®Films for the Humanities & Sciences

I n s t r u c t o r ’s Guide

PHYSICS

TheStandardDeviants®

CoreCurriculum

IntroductionThe Core Curriculum Series is designed to be a complete treatment of major topics covered inan introductory course on a given curricular subject. The series utilizes a unique educationalapproach that combines the best of visual and tutorial elements. Through high-end graphics,animation, and design techniques, academic concepts are broken down, thoroughly explained,and demonstrated through the creative use of presentational devices and examples.

Each individual subject series consists of up to 10 programs of approximately 20 minutes each.These programs can be integrated into the lesson plans for an entire course or as stand-alonesupplements for specific topics, as needed.

Each program is accompanied by an instructor’s guide that contains the following elements:• Program Outline• Key Terms and Formulas• Quiz• Solutions to Quiz• Suggestions for Instructors

The Program Outline lists the main topics covered in each program, and also relates each pro-gram to the overall subject series. For example, if you choose to view Program 3, you’ll alsosee how Program 3 fits into the overall 10-part series.

Note: Occasionally during the show, you may notice a small heading in the corner of the screenthat makes reference to Part and Section divisions in the program. These headings refer to theprogram in its original, extended format and do not correspond to the format of this series.

2Copyright © 2001 Films for the Humanities & Sciences® A Films Media Group Company • P.O. Box 2053 • Princeton, NJ 08543-2053

Program OutlinePROGRAM 1: NUMBERS, UNITS, SCALARS, AND VECTORS . . . . . . . . . . . . . . . . . . . page 6

Introduction to Physics• Using Scientific Notation

Multiplying Exponents Converting Numbers into Scientific Notation

• Using Numbers and Units in PhysicsBasic UnitsConversion and Cancellation of Units

Scalars and Vectors• What are Scalars and Vectors?

MagnitudeDirection

• Adding VectorsHead-to-Tail MethodComponent Method

• Multiplying VectorsDot or Scalar ProductCross or Vector Product

PROGRAM 2: ONE-DIMENSIONAL KINEMATICS . . . . . . . . . . . . . . . . . . . . . . . . . . page 11

One-Dimensional Kinematics• Basic One-Dimensional Quantities

DisplacementVelocityAcceleration

• One-Dimensional Kinematic EquationsInstantaneous VelocityDisplacementFinal Velocity

PROGRAM 3: PROJECTILE MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 15

Two-Dimensional Kinematics• Projectile Motion

First Projectile Motion EquationSecond Projectile Motion EquationThird Projectile Motion EquationFourth Projectile Motion Equation

PROGRAM 4: CIRCULAR AND ROTATIONAL MOTION . . . . . . . . . . . . . . . . . . . . . . page 21

• Uniform Circular MotionCentripetal ForceCentripetal AccelerationPeriod

• Accelerated Circular MotionTangential AccelerationTotal Acceleration

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• Rotational MotionRotating Through an AngleRadiansAngular VelocityLinear Speed

PROGRAM 5: LINEAR MOMENTUM AND NEWTON’S LAWS OF MOTION . . . . . . . page 26

Linear Momentum and Newton’s Laws of Motion• Linear Momentum

The Principle of Conservation of Linear MomentumCollisionsFinal Momentum = Initial Momentum

• Newton’s First Law of MotionObject in Motion Stays in MotionObject at Rest Stays at RestLaw of Inertia

• Newton’s Second Law of MotionForceF = maWeightEquilibriumTension

• Newton’s Third Law of MotionEqual But Opposite ForceMomentum and Newton’s Third Law

PROGRAM 6: FRICTION, WORK, AND ENERGY . . . . . . . . . . . . . . . . . . . . . . . . . . page 32

Friction• What is Friction?• Kinetic and Static Friction

Kinetic Friction Acts Opposite MotionNormal ForceStatic Friction Tries to Prevent MotionCoefficient of Friction

Work and Energy • Work • Power • Kinetic Energy

PROGRAM 7: GRAVITATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 38

Gravitation • The Law of Universal Gravitation • Kepler’s Laws of Planetary Motion

PROGRAM 8: HARMONIC MOTION AND WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . page 43

Harmonic Motion • Simple Harmonic Motion• Behavior of Systems Undergoing Simple Harmonic Motion • The Simple Pendulum

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Waves • Wave Motion • Wave Speed • Types of Waves

PROGRAM 9: HEAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 48

Heat • Heat Transfer • Changes of State • Conduction, Convection, and Radiation • Heat and Work

PROGRAM 10: THERMODYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 54

Thermodynamics • The First Law of Thermodynamics • The Second Law of Thermodynamics • Heat Engines • Entropy

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PROGRAM 1:NUMBERS, UNITS, SCALARS, AND VECTORS

Key Terms and FormulasPhysics is the study of how things work.

Scientific notation is a way of writing really big or really small numbers so they’re easier towork with by using exponents to express the number.

Kinematics is the study of motion.

Magnitude is the quantity of something, or how much of it there is.

Direction is the way something is going, like north, south, east, or west, and is oftendescribed in degrees.

A scalar is a quantity that only has magnitude.

A vector is a quantity that has magnitude AND direction.

The sum of two vectors is called the resultant vector.

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Quiz

1. The density of water is 103 kg/m3 (or 1 gram per cubic centimeter). What is the mass of 1 liter of water? [A liter is equal to 10-3 m3.]

2. The mass of an atom is practically all in its nucleus. The radius of a uranium238 nucleus is8.68 x 10-15 m. Since the atomic mass of 238U is 238 atomic mass units (amu) and 1 amu isequal to a mass of 1.6605 x 10-27 kg, obtain the density of “nuclear matter.”

3. Can the difference of two vectors have a greater magnitude than the sum of the same twovectors? Give an example that shows your answer.

4. Two vectors, 6 and 9 units long, form an angle (a) 0°‚ and (b) 60°. Find the magnitude anddirection of their vector sum.

5. Two vectors, 6 and 9 units long, form an angle (a) 0°, and (b) 60°. Find the magnitude oftheir dot or scalar product.

6. Two vectors, 6 and 9 units long, form an angle (a) 0°, and (b) 60°. Find the magnitude oftheir cross product. If the shorter vector lies along the +X-axis and the other vector is in theXY-plane, determine the direction of the cross product.

7. Find the x and y components of a vector 15 units long when it forms an angle, with respectto the +X-axis, of (a) 50°, (b) 130°, and (c) 230°.

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Solutions to Quiz

1. Since density is mass per unit volume, or ρ = m/V, then m = ρV. Substitution yields:

m = [103 kg/m3][1 liter][10-3 m3/liter] = 1kg.

2. Since density is mass per unit volume, or m/V, we need to find the mass and the volume ofthe 238U nucleus. The mass is found by multiplying the number of nucleons by the atomicmass unit, or:

m = 238 amu{1.6605 x 10-27 kg/amu] = 3.952 x 10-25 kg

Assuming we can think of a nucleus as a sphere, we find the volume from

V = (4/3)πr3 = (4/3) π[8.68 x 10-15m]3 = 2.739 x 10-42 m3

Dividing the mass by the volume, the result is

r = [3.952 x 10-25 kg] / [2.739 x 10-42 m3] = 1.443 x 1017 kg/m3.

3. Yes! The trick here is that we must remember that vectors have d i re c t i o n as well as magnitude.For example, if two vectors have exactly opposite direction and have identical magnitude, theirsum is exactly zero, while their diff e rence is twice that of either one. More involved examplescan be found when the two vectors make an angle other than 180° with respect to each otherand/or do not have the same magnitude.

4. (a) This is easy. Think of the 6 unit vector lying across the +X-axis. Then the 9 unit vectoralso is along the +X-axis and, using head to tail, we see the resultant vector lies along the+X-axis and has a magnitude 15.

(b) Again think of the 6 unit vector lying across the +X-axis. Then draw the 9 unit vectorfrom the head of the 6 unit vector up an angle of 60°. The magnitude of their resultant isless than 15 units and by measuring we could find that length. The angle the resultantmakes with the +X-axis could also be measured with a protractor. To do this problem math-ematically, we can break the 9 unit vector into components as:

9x = 9cos60° = 9[0.500] = 4.500 units

9y = 9sin60° = 9[0.866] = 7.794 units

This then gives Rx = 6 + 4.5 = 10.5 units and Ry = 7.794 units. The square of the magnitudeof the resultant is then

R2 = (Rx)2 + R(y)2 = [10.5]2 + [7.794]2 = 170.996

Giving a magnitude of

R = 13.08 units

To get the angle, the tangent of the angle is given by tanθ= Ry/R = 7.794/10.5 = 0.7422which gives us an angle of θ = 36.6°. We could also get the angle from cosθ = Rx/R =10.5/13.08 = 0.8028, which also gives the angle as 36.6°.

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5. (a) First, we write the basic equation for the product: |A • B|= ABcosθ. For part (a) sincecos0° = 1.00, the dot product is simply 54.

(b) Since cos60° = 0.500, the scalar or dot product has a magnitude (6)(9)(0.500) = 27.

6. (a) First, we write the basic equation for the cross product: |A x B| = ABsinθ. For part (a)since sin0° = 0, the magnitude in this case is zero and there is no vector.

(b) Since sin60° = 0.866, we have a magnitude of (6)(9)(0.866) = 46.76. To find the direction,if we are "turning" the 6 unit vector "into" the 9 unit vector, the right hand rule has thethumb pointing UP, which usually means in the +Z direction.

7. The X component of a vector is given by Ax = Acosθ and the Y component is given by Ay= Asinθ. Knowing this makes this problem quite easy.

(a) Ax = 15cos50° = 15(0.6428) = 9.64 units; Ay = 15sin50° = 15(0.7660) = 11.49 units.

(b) Ax = 15cos130° = 15(-0.6428) = -9.64 units; Ay = 15sin130° = 15(0.7660) = 11.49 units.

(c) Ax = 15cos230° = 15(-0.6428) = -9.64 units; Ay = 15sin230° = 15(-0.7660) = -11.49 units.

These angles were chosen to show you that it is VERY important to consider the angle of a vector.

