calculated; friction forces and drag were addressed conceptually and quantified from force diagrams; and forces required for circular motion were introduced conceptually. In this course, Newton’s laws of motion are applied to mathematically describe and predict the effects of forces on more complex systems of objects and to analyze objects in free fall that experience significant air resistance. Gravitational forces are studied as a universal phenomenon and gravitational field strength is quantified. Elastic forces and a more detailed look at friction are included. At the atomic level, “contact” forces are actually due to the forces between the charged particles of the objects that appear to be touching. These electric forces are responsible for friction forces, normal forces and other “contact” forces. Air resistance and drag are explained using the particle nature


Physics Grade 12


of matter. Projectile motion is introduced and circular motion is quantified. The vector properties of momentum and impulse are introduced and used to analyze elastic and inelastic collisions between objects. Analysis of experimental data collected in laboratory investigations must be used to study forces and momentum. This can include the use of force probes and computer software to collect and analyze data.

Strand Topic Content Statement Days Clear Learning Target Vocabulary Core

Resource Additional Resources Assessment

Motion Chapter 7: Gravitation

1.Planetary Motion and Gravitation

2.Using Law of Universal Gravitation

Chapter 8: Rotational Motion

1.Describing Rotational Motion

2. Rotational Dynamics

3.Equilibrium

WHST.11-‐12.1. Write arrangements focused on Discipline-‐specific content.

WHST.11-‐12.2.

Write informative/explanatory texts, including the narration of historical events, scientific

procedures/experiments, or technical processes.

25 I can participate in and apply the processes of scientific investigation to create models and to design, conduct, evaluate and communicate the results of these investigations. I can describe displacement. I can calculate angular velocity and angular acceleration. I can solve problems involving rotational motion Chapter 8: 1. I can distinguish between Angular velocity of an object and Angular

Chapter 7: Kepler’s First Law Kepler’s Second Law Kepler’s Third Law Gravitational force Law of Universal gravitation Gravitational field Inertial mass Gravitational mass Chapter 8: Radian Angular Displacement Angular velocity Angular acceleration Lever arm Torque Momentum of inertia

Experiment, Lab #7. P.p. 186-‐187

of “Physics , Principles

and Problems

” *Answer the

questions proposed in the

“Analysis, Conclude and Apply and Going Further” on

p.p.187

Error! Hyperlink reference not valid. is a series of videos-‐on-‐demand

produced by Annenberg that show classroom strategies for implementing inquiry into the high school classroom.

While not all of the content is aligned to

physical science, the strategies can be applied to any content.

Use the

“Interactive Chalkboard” using the animations, video, and power point

Review packets (chapters 7 +8) Open Book Quizzes (7+8) Exams (7+8)


Physics Grade 12


acceleration of an object as the mathematical equation on p.p. (198-‐199). 2. I can draw Data Table 8-‐1 on p.p. 199 which supports my understanding to connecting “linear vs. angular; and the relationships between displacement, velocity, and acceleration.”

Newton’s Second Law for rotational motion Center of mass Centrifugal “Force” Coriolis “Force”.

presentation to bring

gravitational theories and laws picture’s closer to

students. Spend at most 20 minutes. Use School Adaptation, video tape to illustrate

“Kepler’s Laws, Navigating in Space, and The Apple and The

Moon.” www.discoveryschools.com

Plato student

Interactive

program,

[email protected]

Error! Hyperlink reference not

valid.

Movie segments can also from:


Physics Grade 12


“Annehttp://www.learner.org/resources/series42.html?pop=yes&pid=582”

https://ieeetv.ieee.org/channels/pels?utm_source=google&utm_medium=cpc&utm_content=Electromagnetism&utm_campaign=IE

EEtv+-‐+Channels+-‐

+Power+Electronics

Content Elaborations: Gravitational Potential Energy: When two attracting masses interact, the kinetic energies of both objects change but neither is acting as the energy source or the receiver. Instead, the energy is transferred into or out of the gravitational field around the system as gravitational potential energy. A single mass does not have gravitational potential energy. Only the system of attracting masses can have gravitational potential energy. When two masses are moved farther apart, energy is transferred into the field as gravitational potential energy. When two masses are moved closer together, gravitational potential energy is transferred out of the field.

Additional Resources or “Examples:”

• the inner lane of a track has a radius of 20 meter and the outer lane has a radius of 30 meter. Compare the centripetal acceleration for two runners in these two lanes that both have constant speeds of 10 m/s. Is the centripetal acceleration less than g (if not, there probably isn't enough static friction to keep the runners on a circular path and their feet will start to slip). If the runners are traveling counterclockwise, what is the direction of the acceleration? Centripetal acceleration of each is:


Physics Grade 12


r=20m: ac = Vc2/r = (10 m/s)2/(20 m) = 5 m/s2 r=30m: ac = Vc2/r = (10 m/s)2/(30 m) = 3.3 m/s2

The acceleration on the inside lane is significantly bigger than for the outside, but both are well under g in magnitude. The acceleration points to the center of the curve, which will be to each runner's left.

• A golf club has a speed just before striking the ball of 40 m/s, if the club head is 1.5 meter from the axis of rotation, what is the magnitude and direction centripetal acceleration? If the club has a mass of 0.2 kg (weight is 7 oz.), what net force is needed to produce this acceleration? centripetal acceleration: just use the formula:

ac = Vc2/r = (40 m/s)2/(1.5 m) = 1067 m/2

The direction is toward the center of rotation (between the golfer's shoulders).

Force: The net force is always given by Newton's Second Law:

F = m*a = (0.20 kg)*(1067 m/2) = 213 N (about 45 lb)

Note that this force (plus the weight of the club) is provided by the golfer's hands on the club (static friction). Since the maximum value of S is related to the normal force strength, the golfer must hold onto the club grip tightly enough [for most people this is no problem with two hands].

• The most extreme example of centripetal acceleration in sport: A hammer (a 16 lb weight, 7.27 kg) on a 1.2 meter wire with a handle) can be thrown 85 meter. To achieve that distance at the ideal launch angle, the launch speed must be at least

R = Vc2/g Vc2 = (D*g) = 840 (m/s)2

Vc =29 m/s

The radius at release is about 1.0 meter (because the thrower is leaning back). What was the centripetal acceleration just at release?


Physics Grade 12


a = Vc2/r = 840 (m/s)2/(1.0 m) = 840 m/s2 = 85* g!

The force at release is tremendous:

F = m*a = (7.27 kg)*(840 m/s2) = 6107 N (= 1360 lb!)

The thrower must exert this force along the cable (which the cable exerts back on the thrower). There is no way the thrower could resist this force using only friction (so the thrower cannot be stationary): the thrower at the other end of the wire is also rotating and the reaction force of the hammer pulling the wire provides the necessary force for the thrower's centripetal acceleration.

End Early 1st Quarter (9 Weeks) District

Short Cycle Assessment

Strand Topic Content Statement Days Clear

Learning Target

Vocabulary Core Resource

Additional Resources Assessment

In regard to launching your own rocket, you may assign points as you wish, as long as you write “a rubric” that would be worth 150 points. Also, make sure you assign a day to launch that weather is a factor.

Content Elaborations: 1. Instructional Strategies and Resources

This section provides additional support and information for educators. These are strategies for actively engaging students with the topic and for providing hands-‐on, minds-‐on observation and exploration of the topic, including authentic data resources for scientific inquiry, experimentation and problem-‐based tasks that incorporate technology and technological and engineering design. Resources selected are printed or Web-‐based materials that directly relate to the particular Content Statement. It is not intended to be a prescriptive list of lessons. • “Collision Lab”: is an interactive simulation that allows students to Investigate collisions on an air hockey table. Students can vary the number of discs, masses, elasticity and initial conditions to see if momentum and kinetic energy are conserved. • “Forces and Motion” : is an interactive simulation that allows students to explore the forces present when a filing cabinet is pushed. Students can create


Physics Grade 12


an applied force and see the resulting friction force and total force acting on the cabinet. Graphs show forces vs. time, position vs. time, velocity vs. time, and acceleration vs. time. A force diagram of all the forces (including gravitational and normal forces) is shown.

2. http://sciencefair.math.iit.edu/projects/pisa/ 3. https://ieeetv.ieee.org/channels/pels?utm_source=google&utm_medium=cpc&utm_content=Electromagnetism&utm_campaign=IEEEtv+-‐

+Channels+-‐+Power+Electronics Have you ever wanted to launch your own rocket?” This project is worth 150 points. The materials needed are easy to obtain. The procedure is also easily manipulated and simple to follow. The math is calculus based, and it is fun. Follow the URL address as the following: http://www.scienceinschool.org/repository/docs/issue22_rockets.pdf

Strand Topic Content Statement Days Clear Learning Target Vocabulary Core Resource Additional

Resources Assessment

Forces, Momentum and Energy

Chapters 9: Momentum and Its

Conservation,

1. Impulse and Momentum

2. Conservation of Momentum

Chapter 10: Energy, Work, and Simple Machines

1. Energy and Work

WHST.11-‐12.1. Write arrangements focused on Discipline-‐specific content.

WHST.11-‐12.2.

Write informative/explanatory texts, including the narration of historical events, scientific

procedures/experiments, or technical processes.

WHST.11-‐12.2. Write informative/explanatory texts, including the narration of historical events, scientific procedures/experiments, or technical processes. (Note: Teachers should be

43 Chapter 9: Momentum and

Its Conservation.

1. I can determine the momentum of different processes as illustarted in the examles below:

a. 60-‐kg halfback moving eastward at 9 m/s. ANS: p=m*v= 540 Kg*m/s, East

b. 1000-‐kg car

Chapter 9: Impulse Momentum Impulse-‐Momentum theorem Angular momentum Angular impulse-‐ angular momentum theorem Closed system Isolated system Law of conservation of momentum Law of

Students need to perform “PHYSICS LAB.” As

suggested on p.p. 302-‐303.

Use this lab as an alternative “Inquiry Lab.” As suggested on p.p. 303 in textbook.

Students must know more about energy by referring to the internet, and log on to “physics Online-‐through

Physicspp.com

www.discoveryschools.co

m

www.physicspp.com/vocabulary_puzzle

maker http://wufs.wustl.edu/lap

is2/

Plato student interactive program, “Work, Power,

Energy and its

conservation.”

Use the conclusion questions from any minilab and Labs as a review session to figure the students’ understandings as well as achievements Re-‐teach by giving additional problems as suggested in the back of each chapter (9-‐11). When the students enter


Physics Grade 12


2. Machines

Chapter 11: Energy and Its Conservation

1. The Many Forms of energy.

2. Conservation of Energy

aware that the common reaction classifications

that are often used in high school physics courses

often lead to misconceptions because they are not based on the actual physics, but on

surface features that may happen when two

substances are mixed may be similar from one

system to another (e.g., exchanging partners), even though the

underlying physics is not the same. However, they may be useful in making predictions about what )

moving northward at 20 m/s. ANS: p=m*v= 20,000 Kg*m/s

c. 40-‐kg freshman moving southward at 2 m/s. ANS: p= m*v= 80 Kg*m/s

2. I can recognize the appropriateness and value of basic questions “What had happened?” “What are the odds?” and “How do scientists and engineers know what will happen?”

3. I can explain how an object’s kinetic energy depends on its mass and its

conservation of angular momentum Chapter 10: Work Energy Kinetic energy Work-‐energy theorem Joule Power Watt Machine Effort force Resistance force Mechanical advantage Ideal mechanical advantage Efficiency Compound machine Chapter 11: Rotational kinetic energy

” Also, Energy flow diagrams picture energy transformatio

ns in an accurate,

quantitative and

conceptually transparent manner.

Energy Flow Diagrams for Teaching Physics

Concepts is a paper that was published in The Physics Teacher that outlines how to use these tools in the classroom. It proceeds from

simple processes to complex socially significant processes

such as global warming.

“The Mechanical Universe, high school adaptation, video tape, “Conservation of Energy”

Distant Learning

Conferences.

More movie segments can be observed

from: “Annehttp://www.learner.org/resources/series42.html?pop=yes&pid=582”

Read and answer questions

proposed for “Technology and Society-‐

Going Further” on p.p. 304 in

text.

classroom, the teacher needs to post a similar problem (either suggested as a review sessions or in the back of each chapter 9-‐11) to students have been learning the concept as a guided practice and reassures students that reviews are always available.

Use study guides such as the ones appear on p.p. 305 in text. Review packets (chapters 9, 10 and 11) Open Book Quizzes (9, 10,


Physics Grade 12


directional speed (KE = ½ mν2).

4. I can

demonstrate that

near Earth’s surface an object’s gravitational potential energy depends upon its weight (m*g*h) where m is the object’s mass and g is acceleration due to gravity) and height (h) above reference surface (PE = mgh)

Chapter 10: Energy, Work, and Simple Machines

Chapter 11:

Gravitational potential energy Reference level Elastic potential energy Law of conservation of energy Mechanical energy Thermal energy Elastic collision Inelastic collision

Assessment Problems Teacher

Work Book-‐Chapter 11

Additional Labs. Such as the one

suggested in text book-‐

“Launch Lab” on p.p. 285.

https://ieeetv.ieee.org/channels/pels?utm_source=google&utm_medium=cpc&utm_content=Electromagnetism&utm_campaign=IE

EEtv+-‐+Channels+-‐+Power+Elec

tronics

and 11) Exams (9, 10, and 11)


Physics Grade 12


Energy and Its Conservation.

1. I can Follow up the discussion on conservation of energy by using “connecting Math to Physics” as illustrated on p.p. 295 in textbook. 2. I can use the Example on Problem # 2, as proposed on p.p. 296; after I make sure I go over the analysis and solution (Step-‐by-‐Step). 3. Also, I can use the Math Handbook, “Square and Cube Roots, p.p. 839-‐840 to assess and evaluate my


Physics Grade 12


understanding.

4. I can learn what was explained in step 26 above, in regard to Example Problem # 3, p.p. 299.

5. I can discuss the suggested misconception on p.p. 298, as well as using models. I can also refer to the diagrams illustrated in Figure 11-‐4 on p.p. 298 to represent the three types of collisions.

6. As a reinforcement of my understanding to collision and


Physics Grade 12


its elasticity, I can use what is suggested on p.p. 299.

Content Elaborations: Energy In physical science, the role of strong nuclear forces in radioactive decay, half-‐lives, fission and fusion, and mathematical problem solving involving kinetic energy, gravitational potential energy, energy conservation and work (when the force and displacement were in the same direction) were introduced. In this course, the concept of gravitational potential energy is understood from the perspective of a field, elastic potential energy is introduced and quantified, nuclear processes are explored further, the concept of mass-‐energy equivalence is introduced, the concept of work is expanded, power is introduced, and the principle of conservation of energy is applied to increasingly complex situations. Energy must be explored by analyzing data gathered in scientific investigations. Computers and probes can be used to collect and analyze data. Gravitational Potential Energy When two attracting masses interact, the kinetic energies of both objects change but neither is acting as the energy source or the receiver. Instead, the energy is transferred into or out of the gravitational field around the system as gravitational potential energy. A single mass does not have gravitational potential energy. Only the system of attracting masses can have gravitational potential energy. When two masses are moved farther apart, energy is transferred into the field as gravitational potential energy. When two masses are moved closer together, gravitational potential energy is transferred out of the field. Energy in Springs The approximation for the change in the potential elastic energy of an elastic object (e.g., a 2spring) is ΔE elastic = ½ k Δx, where Δx is the distance the elastic object is stretched or Compressed from its relaxed length. Work and Power Work can be calculated for situations in which the force and the displacement are at angles to one another using the equation W = F Δx (cos θ) where W is the work, F is the force, Δx is the displacement, and θ is the angle between the force and the displacement. This means when the force and the displacement are at right angles, no work is done and no energy is transferred between the objects. Such is the case for circular motion. The rate of energy change or transfer is called power (P) and can be mathematically represented by P = ΔE / Δt or P = W / Δt. Power is a scalar property. The unit of power is the watt (W), which is equivalent to one Joule of energy transferred in one second (J/s).


Physics Grade 12


Chapters 1-‐>3:

Name ___________________________________________Period _________________

Chapter 1 Test

MULTIPLE CHOICE Identify the letter of the choice that best completes the statement or answers the question. ____1. 1 micrometer (1 µm) equals: a. 10–3 m c. 103 m b. 10–6 m d. 106 m ____2. According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n): a. straight line c. hyperbola b. parabola d. ellipse

____3. The time period T of a simple pendulum is given by the relation , where L is

the length of the pendulum and g is the acceleration due to gravity at the place. From the above relation, we can say that:

a. T ∝ g c. T2 ∝ g b. T ∝ L d. T2 ∝ L

____4. The slope of the graph given below is:

a.

b.

c.

d.

COMPLETE EACH SENTENCE OR STATEMENT: 5. The method of treating the units as algebraic quantities that can be canceled is called ____________________ analysis. 6. The valid digits in a measurement are called ____________________. 7. When the graph of best fit is a straight line, the dependent and independent variables have a(n) ____________________ relationship. 8. The acceleration of the body in the graph given below is ____________________.

SHORT ANSWER ESSAY: 9. Two objects have masses m1 and m2 and their centers are separated by a distance d. The force

of gravitation between these objects is given by the expression .

Here G is the gravitational constant. Find the units of measurement of G. 10. Write the number 13900 using scientific notation. What is the number of significant digits in this number?

11. What is two-point calibration? 12. The position-time graphs of two runners, Ian and Nick, are shown below. How can the faster runner be determined using these graphs?

PROBLEM 13. A body has a velocity of 72 km/hr. Find its value in m/s. 14. The velocity of sound in air is 332 m/s. If the unit of length is km and the unit of time is hour, what is the value of velocity? 15. How many seconds are there in the month of February in a leap year? ESSAY 16. Analyze the graph and answer the following questions.

Ian

Nick

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 t

1

2

3

4

5

6

7

8

9

10d

a. Name the independent and dependent variables in the graph. b. Name the type of relationship between the variables. c. Find the speed of the body. 17. An object has an initial velocity v and uniform acceleration a. If it covers a distance d, then its final velocity, v , is given by the expression, . a. Does this equation give the correct unit of measurement of vf?

b. Is this a base unit or a derived unit?

c. Rewrite this equation to find the distance traveled in terms of initial velocity, final velocity,

and acceleration.

18. Consider the expression for the arithmetic operation, 0.652 km + 195 cm + 1800 cm.

a. Is any conversion required to do this operation?

b. Convert all the measurements to meters and then solve this expression.

c. How many significant digits are present in the obtained answer?

19. Jessica is using a micrometer screw (screw gauge) to measure the diameter of a wire. While

recording the measurements, she views the readings from different angles.

a. Is there any difference in the recorded values of the same measurement reading, when viewed

from different angles?

b. Name the phenomenon that causes this difference.

c. Does this affect the precision of a measurement made by the student?

20. While performing an experiment to find the value of acceleration due to gravity, g, Lee made

observations for the time period of a simple pendulum. He only changed the length of the

pendulum and recorded different values of time period as given in the table.

Length L of pendulum in cm

25 36 49 64 81 100

Time for one oscillation Time period T in s

1.005 1.205 1.405 1.610 1.810 2.01

a. Is there any linear relationship between the values of length and time period of the simple pendulum? b. The L-T graph is a straight line passing through the origin. Find the relationship between the length and time period of the pendulum. c. What are the units of the constant obtained as a ratio of L and T ? CHAPTER 1ANSWER KEY: MULTIPLE CHOICE 1. ANS: B 2. ANS: C 3. ANS: D 4. ANS: B COMPLETION 5. ANS: dimensional 6. ANS: significant digits 7. ANS: linear 8. ANS:

SHORT ANSWER

9. ANS:

10. ANS: Using scientific notation, the number 13900 can be expressed as 1.39 × 104. This number has three significant digits. 11. ANS: Two-point calibration is the method of checking the accuracy of an instrument. In this method, an instrument is checked for zero error and for the correctness of a reading, while measuring an accepted standard. 12. ANS: The position-time graphs of both runners show linear relationship between their positions and time. These straight line graphs show that the runners have uniform speeds. The slopes of these lines of graphs give the value of the speeds of the runners. The steeper the position-time graph of a runner, higher is his speed. Therefore, Ian is the faster runner. PROBLEM 13. ANS:

14. ANS:

15. ANS: 2505600 s ESSAY 16. ANS: a. Time is the independent variable and distance is the dependent variable. b. Linear relationship exists between the variables.

c.

17. ANS:

a. To find the unit of measurement of v , substitute the units of d, a, and v . The unit of is m2/s2. Therefore, the unit of v is m/s, which is the correct unit for the measurement of velocity. b. The unit of measuring velocity is a derived unit. It is created using two basic SI units, meter and second.

c. The equation for finding distance is .

18. ANS: a. To solve any arithmetic operation involving physical quantities, all the quantities involved should have the measurement units of the same system of units. In this expression, 0.615 km should be first converted to cm, or magnitudes in cm should be converted to km. b. 0.652 km = 652 m, 195 cm = 1.95 m, 1800 cm = 18 m Therefore, 0.652 km + 195 cm + 1800 cm becomes 652 m + 1.95 m + 18 m = 672 m c. The number has three significant digits. 19. ANS: a. The recorded values of the same measurement reading differ when the reading is viewed from different angles. b. The phenomenon that causes this difference is called parallax. c. Parallax does not affect the precision of a measurement, but it affects the accuracy of the measurements. 20. ANS: a. The time period of a simple pendulum increases with length. However, from the readings given in the table, it is clear that T is not proportional to L. b. The straight line graph in case of L-T graph shows that L is directly proportional to T .

c. Since , therefore = constant. The units of this constant are m/s .

CHAPTER 1 ANSWER KEY WITH ANALYSIS MULTIPLE CHOICE: 1. ANS: B 10–6 is equivalent to the metric prefix micro (µ). Therefore, 1 micrometer (1 µm) = 10–6 m. DIF: 2 REF: p.p. 6 OBJ: 1.1.2 Use the metric system. TOP: Use the metric system. KEY: SI units MSC: 1 NOT: /a/ The metric prefix milli (m) is equivalent to 10e–3 m. /b/ Correct! /c/ The metric prefix kilo (k) is equivalent to 10e+3 m. /d/ The metric prefix mega (M) is equivalent to 10e+6 m.

2. ANS: C The pressure P is inversely proportional to the volume V. For inverse relation, the graph between the variables is a hyperbola. DIF: 2 REF: p.p. 18 OBJ: 1.3.3 Recognize common relationships in graphs. STO: 12.D.3.2 TOP: Recognize common relationships in graphs. KEY: Inverse relationship MSC: 3 NOT: /a/ P and V do not have a linear relationship. /b/ P and V do not have a quadratic relationship. /c/ Correct! /d/ P and V have an inverse relationship, hence the graph is a hyperbola. 3. ANS: D

. On squaring both sides we get, .

Therefore, T2 ∝ L. DIF: 2 REF: p.p. 17 OBJ: 1.3.3 Recognize common relationships in graphs. STO: 12.D.3.2 TOP: Recognize common relationships in graphs. KEY: Linear relationship MSC: 3 NOT: /a/ T is inversely proportional to the square root of g. /b/ T is directly proportional to the square root of L. /c/ The square of T is inversely proportional to g. /d/ Correct! 4. ANS: B The slope is the ratio of the vertical change to the horizontal change. DIF: 1 REF: p.p. 17 OBJ: 1.3.1 Graph the relationship between independent and dependent variables. TOP: Graph the relationship between independent and dependent variables. KEY: Slope MSC: 2 NOT: /a/ Reciprocate the ratio. /b/ Correct! /c/ PR is neither the vertical nor the horizontal change. /d/ PR is the horizontal change and not the vertical change. COMPLETION 5. ANS: dimensional DIF: 1 REF: p.p. 6 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 1 6. ANS: significant digits DIF: 1 REF: p.p. 7 OBJ: 1.1.4 Perform arithmetic operations using scientific notation. TOP: Perform arithmetic operations using scientific notation. KEY: Significant digits MSC: 1

7. ANS: linear DIF: 1 REF: p.p. 16 OBJ: 1.3.3 Recognize common relationships in graphs. STO: 12.D.3.2 TOP: Recognize common relationships in graphs. KEY: Linear relationship MSC: 2 8. ANS: DIF: 2 REF: p.p. 17 OBJ: 1.3.1 Graph the relationship between independent and dependent variables. TOP: Graph the relationship between independent and dependent variables. KEY: Slope MSC: 2 SHORT ANSWER 9. ANS:

DIF: 2 REF: p.p. 6 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 2 10. ANS: Using scientific notation, the number 13900 can be expressed as 1.39 × 104. This number has three significant digits. DIF: 2 REF: p.p. 7 OBJ: 1.1.4 Perform arithmetic operations using scientific notation. TOP: Perform arithmetic operations using scientific notation. KEY: Significant digits MSC: 2 11. ANS: Two-point calibration is the method of checking the accuracy of an instrument. In this method, an instrument is checked for zero error and for the correctness of a reading, while measuring an accepted standard. DIF: 2 REF: p.p. 13 OBJ: 1.2.2 Determine the precision of measured quantities. STO: 9.D.D.4 TOP: Determine the precision of measured quantities.

KEY: Accuracy MSC: 2 12. ANS: The position-time graphs of both runners show linear relationship between their positions and time. These straight line graphs show that the runners have uniform speeds. The slopes of these lines of graphs give the value of the speeds of the runners. The steeper the position-time graph of a runner, higher is his speed. Therefore, Ian is the faster runner. DIF: 2 REF: p.p. 16 OBJ: 1.3.2 Interpret graphs. TOP: Interpret graphs. KEY: Linear relationship MSC: 2 PROBLEM 13. ANS:

DIF: 2 REF: p.p. 6 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 2 NOT: 1 km = 1000 m and 1 hr = 60*60 s = 3600 s 14. ANS:

DIF: 3 REF: p.p. 7 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 2 NOT: 1 m =0.001 km and 1 s = 1/3600 hr 15. ANS: 2505600 s DIF: 3 REF: p.p. 6 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 2 NOT: In a leap year, in the month of February, there are 29 days. Therefore, the number of seconds = 29*24*60*60 = 2505600 s.

ESSAY 16. ANS: a. Time is the independent variable and distance is the dependent variable. b. Linear relationship exists between the variables.

c.

DIF: 3 REF: p.p. 16 OBJ: 1.3.1 Graph the relationship between independent and dependent variables. TOP: Graph the relationship between independent and dependent variables. KEY: Independent and dependent variables MSC: 3 17. ANS: a. To find the unit of measurement of v , substitute the units of d, a, and v . The unit of is m2/s2. Therefore, the unit of v is m/s, which is the correct unit for the measurement of velocity. b. The unit of measuring velocity is a derived unit. It is created using two basic SI units, meter and second.

c. The equation for finding distance is .

DIF: 3 REF: p.p. 6, p.p. 7 OBJ: 1.1.3 Evaluate answers using dimensional analysis. TOP: Evaluate answers using dimensional analysis. KEY: Dimensional analysis MSC: 2 18. ANS: a. To solve any arithmetic operation involving physical quantities, all the quantities involved should have the measurement units of the same system of units. In this expression, 0.615 km should be first converted to cm, or magnitudes in cm should be converted to km. b. 0.652 km = 652 m, 195 cm = 1.95 m, 1800 cm = 18 m Therefore, 0.652 km + 195 cm + 1800 cm becomes 652 m + 1.95 m + 18 m = 672 m c. The number has three significant digits. DIF: 3 REF: p.p. 6 OBJ: 1.1.4 Perform arithmetic operations using scientific notation. TOP: Perform arithmetic operations using scientific notation. KEY: Significant digits MSC: 1 19. ANS: a. The recorded values of the same measurement reading differ when the reading is viewed from different angles. b. The phenomenon that causes this difference is called parallax.

c. Parallax does not affect the precision of a measurement, but it affects the accuracy of the measurements. DIF: 3 REF: p.p. 13 OBJ: 1.2.2 Determine the precision of measured quantities. STO: 9.D.D.4 TOP: Determine the precision of measured quantities. KEY: Accuracy, Precision MSC: 3 20. ANS: a. The time period of a simple pendulum increases with length. However, from the readings given in the table, it is clear that T is not proportional to L. b. The straight line graph in case of L-T graph shows that L is directly proportional to T .

c. Since , therefore = constant. The units of this constant are m/s .