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Suggestions for Instructors

1. There is considerably more to physics than collections of facts, although students sometimesfeel that is what gets them through a course. Indeed, there are many facts that are necessaryto be known in order to successfully negotiate physics, even at the most basic level. Forinstance, a knowledge of the relative size of "things" is rather important. For this reason, thefirst topic of consideration is units and numbers because, if you don’t measure objects, youare not "doing" physics. The first two problems given are to look at the ordinary world andthen compare it with the subatomic world to show just how very different they are. Othersimilar problems should be assigned: the size of the Sun [e.g., "How many Earths could fit in the volume of the Sun?"], the distance between galaxies [e.g., "if it takes light eight min-utes to travel from the Sun to the Earth, how large is our Milky Way galaxy across?"], or theenormous number of atoms in even a small body [e.g., "Calculate the number of moleculesin one cubic centimeter of water."].

2. By the end of any reasonable course in physics, a student should be aware that there aretwo levels of description of nature. One is macro and thus global and phenomenological,corresponding to the world we perceive directly (sometimes with instruments). The other ismicro, and is structural; it is the domain of atoms and quantum theory. It is not necessary to delve deeply into the micro level; however, the student should be aware that it does existand that "rules" that appear to be different from those followed in the macro world aresometimes developed. Also it is important for the student to recognize that there are twokinds of physical laws. One kind is fundamental, such as the conservation principles or thelaws of gravitation. The other kind is statistical, corresponding to the laws of friction, the gaslaws, Ohm’s law, etc.

3. The need to distinguish quantities that are purely numeric, i.e., scalars, from those physicalquantities that require direction as well, i.e., vectors, may be developed by some rather sim-plistic demonstrations using tools such as a pair of rulers, especially if they are of differentlengths. "Adding" two lengths can easily be shown to have numerous values as one ruler isplaced at the end of another and then rotated. Having a third ruler then extended from the"tail" of the first to the "head" of the second, magnitude and angle change can be seen. It’seven possible to "prove" the Pythagorean theorem this way (as Pythagorus did himself).

4. Introduction of the scalar and vector products is usually difficult for students with little three-dimensional play in their previous experience. Encourage your students to look at some ofthe geometrical relations that are reflected in these products. For instance, the vector producthas a numeric value equal to the area of the parallelogram created when the two vectors areplaced "tail to tail" to form two of the sides, while the remaining sides are a parallel pair. Inthe case of the scalar product, the result is equal to the product of either one by the "verticalshadow" of the other.

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PROGRAM 2:ONE-DIMENSIONAL KINEMATICS

Key Terms and FormulasOne-dimensional kinematics focuses on motion along a straight line.

Displacement, or Δx, is the change in position of an object.

Average velocity, represented by vave, is the change in position, or displacement (Δx) of anobject divided by the change in time (Δt).

Average acceleration, represented by aave, is the change in velocity (Δv) over the change intime (Δt).

Instantaneous velocity is simply the velocity of an object at a given instant in time.

Instantaneous acceleration is the acceleration of an object at a given instant in time.

The kinematic equations can only be used when we have an object traveling in a straightline at a constant acceleration:

vf = vi + atx = vit + 1/2 at2

vf2 = vi

2 + 2axwhere vf is final velocity, vi is initial velocity, a is acceleration, x is position, and t is time.

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Quiz

1. Describe the procedure to measure the average velocity of a body in one-dimensionalmotion.

2. What is the difference between average velocity and instantaneous velocity?

3. A body, at t = 0 seconds, has a velocity of 3 m/s and a constant acceleration of 4 m/s2 in thesame direction as the velocity. (a) What is the velocity of the body at t = 7 s? (b) What is thedistance covered after 7 seconds?

4. A ball is thrown straight up in the air with an initial velocity of 30 m/s. How high will it rise?

5. A car is moving along a street at a rate of 45 km/hr (12.5 m/s) when a red light flashes onat an intersection 50 m ahead. If the driver’s reaction time is 0.7 s, what is the distance tothe intersection when the driver begins to apply the brakes?

6. If the driver in the car above decelerates the car uniformly at 2 m/s2 when the brakes areapplied, will the car stop safely?

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Solutions to Quiz

1. Find the position of the body at two times. Divide the change in position (i.e., final positionminus initial position) by the change in time (i.e., final time minus initial time). In symbols:

vave= [xf – xi]/[tf – ti] = Δx/Δt

2. Average velocity is defined as in answer #1 above; for instantaneous velocity we mean thevelocity at a specific time. Only in the case of uniform rectilinear (one-dimensional) motiondo the two have the same value at all times.

3. (a) a = 4 m/s2 = [vf – vi]/[tf – ti] = [vf – 3 m/s]/[7 s – 0 s]

vf – (3 m/s) = [4 m/s2][7 s] = 28 m/s

vf = 28 m/s + 3 m/s = 31 m/s.

(b) xf = xi + vit + 1/2at2 = 0 m + [3 m/s][7 s] + 1/2[4 m/s2][7 s]2

= 0 m + 21 m + 1/2[4 m/s2][49s2] = 21 m + 98 m = 119 m.

This could also be solved with the equation vf2 – vi

2 = 2a(xf – xi) since we just foundthe final velocity of the body. Using this, we have:

[31 m/s]2 – [3 m/s]2 = 2[4 m/s2][xf – 0 m] = 8 m/s2[xf]

961 m2/s2 – 9 m2/s2 = 952 m2/s2 = 8 m/s2[xf] or xf = [952 m2/s2]/[8 m/s2] = 119 m.

4. For this problem realize that if up is positive, then the acceleration (due to gravity) is down,or negative. The previous equation, vf

2 – vi2 = 2a(xf – xi), now becomes:

0 – vi2 = -2g(xf – 0)

And when we introduce the values of vi = 30 m/s and g = 9.8 m/s2, we have

-[30 m/s]2 = -900 m2/s2 = -2[9.8 m/s2]xf

xf = [-900 m2/s2]/[-19.6 m/s2] = 45.9 m.

5. First, distance equals velocity (uniform, because the driver hasn’t touched the brakes) timestime, or x = (12.5 m/s)(0.7 s) = 8.75 m. This is the distance covered before the driver startsto apply the brakes. So, the distance remaining is 50 m – 8.75 m = 41.25 m.

6. Again, using vf2 – vi

2 = 2a(xf – xi), we have:

0 – [12.5 m/s]2 = 2(-2 m/s2)(Δx)

So, Δx = -[156.25 m2/s2]/[-4 m/s2] = 39.06 m. Whew! Barely 2 meters to spare!13

Suggestions for Instructors

1. The most fundamental and obvious phenomenon we observe around us is motion. Blowingair, waves in the ocean, flying birds, falling leaves—all of these are examples of motion.Practically all imaginable processes can be traced back to the motion of certain particles orobjects. The Earth and other planets move around the Sun. The Sun, in turn, carries thesolar system around the center of our galaxy, the Milky Way. Electrons move within atoms,giving rise to the absorption and emission of electromagnetic radiation. Electrons move with-in metal, producing an electric current. Gas molecules move in a random fashion, giving riseto pressure and diffusion processes. Our everyday experience tells us that the motion of abody is influenced by the bodies around it; that is, by its interactions with them. One roleof the physicist, and the engineer as well, is to discover the relation between motions andthe interactions responsible for them and then to arrange things so that useful motions areproduced. The rules (or principles) that apply to the analysis of all kinds of motion arecalled mechanics and this is why most physics courses begin with its study. To carry out aprogram in mechanics, however, we must begin by learning how to describe the motions weobserve; hence, one-dimensional kinematics (from the Greek, kinema) as our first step.

2. An object is in motion re l a t i v e to another when its position, measured re l a t i v e to the secondbody, is changing with time [when the position doesn’t change, the object is at relative re s t] .Both rest and motion are relative concepts, so it is rather necessary to define a frame of re f e re n c e relative to which motion (or rest) is analyzed. Usually we choose the "fixed, flat"Earth as that frame, and this works for many cases. However, this frame fails, miserably insome outstanding instances, and such cases should be discussed with students. For instance,using the Sun as the frame of re f e rence, planetary motion becomes quite simple—which intro-duces such wonderful characters as Copernicus and Kepler. Also, Coriolis motion is "magic"until students recognize that it’s simply a case of ignoring the rotation of the Earth on its axis.

3. Graphing position (displacement) versus time is a useful exercise for students. More of intere s t ,h o w e v e r, is graphing velocity as a function of time for a number of reasons: A uniform veloci-ty graphs as a straight horizontal line; a uniform acceleration results in a graph of velocity vs.time as a straight line, whose s l o p e is the value of the uniform acceleration; the a re a under thecurve between any two times is the displacement during that time interval. All the basic equa-tions developed in the program can be "proven" from the geometry of these graphs—even thedisplacement equation under uniform acceleration: xf = xi + vit + 1/2a(tf – ti)

2.

4. Simple 1-D problems, where gravity is the uniform acceleration, help the student becomefamiliar with breaking down a problem into parts. This is a skill instructors should encour-age students to develop because it can carry over to many other fields of endeavor.

5. To spark interest, get the latest data on the expanding universe from the Hubble telescopeinformation available at www.nasa.gov.

6. Even though the work here is in 1-D, continue to stress the vector nature of displacement,velocity, and acceleration.

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PROGRAM 3:PROJECTILE MOTION

Key Terms and FormulasTwo-dimensional kinematics is motion in a plane, as opposed to motion along a straightline, like one-dimensional kinematics.

Projectile motion is when an object is tossed into the air. The object’s flight follows a curved path.

Gravity is the force pulling down on the object (in the -y direction). The acceleration of any object due to gravity, near the surface of the earth, is 9.8 m/s2.

The initial magnitude of the horizontal or x-component of a projectile is found using the equationvix = vi cosθ

and the initial magnitude of the vertical or y-component of a projectile is found using theequation viy = vi sinθ

To find the velocity in the y direction, use vy = viy – gt

and to find the horizontal velocity, use vx = vix

To find the distance that an object has traveled in the x direction, use xf = xi + vixt

To determine the displacement of an object in the y direction, use yf = yi + viyt – 1/2 gt2

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Quiz

1. In projectile motion the horizontal component of the projectile’s velocity remains constant.Why?

2. For the ideal case of no air resistance, a projectile follows the path of a parabola. What happens to the path when there is some air resistance?