DIF: 3 REF: p.p. 16, p.p. 17 OBJ: 1.3.3 Recognize common relationships in graphs. STO: 12.D.3.2 TOP: Recognize common relationships in graphs. KEY: Linear relationship, Nonlinear relationship MSC: 3

Name ___________________________________________________ Period _______________

Chapter 2 Test MULTIPLE CHOICE Identify the letter of the choice that best completes the statement or answers the question. ____1. A car is moving with a uniform speed of 15.0 m/s along a straight path. What is the distance covered by the car in 12.0 minutes? a. 1.02 × 10–3 km c. 8.00 × 10–5 km b. 1.80 × 10–1 km d. 1.08 × 101 km ____2. Which of the following is a pair of vector quantities? a. Speed — Distance c. Velocity — Displacement b. Velocity — Distance d. Speed — Displacement ____ 3. What is the resultant of two displacement vectors having the same direction? a. The resultant is the sum of the two displacements having the same direction as the

original vectors. b. The resultant is the difference of the two displacements having the same direction

as the original vectors. c. The resultant is the sum of the two displacements having the direction opposite to

the direction of the original vectors. d. The resultant is the sum of the two displacements having the direction

perpendicular to the direction of the original vectors. COMPLETION: Complete each sentence or statement. 4. A(n) ____________________ model is a simplified version of a motion diagram that represents the object in motion by a series of single points. 5. The vector that represents the sum of the other two vectors is called the ____________________. SHORT ANSWER: 6. What is a motion diagram? 7. What is the significance of defining a coordinate system to study the motion of an object? 8. How is displacement different from distance? Under what conditions can an object travel a certain distance and yet its resultant displacement be zero? 9. What is the equation of motion for average velocity?

10. Define average velocity. 11. The position-time graph of an object is found to be a straight line passing through the origin. What information about the motion of the object is provided by the graph? 12. Given below is the position-time graph representing motions of two runners, Nick and Ian. Use this graph to determine which runner has greater average velocity.

13. Given below is the particle model of a boy skating on a smooth, pedestrian-free sidewalk. The time interval between successive dots is 2 s.

Plot a position-time graph to represent the motion of the boy.

Ian

Nick

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

14. The position-time graph given below represents the motion of Ted returning home from the market on his bike. What is the similarity between his displacement and the average velocity?

15. Given below is the motion diagram of a small ball rolling straight on a frictionless surface.

The time interval between successive positions is 1 s.

a. What is the displacement of the ball after 3 s?

b. Where does the ball reach after 5 s?

c. After rolling for 7 s, the direction of motion of the ball is changed. It starts rolling toward

its starting point. Assume the coordinate system and the speed of the ball to remain unchanged.

What is the displacement of the ball between 7 s and 9 s?

Time (s)

Posi

tion

(m)

1 2 3 4 5 6 7 8 9 10 11 12 13–1

5

10

15

20

25

30

35

40

45

50

–5

–10

16. Given below is the graph representing the position-time graphs of two swimmers (A and B), swimming in a pool along a straight line. Both the swimmers start from two different positions. Use the graph to find when and where swimmer B passes swimmer A.

17. The position-time graph of a pedestrian is given below. What is his displacement after 2.5 s?

18. What information is provided by the points on the line of a position-time graph of an object? 19. A boy starts from point A and moves 5 units toward the east, then turns back and moves 3 units toward the west. What is the displacement in the position of the boy? 20. What are the two attributes of the coordinate system chosen for a motion diagram? 21. How is the time interval affected when the origin of the coordinate system of a motion diagram is changed?

Swimmer A

Swimmer B

Posi

tion

(m)

Time (s)10 20 30 40 50 60 70 80

50

100

150

200

250

300

350

–50

–100

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 10

2

4

6

8

10

12

–2

–4

–6

–8

22. What is the distance traveled by a vehicle in 12 minutes, if its speed is 35 km/h? 23. Given below is the position-time graph representing the motion of two friends, A and B, jogging in a park. Use this graph to find their displacements after 4 s.

ESSAY 24. A boy is cycling at a constant speed along a straight road to his school. The positions of the

boy after fixed time intervals are observed.

a. What type of position-time graph can be expected for the motion of the bicycle?

b. How can the average velocity of the bicycle be calculated?

c. How is the average speed of the bicycle related to its average velocity?

25. Linda starts for her school at eight in the morning. There is a straight road that connects her

house to the school. She reaches school after 3.5 minutes. Her physics teacher has asked her to

note her position after every 30 seconds on her way to school.

The values for her position (distance from her home) are listed with each time interval in the

table below:

Time (in seconds)

0 30 60 90 120 150 180 210

Position (in meters)

0 45 90 135 180 225 270 315

A

B

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

Using this position time table, a. Draw a position-time graph to represent Linda’s motion from home to school.

b. Calculate the average velocity of Linda using the position-time graph.

c. What is the velocity of Linda on a return trip home if she reaches back home in the same

time of 3.5 minutes? Assume Linda’s house to be the fixed origin of the graph.

26. A mouse is running at a high speed between the kitchen and storeroom, situated along the

same line in a building. A boy observes the mouse picking up pieces of bread from the kitchen

and leaving them in a box in the storeroom. The boy observes the positions of the mouse after

fixed intervals. He finds that the mouse travels the same displacement of 2.0 m in every half

second. Starting from the door of the kitchen, the mouse reaches the door of the storeroom in 3.0

s.

a. Make a position-time table for the motion of the mouse from the kitchen to the

storeroom.

b. Draw the particle model of this motion.

c. Convert this model into the corresponding position-time graph.

27. Thomas cycles on a straight road near his house. After sometime, his sister Anna also starts

cycling. Their motions are represented by the position-time graphs given below.

a. How long had Thomas been cycling when Anna started cycling?

b. What is the position at which Anna crosses Thomas?

c. What is the difference between their positions after 5 s?

28. Three railway stations, A, B, and C, are situated on a straight railway route. Station B is at 30

km in the east from station A and station C is at 60 km in east from station A. A train, with a

constant average velocity, starts from A and reaches B in 15 minutes, and reaches C in 30

minutes. Assuming that station A is the origin, find the following:

a. The position of the train moving from station A to station B after 10 minutes.

b. The position of the train moving from station A to station B after 10 minutes, if station B

is assumed to be the origin.

c. The position of the train moving from station A to station B after 10 minutes, if station C

is assumed to be the origin.

29. Tom drives his car at a constant speed on a straight road for two hours while going to meet

his friend. He noted the position of his car after every 15 minutes, assuming his house to be the

origin. On his way back home, he again noted the positions of the car after every 15 minutes,

without changing the origin. If on both of the trips, his car covers 25 km in every 15 minutes,

then:

a. Draw the position-time graphs of the car for the two sides of the journey.

b. Find the difference between the average velocities of the two sides of the journey.

c. Find what happens if during the return journey, the friend’s house is taken as the origin.

CHAPTER 2 ANSWER KEY MULTIPLE CHOICE 1. ANS: D 2. ANS: C 3. ANS: A COMPLETION 4. ANS: particle 5. ANS: resultant SHORT ANSWER 6. ANS: Motion diagram is a series of images showing the positions of a moving object at equal time intervals. 7. ANS: A coordinate system tells about the location of the zero point of the variables defining the motion of the object to be studied. The coordinate system also explains the direction in which the values of the variables increase. 8. ANS: Displacement is a vector quantity, while distance is a scalar quantity. Displacement has both the magnitude and direction. Distance has only magnitude. The displacement of an object is equal to the difference between its final position and initial position. When an object starts moving from a certain point and after covering certain distance returns to its starting position, its displacement becomes zero. 9. ANS: The equation of motion for average velocity is . In this equation, d is the position of an object, is the average velocity of the object, t is the time and di is the initial position of the object. According to this equation, an object’s position is equal to the average velocity multiplied by time plus the initial position. 10. ANS: Average velocity of an object is defined as the change in position of the object moving along a straight line, divided by the time during which the change occurred. In symbolic form, the average velocity

.

11. ANS: The straight line position-time graph of a moving object gives the information about the nature of the motion of the object. It is also used to calculate the average velocity of the object. Since the graph is a straight line, the object is undergoing uniform motion. The slope of this line gives the value of the average velocity of the object. 12. ANS: The average velocity of an object is equal to the slope of the line representing the position-time graph of the object. In this graph, the slope of Ian’s line is steeper than the slope of Nick’s line. A steeper slope indicates a greater change in displacement during each time interval. Average velocity for a given time interval is proportional to the change in the displacement. Therefore, the average velocity of Ian is more than the average velocity of Nick. 13. ANS: The position-time graph corresponding to this particle model is plotted below.

14. ANS: In this case, motion is in the negative direction. Therefore, displacement and velocity of the bike are negative. Hence, both the displacement and the velocity have a negative sign. 15. ANS: a. The value of displacement after 3 s is 12 m. b. After 5 s, the ball reaches at 20 m from its starting point. c. The position of the ball after 7 s = 28 m. The position of the ball after 9 s = 20 m. The direction of the motion of the ball is reversed after 7 s. Therefore, the displacement of the ball between 7 s and 9 s = Δd = 20 – 28 = –8 m. The negative sign shows that the direction of displacement in this case is negative, toward the starting point.

Time (s)

Posi

tion

(m)

2 4 6 8 10 12 14 16 18 20 22 24

5

10

15

20

25

30

35

40

45

50

16. ANS: The point of intersection of the two graphs is 150 m at about 38 s. Swimmer B passes Swimmer A about 150 m beyond the origin 38 s after A has passed the origin. 17. ANS: The position-time graph of the pedestrian intersects the x-axis at 2.5 s. Thus, the displacement in 2.5 s is –10 m. 18. ANS: The points on the line of a position-time graph of an object represent the most likely positions of the moving object at the times between the recorded data points. 19. ANS:

The resultant displacement is 2 units toward the east, and his position is at 2. 20. ANS: The two attributes of the coordinate system chosen for a motion diagram are the origin and the axis of the coordinate system. 21. ANS: The time interval is a scalar quantity. Its value is not affected by the change in the relative position of the origin of the coordinate system. 22. ANS: Speed is given in km/h, therefore convert time in minutes to hours.

Distance traveled = (speed)(time taken) = (35 km/h)(12 min)(1 h/60 min) = 7.0 km 23. ANS: The displacements of A and B are the coordinates of the vertical line with their respective position-time graphs at 4 s.

The value of displacement of A is approximately 4 m The value of displacement of B is approximately 3 m. ESSAY 24. ANS: a. The bicycle is moving with a constant speed, and motion is in the straight line, without any change in direction. The velocity of the bicycle is also constant. Therefore, the position-time graph will be a straight line. b. The average velocity of the bicycle is calculated as the slope of the position-time graph. c. The average speed of the bicycle is the absolute value of its average velocity. 25. ANS: a. The position-time graph of Linda’s motion from home to school is as follows:

b. Average velocity of Linda = Slope of the above graph =

c. During her return journey from school, the average velocity of Linda is –1.5 m/s.

A

B

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8Po

sitio

n (m

)

Time (s)30 60 90 120 150 180 210 240

45

90

135

180

225

270

315

26. ANS: a. The position-time table for the motion of the mouse from the kitchen to the store is:

Position vs Time Time (s) Position

(m) 0.0 0.0 0.5 2.0 1.0 4.0 1.5 6.0 2.0 8.0 2.5 10.0 3.0 12.0

b. The following particle model represents the motion of the mouse from the kitchen to the store. Kitchen Store

The time interval between two successive dots is of 0.5 s. c. The position-time graph corresponding to the position-time table formed in part a, is plotted below.

27. ANS: a. Thomas had been cycling for 1.0 s, when Anna started cycling. b. Anna crosses Thomas at a point 12.0 m from the starting point. c. Displacement of Thomas after 5.0 s = 20.0 m – 0.0 m = 20.0 m. Displacement of Anna after 5.0 s = 24.0 m – 0.0 m = 24.0 m. The difference in their positions after 5.0 s is 4.0 m with Anna being ahead. 28. ANS: a. The position of the train is 20 km in the east to station A. b. The position of the train becomes –10 km. c. The position of the train becomes –40 km.

29. ANS: a.

b. There is the difference of sign in the two velocities. During the return journey, the displacement is negative. Therefore, the average velocity is also negative. c. The average velocity during the return journey also becomes positive. CHAPTER 2 ANSWER KEY WITH ANALYSIS MULTIPLE CHOICE 1. ANS: D Distance = speed × time Substituting the values in the relation for distance, we get, d = (15.0 m/s)(12.0 min)(60 s/1 min) = 10,800 m = 1.08 × 101 km DIF: 2 REF: p.p. 44 OBJ: 2.4.2 Differentiate between speed and velocity. TOP: Differentiate between speed and velocity. KEY: Average speed MSC: 2 NOT: /a/ Speed is multiplied with time to calculate the distance. All the quantities should have units of the same system. /b/ Time must be converted to seconds. /c/ Time should be multiplied with speed, not divided, to calculate the distance. /d/ Correct! 2. ANS: C Velocity and displacement are vector quantities. DIF: 1 REF: p.p. 35 OBJ: 2.2.3 Define displacement. TOP: Define displacement. KEY: Vectors MSC: 1 NOT: /a/ Speed and distance are scalar quantities. /b/ Velocity is a vector, but distance is scalar. /c/ Correct! /d/ Displacement is a vector, but speed is scalar.

3. ANS: A When two displacement vectors have the same direction, their resultant has magnitude equal to the sum of their magnitudes. The direction of the resultant is the same as the direction of the original vectors. DIF: 2 REF: p.p. 37 OBJ: 2.2.3 Define displacement. TOP: Define displacement. KEY: Displacement, Resultant MSC: 2 NOT: /a/ Correct! /b/ The resultant is the sum of the vectors, and not the difference of the vectors. /c/ The resultant of two vectors, having the same direction, also has the same direction as the original vectors. /d/ The resultant is not perpendicular, but is in the direction of the original vectors. COMPLETION 4. ANS: particle DIF: 1 REF: p.p. 33 OBJ: 2.1.2 Develop a particle model to represent a moving object. STO: 9.D.D.3 TOP: Develop a particle model to represent a moving object. KEY: Particle model MSC: 2 5. ANS: resultant DIF: 1 REF: p.p. 35 OBJ: 2.2.5 Use a motion diagram to answer questions about an object's position or displacement. TOP: Use a motion diagram to answer questions about an object's position or displacement. KEY: Vectors, Resultant MSC: 1 SHORT ANSWER 6. ANS: Motion diagram is a series of images showing the positions of a moving object at equal time intervals. DIF: 1 REF: Pp.p. 33 OBJ: 2.1.1 Draw motion diagrams to describe motion. TOP: Draw motion diagrams to describe motion. KEY: Motion diagram MSC: 2 7. ANS:A coordinate system tells about the location of the zero point of the variables defining the motion of the object to be studied. The coordinate system also explains the direction in which the values of the variables increase. DIF: 1 REF: p.p. 34 OBJ: 2.2.1 Define coordinate systems for motion problems. TOP: Define coordinate systems for motion problems. KEY: Coordinate system, Origin

MSC: 2 8. ANS:Displacement is a vector quantity, while distance is a scalar quantity. Displacement has both the magnitude and direction. Distance has only magnitude. The displacement of an object is equal to the difference between its final position and initial position. When an object starts moving from a certain point and after covering certain distance returns to its starting position, its displacement becomes zero. DIF: 2 REF: p.p. 37 OBJ: 2.2.3 Define displacement. TOP: Define displacement. KEY: Distance, Displacement, Scalars, Vectors MSC: 2 9. ANS:The equation of motion for average velocity is . In this equation, d is the position of an object, is the average velocity of the object, t is the time and di is the initial position of the object. According to this equation, an object’s position is equal to the average velocity multiplied by time plus the initial position. DIF: 1 REF: p.p. 47 OBJ: 2.4.3 Create pictorial, physical, and mathematical models of motion problems.; STO: 9.D.D.3 TOP: Create pictorial, physical, and mathematical models of motion problems.; KEY: Average velocity MSC: 2 10. ANS: Average velocity of an object is defined as the change in position of the object moving along a straight line, divided by the time during which the change occurred. In symbolic form, the average velocity

.