3. Since a thrown baseball is a projectile, how is it possible to throw a curve?

4. A projectile is shot with a velocity of 70 m/s at an angle of 60° with the horizontal. Calculate(a) the maximum height; (b) the time to reach the maximum height; (c) the total time toagain reach the ground; (d) the horizontal range.

5. A helicopter is flying at an altitude of 150 m with a horizontal velocity of 100 km/hr (27.78 m/s). It drops a package of supplies to hit the blanket of a stranded hiker. Show that the package should be released when the copter is a horizontal distance of 154 m from the blanket.

6. The professor sits on a platform 15 m high. A student stands 30 m away from the base ofthe platform with a ripe, red tomato. The student can toss the tomato with a speed of 50miles per hour (22.3 m/s). Calculate the angle the tomato should be thrown if the studentREALLY wants to hit the professor.

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Solutions to Quiz

1. The "force of gravity" is straight down, so the only acceleration is straight down. If there’s no acceleration in the horizontal direction, then there’s no acceleration [i.e., CHANGE invelocity] in that direction. No acceleration, no change in velocity.

2. Lots of interesting things happen. Thinking of the components of the velocity, the verticalcomponent will slow down MORE than just due to gravity and so the projectile will not goas high as in the ideal case; on the way down the vertical component will not speed up asfast as it would without friction, so its value is considerably different than its original valuewhen it gets back to the ground; now the horizontal component slows down all during theflight. So if you first draw the ideal parabola case, the real path will not go as high, willpeak at a shorter horizontal value and then curve down more sharply than the original, butnever vertical.

3. The only way possible is for a force to be applied to the ball in a direction perpendicular tothe plane that the ball would normally follow. A pitcher is able to have such a force be pro-duced by spinning the ball as it is released. The aerodynamics of this is rather involved andwill take more analysis than we can spend time on here. Nevertheless, go throw curve balls!

4. Before we begin, break the velocity in the horizontal (x) and vertical (y) components:

vx = v0cos60° = (70 m/s)0.5 = 35 m/s

vy = v0sin60° = (70 m/s)0.866 = 60.62 m/s

(a) This is just like throwing a ball straight up with an initial velocity of 60.62 m/s. Gravity isdown and so we use vf

2 = vi2 – 2gy or 0 = (60.62 m/s)2 – 2(9.8 m/s2)h where we have

used h instead of y because we are after the height when the projectile has no morepositive velocity in the vertical direction. Solving, we get:

h = (60.62 m/s)2/[2(9.8 m/s2)] = (3675 m2/s2)/(19.6 m/s2) = 187.5 m.

(b) To find the time, let’s use the equation vf = vi – gt or 0 = 60.62 m/s – (9.8 m/s2)t.This gives t = [60.62 m/s]/(9.8 m/s2) = 6.18 s.

(c) If it takes 6.18 seconds to get up to the highest point, it will take the same amount of time to go back down. The total time then is simply T = 2t = 2(6.18 s) = 12.36 s.

(d) This is really easy now because we know the horizontal velocity, 35 m/s, and the totaltime the projectile is in the air, and so: Range = vxT = (35 m/s)(12.36 s) = 432.6 m.

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5. If we knew how long it would take the package to fall, we could calculate the horizontaldistance it would travel in that time. That is, D = vxt. We know vx equals 27.78 m/s andfrom the "answer" given, we see the time needed is D/vx = (154 m)/(27.78 m/s) = 5.54 s. Sowe’ve turned the problem into "proving" that it takes 5.54 seconds to fall the 150 m and thatmakes us realize the package WILL fall 150 m, starting from a zero vertical velocity. Thebasic equation is yf = yi + vit – 1/2 gt2 where yf = 0 m (ground), yi = 150 m, vi = 0 m/s, andg = 9.8 m/s2. This gives: 0 = 150 m + (0m/s)(t) – 1/2(9.8 m/s2)t2 or, rearranging:

t2 = [150 m]2/(9.8 m/s2) = 300 m/(9.8 m/s2) = 30.61 s2

The square root of that gives t = 5.53 s, and I guess we’ve proven it.

6. This could turn out to be a real messy problem (pun intended). We have to figure out theangle the student should throw the tomato, so we’ll just call it θ for now. Then we have ourtwo equations: vx = v0cosθ and vy = v0sinθ where v0 = 22.3 m/s. Since we know the hori-zontal distance to the target (har har!) equals 30 m, using the horizontal velocity we can findthe time to target, namely: D = vxt, or

t = (30 m)/([22.3 m/s]cosθ) = (1.345/cosθ) s

Using the height equation, yf = yi + vit – 1/2 gt2 where yf = 15 m, yi = 0 m, vi is v0sinθ,and t is given in the above equation, we can write:

15 m = 0 m + ([22.3 m/s]sinθ){(1.345/cosθ)s} – 1/2 (9.8 m/s2){(1.345/cosθ)s}2.

Now this IS a mess! Let’s see how to make it look better. Each term has the unit of meters,so we can drop having to write units. Then, doing all multiplications in each term and multi-plying by cos2θ, the above equation reads:

15cos2θ = 30sinθcosθ – 8.8680.

Such an equation is called transcendental, which means "we haven’t a clue how to solve itin a straightforward manner." These equations appear in real life more often than the mathe-maticians and other scientists would lead you to believe, but having calculators and comput-ers available allows us to get very close to a solution rather quickly. Here’s one way: First,just for ease in calculation, rewrite the equation as

15 cos2θ – 30sinθcosθ – 8.868 = 0

Then we guess a value for θ [some people would say "approximate"] and see how close tozero you get when you evaluate. For instance, guess θ = 30°. Substitution gives

15(cos30°)2 – 30sin30°cos30° + 8.868 = 15(0.866)2 – 30(0.5)(0.866) + 8.868

= 11.25 – 12.990 + 8.868 = 7.128

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And that’s not too good a guess. Now try 40°:

15(cos40°)2 – 30sin40°cos40° + 8.868 = 15(0.766)2 – 30(0.643)(0.766) + 8.868

= 8.802 – 14.772 + 8.868 = 2.898

This is closer to zero, so let’s try 60°:

15(cos60°)2 – 30sin60°cos60°+ 8.868 = 15(0.5)2 – 30(0.866)(0.5) + 8.868

= 3.75 – 12.990 + 8.868 = -0.372

And now we’re pretty close. Notice also that the result has changed sign from a positiveresult to a negative one; that means the solution is between 40° and 60° and a lot closer to 60°. When we try 50°, we get 15(0.643)2 – 30(0.766)(0.643) + 8.868 = +0.294. So you have the idea now. The correct angle is between 50° and 60°. By this iterative technique, it is possible to get the angle down to within a tenth of a degree on only a few more tries.

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Suggestions for Instructors

1. Projectile motion is a beginning step toward general curvilinear motion and starts the studenttoward visualizing in a 3-D world. The reason projectile motion appears is that it allows discussion of acceleration beyond the 1-D case; whenever the acceleration (recall it’s a vec-tor) is constant, the resultant motion MUST be in a single plane. Not only that, the trajectoryof the motion is a parabola.

2. Many texts produce a series of rather involved equations for projectile motion, such as theone for the range of the projectile. Unfortunately, this type of equation results in the studentconsidering physics as a collection of specialized "formulas" that must be memorized for suc-cess in the course. Try to emphasize that the equations developed are all based on the firstfew they learned in the definitions for velocity and acceleration.

3. In today’s world of long-range missiles, space shuttles, and satellites, help the student realizethe basic projectile equations only "work" for a flat Earth where the acceleration of gravity isconstant, both in magnitude and direction. Ask your students what happens when youthrow a rock "really hard" horizontally from the top of a building—so hard the Earth beginsto "fall away" as the rock also falls. This was a concept Galileo had begun to work on andNewton was very good at.

4. The last problem (#6) was chosen because it is not the standard kind you find in textbooks.Because our students now have truly powerful tools at their disposal, they should be giventhe opportunity to use them. After all, when a spaceship was sent to the Moon, the dynam-ics of the situation required a solution of the three-body problem, for which there is no sat-isfactory analytic solution. Let’s have students work in the real world.

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PROGRAM 4:CIRCULAR AND ROTATIONAL MOTION

Key Terms and FormulasUniform circular motion occurs when an object moves around a circle, with a radius, r,at a constant velocity, v—like when you swing your key chain around your finger.

The force that allows an object to follow this curved path is called the centripetal force, Fc.Centripetal force is directed towards the center of the circle.

The object’s acceleration is directed towards the center of the circle and is called the centripetal acceleration, ac. It can be found using the equation

ac = v2

r

Period, T, is the time it takes for the object to complete one orbit.

The velocity of an object traveling in uniform circular motion can be found using the equation

v = 2πrT

Accelerated circular motion happens when an object moving around a circle changes themagnitude of its velocity. This type of acceleration is called tangential acceleration, at.Tangential acceleration is always in the direction that the object is moving and is always tan-gent to the circle. The total acceleration of the object is the vector sum of the tangentialacceleration and the centripetal acceleration. Its magnitude can be found using the equation

a2 = at2 + ac

2

The term rotational motion describes objects that rotate about an axis, like a top.

The angular velocity of a rotating object, represented by the Greek letter little omega, ω,is the angle, θ, through which it turns, over the time, t, it takes the object to turn. It can befound using the equation

ω = θt

The axis of rotation is the axis about which all parts of the rotating body are moving.

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Quiz

1. Why MUST the acceleration point toward the concave side of circular motion?

2. Why is the acceleration in uniform circular motion called "centripetal"?

3. Is it possible for a body to have centripetal acceleration but no tangential acceleration? Is it possible to have tangential acceleration but no centripetal acceleration?

4. There is a relation in motion in a circle between angular velocity, ω, linear velocity, v, andthe radial distance, r, of the point with velocity v from the axis of rotation. But velocity anddisplacement are vectors. Can you figure out a vector equation that could relate these quan-tities based on the equation ω = v/r ? [Hint: you didn’t learn any vector equation for dividingbut you did for multiplying.]

5. A large flywheel, whose diameter is 3 meters, is rotating at 120 rpm. Calculate: (a) the periodof rotation; (b) the angular velocity; (c) the linear velocity of a point on the rim; (d) the centripetal acceleration of a point on the rim.