DIF: 2 REF: p.p. 44 OBJ: 2.4.1 Define velocity. TOP: Define velocity. KEY: Average velocity MSC: 2 11. ANS: The straight line position-time graph of a moving object gives the information about the nature of the motion of the object. It is also used to calculate the average velocity of the object. Since the graph is a straight line, the object is undergoing uniform motion. The slope of this line gives the value of the average velocity of the object. DIF: 2 REF: p.p. 43 OBJ: 2.4.1 Define velocity. TOP: Define velocity. KEY: Average velocity MSC: 12. ANS: The average velocity of an object is equal to the slope of the line representing the position-time graph of the object. In this graph, the slope of Ian’s line is steeper than the slope of Nick’s line. A steeper slope indicates a greater change in displacement during each time interval. Average velocity for a given time interval is proportional to the change in the displacement.

Therefore, the average velocity of Ian is more than the average velocity of Nick. DIF: 3 REF: p.p. 43 OBJ: 2.4.1 Define velocity. TOP: Define velocity. KEY: Average velocity MSC: 2 13. ANS: The position-time graph corresponding to this particle model is plotted below.

DIF: 3 REF: p.p. 40 OBJ: 2.3.1 Develop position-time graphs for moving objects. TOP: Develop position-time graphs for moving objects. KEY: Position-time graph MSC: 2 14. ANS: In this case, motion is in the negative direction. Therefore, displacement and velocity of the bike are negative. Hence, both the displacement and the velocity have a negative sign. DIF: 1 REF: p.p. 44 OBJ: 2.4.1 Define velocity. TOP: Define velocity. KEY: Displacement, Average velocity MSC: 1 15. ANS: a. The value of displacement after 3 s is 12 m. b. After 5 s, the ball reaches at 20 m from its starting point. c. The position of the ball after 7 s = 28 m. The position of the ball after 9 s = 20 m. The direction of the motion of the ball is reversed after 7 s. Therefore, the displacement of the ball between 7 s and 9 s = Δd = 20 – 28 = –8 m. The negative sign shows that the direction of displacement in this case is negative, toward the starting point. DIF: 3 REF: p.p. 36 OBJ: 2.2.5 Use a motion diagram to answer questions about an object's position or displacement. TOP: Use a motion diagram to answer questions about an object's position or displacement.

Time (s)

Posi

tion

(m)

2 4 6 8 10 12 14 16 18 20 22 24

5

10

15

20

25

30

35

40

45

50

KEY: Displacement MSC: 3 16. ANS: The point of intersection of the two graphs is 150 m at about 38 s. Swimmer B passes Swimmer A about 150 m beyond the origin 38 s after A has passed the origin. DIF: 2 REF: p.p. 38 OBJ: 2.2.4 Determine a time interval. TOP: Determine a time interval. KEY: Time interval, Origin MSC: 2 17. ANS: The position-time graph of the pedestrian intersects the x-axis at 2.5 s. Thus, the displacement in 2.5 s is –10 m. DIF: 2 REF: p.p. 38 OBJ: 2.3.2 Use a position-time graph to interpret an object's position or displacement. TOP: Use a position-time graph to interpret an object's position or displacement. KEY: Position-time graph, Displacement MSC: 1 18. ANS: The points on the line of a position-time graph of an object represent the most likely positions of the moving object at the times between the recorded data points. DIF: 2 REF: p.p. 38 OBJ: 2.3.2 Use a position-time graph to interpret an object's position or displacement. TOP: Use a position-time graph to interpret an object's position or displacement. KEY: Position-time graph, Position MSC: 1 19. ANS:

The resultant displacement is 2 units toward the east, and his position is at 2. DIF: 2 REF: p.p. 34 OBJ: 2.2.5 Use a motion diagram to answer questions about an object's position or displacement. TOP: Use a motion diagram to answer questions about an object's position or displacement. KEY: Displacement MSC: 2

20. ANS:The two attributes of the coordinate system chosen for a motion diagram are the origin and the axis of the coordinate system. DIF: 2 REF: p.p. 34 OBJ: 2.2.5 Use a motion diagram to answer questions about an object's position or displacement. TOP: Use a motion diagram to answer questions about an object's position or displacement. KEY: Coordinate system, Origin MSC: 2 21. ANS:The time interval is a scalar quantity. Its value is not affected by the change in the relative position of the origin of the coordinate system. DIF: 2 REF: p.p. 36 OBJ: 2.2.4 Determine a time interval. TOP: Determine a time interval. KEY: Time interval MSC: 2 22. ANS:Speed is given in km/h, therefore convert time in minutes to hours.

distance traveled = (speed)(time taken) = (35 km/h)(12 min)(1 h/60 min) = 7.0 km DIF: 2 REF: p.p. 36 OBJ: 2.2.3 Define displacement. TOP: Define displacement. KEY: Distance MSC: 2 23. ANS:The displacements of A and B are the coordinates of the vertical line with their respective position-time graphs at 4 s.

The value of displacement of A is approximately 4 m The value of displacement of B is approximately 3 m. DIF: 3 REF: p.p. 36 OBJ: 2.2.3 Define displacement. TOP: Define displacement. KEY: Displacement MSC: 1

A

B

Posi

tion

(m)

Time (s)1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

ESSAY: 24. ANS: a. The bicycle is moving with a constant speed, and motion is in the straight line, without any change in direction. The velocity of the bicycle is also constant. Therefore, the position-time graph will be a straight line. b. The average velocity of the bicycle is calculated as the slope of the position-time graph. c. The average speed of the bicycle is the absolute value of its average velocity. DIF: 3 REF: p.p. 44 OBJ: 2.4.2 Differentiate between speed and velocity. TOP: Differentiate between speed and velocity. KEY: Average speed, Average velocity MSC: 2 25. ANS: a. The position-time graph of Linda’s motion from home to school is as follows:

b. Average velocity of Linda = Slope of the above graph =

c. During her return journey from school, the average velocity of Linda is –1.5 m/s. DIF: 3 REF: p.p. 43 OBJ: 2.4.1 Define velocity. TOP: Define velocity. KEY: Average velocity MSC: 3 26. ANS: a. The position-time table for the motion of the mouse from the kitchen to the store is:

Position vs Time Time (s) Position

(m) 0.0 0.0 0.5 2.0 1.0 4.0

Posi

tion

(m)

Time (s)30 60 90 120 150 180 210 240

45

90

135

180

225

270

315

1.5 6.0 2.0 8.0 2.5 10.0 3.0 12.0

b. The following particle model represents the motion of the mouse from the kitchen to the store. Kitchen Store

The time interval between two successive dots is of 0.5 s. c. The position-time graph corresponding to the position-time table formed in part a, is plotted below.

DIF: 3 REF: p.p. 40 OBJ: 2.3.3 Make motion diagrams, pictorial representations, and position-time graphs that are equivalent representations describing an object's motion.; TOP: Make motion diagrams, pictorial representations, and position-time graphs that are equivalent representations describing an object's motion.; KEY: Position-time graph MSC: 2 27. ANS: a. Thomas had been cycling for 1.0 s, when Anna started cycling. b. Anna crosses Thomas at a point 12.0 m from the starting point. c. Displacement of Thomas after 5.0 s = 20.0 m – 0.0 m = 20.0 m. Displacement of Anna after 5.0 s = 24.0 m – 0.0 m = 24.0 m. The difference in their positions after 5.0 s is 4.0 m with Anna being ahead. DIF: 3 REF: p.p. 42 OBJ: 2.3.2 Use a position-time graph to interpret an object's position or displacement. TOP: Use a position-time graph to interpret an object's position or displacement. KEY: Position-time graph, Instantaneous position MSC: 2

28. ANS: a. The position of the train is 20 km in the east to station A. b. The position of the train becomes –10 km. c. The position of the train becomes –40 km. DIF: 3 REF: p.p. 35 OBJ: 2.2.2 Recognize that the chosen coordinate system affects the sign of objects' positions. TOP: Recognize that the chosen coordinate system affects the sign of objects' positions. KEY: Coordinate system, Position, Origin MSC: 3 29. ANS: a.

b. There is the difference of sign in the two velocities. During the return journey, the displacement is negative. Therefore, the average velocity is also negative. c. The average velocity during the return journey also becomes positive. DIF: 3 REF: p.p. 38 OBJ: 2.3.1 Develop position-time graphs for moving objects. TOP: Develop position-time graphs for moving objects. KEY: Displacement, Position-time graph, Average velocity MSC: 2

Name ___________________________________________________ Period _______________

Chapter 3 Test MULTIPLE CHOICE Identify the letter of the choice that best completes the statement or answers the question. ____1. A man starts his car from rest and accelerates at 1 m/s2 for 2 seconds. He then continues at a constant velocity for 10 seconds until he sees a tree blocking the road and applies brakes. The car, decelerating at 1 m/s2, finally comes to rest. Which of the following graphs represents the motion correctly? a.

c.

b.

d.

v

t5 10 15 20 25 30 35 40 (s)

2468

1012141618202224262830

(m/s) v

t5 10 15 20 25 30 35 (s)

2468

101214161820222426283032343638

(m/s)

v

t5 10 15 20 25 (s)

2

4

6

8

10 (m/s) v

t5 10 15 20 25 (s)

2

4

6

8

10 (m/s)

SHORT ANSWER 2. The velocity-time graph of a car’s motion is given below. Plot the corresponding acceleration-time graph.

PROBLEM 3. The velocity-time graph of the motion of a particle is shown below. Calculate the total displacement of the particle from 0 to 29 seconds.

4. A car starts from rest with an acceleration of 2.84 m/s2 at the instant when a second car moving with a velocity of 25.7 m/s passes it in a parallel line. How far does the first car move before it overtakes the second car? 5. A ball is thrown vertically upward with a speed of 1.53 m/s from a point 4.21 m above the ground. Calculate the time in which the ball will reach the ground.

v

t5 10 15 (s)

5

10

–5

–10

(m/s)

v (m/s)

t (s)2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

2468

10121416182022242628303234

6. A ball is thrown vertically upward with a speed of 1.86 m/s from a point 3.82 m above the ground. Calculate the time in which the ball will reach the ground. 7. Julia throws a ball vertically upward from the ground with a speed of 5.89 m/s. Andrew catches it when it is on its way down at a height of 1.27 m from the ground. After how much time does Andrew catch the ball? 8. The graph below shows how the velocity of a rolling ball changes with time. Calculate the acceleration of the ball.

9. A car accelerates from rest at 5 m/s2 for 5 seconds. It moves with a constant velocity for some time, and then decelerates at 5 m/s2 to come to rest. The entire journey takes 25 seconds. Plot the velocity-time graph of the motion.

(m/s)v

t2 4 6 8 10 12 14 16 18 (s)

1

2

3

4

5

6

7

8

9

10

10. The graph below represents the velocity-time variation of a car’s motion.

Use the graph to find: a) The acceleration of the car between t = 0 s and t = 5 s.

b) The acceleration of the car between t = 5 s and t = 10 s.

c) The acceleration of the car between t = 10 s and t = 20 s.

d) The acceleration of the car between t = 20 s and t = 30 s.

11. A boy throws a ball vertically upward with a speed of 19 m/s. Calculate the speed of the ball

when it is at a height equal to 0.77 times the maximum height reached by the ball.

v

t5 10 15 20 25 30 (s)

5

10

15

(m/s)

Chapter 3 ANSWER KEY MULTIPLE CHOICE 1.ANS: B SHORT ANSWER 2. ANS:

PROBLEM 3. ANS: 532.5 m 4. ANS: 465 m 5. ANS: 1.10 s 6. ANS: 1.093 s 7. ANS: 0.92 s 8. ANS: 4 m/s2 9. ANS:

10. ANS: a) 0.80 m/s2 b) 0 m/s2 c) 1 m/s2 d) –1.6 m/s2 11. ANS: 9.1 m/s

a

t1 2 3 4 5 6 7 8 9 (s)

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

(m/s^2)

v

t5 10 15 20 25 30 (s)

5

10

15

20

25

30

(m/s)

ANSWER KEY ANALYSIS MULTIPLE CHOICE 1. ANS: B DIF: 3 REF: p.p. 57 OBJ: 3.1.3 Create velocity-time graphs. TOP: Create velocity-time graphs. KEY: Velocity-time graph MSC: 3 NOT: /A/ This graph does not show a period of constant velocity. /B/ Correct! /C/ The car starts from rest instead of at constant speed and the slopes of the different parts of the graph are incorrect./D/ The car moves at a constant velocity for 12 seconds. SHORT ANSWER 2. ANS:

DIF: 3 REF: p.p. 66 OBJ: 3.1.2 Relate velocity and acceleration to the motion of an object. TOP: Relate velocity and acceleration to the motion of an object. KEY: Relate velocity and acceleration MSC: 3 PROBLEM 3. ANS: 532.5 m DIF: 3 REF: p.p. 65 OBJ: 3.2.1 Interpret position-time graphs for motion with constant acceleration. TOP: Interpret position-time graphs for motion with constant acceleration. KEY: Position-time graph MSC: 3 NOT: The area under the graph gives the distance traveled. 4. ANS: 465 m DIF: 3 REF: p.p. 65

a

t1 2 3 4 5 6 7 8 9 (s)

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

(m/s^2)

OBJ: 3.2.2 Determine mathematical relationships among position, velocity, acceleration, and time.; TOP: Determine mathematical relationships among position, velocity, acceleration, and time.; KEY: Position, Velocity, Acceleration, Time MSC: 3 NOT: The distance covered by both cars is the same when the second car just overtakes the first car. 5. ANS: 1.10 s DIF: 3 REF: p.p. 72 OBJ: 3.3.2 Solve objects involving objects in free fall. TOP: Solve objects involving objects in free fall. KEY: Free fall MSC: 3 NOT: The total time is the sum of the time taken by the ball to reach its topmost point and the time to come down from the topmost point to the ground. 6. ANS: 1.093 s DIF: 3 REF: p.p. 72 OBJ: 3.3.2 Solve objects involving objects in free fall. TOP: Solve objects involving objects in free fall. KEY: Free fall MSC: 3 NOT: The total time is the sum of the time taken by the ball to reach its topmost point and the time to come down from the topmost point to the ground. 7. ANS: 0.92 s DIF: 3 REF: p.p. 72 OBJ: 3.3.2 Solve objects involving objects in free fall. TOP: Solve objects involving objects in free fall. KEY: Free fall MSC: 3 NOT: The total time is the sum of the time taken by the ball to reach its topmost point and the time to come down from the topmost point to the point where the ball is caught. 8. ANS: 4 m/s2 DIF: 3 REF: p.p. 59 OBJ: 3.1.1 Define acceleration. STO: 9.C.C.23 TOP: Define acceleration. KEY: Acceleration MSC: 3 NOT: Acceleration is the rate of change of velocity. 9. ANS:

DIF: 3 REF: p.p. 59 OBJ: 3.1.3 Create velocity-time graphs.

v

t5 10 15 20 25 30 (s)

5

10

15

20

25

30

(m/s)

TOP: Create velocity-time graphs. KEY: Velocity-time graph MSC: 3 NOT: The velocity-time graph is a straight line making an acute angle with the positive x-axis for accelerated motion, a horizontal line for uniform motion, and a straight line making an obtuse angle with the positive x-axis for decelerated motion. 10. ANS: a) 0.80 m/s2 b) 0 m/s2 c) 1 m/s2 d) –1.6 m/s2 DIF: 3 REF: p.p. 66 OBJ: 3.2.3 Apply graphical and mathematical relationships to solve problems related to constant acceleration. TOP: Apply graphical and mathematical relationships to solve problems related to constant acceleration. KEY: Problems related to constant acceleration MSC: 3 NOT: The acceleration can be obtained by calculating the slope of the velocity-time graph. 11. ANS: 9.1 m/s DIF: 3 REF: p.p. 72 OBJ: 3.3.1 Define acceleration due to gravity. STO: 9.C.B.13 TOP: Define acceleration due to gravity. KEY: Acceleration due to gravity MSC: 3 NOT: Calculate the maximum height reached and then use the starting point as the highest point.

*END OF Exams 1-‐> 3*

Chapter 4 Test Multiple Choices Identify the letter of the choice that best completes the statement or answers the question. ____1. Which of the following system of forces provides the block the highest net force?

a.

11 N 71 N

c.

227 N 153 N

b.

405 N 403 N

d.

22.7 N 15.3 N PROBLEM 2. A car of mass 1330 kg is traveling at 28.0 m/s. The driver applies the brakes to bring the car

to rest over a distance of 79.0 m. Calculate the retarding force acting on the car. 3. Two men pull a 31-kg box with forces 9.7 N and 7.6 N in the directions shown below. Find

the resultant acceleration of the box and the direction in which the box moves.

9.7 N 7.6 N 4. An elevator is moving down with an acceleration of 3.36 m/s2. What would be the apparent

weight of a 64.2-kg man in the elevator? 5. An elevator is moving down with an acceleration of 1.40 m/s2. A 14.5-kg block hangs from a Spring balance fixed to the roof of the elevator. What is the apparent weight of the block? 6. The blocks shown below are placed on a smooth horizontal surface and connected by a piece

of string. If a 8.8-N force is applied to the 8.8-kg block, what is the tension in the string? 8.8 N

19.4 kg 8.8 kg

7. Three blocks A, B, and C are connected by two massless strings passing over smooth pulleys as shown below, with the 3.4-kg block on a smooth horizontal surface. Calculate the tension in the strings connecting A and B, and, B and C.

3.4 kg

6.9 kg 4.1 kg 8. A 2.1-kg block is kept on a 3.5-kg block resting on the floor of an elevator. If the elevator is

moving up at 1.1 m/s2, calculate the following: a. Force exerted by the 2.1-kg block on the 3.5-kg block. b. Force exerted by the floor of the elevator on the 3.5-kg block. 9. Two blocks A and B are kept in contact with each other on a smooth horizontal surface and a

3.9-N force is applied to the block A as shown below. Calculate the force exerted by the block A on the block B, assuming that they do not separate at any time. The mass of block A is 4.4 kg and that of block B is 5.5 kg.

3.9 N

10. Raindrops fall on Brian’s head at the rate of 4 drops per second. Each raindrop has a mass of

1.6 mg and falls with a speed of 25 m/s. assuming that on making contact with Brian’s head the drops come to rest and do not rebound, calculate the force felt by Brian.

Chapter4 ANSWER KEY

MULTIPLE CHOICES 1. ANS: A PROBLEM

2. ANS: N

3. ANS: m/s2 to left

4. ANS: 413 N

5. ANS: 12.4 N

6. ANS: 6.1 N

7. ANS: Tension in the string connecting A and B = 54 N Tension in the string connecting B and C = 48 N

8. ANS:

a. Force exerted by the 2.1-kg block on the 3.5-kg block = 23 N b. Force exerted by the floor of the elevator on the 3.5-kg block = 61 N

9. ANS 2.2 N 10. ANS: N

ANSWER KEY ANALYSIS MULTIPLE CHOICES

1. ANS: A DIF: 1 REF: p.p. 87 OBJ: 4.1.1 Define force. TOP: Define force. KEY: Force MSC: 1 NOT: /A/ Correct! /B/ the difference between the forces is not the maximum./C/ the net force is the difference between the two forces. /D/ this system does not give the highest net force. PROBLEM

2. ANS: N DIF: 3 REF: p.p. 87 OBJ: 4.1.2 Apply Newton's second law to solve

6.60× 103

6.8× 10−2

1.6× 10−4

6.60× 103

problems. STO: C.11-12.D, 12.C.2.5 TOP: Apply Newton's second law to solve problems. KEY: Newton's second law MSC: 3 NOT: Calculate the acceleration using the initial and final velocities. The product of the mass and acceleration gives the force.

3. ANS: m/s2 to left DIF: 3 REF: p.p. 87 OBJ: 4.1.2 Apply Newton's second law to solve problems. STO: C.11-12.D, 12.C.2.5 TOP: Apply Newton's second law to solve problems. KEY: Newton's second law MSC: 3 NOT: The resultant acceleration is obtained on dividing the net force by the mass of the box.

4. ANS: 413 N DIF: 3 REF: p.p. 97 OBJ: 4.2.2 Differentiate between actual weight and apparent weight. TOP: Differentiate between actual weight and apparent weight. KEY: Apparent weight MSC: 3 NOT: Use Newton's second law to find the apparent weight of the man.

5. ANS: 12.4 N DIF: 2 REF: p.p. 96 OBJ: 4.2.2 Differentiate between actual weight and apparent weight. TOP: Differentiate between actual weight and apparent weight. KEY: Apparent weight MSC: 3 NOT: Use Newton's second law to find the apparent weight of the block.

6. ANS: 6.1 N DIF: 3 REF: p.p. 102 OBJ: 4.3.2 Explain the tension in ropes and strings in terms of Newton's third law. TOP: Explain the tension in ropes and strings in terms of Newton's third law. KEY: Tension MSC: 3 NOT: Draw the free-body diagram for each block and then use Newton's second law to find the tension.

7. ANS: Tension in the string connecting A and B = 54 N Tension in the string connecting B and C = 48 N DIF: 3 REF: p.p. 102 OBJ: 4.3.2 Explain the tension in ropes and strings in terms of Newton's third law. TOP: Explain the tension in ropes and strings in terms of Newton's third law. KEY: Tension MSC: 3 NOT: Draw the free-body diagram for each block and then use Newton's second law to find the

6.8× 10−2

tensions.

8. ANS: a. Force exerted by the 2.1-kg block on the 3.5-kg block = 23 N b. Force exerted by the floor of the elevator on the 3.5-kg block = 61 N DIF: 3 REF: p.p. 102 OBJ: 4.3.4 Determine the value of the normal force by applying Newton's second law. STO: C.11-12.D, 12.C.2.5 TOP: Determine the value of the normal force by applying Newton's second law. KEY: Normal force MSC: 3 NOT: Draw the free-body diagram for each block and then use Newton's second law to find the interaction forces.

9. ANS 2.2 N DIF: 3 REF: p.p. 102 OBJ: 4.3.4 Determine the value of the normal force by applying Newton's second law. STO: C.11-12.D, 12.C.2.5 TOP: Determine the value of the normal force by applying Newton's second law. KEY: Normal force MSC: 3 NOT: Draw the free-body diagram for each block and then use Newton's second law to find the interaction force.

10. ANS: N DIF: 2 REF: P.P. 87 OBJ: 4.1.1 Define force. TOP: Define force. KEY: Force MSC: 3 NOT: The rate of change of momentum is equal to the force.

1.6× 10−4

Name ___________________________________________________ Period _________________

Chapter 5 Test PROBLEM 1. The magnitude of the force is 57 N, of force is 57 N, and of is 57 N. The angles

and are 30° each. Use the graphical method to find the resultant of the forces , and .

2. A vector has magnitude 58.0 units and is inclined to the positive axis of x at 42.0°. Vector

has magnitude 36.0 units and is inclined to the positive axis of x at 121°. Vector has magnitude 40.0 units and is inclined to the positive axis of x at 53.0°. Use the graphical method to find the resultant of the vectors , and .

3. A river flows at a speed of 4.60 m/s. A boat, capable of moving with a speed of 5.80 m/s in

still water is rowed across the river at an angle of 53.0° to the river flow. Calculate the resultant velocity with which the boat moves and the angle that its resultant motion makes to the river flow.

4. What should be the angle between two vectors of magnitudes 3.20 and 5.70 units, so that their

resultant has a magnitude of 6.10 units? 5. An airplane has to fly eastward to a destination 856 km away. If wind is blowing at 18.0 m/s

northward and the air speed of the plane is 161 m/s, in what direction should the plane head to reach its destination?

6. A 2.7-kg box is released on a horizontal surface with an initial speed of 2.9 m/s. It moves on

the surface with a deceleration of 0.27 m/s2. Calculate the kinetic friction force on the box.

7. A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0°. Once it has been set into motion, the angle is reduced to 30.0° to keep the block moving at constant speed. Calculate the maximum force of static friction and the force of kinetic friction for the surfaces in contact.

8. The system shown below is in equilibrium. Calculate the force of friction acting on the block

A. The mass of block A is 4.10 kg and that of block B is 4.30 kg. The angle q is 33.0°.

9. The blocks A and B are connected by a piece of spring. Block B rests on an inclined plane of

40° and blocks a hangs vertically. The coefficient of friction between block A and the inclined plane is 0.29. Calculate the acceleration of the system if the mass of block A is 0.15 kg and that of block B is 7.5 kg.

10. The blocks A and B are connected by a piece of spring. Block B rests on a smooth inclined

plane of 35° and blocks a hangs vertically. Calculate the acceleration of the system if the mass of block A is 1.06 kg and that of block B is 6.4 kg.

ANSWER KEY PROBLEM

1. ANS: Zero 0

2. ANS: The resultant is 113 units at an angle of 64.4° to the positive axis of x.

3. ANS: The boat moves with 9.32 m/s at 29.8° to the river flow.

4. ANS: 98.7°

5. ANS: 6.42° south of east 6. ANS: 0.73 N

7. ANS: The maximum force of static friction is 43.8 N and the force of kinetic friction is 38.2 N.

8. ANS: 64.9 N

9. ANS: 3.8 m/s2

10. ANS: 3.4 m/s2

ANSWER KEY ANALYSIS PROBLEM 1. ANS: Zero 0 DIF: 3 REF: p.p. 120 OBJ: 5.1.1 Evaluate the sum of two or more vectors in two dimensions graphically. TOP: Evaluate the sum of two or more vectors in two dimensions graphically. KEY: Sum of vectors MSC: 3 NOT: Draw the vectors to scale and use the protractor to draw them tip to tail. The closing side of the polygon in reverse order is the resultant.

2. ANS: The resultant is 113 units at an angle of 64.4° to the positive axis of x. DIF: 3 REF: p.p. 120 OBJ: 5.1.1 Evaluate the sum of two or more vectors in two dimensions graphically. TOP: Evaluate the sum of two or more vectors in two dimensions graphically. KEY: Sum of vectors MSC: 3 NOT: Draw the vectors to scale and use the protractor to draw them tip to tail. The closing side of the polygon in reverse order is the resultant. Measure the length of the resultant to get the magnitude of the vector, and use a protractor to find the direction of the resultant.

3. ANS: The boat moves with 9.32 m/s at 29.8° to the river flow. DIF: 3 REF: p.p. 122 OBJ: 5.1.3 Solve for the sum of two or more vectors by adding the components of the vectors. TOP: Solve for the sum of two or more vectors by adding the components of the vectors. KEY: Sum of vectors MSC: 3 NOT: Resolve the velocity of the boat into components along and perpendicular to the river flow.

4. ANS: 98.7° DIF: 3 REF: p.p. 122 OBJ: 5.1.3 Solve for the sum of two or more vectors by adding the components of the vectors. TOP: Solve for the sum of two or more vectors by adding the components of the vectors. KEY: Sum of vectors MSC: 3 NOT: Resolve one of the vectors into components along and perpendicular to the second vector.

5. ANS: 6.42° south of east DIF: 3 REF: p.p. 122 OBJ: 5.1.3 Solve for the sum of two or more vectors by adding the components of the vectors.

TOP: Solve for the sum of two or more vectors by adding the components of the vectors. KEY: Sum of vectors MSC: 3 NOT: Resolve the velocity of the plane in and perpendicular to the direction of the wind. 6. ANS: 0.73 N DIF: 2 REF: p.p. 127 OBJ: 5.2.1 Define the friction force. TOP: Define the friction force. KEY: Friction MSC: 2 NOT: The only force on the box is the kinetic friction, which should be equal to the product of mass and acceleration of the box.

7. ANS: The maximum force of static friction is 43.8 N and the force of kinetic friction is 38.2 N. DIF: 3 REF: p.p. 127 OBJ: 5.2.2 Distinguish between static and kinetic friction. TOP: Distinguish between static and kinetic friction. KEY: Inclined plane MSC: 3 NOT: The maximum force of static friction is equal to the force that just moves a static object. The kinetic friction force is equal to the force that keeps it moving at constant speed.

8. ANS: 64.9 N DIF: 2 REF: p.p. 131 OBJ: 5.3.1 Determine the force that produces equilibrium when three forces act on an object. TOP: Determine the force that produces equilibrium when three forces act on an object. KEY: Equilibrant MSC: 3 NOT: The tension in the three strings keeps the knot in equilibrium. The tension in the horizontal string is equal to the friction on the block A.