6. Find the magnitude of the velocity and the centripetal acceleration of the Earth around theSun. The radius of Earth’s orbit about the Sun is 1.49 x 1011 m and the period of its motion is 3.16 x 107 s [i.e., 365.25 days].

7. The angular velocity of a flywheel increases uniformly from 20 rad/s to 30 rad/s in 5 s.Calculate: (a) the tangential acceleration of a point 1.5 m from the axis of rotation; (b) thecentripetal acceleration at the start and at the end of the 5 seconds.

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Solutions to Quiz

1. The acceleration must always point "in" because that is the DIRECTION in which the velocityis changing. When the motion is uniform circular motion this is the easiest case to see; buteven when the motion is not, a few simple drawings of the v e c t o r s for the two velocitiesp retty close together show you that final velocity vector MINUS first velocity vector alwaysgives a vector that points in.

2. The answer here is obvious for anyone who knows some Greek. For those of us who arenot up on our Greek, centri- clearly means center and the petal part is associated with theconcept of "seeking".

3. Yes, for the first part. Uniform circular motion is the case where this is possible. As long asthe speed of the body is unchanged, there is no tangential acceleration. For the second ques-tion, the answer is yes but only for the trivial case when the body moves in a straight line;that is, one-dimensional motion.

4. This is kind of difficult, except that we can make up a direction for ω and so turn it into a vector ω. Think of a circle or wheel rotating about an axis through its center. Now grab—not really—the wheel with your right hand pointing the direction of rotation and point outyour thumb. That’s the direction we’ll give to vector ω. Now we have to take our relation ω = v/r or rω = v and write it as a vector product equation. To get the magnitude correct, r and ω must be perpendicular to each other. Great! The vector r is the one that goes fromthe axis of rotation in the plane of the wheel out to the point on the rim, and the vector ωpoints along the axis. All we need now is to make sure the direction of v comes out right.And that can be done by making the equation v = ω x r. Not the other way around, becausethen the velocity vector would point OPPOSITE to the direction of rotation [look at thosefingers on your right hand when they wrapped around the wheel].

5. Rather than talking through the solution of numerical problems, they will just be written outas we would expect a student to do them. For instance:

120 rpm = [120rev/min]/[60sec/min] = 2 rev/sec = frequency

(a) But Period = 1/frequency, or Period = 1/[2 rev/sec] = 1/2 sec (per revolution)

(b) 120 rpm = [(120 rev/min)(2πrad/rev)]/[60 sec/min] = 4πrad/s = ω

(c) v = ωr = [4πrad/s][1.5m] = 6π m/s (= 18.85 m/s)

(d) ac = v2/r = [36π2 m2/s2]/[1.5 m] = 24π2 m/s2 (= 236.9 m/s2)

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6. v = 2πR/T = 2π(1.49 x 1011 m)/(3.16 x 107 s) = 2.96 x 104 m/s

(= 29.6 km/s = 1.07 x 105 km/h = 66,300 mph)

ac = v2/R = [2.96 x 104 m/s]2/[1.49 x 1011 m] = 5.89 x 10-3 m/s2

7. ωf = 30 rad/s; ωi = 20 rad/s; r = 1.5 m; then:

vf = ωf r = [30 rad/s][1.5 m] = 45 m/s; vi = ωi r = [20 rad/s][1.5 m] = 30 m/s

(a) aT = Δv/Δt = [45 m/s – 30 m/s]/5 s = [15 m/s]/5 s = 3 m/s2

(b) ac(initial) = vi2/R = [30 m/s]2/1.5 m = 600 m/s2 (= 61g)

ac(final) = vf2/R = [45 m/s]2/1.5 m = 1350 m/s2 (= 140g)

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Suggestions for Instructors

1. The program material may be extended to include angular acceleration, α, where α =Δω /Δt just as linear acceleration, a, is defined as a = Δv/Δt. A comparable set of equationsmay then be written with θ equivalent to x, ω equivalent to v, and α equivalent to a. Forinstance, the equation ωf

2 = ωi2 + 2αθ is correct, as is θf = θi + ωi t + 1/2 α t2.

2. So many "accelerations" may be confusing to the student. There is linear acceleration, a;but in curvilinear motion we may have centripetal acceleration, ac, and tangential accelera-tion, aT, which are perpendicular to each other [therefore, ac is sometimes called "normal"acceleration, aN—in circular motion it is easier to see this is true]. Most importantly, don’tbreathe "centrifugal acceleration"!

3. Rigid body rotation can be quite involved. Try to keep all discussion with rotation about a principal axis. If students—or the text—extend you toward moment of inertia, be sure toinclude a discussion of torque—τ = r x F and then τ = I α when you start dynamics (seeProgram 5).

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PROGRAM 5:LINEAR MOMENTUM AND NEWTON’S LAWS OF MOTION

Key Terms and FormulasAccording to the Principle of Conservation of Momentum, the final momentum of the sys-tem equals the initial momentum of the system. When two particles collide, the momentum ofthe whole system, that is—of what happens before, during, and after the collision—is the samebefore and after the collision. This is called the principle of the conservation of linear momen-tum, and gives us the relation p = mv.

Newton’s First Law of Motion states that an object at rest will remain in its state of restunless acted upon by an external force, and an object moving with uniform motion will remainin uniform motion unless acted upon by an external force.

Newton’s Second Law of Motion states that the net force acting on an object equals the massof the object times its acceleration, or F = ma.

The unit of force is the Newton. One newton equals one kilogram meter per seconds squared.

Weight, w, is the force of gravity acting upon an object’s mass. So, w = mg.

Newton’s Third Law of Motion states that when an object exerts a force on another object,the second object exerts an equal but opposite force on the first object.

When two particles collide,

m1v1 (after) + m2v2 (after) = m1v1 (before) + m2v2 (before)

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Quiz

1. Is linear momentum a vector or scalar quantity? Why?

2. The principle of conservation of momentum says the momentum of a system remains constant (or fixed). How is it then possible for the momentum of a body to change?

3. When the force on a body is zero, what is the change in its momentum?

4. What is the direction of the change in momentum of a body when there is a force on it?

5. If I apply a force F on a body, does the body exert a force on me? If so, how large is it and in what direction? Explain then how a body can possibly move.

6. A log, with a mass of 45 kg, is floating downstream at 2.5 m/s. A 10 kg swan attempts toland on the log, flying upstream at 2.5 m/s, but slides the length of the log and falls off theend with a velocity of 0.5 m/s (upstream). Calculate the final velocity of the log downstreamafter this weird interaction.

7. (a) Calculate the time needed for a constant force of 80 N to act on a body of 12.5 kg inorder to take it from rest to a speed of 72 km/h. (b) How long, and in what direction, mustthe same constant force act on the body to return it to rest?

8. A cart having a mass of 1.5 kg moves along its track at 0.2 m/s until it runs into a fixedbumper at the end of a track. What is its change in momentum and the average force exerted on the cart if, in 0.1 s, it: (a) is brought to rest; (b) rebounds with a speed of 0.1 m/s?(c) Discuss the conservation of momentum in these collisions.

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Solutions to Quiz

1. Since linear momentum, p, is the product of a scalar, m, and a vector, v, it must be a vectorquantity. That means it has direction as well as magnitude.

2. The momentum any body gains (with respect to someone watching!!) must be at the"expense" of some other body having its momentum changed by an equal AND OPPOSITEamount (again with respect to the same someone!!). When a ball drops toward the Earth andgains momentum, the Earth also gains momentum, equal and opposite to that gained by theball; but the mass of the Earth is so immense compared to that of the ball, its velocity ishardly changed.

3. By the very definition of force, it is the change in momentum of a body per unit time. So, ifthere is no force on a body, its momentum doesn’t change—in magnitude OR direction. Togo round in a circle, then, a force must be present.

4. Again, from the definition of force, F = Δp/Δt, the direction of the change in momentummust be in exactly the same direction as the direction of the force. For instance, in circularmotion, the direction of the change in momentum is "centripetal"; so then is the direction of the force.

5. The body must exert a force equal and opposite to the force I exert; otherwise it would beimpossible for us to remain at relative rest. If my hand is not in constant contact with thebody, I can’t apply the force. Besides, the force the body exerts on me is exerted ON ME,not on the body, and so F = ma still works because F is all the forces on the mass m, notthe forces m exerts on other bodies.

6. Before the "collision":

plog = (45 kg)(2.5 m/s downstream) = 112.5 kg-m/s downstream;

pswan = (10 kg)(2.5 m/s upstream) = 25 kg-m/s upstream;

ptotal = plog + pswan = (112.5 – 25) kg-m/s downstream = 87.5 kg-m/s downstream

After the collision:

pswan = (10 kg)(0.5 m/s) upstream = 5 kg-m/s upstream;

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Since ptotal = plog + pswan = 87.5 kg-m/s downstream, and this remains constant, we can write that after the collision:

plog = ptotal – pswan = 87.5 kg-m/s downstream – 5 kg-m/s upstream

= (87.5 + 5) downstream = 92.5 kg-m/s downstream

With momentum equal to mass times velocity, we get the velocity of the log as:

vlog = plog /m = [92.5 kg-m/s downstream]/45 kg = 2.056 m/s downstream.

7. (a) 72 km/h = (72 x 103 m)/(3.6 x 103 s) = 20 m/s;

F = ma = 80 N = (12.5 kg)a

a = (80 N)/(12.5 kg) = 6.4 m/s2

vf = vi + at

t = (20 m/s – 0 m/s)/6.4 m/s2 = 3.125 s

(b) If the same constant force is applied in the opposite direction for 3.125 seconds, thebody will return to rest.

8. (a) Δp = mΔv = (1.5 kg)(0 m/s – 0.20 m/s) = -0.3 kg-m/s;

F = Δp/Δt = [-0.3 kg-m/s]/0.1 s = -3.0 N

[The minus sign means the force is opposite to the original momentum direction]

(b) Δp = mΔv = (1.5 kg)(-0.10 m/s – 0.20 m/s) = -0.45 kg-m/s;

F = Δp/Δt = [-0.45 kg-m/s]/0.1 s = -4.5 N

(c) The momentum of the cart is not conserved because the cart is not the entire system. It interacted with the bumper in both cases and we would have to consider what hap-pened to the whole rest of the system—bumper, Earth, etc.