9. ANS: 3.8 m/s2 DIF: 3 REF: p.p. 132 OBJ: 5.3.2 Analyze the motion of an object on an inclined plane with and without friction. TOP: Analyze the motion of an object on an inclined plane with and without friction. KEY: Inclined plane MSC: 3 NOT: Resolve all forces parallel and perpendicular to the inclined plane. The friction force opposes the motion of the block B.

10. ANS: 3.4 m/s2 DIF: 3 REF: p.p. 132 OBJ: 5.3.2 Analyze the motion of an object on an inclined plane with and without friction. TOP: Analyze the motion of an object on an inclined plane with and without friction. KEY: Inclined plane MSC: 3 NOT: Resolve all forces parallel and perpendicular to the inclined plane.

Name ____________________________________________Period _________________

Chapter 6 Test MULTIPLE CHOICES Identify the letter of the choice that best completes the statement or answers the question.

____1. The path of a projectile through space is called its: a. equilibrant c. range b. torque d. trajectory ____2. A soldier throws a grenade horizontally from the top of a cliff. Which of the following

curves best describes the path taken by the grenade? a. Circle c. Hyperbola b. Ellipse d. Parabola ____3. A stone is thrown horizontally from the top of a 25.00-m cliff. The stone lands at a

distance of 40.00 m from the edge of the cliff. What is the initial horizontal velocity of the stone?

a. 2.260 m/s c. 17.70 m/s b. 15.60 m/s d. 22.05 m/s ____4. A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from

the base of the cliff, would the ball hit the ground? a. 23.6 m c. 31.9 m b. 26.4 m d. 45.0 m ____5. A ball is thrown horizontally from a hill 29.0 m high at a velocity of 4.00 m/s. Find the

distance between the base of the hill and the point where the ball hits the ground. a. 2.43 m c. 10.06 m b. 9.73 m d. 3.28 m ____6. A player kicks a football at an angle of 30.0° above the horizontal. The football has an

initial velocity of 20.0 m/s. Find the horizontal component of the velocity and the maximum height attained by the football.

a. 10.0 m/s, 17.6 m c. 25.1 m/s, 7.40 m b. 17.3 m/s, 5.10 m d. 30.3 m/s, 9.50 m

____7. A missile launches at a velocity of 30.0 m/s at an angle of 30.0° to the normal. What is

the maximum height the missile attains? a. 11.5 m c. 34.4 m b. 27.5 m d. 45.9 m ____8. A player kicks a football with an initial velocity of 3.00 m/s at an angle of 60.0° above the horizontal. What is the horizontal distance traveled by the football? a. 0.312 m c. 0.673 m b. 0.397 m d. 0.795 m ____9. The movement of an object or a point mass at a constant speed around a circle that has a

fixed radius is called uniform: a. circular motion c. elliptical motion b. parabolic motion d. rotational motion ____10. A sprinter runs at a speed of 3.00 m/s on a circular track that has a radius of 40.00 m.

Find the centripetal acceleration of the sprinter. a. 0.225 m/s2 c. 0.750 m/s2 b. 4.44 m/s2 d. 0.0750 m/s2 ____11. A 0.50-kg ball is attached to a string of 0.50 m and swung in a horizontal circle with a

velocity of 1.0 m/s. Find the centripetal force of the ball. a. 0.50 N c. 2.0 N b. 1.0 N d. 2.5 N ____12. A 1.00-kg ball is attached to a string of 0.50 m and swung in a horizontal circle with a

velocity of 2.00 m/s. Find the centripetal acceleration. a. 0.25 m/s2 c. 4.0 m/s2 b. 2.0 m/s2 d. 8.0 m/s2 SHORT ANSWERS 13. An object in uniform circular motion moves at a constant speed around a circle with a fixed

radius. Why is the object said to be accelerating though it has a constant speed? PROBLEMS: 14. A hiker throws a ball at an angle of 21.0° above the horizontal from a hill 21.0 m high. The

hiker’s height is 1.750 m. The magnitudes of the horizontal and vertical components of the velocity are 14.004 m/s and 5.376 m/s, respectively. Find the distance between the base of the hill and the point where the ball hits the ground. (Consider the hiker’s height while calculating the answer.)

15. A ball is tied to an elastic string of length 8.0 m and swung in a horizontal circle with a velocity of 0.8 m/s. When a metallic object is tied to a rope of length 2.75 m and swung in a horizontal circle, it makes one revolution in 2.9 s. The ratio of the centripetal force in the

string to the centripetal force in the rope is . Find the mass of the metallic object attached

to the rope, if the centripetal force in the string is 0.20 N.

16. Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m/s and the speed of B is 3.40 m/s. Both grenades remain

in air for 3.77 s. Assume that the acceleration due to gravity is 9.86 m/s2. What is the distance between A and B if they are thrown along the same straight line?

17. A boat traveling east covers a distance of 40.0 m in 20.0 s. It encounters a current moving at

a speed of 2.50 m/s traveling north. Find the resultant velocity of the boat. 18. A riverboat travels with a velocity of 4.60 m/s from one shore to another. The velocity of the

river is 2.30 m/s. If the width of the river is 72.0 m, how far does the boat travel downstream to reach the other shore?

19. A bus travels from west to east with a velocity of 11.5 m/s. A marble rolls on the surface of

the bus floor with a velocity of 0.200 m/s north. What is the velocity of the marble relative to the road?

20. Two boats, A and B, travel with a velocity of 4.90 m/s across a river of width 72.0 m. The

river flows with a velocity of 2.50 m/s. Boat A travels the shortest distance and boat B travels in the shortest time. If both start at the same time, how much time will they take to cross the river?

21. Tracy swims across a stream of width 40.0 m in 33.0 s when there is no current. She takes

59.0 s to cover the same distance when there is a current. Find the speed of the river current.

Chapter 6 ANSWER KEY

MULTIPLE CHOICES

1. ANS: D 2. ANS: D

3. ANS: C

4. ANS: C

5. ANS: B

6. ANS: B

7. ANS: C

8. ANS: D

9. ANS: A 10. ANS: A

11. ANS: B

12. ANS: D SHORT ANSWER 13. ANS:

An object in uniform circular motion has a constant speed, but its velocity keeps changing. Since velocity is a vector quantity, a change in direction indicates a change in velocity. Since the velocity changes, the object is said to be accelerating.

PROBLEM 14. ANS: 38.8 m

15. ANS: 0.05 kg

16. ANS: 3.39 m.

17. ANS: The boat is traveling 3.20 m/s at 51.3 m/s north of east.

18. ANS: 36.0 m

19. ANS: The marble is traveling 11.5 m/s at 1.00° north of east.

20. ANS: The time taken by boat A is 17.1 s. The time taken by boat B is 14.7 s.

21. ANS: 1.005 m/s

Chapter 6 ANSWER KEY ANALYSIS MULTIPLE CHOICES

2. ANS: D A projectile is any object that has been given an initial thrust and moves through air. Its path through space is called its trajectory. DIF: 1 REF: p.p. 147 OBJ: 6.1.1 Recognize that the vertical and horizontal motions of a projectile are independent. TOP: Recognize that the vertical and horizontal motions of a projectile are independent. KEY: Projectile, Trajectory MSC: 1 NOT: /a/ Equilibrant is a type of force. /b/ Torque is the product of the force and length of the lever arm. A projectile the horizontal distance travels /c/ Range. /d/ Correct!

3. ANS: D Draw a motion diagram for the trajectory showing the downward acceleration. The velocity will have two components, a horizontal and a vertical component. The combination of constant horizontal velocity and uniform vertical acceleration produces a distinct trajectory. DIF: 1 REF: p.p. 148 OBJ: 6.1.3 Explain how the shape of the trajectory of a projectile depends upon the frame of reference from which it is observed. TOP: Explain how the shape of the trajectory of a projectile depends upon the frame of reference from which it is observed. KEY: Projectile, Trajectory MSC: 1 NOT: /a/ a circular path would mean that the grenade falls on the soldier! /b/ the grenade cannot take an elliptical path, as gravity would act on it. /c/ the trajectory of the grenade cannot be a hyperbola as Earth would exert a gravitational pull on it. /d/ Correct!

4. ANS: C Solve the equation for the time the stone is in the air. Then solve the equation for the initial horizontal velocity by substituting the values of time and the range. DIF: 3 REF: p.p. 149

OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Projectiles launched horizontally MSC: 3 NOT: /a/ the time is 2.260 s. Divide the range by the time to get the initial horizontal velocity. /b/ Divide the range by the period, instead of multiplying the range by the period. /c/ Correct! /d/ you calculated the vertical velocity instead of the horizontal velocity.

5. ANS: C Use the equation for the y-position, and solve the equation for the time the stone is in the air. Multiply the time with the horizontal velocity to obtain the horizontal distance (range). DIF: 2 REF: p.p. 149 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Projectiles launched horizontally MSC: three NOT: /a/ Use the equation for the y-position to find the time period. /b/ Multiply the period with the horizontal velocity. /c/ Correct! /d/ the magnitude of the acceleration due to gravity is 9.80 m/s2.

6. ANS: B Use the equation for the y-position, and solve the equation for the time the stone is in the air. Multiply the time with the horizontal velocity to obtain the distance. DIF: 2 REF: p.p. 149 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Projectiles launched horizontally MSC: 3 NOT: /a/ you calculated the time the stone is in the air. Multiply this value with the initial horizontal velocity to find the range. /b/ Correct! /c/ Use the equation for the y-position to calculate how long the stone is in the air. /d/ to calculate the time the stone is in the air, use the formula t2 = –2 y/g.

7. ANS: B Write expressions for the vertical and horizontal velocity components. Solve the velocity equation for the time of maximum height. Substitute this time into the equation for vertical position to find the height.

DIF: 3 REF: p.p. 150 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Maximum range MSC: 3 NOT: /a/you calculated the vertical component instead of the horizontal component. /b/ Correct! /c/ the actual velocity is always more than its horizontal component. /d/ Can the horizontal component of the velocity be more than the actual velocity?

8. ANS: C Write the equations for the initial velocity components, the velocity components at time t, and the position in both directions. DIF: 2 REF: p.p. 150 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Maximum height MSC: 2 NOT: /a/ did you multiply the velocity with the cosine of the angle instead of the sine? /b/ the time taken to reach maximum height can be calculated using the formula t = vyo/g, where vyo is the vertical component of the velocity. /c/ Correct! /d/ Substitute the vertical component of the velocity in the equation for the maximum height.

9. ANS: D Solve the vertical-position equation for the time at the end of the flight, when y = 0. Substitute that value of time into the equation for horizontal distance to get the range. DIF: 2 REF: p.p. 150 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Maximum range MSC: 2 NOT: /a/ Solve the vertical-position equation for the time at the end of the flight when y = 0. /b/ to find the range, multiply the horizontal component of velocity with the time obtained when y = 0. /c/ the vertical component of velocity is the product of the velocity and the sine of the angle the football makes with the horizontal. /d/ Correct!

10. ANS: A Uniform circular motion is the movement of an object or point mass at a constant speed around a circle with a fixed radius.

DIF: 1 REF: p.p. 153 OBJ: 6.2.1 Explain why an object moving in a circle at constant speed is accelerated. TOP: Explain why an object moving in a circle at constant speed is accelerated. KEY: Uniform circular motion MSC: 1 NOT: /a/ Correct! /b/ a parabola is not a circle. /c/ an elliptical orbit does not have a fixed radius. /d/ a rotating object moves around its own vertical axis.

11. ANS: A

Substitute the values of velocity and radius in the equation for centripetal acceleration. DIF: 1 REF: p.p. 153 OBJ: 6.2.2 Describe how centripetal acceleration depends upon the object's speed and the radius of the circle. TOP: Describe how centripetal acceleration depends upon the object's speed and the radius of the circle. KEY: Centripetal acceleration MSC: 2 NOT: /a/ Correct! /b/ Divide the square of the velocity by the radius to calculate centripetal acceleration. /c/ Square the velocity in the formula for centripetal acceleration. /d/ Divide the square of the velocity by the radius.

12. ANS: B

Use Newton’s second law, .

DIF: 2 REF: p.p. 154 OBJ: 6.2.3 Identify the force that causes centripetal acceleration. TOP: Recognize the direction of the force that causes centripetal acceleration. KEY: Centripetal force MSC: 2 NOT: /a/ did you include the value of radius in the formula? /b/ Correct! /c/ Use Newton's second law. /d/ multiplies the mass with the square of the velocity and divides it by the radius.

13. ANS: D Substitute the value of velocity and radius in the equation for centripetal acceleration. DIF: 2 REF: p.p. 154 OBJ: 6.2.2 Describe how centripetal acceleration depends upon the object's speed and the radius of the circle. TOP: Describe how centripetal acceleration depends upon the object's speed and the radius of the circle. KEY: Centripetal acceleration MSC: 2 NOT: /a/ Divide the square of the velocity by the radius to get the centripetal acceleration. /b/ did you include the value of radius in the formula? /c/ Square the velocity in the formula. /d/ Correct!

SHORT ANSWER 14. ANS: An object in uniform circular motion has a constant speed, but its velocity keeps changing. Since velocity is a vector quantity, a change in direction indicates a change in velocity. Since the velocity changes, the object is said to be accelerating. DIF: 3 REF: p.p. 153 OBJ: 6.2.1 Explain why an object moving in a circle at constant speed is accelerated. TOP: Explain why an object moving in a circle at constant speed is accelerated. KEY: Uniform circular motion MSC: 2 PROBLEM 15. ANS: 38.8 m DIF: 3 REF: p.p. 150 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Projectiles launched at an angle MSC: 3 NOT: Solve the y-equation for y = 0 to find the time the ball is in the air. Multiply the time with the horizontal component of the velocity to obtain the range.

16. ANS: 0.05 kg DIF: 3 REF: p.p. 154 OBJ: 6.2.3 Identify the force that causes centripetal acceleration. TOP: Identify the force that causes centripetal acceleration. KEY: Centripetal force MSC: 3 NOT: Find the velocity of the metallic object. The centripetal force for the metallic object is the product of the ratio and the centripetal force. Substitute this value in the relation F = mv^2/r to get the mass.

17. ANS: 3.39 m. DIF: 3 REF: p.p. 149 OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion.; TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range using the horizontal motion. KEY: Projectile, Trajectory MSC: 3 NOT: Use the equation for the y-position to solve the equation for the time the objects are in the air.

18. ANS: The boat is traveling 3.20 m/s at 51.3 m/s north of east. DIF: 2 REF: p.p. 157 OBJ: 6.3.2 Solve relative-velocity problems. TOP: Solve relative-velocity problems. KEY: Relative velocity MSC: 3 NOT: Use the equation for the relative velocity of an object.

19. ANS: 36.0 m DIF: 3 REF: p.p. 157 OBJ: 6.3.2 Solve relative-velocity problems. TOP: Solve relative-velocity problems. KEY: Relative velocity MSC: 3 NOT: While calculating the time taken to reach the other shore, divide the width of the river by the boat velocity. While calculating the distance, multiply the time by the river velocity.

20. ANS: The marble is traveling 11.5 m/s at 1.00° north of east.

DIF: 2 REF: Page 157 OBJ: 6.3.2 Solve relative-velocity problems. TOP: Solve relative-velocity problems. KEY: Relative velocity MSC: 3 NOT: Use the equation for the relative velocity of an object.

21. ANS: The time taken by boat A is 17.1 s. The time taken by boat B is 14.7 s.

DIF: 3 REF: Page 157 OBJ: 6.3.2 Solve relative-velocity problems. TOP: Solve relative-velocity problems. KEY: Relative velocity MSC: 3 NOT: Use the equation for relative velocity of an object.

22. ANS: 1.005 m/s DIF: 3 REF: Page 157 OBJ: 6.3.2 Solve relative-velocity problems. TOP: Solve relative-velocity problems. KEY: Relative velocity MSC: 3 NOT: Use the equation to calculate the relative velocity of an object.

*END OF EXAMS 4 ->6*

Name ___________________________________________________ Period _________________

Chapter 7 Test

PROBLEM 1. The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun

and Earth. Calculate the time taken by Pluto to orbit the Sun in Earth days. 2. The Moon has an orbital period of 27.3 days around Earth and a mean distance of 3.85 x 105

km from Earth’s center. Use Kepler’s laws to find the orbital period of an artificial satellite orbiting Earth at a distance of km from the center of Earth.

3. Venus orbits the Sun with an orbital radius of m. Given that the mass of the Sun is

2.0 x 1030 kg, calculate the period of Venus’s orbit. 4. Jupiter orbits the Sun with an orbital radius of m. Given that the mass of the Sun is 2.00 x 1030 kg, calculate the orbital velocity of Jupiter. 5. Calculate the force of gravitational attraction between two spheres of mass 10.1 kg and 45.4

kg that are 38.5 m apart. 6. Venus has radius m and mass kg. Calculate the value of acceleration

due to gravity on Venus’s surface. 7. If Earth shrinks in size such that its shape and mass remain the same, but the radius

decreases to 0.21 times its original value, find the acceleration due to gravity on its surface. 8. A satellite orbits Neptune 4000 km above its surface. Given that the mass of Neptune is

kg and the radius of Neptune is m, calculate the orbital speed of the satellite.

9. A satellite orbits Jupiter km above its surface. Given that the mass of Jupiter is

kg and the radius of Jupiter is m, calculate the period of orbit of the satellite.

10. At what height above Earth’s surface does the gravitational intensity becomes 0.67 times its

value on the surface of Earth? Given the radius of the Earth is 6.40 x 106 m.

3.03× 104

1.08× 1011

7.78× 1011

6.05× 106 4.87× 1024

1.02× 1026 2.48× 107

6.40× 103

1.90× 1027 7.15× 107

Chapter7 ANSWER KEY PROBLEM

1. ANS: 89300 Earth days

2. ANS: 0.603 days

3. ANS: s

4. ANS: 13.1 km/s

5. ANS: N 6. ANS: 8.9 m/s2

7. ANS: 220 m/s2

8. ANS: 15 km/s

9. ANS: s

10. ANS: 1400 km

1.9× 107

2.06× 10−11

1.21× 104

Chapter7 ANSWER KEY ANALYSIS PROBLEM

1. ANS: 89300 Earth days DIF: 3 REF: p.p. 173 OBJ: 7.1.1 Relate Kepler's laws to the law of universal gravitation. TOP: Relate Kepler's laws to the law of universal gravitation. KEY: Kepler's laws MSC: 3 NOT: The square of the period of a planet's orbit around the Sun is proportional to the cube of its mean distance from the Sun.

2. ANS: 0.603 days DIF: 3 REF: p.p. 173 OBJ: 7.1.1 Relate Kepler's laws to the law of universal gravitation. TOP: Relate Kepler's laws to the law of universal gravitation. KEY: Kepler's laws MSC: 3 NOT: The square of the period of a planet's orbit around the Sun is proportional to the cube of its mean distance from the Sun.

3. ANS: s DIF: 3 REF: p.p. 176 OBJ: 7.1.2 Calculate orbital speeds and periods. TOP: Calculate orbital speeds and periods. KEY: Orbital period MSC: 3 NOT: The square of the period of a planet's orbit around the Sun is proportional to the cube of its mean distance from the Sun.

4. ANS: 13.1 km/s DIF: 3 REF: p.p. 176 OBJ: 7.1.2 Calculate orbital speeds and periods. TOP: Calculate orbital speeds and periods. KEY: Orbital speed MSC: 3 NOT: The orbital velocity of a planet is inversely proportional to the square root of its radius of orbit around the Sun.

5. ANS: N DIF: 3 REF: p.p. 177, p.p. 178 OBJ: 7.1.3 Describe the importance of Cavendish's experiment.

1.9× 107

2.06× 10−11

TOP: Describe the importance of Cavendish's experiment. KEY: Gravitational force MSC: 3 NOT: Use the mathematical form of Newton's law of gravitation.

6. ANS: 8.9 m/s2

DIF: 3 REF: p.p. 177, p.p. 178 OBJ: 7.1.3 Describe the importance of Cavendish's experiment. TOP: Describe the importance of Cavendish's experiment. KEY: Acceleration due to gravity MSC: 3 NOT: Acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of the planet's radius.

7. ANS: 220 m/s2 DIF: 3 REF: p.p. 177, p.p. 178 OBJ: 7.1.3 Describe the importance of Cavendish's experiment. TOP: Describe the importance of Cavendish's experiment. KEY: Acceleration due to gravity MSC: 3 NOT: Acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of the planet's radius.

8. ANS: 15 km/s DIF: 3 REF: p.p. 180 OBJ: 7.2.1 Solve orbital motion problems. TOP: Solve orbital motion problems. KEY: Orbital velocity MSC: 3 NOT: The orbital velocity of a satellite is directly proportional to the square root of the planet's mass and inversely proportional to the square root of the satellite's distance from the center of the planet.

9. ANS: s DIF: 3 REF: p.p. 180 OBJ: 7.2.1 Solve orbital motion problems. TOP: Solve orbital motion problems. KEY: Orbital period MSC: 3 NOT: The orbital period of a satellite is directly proportional to the square root of the cube of the satellite's distance from the center of the planet and inversely proportional to the square root of the planet's mass.

10. ANS: 1400 km DIF: 3 REF: p.p. 182, p.p. 183 OBJ: 7.2.3 Describe gravitational fields. TOP: Describe gravitational fields. KEY: Gravitational field

1.21× 104

MSC: 3 NOT: The gravitational intensity of a planet is inversely proportional to the square of the distance from Earth's center.

Name ___________________________________________________ Period _________________

Chapter 8 Test

PROBLEM 1. A disc of radius 5.70 cm rotates about its axis and a point 1.90 cm from the center of the disc

moves 34.5 cm in 12.2 s. Calculate the angular speed of the disc. 2. A car wheel turns through 277° in 10.7 s. Calculate the angular speed of the wheel.

3. A torque acts on a wheel rotating at 19.8 rad/s and increases its angular speed to 23.5 rad/s in 11.2 s. Find the angle through which the wheel turns during this time. 4. A disc rotates at 5.60 rad/s. An angular acceleration acts on the disc to change its angular

speed to 16.8 rad/s. If the disc turns through 12.2 rad during this time, calculate the angular acceleration of the disc.

5. A car travels at 1.70 m/s. The driver accelerates and increases the speed to 10.7 m/s in 2.20 s.

If the radius of its wheel is 0.590 m, calculate the angle turned by the wheel during this time. 6. A beam that weighs 10.0 N/m is 2.5 m long. It is supported at a point 0.78 m from one end.

Find the weight of the object that must be placed on the other end of the beam to balance it. 7. A worker sits at one end of a 183-N uniform rod that is 2.80 m long. A weight of 107 N is

placed at the other end of the rod. The rod is balanced when the pivot is 0.670 m from the worker. Calculate the weight of the worker.

8. A rod of length 2.4 m is pivoted at its center and masses of 2.3 kg and 3.1 kg are hung from

its two ends. Find the initial angular acceleration of the rod when the system is released from a starting position of the rod that is horizontal.

9. Calculate the net torque on the rectangular plate about the point O. Force F1 is 12.0 N, F2 is

21.0 N, F3 is 12.0 N, F4 is 16.0 N, and F5 is 19.0 N.

10. A 1.29-m long ladder weighing 200 N rests against a vertical wall so that the top of the

ladder is at a height of 0.590 m. A 603-N man stands on the ladder at a distance of 0.390 m along the ladder. Calculate the force exerted by the wall on the ladder. Assume that the wall is perfectly smooth.


PROBLEM

1. ANS: 1.49 rad/s

2. ANS: 0.452 rad/s

3. ANS: 243 rad

4. ANS: 10.3 rad/s2

5. ANS: 23.1 rad 6. ANS: 15 N

7. ANS: 549 N

8. ANS: 0.50 rad/s2

9. ANS: 73 N·m

10. ANS: 617 N

Chapter 8

ANSWER KEY ANALYSIS PROBLEM

1. ANS: 1.49 rad/s DIF: 3 REF: p.p., p.p. 199 OBJ: 8.1.2 Calculate angular velocity. TOP: Calculate angular velocity. KEY: Angular velocity MSC: 3 NOT: The angular speed of a rotating body is equal to the linear speed of a point divided by the distance of the point from the axis of rotation.

2. ANS: 0.452 rad/s DIF: 3 REF: p.p. 198, p.p. 199 OBJ: 8.1.2 Calculate angular velocity. TOP: Calculate angular velocity. KEY: Angular velocity MSC: 3 NOT: The angular speed of a rotating body is equal to the linear speed of a point divided by the distance of the point from the axis of rotation.

3. ANS: 243 rad DIF: 3 REF: p.p. 198, p.p. 199 OBJ: 8.1.4 Solve problems involving rotational motion. TOP: Solve problems involving rotational motion. KEY: Rotational motion MSC: 3 NOT: The angle turned is the product of the average angular speed and the time.

4. ANS: 10.3 rad/s2 DIF: 3 REF: p.p. 198, p.p. 199 OBJ: 8.1.4 Solve problems involving rotational motion. TOP: Solve problems involving rotational motion. KEY: Rotational motion MSC: 3 NOT: The angular acceleration is equal to the difference between the squares of the final and initial angular speeds divided by twice the angle turned.

5. ANS: 23.1 rad DIF: 3 REF: p.p. 198, p.p. 199 OBJ: 8.1.4 Solve problems involving rotational motion. TOP: Solve problems involving rotational motion. KEY: Rotational motion

MSC: 3 NOT: The angle turned is equal to the distance traveled by the car divided by the radius of the wheel. 6. ANS: 15 N DIF: 3 REF: p.p. 202, p.p. 203 OBJ: 8.2.2 Calculate net torque. TOP: Calculate net torque. KEY: Net torque MSC: 3 NOT: The torque due to the weight of the beam is equal to the torque due to the weight required to balance the beam.

7. ANS: 549 N DIF: 3 REF: p.p. 202, p.p. 203 OBJ: 8.2.2 Calculate net torque. TOP: Calculate net torque. KEY: Net torque MSC: 3 NOT: The torque due to the weight of the beam and the required weight is equal to the torque due to the man's weight.

8. ANS: 0.50 rad/s2 DIF: 3 REF: p.p. 202, p.p. 203 OBJ: 8.2.2 Calculate net torque. TOP: Calculate net torque. KEY: Net torque MSC: 3 NOT: The net torque is equal to the difference between the torque due to the weight of each mass. The angular acceleration is equal to the torque divided by the total moment of inertia.

9. ANS: 73 N·m DIF: 3 REF: p.p. 202, p.p. 203 OBJ: 8.2.2 Calculate net torque. TOP: Calculate net torque. KEY: Net torque MSC: 3 NOT: The net torque is the algebraic sum of the product of each force and its perpendicular distance from O.

10. ANS: 617 N DIF: 3 REF: p.p. 213 OBJ: 8.3.3 Define the conditions for equilibrium. TOP: Define the conditions for equilibrium. KEY: Equilibrium MSC: 3 NOT: Because the ladder is in equilibrium, the net force must be equal to zero and the net torque must be equal to zero.

*End of exams 7+8*

Name ___________________________________________________ Period _________________

Chapter 9 Test PROBLEM 1. A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes.

If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?

2. A 0.140-kg baseball is pitched horizontally at 36.7 m/s. When a player hits the ball, it moves

at the same speed, but in the opposite direction. If the bat and the ball are in contact for 0.450 ms, calculate the average force the bat exerts on the ball.

3. Candona strikes a 0.055-kg golf ball with a force of 260 N. If the ball moves with a velocity

of 65 m/s, calculate the time the ball is in contact with the club. 4. A force of 200 N acts on a 7.20-kg bowling ball for 0.350 s. Calculate its change in velocity. 5. The moment of inertia of an asteroid rotating about its own axis is kg m2. Its angular

velocity is 40.0 rad/s. If a force acts on the asteroid for 0.100 s, increasing the angular velocity to 48.0 rad/s, find its magnitude.

6. A toy car X of mass 0.200 kg moves along a frictionless surface with a velocity of 0.180 m/s.

It collides with another toy car Y, with a mass of 0.250 kg and a speed of 0.130 m/s in the same direction. After the collision, toy car X continues to move in the same direction with a velocity of 0.177 m/s. Calculate the speed of toy car Y after the collision.

7. A marksman at rest fires a 4.00-kg gun that expels a bullet of mass 0.0140 kg with a velocity

of 181 m/s. The marksman’s mass is 81.0 kg. What is the marksman’s velocity after firing the gun?

8. A rocket expels gases at a rate of kg/s with a speed of m/s. What is the

force exerted on the rocket? 9. Ball A, with a mass of 1.75 kg, moves with a velocity 3.50 m/s. It collides with a stationary

ball B, with a mass of 2.50 kg. After the collision, ball A moves in a direction 55.0 north to the left of its original direction, while ball B moves in a direction 35.0 south to the right of ball A’s original direction. Calculate the velocity of each ball after the collision.