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Suggestions for Instructors

1. This program begins the investigation into why particles move the way they do. Why do bodies near the surface of the Earth fall with constant acceleration? Why does the Earth orbitthe Sun along an elliptical path? Why do atoms bind together to form molecules? Why does aspring oscillate when it is stretched? We want to understand these and many other motionsthat we constantly observe around us. This understanding is not only important to our basicknowledge of nature, but also for engineering and other practical applications. When weunderstand how motions in general are produced, we are able to design machines and otherpractical devices that move as we desire. The study of the relationship between the motion of a body and the causes for this motion is called d y n a m i c s.

2. F rom daily experience, we know that the motion of a body is a direct result of its i n t e r a c t i o n swith other bodies around it. When a batter hits a ball, there is an interaction between bat andball. The path of a projectile is but a result of its i n t e r a c t i o n with the Earth. The motion of ane l e c t ron around a nucleus is the result of its i n t e r a c t i o n with the nucleus, and perhaps withother electrons. Interactions are conveniently expressed quantitatively in terms of a conceptcalled f o rc e, where we have this intuitive notion of force, and of its strength, in terms of apush or a pull such as the weight of a body or the pull of a magnet on iron fillings.

3. T h e re are some easily perf o rmed experiments that support the law of inertia. A spherical ballresting on a smooth horizontal surface will remain at rest unless acted upon. [If the surface isnot quite smooth, but has small variations in its surface, the ball will be found to roll about—something that attracts the students’ attention when you assert initially that the tabletop is flat.]We assume that the surface on which the ball is resting balances the i n t e r a c t i o n between balland Earth and hence the ball is essentially free of interactions. When the ball is struck, as inbilliards, it momentarily suffers an i n t e r a c t i o n and gains velocity (and there f o re momentum),but afterward is free again, moving in a straight line with the constant velocity it obtainedwhen it was struck. Since the ball does not go on indefinitely, because the surface is finite andbecause of f r i c t i o n, we can begin to anticipate a discussion of this additional i n t e r a c t i o n, whichwill be introduced in the next pro g r a m .

4. The principle of the conservation of momentum holds for any number of particles that forman isolated system, i.e., particles which are subject only to their own mutual interactions andnot to interactions with other parts of the world. For example, consider a hydrogen moleculecomposed of two hydrogen atoms (there f o re, of two protons and two electrons). If the mole-cule is isolated so that only the interactions among these four particles have to be considere d ,the sum of their momenta relative to an inertial frame of re f e rence will be constant. Similarly,consider the Solar System, composed of our Sun, the planets, and their satellites. If we couldneglect the interactions with all other heavenly bodies, the total momentum of the SolarSystem, relative to an inertial frame of re f e rence, would be constant.

No exceptions to this general principle (of conservation of momentum) are known. In fact,whenever this principle seems to be violated in an experiment, the experimenter immediatelylooks for some unknown or hidden particle, which remained unnoticed and which may beresponsible for the apparent lack of momentum conservation. It was this type of search thatled physicists to identify ("discover") the neutron, the neutrino, the photon, and many otherelementary particles.

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5. Because of the existence of the principle of conservation of momentum, all three of Newton’sLaws exist. It would be interesting to investigate with the student the rationale behind this conservation principle—it’s not something we have worked out, like the gas laws or evenNewton’s laws. Rather, the conservation of linear momentum is due to the invariance of spacetranslation, which means that an isolated system, no matter where in the universe it may beplaced, will behave in the exact same manner. The other two dynamic conservation principles,that of angular momentum conservation and energy conservation, are the result of similari n v a r i a n c e s .

6. When we look around, we see many diff e rent kinds of force. We apply a force on theg round when we walk. We push and lift diff e rent objects. To stretch a string, we must applya force. The wind upsets a tree or pushes a sailboat by applying a force. The expansion ofgases in an internal combustion engine produces a force that causes an automobile, a boat,or an airplane to move. Electric motors produce a force that moves objects. The above exam-ples of physical forces occur in complicated systems, such as our bodies, a machine, theEarth’s atmosphere, and on wires carrying an electrical current. We may say that these are s t a t i s t i c a l f o rces because, when we analyze where these forces appear, we notice that theyinvolve bodies composed of large numbers of atoms. The question is then, if we break thesephenomena down into microscopic components, namely, molecules, atoms, electrons, andtheir interactions, what do we discover as the i n t e r a c t i o n s or forces? One of the very gre a tachievements of the last few decades has been to reduce all forces observed in nature to avery few b a s i c or f u n d a m e n t a l i n t e r a c t i o n s. This means that statistical forces are simply mani-festations of the fundamental forces when a very large number of particles are involved. Thefour fundamental forces that we currently recognize are gravitational, electromagnetic, stro n g(or nuclear), and weak.

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PROGRAM 6:FRICTION, WORK, AND ENERGY

Key Terms and FormulasFriction is a phenomenon that impedes motion between two surfaces in contact.

There is kinetic friction when an object slides on a surface, and the interaction between theobject and the surface results in a force of friction that acts in a direction opposite to that ofthe motion.

Static friction is the force between two stationary surfaces that inhibits motion.

The normal force is the force perpendicular to the touching surfaces, and is opposite indirection to the object’s weight when the surface is horizontal.

The coefficient of friction, µ, is a measure of the amount of interaction between two surfacesthat results in a frictional force when an attempt is made to move an object.

Kinetic friction equals the coefficient of friction times the normal force, or, Ff = µkFN.

Static friction has a maximum value equal to the coefficient of friction times the normal force,or Ff = µsFN.

Work is the dot scalar product of force and displacement, W = F • r. The unit of work is theJoule, which is one Newton-meter.

Power is the rate at which work is done.

The Work-Kinetic Energy Principle encompasses the relationship between work and kinetice n e rgy, where net work on a body equals its final kinetic energy minus its initial kinetic energy.

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Quiz

1. Is there a specific relation between the direction of the force of friction and the direction of the velocity of a body? Is the direction of the force of friction on a body related to itsacceleration?

2. A body moves across a horizontal surface under an applied force. Describe the kind ofmotion that results when the applied force is: (a) larger than; (b) equal to; (c) smaller thanthe force of friction.

3. A body has a mass of 10 kg. The coefficient of friction between the body and a horizontalsurface is 0.35. Describe the motion when a horizontal force of 78 N is applied to the body.

4. Aristotle said different bodies fell at different rates. Galileo proved that all bodies fall withthe same acceleration. Under what conditions might both assertions be correct?

5. State the conditions under which the work done by a force is (a) zero; (b) positive; (c) negative.

6. What happens to the kinetic energy of a particle when the work of the applied force is (a) zero; (b) positive; (c) negative.

7. When I raise up a 10 kg (980 N) weight through a distance of 2 m, I perform (980 N)(2 m)= 1960 N-m [1960 J] of work. But the kinetic energy of the body is still zero! Where did mywork go?

8. Calculate the work of a constant force of 12 N when the particle on which it acts moves 7 m if the angle between the directions of the force and the displacement is (a) 0°; (b) 60°;(c) 90°; (d) 135°; (e) 180°.

9. In each of the above, calculate the power generated if the work is done in 5 s.

10. (a) Calculate the constant force required by the motor of an automobile, whose mass is1500 kg, in order to increase its velocity from 4.0 km/hr (1.11 m/s) to 40 km/hr (11.11 m/s)in 8 s. (b) Calculate the change in momentum and kinetic energy. (c) Calculate the workdone by the force. (d) Compute the average power of the motor.

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Solutions to Quiz

1. To the best of our measuring ability, and in the basic physics class, the direction of the forceof friction is always opposed to the direction of motion—even opposed to the attempteddirection of motion for the case where the applied force isn’t large enough to overcome themaximum force of static friction. Acceleration, as far as we can measure, has no effect onthe direction of the friction force. Incidentally, fluid or viscous friction, where the resistingmaterial more or less surrounds the object in relative motion, not only is always opposite tothe direction of motion but is usually some function of the velocity—and not just to the firstpower (i.e., viscous friction is a nonlinear phenomenon).

2. (a) Since the applied force is greater than the force of friction, there will be a net force inthe direction of motion and so the body will accelerate in the direction of the net force. (b) When the applied force is equal to the force of friction, the net force is zero and thebody will have constant motion [F = 0, a = 0, or v = a constant]. (c) If the applied force issmaller than the force of friction, the net force will have a direction opposite to the directionof motion and so the body will decelerate, finally stopping. Since the force of static frictionis always (well, almost always) larger than the force of kinetic friction, the body really stops!

3. The normal force pressing the body against the horizontal surface is its weight, mg, and the force of friction—assuming the body will be moving—is

Ff = µmg = 0.35(10 kg)(9.8 m/s2) = 34.3 kg-m/s2 = 34.3 N

Since the horizontal force being applied is 78 N, we get a net positive force of (78 – 34) N = 44 N and then using F = ma, we have a = (44 N)/(10 kg) = 4.4 m/s2.

4. Both assertions cannot be correct at the same time. However, the feather and penny in thetube is a very good experiment to show how both Aristotle and Galileo can be right. Whenwe can ignore the frictional effects of surrounding materials, Galileo is shown to be correctin that ALL objects will accelerate downward with the acceleration of gravity. But when theair is let into the tube, we see that in the "real" world Aristotle was right. So it’s a case ofcarefully stating the conditions of your experiment as well as the results. Too often what weread in today’s newspapers or see and hear on television doesn’t completely give all thefacts, and so two sides have conflicting "results."

5. Work is a scalar pro d u c t of force (a vector) and displacement (also a vector) and is there f o rewritten as W = F • r or W = Frc o sθw h e re θ is the angle between the direction of the force andthe direction of the displacement of the body on which the force acts [that’s a lot of words, butlet’s be v e r y c a reful before we start using concepts!]. (a) The work done by a [n o n z e ro] forc ewill be zero e i t h e r when the body does not move, which means there must be some otheropposing force such as the force of static friction, o r the value of cosθ is zero, which happensonly when the force and the displacement are perpendicular to each other; (b) The work ispositive when there is an actual displacement of the body and t h e value of cosθ is positive; (c) The work is negative only when there is a displacement and c o sθ is negative—this usuallymeans that the force and the displacement are in somewhat opposite dire c t i o n s .

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6. Since kinetic energy depends on the velocity [squared] of the body: (a) zero work results inzero change in the kinetic energy; (b) positive work results in a positive change (increase)in the body’s kinetic energy; (c) negative work produces a negative change (decrease) in thekinetic energy of a body.

7. You performed 1960 J of positive work on the body because the direction of your force andthe direction of displacement were both up. Since the kinetic energy of the body did notchange, the net work done on the body must have been zero! What you forgot was thatthere is a negative force due to "gravity" down, equal to mg. So, you did work on the bodyand so did the Earth, or gravity, or whatever you want to call it. One way we get out of thisseeming paradox is to say your positive work increased the [gravitational] potential energyof the body. This then lets us talk about potential energy and kinetic energy and the sum of the two, which we call total energy.

8. W = F • r or W = Frc o sθ where θ is the angle between the direction of the force and the direction of the displacement of the body on which the force acts. For all these cases, F = 12 N and r = 7 m, so Fr = (12 N)(7 m) = 84 N-m = 84 J. Then:

(a) W = (84 J)cos0° = (84 J)(1.0) = 84 J

(b) W = (84 J)cos60° = (84 J)(0.5) = 42 J

(c) W = (84 J)cos90° = (84 J)(0.0) = 0 J

(d)W = (84 J)cos135° = (84 J)(-0.707) = -59.4 J

(e) W = (84 J)cos180° = (84 J)(-1.0) = -84 J

9. P = W/t, where P is in watts if work is in joules and time in seconds. So:

(a) P = (84 J)/(5 s) = 16.8 W

(b) P = (42 J)/(5 s) = 8.4 W

(c) P = (0 J)/(5 s) = 0 W

(d) P = (-59.4 J)/(5 s) = -11.9 W

(e) P = (-84 J)/(5 s) = -16.8 W

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10. (a) F = ma = m(Δv/Δt) = (1500 kg)([11.11 m/s] – [1.11 m/s])/(8 s) = 1500 kg(10 m/s)/8s = 1875 kg-m/s2 = 1875 N. For later: a = F/m = 1875 N/1500 kg = 1.25 m/s2.

(b) Δp = m(Δv) = 1500 kg([11.11 m/s] – [1.11 m/s]) = 1500 kg(10 m/s) = 1.5 x 104 kg-m/s.

ΔEk = 1/2mΔv2 = 1/2(1500 kg)([11.11 m/s]2 – [1.11 m/s]2)

= 1/2(1500 kg)(123.456 m2/s2 – 1.234 m2/s2) = 750 kg(122.222 m2/s2) = 9.17 x 104 J.

(c) W = Fs, but we don’t have the distance covered. So, let’s calculate that using

vf2 = v0

2 + 2as, or s = [vf2 – v0

2]/2a

This gives

s = ([11.11 m/s]2 – [1.11 m/s]2)/2[1.25 m/s2] = (123.456 m2/s2 – 1.234 m2/s2)/2.5 m/s2

= (122.222 m2/s2)/2.5 m/s2 = 49.0 m.

So now, we use W = Fs = (1875 N)(49.0 m) = 9.17 x 104 N-m = 9.17 x 104 J. Please notethat this is exactly the change in the kinetic energy of the body—as we could have real-ized from the beginning and simply invoked the work–kinetic energy theorem!

(d) Power is work per unit time, P = W/t = [9.17 x 104 J]/8 s = 1.146 x 104 W (11.5 kW).

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Suggestions for Instructors

1. When it comes to friction, it is important to stress the statistical nature of the friction force.Simple experiments, such as placing a block on a relatively rough board and then raising theboard until the block slides, illustrate how the force of static friction increases until it reachesits maximum value, which allows the coefficient of [static] friction to be experimentallymeasured [= tanθ]. Also, by placing the block on the board in different manners, such assliding it back and forth initially or pressing down hard first, it becomes pretty clear that thisis a difficult property to determine with any accuracy. Kinetic friction can be studied—andmeasured—by pulling a block along a horizontal surface using a spring calibrated to readforce (rather than grams, as so many are labeled!) and attempting to keep the velocity a constant, which should keep the force constant.

2. The spring measuring device is also useful in illustrating work done on a body. You maypull horizontally, at an angle, while the block slides horizontally, or pull up an inclinedboard. The only real way to show work resulting in a change in kinetic energy is to drop abody and try to measure the change in velocity [therefore, kinetic energy]. In the laboratorya more sophisticated apparatus must be set up where first the kinetic coefficient of friction isevaluated by having the block slide at a constant velocity [due to a set of "weights" that pullhorizontally on the block as they "fall"—the string over a pulley]; then a larger amount ofmass is added and the resulting acceleration measured [displacement from zero velocity in a measured time interval], allowing final velocity to be calculated.

3. Have students perform a task that involves real physics-type work in differing time intervalsto have them understand the concept of power. Relate their actual values to sources ofpower—particularly the "horsepower" of their automobiles and kilowatts of electric lights.

4. Potential energy only appears peripherally in this program. It would be useful to introducemore information and some cases into class discussion.

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PROGRAM 7:GRAVITATION

Key Terms and FormulasGravitation is the attraction between every mass in the universe.

The Law of Universal Gravitation states that the attractive force between two masses is proportional to the product of the masses of the two objects, over the square of the distancebetween the objects. Or,

F = G m1m2r2

The G in this equation is the gravitational constant and has a value of 6.67 x 10- 1 1 N - m2/ k g2.

Kepler’s First Law of Planetary Motion states that every planet has an elliptical orbit, withthe Sun at one focus.

Kepler’s Second Law of Planetary Motion states that an imaginary line that connects theSun to a planet sweeps out equal areas in equal times.

Kepler’s Third Law of Planetary Motion states that the square of a planet’s orbital period isproportional to the cube of its semi-major axis, or

T2 = 1 4 π2

2 a3

GM

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Quiz

1. Why do all [massive] bodies attract each other the way described by Newton’s law of universal gravitation?

2. If you were in a tightly enclosed, very well equipped laboratory, devise an experiment thatcould prove you were on the Earth’s surface where the "acceleration of gravity" was 9.8m/s2, rather than in a rocket ship in empty space accelerating at 9.8 m/s2.

3. Gravitation force falls off as 1/r2 and so is called an inverse square force. If the radius of theEarth is 6.38 x 106 m, where the gravitational force on a 1 kg mass is 9.8 N, how far from[the center of the] Earth must we be for the gravitational force on the 1 kg mass to equal 9.8 x 10-10 N?

4. How hard would you have to throw a stone straight up so that it would never come back?(This speed is called the escape velocity from Earth.)

5. A satellite of mass m is going around the Earth in a circular orbit, with a radius r and aspeed v. [The centripetal force needed to keep a mass moving along a circular path is FC =mv2/r.] If the entire centripetal force is supplied by the gravitational force, FG = GmM/r2,where M is the mass of the Earth (i.e., FC = FG = mv2/r = GmM/r2), calculate the radius["height"] such that the period of the satellite is exactly 24 hours.

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Solutions to Quiz

1. There is no obvious reason why masses attract although many models, or ideas, have beenput forward. Kepler, with the help of Tycho Brahe, developed what we call Kepler’s Laws inthe late 1500’s and Newton collapsed them into a single idea in the late 1600’s. This conceptappears to be quite valid, with the possibility that it must be modified when Einstein’s ideasconcerning General Relativity are introduced, which says that a mass "warps" space, or viceversa.

2. If you can come up with such an experiment, you are guaranteed a Nobel Prize becauseyou will have proven Einstein’s statement of the Principle of equivalence to be incorrect. Theproposed experiment is exactly what Einstein proposed to prove his point that gravitationalattraction caused by a mass is identical in all ways with an acceleration.

3. Using the law of gravitational force, we write:

F1 = G(ME[1 kg])/RE2 = 9.8 N, and

FX = G(ME[1 kg])/RX2 = 9.8 x 10-10 N

where the X stands for the new distance. There are a lot of common symbols in both equations; so to solve with the least amount of math, let’s equate the same batch of symbolsfrom both equations:

G(ME[1 kg]) = RE2 [9.8 N] = RX

2[9.8 x 10-10 N], or

RE2[9.8 N] = RX

2[9.8 x 10-10 N]

Then, canceling the common 9.8 N on each side, we have RE2 = RX

2[10-10] and taking squareroots gives RE = RX[10-5]. Therefore, RX = RE[105] = 6.38 x 1011 m, which is further than thedistance to the Sun.

4. To do this problem, we have to know what the gravitational potential energy of a mass is.We can’t use PE = mgh because g is not a constant as we move away from Earth (or if wego inside the Earth!). When two masses are infinitely far apart, the force of gravity is zero; as they get closer to each other the force [of attraction] increases and so, to keep them apartthe direction of the force is opposite to their distance apart. This means that if we call gravi-tational potential energy zero when two masses are infinitely far apart, the value of theirpotential energy is negative for any value less than infinity. What we discover is that therelation is PE = -GM1M2/R; that is, it’s negative always and inversely proportional to the distance—to the first power. So, with M1 = Earth’s mass, ME, and M2, the mass of what wethrow, we can write the total energy of a mass at the Earth’s surface:

1/2M2v2 – GMEM2/RE = E

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Then, when the mass reaches infinity, the potential energy will be zero. Let the kinetic energ ybe zero also at that point. There f o re, if the total energy of the mass at the Earth’s surface isz e ro and we throw the mass directly "up", it will reach infinity. So, let E = 0 and we can write:

1/2M2v2 = GMEM2/RE

or, canceling out the common M2, multiplying by 2, and taking the square root, we have:

v = [2GME/RE]1/2

= [2(6.67 x 10-11 Nm2kg-2)(5.97 x 1024 kg)/(6.38 x 106 m)]1/2

v = [1.248 x 108 Nm/kg]1/2

= 1.12 x 104 m/s [= about 40,000 km/hr].