5.00× 104

1.30× 103 3.00× 104


PROBLEM 1. ANS:

N

N

2. ANS: N

3. ANS: 0.014 s

4. ANS: 9.72 m/s

5. ANS: N

6. ANS: 0.132 m/s

7. ANS: 0.0298 m/s

8. ANS: N

9. ANS: The velocity of ball A is 2.01 m/s. The velocity of ball B is 2.01 m/s.

5.09× 103

3.13× 104

2.28× 104

4.00× 106

3.90× 107

Chapter 9

ANSWER KEY ANALYSIS PROBLEM 2. ANS:

N

N

DIF: 3 REF: p.p. 230 OBJ: 9.1.2 Determine the impulse given to an object. TOP: Determine the impulse given to an object. KEY: Impulse MSC: 3 NOT: Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle.

2. ANS: N DIF: 3 REF: p.p. 230 OBJ: 9.1.2 Determine the impulse given to an object. TOP: Determine the impulse given to an object. KEY: Impulse MSC: 3 NOT: Apply the impulse-momentum theorem to obtain the force the bat exerts on the ball.

3. ANS: 0.014 s DIF: 3 REF: p.p. 230 OBJ: 9.1.2 Determine the impulse given to an object. TOP: Determine the impulse given to an object. KEY: Impulse MSC: 3 NOT: Apply the impulse-momentum theorem to obtain the time the ball is in contact with the club.

4. ANS: 9.72 m/s DIF: 3 REF: p.p. 230 OBJ: 9.1.2 Determine the impulse given to an object. TOP: Determine the impulse given to an object. KEY: Impulse MSC: 3 NOT: Apply the impulse-momentum theorem to obtain the change in velocity.

5. ANS: N DIF: 3 REF: p.p. 234 OBJ: 9.1.3 Define the angular momentum of an object. TOP: Define the angular momentum of an object. KEY: Angular momentum MSC: 3 NOT: Apply the angular impulse-angular momentum theorem to find the angular impulse.

6. ANS: 0.132 m/s

5.09× 103

3.13× 104

2.28× 104

4.00× 106

DIF: 3 REF: p.p. 237 OBJ: 9.2.1 Relate Newton's third law to conservation of momentum. TOP: Relate Newton's third law to conservation of momentum. KEY: Conservation of momentum, Newton's third law MSC: 3 NOT: Apply Newton's third law and the law of conservation of momentum.

7. ANS: 0.0298 m/s DIF: 3 REF: p.p. 240 OBJ: 9.2.3 Solve conservation of momentum problems. TOP: Solve conservation of momentum problems. KEY: Conservation of momentum MSC: 3 NOT: Apply the law of conservation of momentum.

8. ANS: N DIF: 3 REF: p.p. 237 OBJ: 9.2.1 Relate Newton's third law to conservation of momentum. TOP: Relate Newton's third law to conservation of momentum. KEY: Conservation of momentum MSC: 3 NOT: Apply the law of conservation of momentum.

9. ANS: The velocity of ball A is 2.01 m/s. The velocity of ball B is 2.01 m/s. DIF: 3 REF: p.p. 242 OBJ: 9.2.3 Solve conservation of momentum problems. TOP: Solve conservation of momentum problems. KEY: Conservation of momentum MSC: 3 NOT: Apply the law of conservation of momentum in two dimensions.

3.90× 107

Name ___________________________________________________ Period _________________

Chapter 10 Test PROBLEM 1. Carol and Bruno drag a box of mass 58.0 kg along a frictionless floor. Carol pushes the box

with a force of 11.4 N at an angle of 40.0° downward from the horizontal. Bruno pulls the box from the other side with a force of 11.0 N at an angle of 40.0° above the horizontal. What is the net work done on the box if the displacement of the box is 14.5 m?

2. A cable pulls a stationary crate of mass 19.0 kg over a frictionless ramp at an angle 20.1°

above the ground. If the total distance traveled is 5.40 m, find the work done by the cable on the crate.

3. Raul pushes a stalled car with a force of 204 N. If the required force decreases at a constant rate from 204 N to 44.0 N for a distance of 16.3 m in 16.0 s, calculate the average power required to move the car.

4. A 1600-kg vehicle moves with a velocity of 19.5 m/s. Calculate the power required to reduce

the velocity to 3.20 m/s in 11.0 s. 5. Ayesha exerts a force of 186 N on a lever to raise a 2.70 103-N object to a height of 29.0

cm. If the efficiency of the lever is 83.3 percent, how far does Ayesha move her end of the lever?

6. A bicycle has a wheel of radius 43.4 cm and a gear of radius 5.20 cm. When the chain is

pulled with a force of 152.4 N, the wheel rim moves 15.7 cm. If the efficiency of the bicycle is 92.6 percent, how far was the chain pulled to move the rim 15.7 cm? Also, find the resistance force.

7. Aryton uses a pulley system to raise a 23.1-kg block to a height of 18.0 m. If a force of 124 N

is exerted and the rope is pulled 34.2 m, find the efficiency of the system.

8. A worker exerts a force of 119 N on a lever to raise a block. The efficiency of the lever is 91.6 percent. If the lever is moved 0.770 m, the block rises to a height of 0.180 m. What is the mass of the block?

Chapter10 ANSWER KEY PROBLEM

1. ANS: J

2. ANS: J

3. ANS: 126 W 4. ANS: W

5. ANS: 5.05 m

6. ANS: The distance is 1.88 cm. The resistance force is 16.9 N.

7. ANS: 95.8 percent

8. ANS: 47.6 kg

2.49× 102

3.46× 102

2.69× 104

Chapter 10 ANSWER KEY ANALYSIS PROBLEM

1. ANS: J DIF: 3 REF: p.p. 260 OBJ: 10.1.2 Calculate work. TOP: Calculate work. KEY: Work MSC: 3 NOT: Work is equal to the product of force and displacement times the cosine of the angle between the force and the direction of the displacement.

2. ANS: J DIF: 3 REF: p.p. 260 OBJ: 10.1.2 Calculate work. TOP: Calculate work. KEY: Work MSC: 3 NOT: Use the equation for work when a constant force is exerted in the same direction as the object's displacement. Since the object is moving against gravity, find the height of the object and substitute in the equation.

3. ANS: 126 W DIF: 3 REF: p.p. 263 OBJ: 10.1.3 Calculate the power used. TOP: Calculate the power used. KEY: Power MSC: 3 NOT: Power is equal to the work done divided by the time taken to do the work. 4. ANS: W DIF: 3 REF: p.p. 263 OBJ: 10.1.3 Calculate the power used. TOP: Calculate the power used. KEY: Power MSC: 3 NOT: Work is equal to the change in kinetic energy. Power is equal to the work done divided by the time taken to do the work.

5. ANS: 5.05 m DIF: 3 REF: p.p. 271 OBJ: 10.2.4 Calculate efficiencies for simple and compound machines. TOP: Calculate efficiencies for simple and compound machines. KEY: Efficiency MSC: 3 NOT: Use the formula for efficiency of a machine.

6. ANS: The distance is 1.88 cm. The resistance force is 16.9 N.

2.49× 102

3.46× 102

2.69× 104

DIF: 3 REF: p.p. 271 OBJ: 10.2.4 Calculate efficiencies for simple and compound machines. TOP: Calculate efficiencies for simple and compound machines. KEY: Mechanical advantage, Ideal mechanical advantage, Efficiency MSC: 3 NOT: For a wheel-and-axle machine, IMA is equal to the ratio of radii.

7. ANS: 95.8 percent DIF: 3 REF: p.p. 271 OBJ: 10.2.4 Calculate efficiencies for simple and compound machines. TOP: Calculate efficiencies for simple and compound machines. KEY: Efficiency MSC: 3 NOT: The efficiency of a machine is equal to the output work, divided by the input work, multiplied by 100.

8. ANS: 47.6 kg DIF: 3 REF: p.p. 271 OBJ: 10.2.4 Calculate efficiencies for simple and compound machines. TOP: Calculate efficiencies for simple and compound machines. KEY: Efficiency MSC: 3 NOT: Use the formula for efficiency of a machine.

Name ___________________________________________________ Period _________________

Chapter 11 PROBLEM 1. Andrew throws a 0.11-kg ball toward Donald, who is standing on a ledge. The ball leaves

Andrew’s hands at a height of 0.24 m and Donald catches it at a height of 0.82 m. Calculate the gravitational potential energy of the ball relative to the ground before being thrown.

2. A warehouse worker pushed a cart weighing 4.50 kg to the top of an inclined plane. Initially,

the cart was 0.670 m above the floor. If the top of the inclined plane is 2.70 m above the floor,

calculate the work done by gravity as the worker pushed the cart to the top of the plane. 3. A student lifts a 1.2-kg bag from her desk, which is 0.59-m high, to a locker that is 2.9-m

high. What is the gravitational potential energy of the bag relative to the desk? 4. A hemispherical bowl of radius 0.14 m is kept inverted on a tabletop. A 4.3-g matchbox slides

off from rest from the top of the hemisphere. What is the speed of the matchbox when it reaches a point on the bowl at an angle of 16.0° to the horizontal?

5. A skier is pushed from the top of a hill so that he starts moving down the hillside sloped at

27.6° to the horizontal with an initial speed of 0.434 m/s. After traveling 80.4 m, he reaches the bottom of the valley. Due to inertia, he then continues 70.4 m up another hillside sloped at 20.7° to the horizontal. What is the skier’s speed when he reaches the top of the hill? Assume that you can neglect friction.

6. A roller coaster car starts with an initial speed of 0.375 m/s from the top of a section of a

track that is sloping down at 26.2° to the horizontal. After traveling 80.3 m, it reaches the end of that section of the track and continues up the next section of the track that slopes up at 22.4° to the horizontal. If this section of the track has a length of 73.2 m, will the car reach the top of the section? Assume that you can neglect friction and that the car has no source of power.

7. A -kg bullet is fired with a velocity of 154 m/s toward a 5.44-kg stationary solid

block resting on a surface that has a coefficient of friction 0.215. The bullet emerges with a reduced velocity of 20.2 m/s after passing through the block. What distance will the block slide before coming to rest? Assume that the block does not lose any mass.

2.90× 10−3

8. A -kg bullet is fired with a velocity of 108.6 m/s toward a 5.99-kg block moving at a speed of 0.280 m/s in the same direction. The bullet emerges with a speed of 20.7 m/s and with a small piece of the block with a mass of 0.0016 kg sticking to it. What is the kinetic energy lost in the collision?

9. A 0.390-kg piece of wood is at rest on a frictionless table. A 8.30-g bullet, moving with a

speed of 495 m/s, strikes the piece of wood, and is embedded in it. After the collision, the piece of wood and the bullet move slowly down the table. What percentage of the system’s original kinetic energy was lost?

2.10× 10−3


PROBLEM

1. ANS:

2. ANS: –89.5 J

3. ANS: 27 J

4. ANS: Speed of matchbox = 1.4 m/s 5. ANS: Speed of skier = 15.6 m/s

6. ANS: Yes. The car will reach the top of the next hill.

7. ANS: m

8. ANS: 12 J

9. ANS: 97.9 %

1.21× 10−3

Chapter 11

ANSWER KEY ANALYSIS PROBLEM

1. ANS: DIF: 2 REF: p.p. 285 OBJ: 11.1.3 Determine the gravitational potential energy of a system. TOP: Determine the gravitational potential energy of a system. KEY: Gravitational potential energy MSC: 3 NOT: The gravitational potential energy of an object is equal to the product of its mass, the acceleration due to gravity, and its height from the reference level.

2. ANS: –89.5 J DIF: 2 REF: p.p. 285 OBJ: 11.1.3 Determine the gravitational potential energy of a system. TOP: Determine the gravitational potential energy of a system. KEY: Gravitational potential energy MSC: 3 NOT: The work done by gravity is the weight of the cart times the vertical distance through which the cart is pushed.

3. ANS: 27 J DIF: 2 REF: p.p. 285 OBJ: 11.1.3 Determine the gravitational potential energy of a system. TOP: Determine the gravitational potential energy of a system. KEY: Gravitational potential energy MSC: 3 NOT: The gravitational potential energy of an object is equal to the product of its mass, the acceleration due to gravity, and its height from the reference level.

4. ANS: Speed of matchbox = 1.4 m/s DIF: 3 REF: p.p. 293 OBJ: 11.2.1 Solve problems using the law of conservation of energy. TOP: Solve problems using the law of conservation of energy. KEY: Law of conservation of energy MSC: 3 NOT: The kinetic energy gained by an object is equal to the gravitational potential energy lost by the object.

5. ANS: Speed of skier = 15.6 m/s

DIF: 3 REF: p.p. 293 OBJ: 11.2.1 Solve problems using the law of conservation of energy. TOP: Solve problems using the law of conservation of energy. KEY: Law of conservation of energy MSC: 3 NOT: The kinetic energy gained by an object is equal to the gravitational potential energy lost by the object.

6. ANS: Yes. The car will reach the top of the next hill. DIF: 3 REF: p.p. 293 OBJ: 11.2.1 Solve problems using the law of conservation of energy. TOP: Solve problems using the law of conservation of energy. KEY: Law of conservation of energy MSC: 3 NOT: The gravitational potential energy gained by an object is equal to the kinetic energy lost by the object.

7. ANS: m DIF: 3 REF: p.p. 293 OBJ: 11.2.2 Analyze collisions to find the change in kinetic energy. TOP: Analyze collisions to find the change in kinetic energy. KEY: Collisions, Change in kinetic energy MSC: 3 NOT: The principle of conservation of momentum can be used to find the final velocity of the block. The kinetic energy of the block is used in doing work against friction.

8. ANS: 12 J DIF: 3 REF: p.p. 293 OBJ: 11.2.2 Analyze collisions to find the change in kinetic energy. TOP: Analyze collisions to find the change in kinetic energy. KEY: Collisions, Change in kinetic energy MSC: 3 NOT: The principle of conservation of momentum can be used to find the final velocity of the block without the missing part. The loss in kinetic energy is the difference between the initial total kinetic energy and the final total kinetic energy.

9. ANS: 97.9 % DIF: 3 REF: p.p. 293 OBJ: 11.2.2 Analyze collisions to find the change in kinetic energy. TOP: Analyze collisions to find the change in kinetic energy. KEY: Collisions MSC: 3 NOT: Use the conservation of momentum equation to find the final velocity. Determine the

1.21× 10−3

change in kinetic energy of the system. Use the change in the kinetic energy to find the fraction of the original kinetic energy lost.

*End of Exams 9 -> 11*

Physics%–%1st%Quarter% - Lorain City School District · PDF file11 L12text’ ......

Documents

Transcript of Physics%–%1st%Quarter% - Lorain City School District · PDF file11 L12text’ ......