5. Using the relation FC = FG = mv2/r = GmM/r2 and canceling the common m and one r,we have v2 = GM/r. This is a relationship between any satellite orbiting a massive body in a circular orbit. What we are looking for is a circular orbit with a period of 24 hours, whichequals (24 hours)(60 minutes/hour)(60 seconds/minute) = 8.64 x 104 seconds. Well, since v = d/t, if we substitute 2πr for d and let t be the 24 hours, we have:

[2πr/t]2 = GM/r or [4π2r3]/GM = t2

Darn if that doesn’t look like one of Kepler’s Laws! So, we solve for r and get:

r = [GMt2/4π2]1/3

r = [(6.67 x 10-11 Nm2kg-2)(5.97 x 1024 kg)(8.64 x 104 s)2/4π2]1/3

= 4.22 x 107 m.

If you subtract the Earth’s radius, you then find how high this stationary satellite will sit in the sky.

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Suggestions for Instructors

1. When there is time, the introduction of angular momentum, L = r x p, when discussing plan-etary motion is of help in explaining why the motion is in a plane. Since the force of attrac-tion between masses has a direction along the line joining them [a central forc e], it can beshown that d L / d t must be zero, which results in a constant angular momentum, both in mag-nitude AND direction. If then the planets follow elliptical orbits, their distance to the Sun isvarying; there f o re, their speed must as well, such that the magnitude of L remains constant.

2. Students will become interested in this subject when artificial satellites, such as the orbitingspacecraft of the U.S. and Russia, are used as examples. The Smithsonian Institution has a goodcollection of such craft, and a Web site they may wish to visit http://www.nasm.si.edu/museum.

3. The first time the student becomes aware of the fact that the "interaction" between twomasses exists throughout all space, it may be possible to start a discussion about the gravita-tional field of a single mass. That is, we may say that a mass creates a gravitational fieldabout itself whose strength [a vector] is measured in terms of the force [a vector] per unitmass at any point in space. That is, G = FG/m = -[GM/r2]ur, where ur is a unit vector in theradial direction. When the electric interaction is studied, the analogy is usually not lost onthe student between electric field and gravitational field. Such a comparison allows for anunderstanding of the relationship between electric and gravitational potential as well as elec-tric and gravitational potential energy.

4. Here is it appropriate to mention that there are only four known fundamental forces: gravita-tional, electromagnetic, nuclear [also called "strong"], and weak. We know "everything" aboutthe gravitational case, a lot about the electromagnetic case, little about the strong interaction,and hardly anything about the weak "force". There is a lot of room for new studies, ideas,theories, and experimentation. That’s why physicists keep trying to build these very-high-energy machines.

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PROGRAM 8:HARMONIC MOTION AND WAVES

Key Terms and FormulasSimple harmonic motion is the basic vibratory motion an object undergoes when it is subjected to a restoring force that’s directly proportional to its displacement from equilibrium.

Hooke’s Law states that F = -kx, where k is the constant of proportionality, particular to everyobject undergoing simple harmonic motion.

The period, T, of an object undergoing simple harmonic motion is the time it takes to com-plete one cycle.

Frequency, f, is the inverse of period, and is measured in Hertz.

Amplitude is the maximum displacement of an object undergoing simple harmonic motionfrom its equilibrium position.

mFor an object oscillating on the end of a spring, T = 2π!k

LFor a simple pendulum, T = 2π!g

A wave is a periodic disturbance.

The bump which is sent in a wave is called the pulse.

The top of each wave is the peak; the bottom is called the trough.

Where the wave originally begins is called the equilibrium position, and the top of the peakto the equilibrium position is the amplitude of the wave.

The distance between peaks is the wavelength, λ.

The frequency, f, is the number of waves that pass a certain point per second.

Periodic waves are waves that move ceaselessly and regularly.

The speed of a wave equals its frequency times its wavelength, or v = f λ.

Transverse waves occur when the particle motion is perpendicular to the wave motion.

Longitudinal waves occur when the particle motion is parallel to the wave motion.

Surface waves occur when the particles move in a circular motion, while the circle centerstays in one place.

For transverse waves on a stretched string, wave speed equals the square root of the tensionover the mass of the string divided by its length, or T

v =!m/L .

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Quiz1. How is the period of Simple Harmonic Motion [SHM] changed: (a) when the mass of the

particle is increased without changing the elastic constant; (b) when the elastic constant isincreased without changing the mass; (c) when the mass and the elastic constant arechanged by the same ratio?

2. Describe an experiment that will measure the spring [or elastic] constant of a spring.

3. The length of a pendulum clock is adjusted so that it gives the correct time at the Earth’ssurface. How would the length have to be adjusted if the clock were taken to the top of a very high mountain?

4. When a man of mass 60 kg gets into a car, the center of gravity of the car lowers 0.3 cm [3 x 10-3 m]. What is the net elastic constant of the springs of the car? Given that the mass of the car is 500 kg, what is the period of vibration when it is empty and when the man isinside?

5. By rocking a boat, surface waves are produced on a quiet lake. The boat performs 12 oscillations in 20 seconds, each oscillation producing a wave crest. It takes 6 seconds for a given crest to reach the shore 12 m away. Calculate the wavelength of the surface waves.

6. A spring having a normal length of 1 m and a mass of 0.2 kg is stretched 0.04 m by a forceof 10 N. Calculate the velocity of propagation of longitudinal waves along the spring.

7. A steel wire having a diameter of 0.2 mm [2 x 10-4 m] is subject to a tension of 200 N.Determine the velocity of propagation of transverse waves along the wire. Steel has a mass density of 7.86 x 103 kg/m3.

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Solutions to Quiz1. The basic equation for the period of motion for a mass on a spring is T = 2π !m/k, where

m is the mass on the end of the spring and k is the elastic or spring constant. So, (a) if themass is increased, the period increases, but only by the square root, while (b) if the elasticconstant increases, the period decreases again by the square root. Then (c) if both changeby the same ratio, nothing happens to the period.

2. One experiment would be to hang the spring vertically with no mass attached. Then, afternoting (i.e., measuring) the position of the lower end, add a small mass and slowly allowthe spring to expand. Measure the new position of the lower end and then calculate theextension of the spring, x, due to the addition of a force, F (= mg). The ratio of F/x will givethe spring constant. Of course, a good scientist would add further "weights" and create agraph of total added force versus extension and get the value for k from the slope of thecurve. Alternatively, a computer program could be used to evaluate the least-squares fit ofthe experimental data.

A second experiment could be done by using the SHM equation: add a mass and allow thesystem to oscillate; measure the period of oscillation. Repeat for other masses and plot oruse your computer program.

3. The period of a pendulum clock is given by T = 2π!L/g, w h e re L is the length of the pendu-lum and g is the acceleration of gravity. At the top of a mountain the value of g is somewhatlower than at sea level. There f o re, recalling Problem 1 above, we should re d u c e the length ofthe pendulum by the same ratio that g has been re d u c e d .

4. F = -kx. Then k = [60 kg(9.8 m/s2)]/[3 x 10-3 m] = [588 N]/[3 x 10-3 m] = 1.96 x 105 N/m wherewe have dropped the minus sign because we are not "doing" vectors. Using the periodequation:

T = 2π !m/k, for the two cases we get

T = 2π!500kg/1.96 x 105 N/m) = 2π!2.55 x 10-3 s2 = 0.317 s, and

T = 2π!(560kg)/(1.96 x 105 N/m) = 2π!2.86 x 10-3 s2 = 0.336 s.

5. If the boat makes 12 oscillations in 20 seconds, its f re q u e n c y (of oscillation) is 12/[20 s] = 0.6 Hz. Since the waves take 6 seconds to reach a shore 12 m away, their v e l o c i t y is [12 m/6 s] = 2 m/s. Then since v = f λ, the wavelength of these is

v /f = [2 m/s]/0.6 Hz = 3.33 m

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6. This spring has a mass per unit length, m/L, of [0.2 kg]/1 m] = 0.2 kg/m. Its spring constantis k = F/x = [10 N]/[4 x 10-2 m] = 2.5 x 102 N/m. Longitudinal waves on a spring must have avalue related to these properties. Using dimensional analysis, breaking newtons into its basicunits, we get N = kg[m/s2] and then k is 2.5 x 102 {kg[m/s2]}/m = 2.5 x 102 kg/s2. Just lookingat units, k divided by the mass per unit length gives units of {kg/s2}/{kg/m} = m/s2. If wemultiplied the length of the spring (one more time), we would get units of m2/s2 and wecould write an equation:

vlongitudinal = !kL/(m/L) = !kL2/m = !2.5(102)(1)2/0.2 = !1250 = 35.4 m/s.

7. The velocity of propagation of transverse waves along a stretched wire is given by

vtransverse = !T/(m/L)

Where T is the tension in the wire and m/L is the mass per unit length of the wire. We’regiven the tension but not the length density of the wire, so we have some work to do.Suppose the wire is L long. It has a cross-sectional area of πr2 and so a total volume of V = πr2L. Since we are given the mass per unit volume, ρ, we have the mass as m = ρV= ρπr2L and so m/L = ρπr2. Remember, radius is half the diameter. So

vtransverse = !200N/(7.86 • 103kg/m3) π(1 • 10−4m)2 = !200N/π(7.86 • 10-5 kg/m)

=!8.099 • 105m2/s2 = 900 m/s.

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Suggestions for Instructors 1. Some students find the use of a point in uniform circular motion and its projection [onto the

X-axis or, better, onto a screen] helpful in understanding SHM. Of course, if you are findingthat using trig functions slows things down it might be better not to introduce SHM as x = Asiθt.

2. Adding acoustic SHMs to produce beats is a demonstration that students appreciate. Also,adding SHMs on an oscilloscope screen so they are perpendicular gives patterns that beginto uncover some of the mysteries of what a CRT can produce. Coupled oscillators are alsolots of fun to observe in demonstration.

3. Since real oscillators do not keep vibrating forever, some discussion of resistive terms is useful. Most often the resistive term is chosen to be directly proportional to the velocity [and opposed to the direction of motion] because the "second order differential equation" is then linear (only composed of first-order terms, hence "linear") and is solvable by stan-dard techniques taught in standard calculus courses. Of course, if students [and/or instruc-tors] can program iterative equations, ANY type of resistive term can be added. For example,with no resistance, SHM is:

F = ma = -kx, or m[dv/dt] + kx = 0, or m[d2x/dt2] + kx = 0

which can be "programmed." Then a resistive force term is added so the equation is:

F = ma = -kx – Rv, or m[dv/dt] + Rv + kx = 0, or m[d2x/dt2] + R[dx/dt] + kx = 0

or any other such term to play with. Give the system some initial energy [by letting the initial values of x and d2x/dt2 be zero and dx/dt some number]. Let the system "go"!

4. Experiments with wave interference, or superposition, intrigue students. In particular, settingup standing waves in the laboratory gives students a good handle on the relationshipbetween frequency, wavelength, and velocity of propagation.

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PROGRAM 9:HEAT

Key Terms and FormulasHeat is the energy that’s transferred from one object to another due to a difference in tempera-ture; heat is measured in Joules.

The specific heat capacity of a substance is how much heat it takes to change one kilogramof it by one degree Celsius.

Heat added or removed equals the mass of a substance times its specific heat capacity timesthe change in temperature, or Q = mcΔT

The equilibrium temperature is the temperature at which, in an isolated system, no heat islost or gained by the system.

The latent heat, L, of a substance is the amount of heat needed to turn one kilogram of itfrom one phase to another, or Q = mL.

The heat of fusion is the amount of heat required to change a substance from a solid into aliquid.

The heat of vaporization is the amount of heat required to change a substance from a liquidinto a gas.

Conduction is heat transfer via the collision of molecules.

Convection is the transfer of heat due to the mass movement of a heated substance.

Radiation is the transfer of energy through space.

The change in work done by gases is equal to the product of the pressure and the change involume of the gas, or ΔW = pΔV.

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Quiz

1. Why do diff e rent materials have diff e rent specific heat capacities? Why does the same materialhave diff e rent specific heat capacities when it is in diff e rent states?

2. How do you measure temperature? Does the temperature of a molecule make any sense? Is temperature a statistical concept?

3. How do you measure pressure? Does the pressure from a molecule make any sense? Is pressure a statistical concept?

4. How does a microwave oven heat anything since there is no fire present? When you grill asteak are you using conduction, convection, or radiation?

5. Assuming no heat is lost or gained to the outside world, will a 10 g [10-2 kg] ice cube at -10°C melt completely in 100 cm3 [10-4 m3] of water originally at +10°C? If yes, what’s the finaltemperature?

6. The rate at which a hot body cools depends directly on the temperature difference betweenthe hot body and the medium surrounding it (e.g., the room temperature). Assume an initialtemperature difference of 100°C and a rate constant such that in 1 minute the temperaturedifference becomes 95°C. Estimate the temperature difference after 1 minute more; after 2minutes more. How long would it take—make a rough guess—before the temperature dif-ference is less than 1°C. [This phenomenon is called Newton’s law of cooling and, more orless, answers the question about when you should add cream to your hot cup of coffee.]

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Solutions to Quiz

1. Heat is a form of energy. There f o re, when a substance absorbs heat, it must increase its ener-gy. Since the substance is "not moving" this energy must be within itself. This means that thepotential energy and the kinetic energy w i t h i n the substance must increase. Since potentiale n e rgy is associated with change in position, as long as a substance doesn’t "grow" [ALL re a lsubstances DO], then the i n t e rn a l kinetic energy must go up and that will be associated withthe molecular structure of the substance—and they’re all diff e rent. Ergo[??], the amount ofe n e rgy per unit mass and per temperature unit change is diff e rent for diff e rent substances.For the same substance in diff e rent states the same kind of argument: diff e rent moleculars t r u c t u re, diff e rent specific heat capacities.

2. Very interesting question. Simple answer: with a thermometer. But what’s a thermometer?A thermometer is a device that uses some temperature-dependent property of some sub-stance to allow for observation of changes in the internal kinetic energy of the substance.For instance mercury [or red-tinted alcohol] expands when heated; so place some mercury ina very thin tube and watch it expand. Stick the tube in a mixture of ice and water and whenthe mercury level settles, mark the tube and call the temperature 0°C. Do the same thing asthe water boils and then not only do you have a thermometer, you also have a definedtemperature scale, arbitrary though it may be. No way are you going to be able to measurethe temperature of one molecule because it doesn’t have any internal energy—it’s all exter-nal kinetic energy [unless you look at it REAL close]. Since there is no way each and everymolecule of a substance will have identically the same amount of kinetic energy, the temper-ature of a sample of a material is an overall average of some kind. You are going to needstatistics of some kind to figure it out "exactly."

3. Pressure is a measure of force per unit area. So if you have a device with a known area onwhich you can measure the force of the gas, you can divide the size of the force by thatarea and measure pressure, by marking off the scale appropriately. No way will one mole-cule give you pressure; it’s just too small and will strike the device too infrequently. As amatter of fact, if you think of a few million molecules [only] within the device, you can thinkof them peppering the area like a bunch of "little balls" [bad analogy, but it helps] and youcan see the scale jiggle like crazy. More statistics are needed to make sense out of pressure.Outer space doesn’t have a pressure—there’s only one molecule per cubic meter, on theaverage!

4. A microwave oven sends out "microwaves," that’s how. Really, there’s an electronic tubegenerating very short electromagnetic waves—wavelengths much longer than those of visiblelight, however—of just the right size that they resonate the water molecules within the mate-rial to be heated. The resonance causes the molecules to increase their internal kinetic ener-gy, which is then an increase in the temperature of the material. Voila, we’re cooked! Whenyou grill a steak, inside or out, you are using all three methods for heat transfer. The surfaceis hot, the coals or fire radiate, and there is a bit—not much—of moving warm air to con-vect some heat.

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5. To raise the temperature of the 10 g of ice to the melting point will require

Q[-10 to 0] = mciceΔT = [10-2 kg][1.946 x 103 J/kg°C][10°C] = 1.946 x 102 J

I looked up the heat capacity of ice so this part could be done. Next, to melt the ice weneed the heat of fusion for ice and that’s equal to 3.335 x 103 J/kg. For our ice cube:

Q[melt] = mL = [10-2 kg][3.335 x 105 J/kg] = 3.335 x 103 J

So just to get the ice cube melted requires 3335 + 194.6 = 3530 J. To reduce the temperatureof 100 g [100 cm3 has a mass of 100g] to 0° would require

Q[+10 to 0] = mcwaterΔT = [10-1 kg][4.186 x 103 J/kg°C][10°C] = 4.186 x 103 J

Therefore, since it would take only 3530 J to "heat up" and then completely melt the icecube, there was more than enough heat in the water to give up. So, we have to go backand see what the temperature of the water was at the time the ice was all melted; that is,when the water had given up 3530 J. We use the above equation but solve for ΔT instead:

ΔT = +10°C – T° = Q[+10 to T°]/[mcwater] = [3530 J]/[10-1 kg][4.186 x 103 J/kg°C] = 8.43°C

or

T° = 10 – 8.43 = 1.57°C. We are now at the point where 100 g of water is at 1.57°C and 10 gof water is at 0°C. The 10 g will "warm up" a bit and the 100 g will cool down so that bothhave the same final temperature. In other words, there will be a heat loss Q" by the 100 gequal to [100 g]cwater[1.57°C – Tfinal] and an equal heat gain by the 10 g equal to [10 g]cwater[Tfinal – 0°C]

or

[100 g]cwater[1.57°C – Tfinal] = [10 g]cwater[Tfinal – 0°C]

or

10[1.57°C – Tfinal] = [Tfinal – 0°C], which gives 11Tfinal = 15.7°C or Tfinal = 1.43°C.

6. Newton’s law of cooling has to be put into an equation form so we can use it. What it saysis that the heat loss per second is directly proportional to the temperature difference. So wewrite:

ΔQ/Δt = K(T – Troom)

The information given was an initial temperature diff e rence of 100°C, so let’s say the ro o mt e m p e r a t u re, which will remain constant, is 20°C. Then the initial temperature of the sub-stance to cool is 120°C. This is so you can develop a plot of temperature versus time to getsome idea of what’s going on. Now with all these numbers, we can calculate the value of theconstant K f ro m

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K = (T – Troom)/[ΔQ/Δt] = (100°C)/[(5°C/1 min)] = 20 min,

where we are just using the temperature drop as the amount of heat loss per unit time,ΔQ/Δt, because we aren’t given any particulars about what’s cooling. This is what we know:Something is cooling and starts at a temperature of 120°C and after 1 minute its temperatureis 115°C. During that initial minute, the substance lost heat at the rate of [20 min][100°C] and,all things being equal, will lose heat at the rate of [20 min][95°C] during the second minute,or only 95% of what it lost the first minute. This means the temperature drop of the sub-stance should be about 0.95[5°C] = 4.75°C and the new temperature will be 115°C – 4.75°C= 110.25°C, giving a new temperature difference of 90.25°C. Using the same reasoning, dur-ing the next (third) minute the rate of heat loss will be [20 min][90.25°C], or 0.9025[4.75°C] =4.287°C and temperature 110.25°C – 4.287°C = 105.96°C. This game [process] can be contin-ued to find new lower temperatures. Plotting these values will prove very interesting. As youcan clearly see, it is going to take a lot longer than 20 minutes for the substance to cool toroom temperature because it loses less and less heat each unit of time.

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Suggestions for Instructors

1. Calorimetry, the calculation of heat transfer from one body to another, is usually a simplealgebraic process and as such is useful to help students improve mathematical skills.However, there is not too much physics involved. More time can be spent on more difficultideas, such as what is a temperature scale and how temperature [absolute] is a measure ofthe internal energy of a substance.

2. Once you have introduced the absolute temperature scale, your text [or curriculum] shoulddiscuss internal energy of materials. Without necessarily going into the "kinetic theory ofmatter" and a derivation, students should know that U = (3/2)kNT where k, the Boltzmannconstant, has been chosen to have its value so 3/2 remains as does N, the number of mole-cules of material. The other form, U = (3/2)nRT, where n is the number of moles of thematerial and R is the gas constant, is sometimes useful. The critical point is that studentsmust understand that, under usual circumstances, the internal energy of a substance is direct-ly proportional to its absolute temperature, whatever form the constant of proportionalitytakes.

3. The reason we can write any "laws" in Heat and Thermodynamics is because of the statisti-cal nature of these laws. When you get down to the level of being able to count individualmolecules, the best we can do is attempt to introduce some kind of statistical mechanicsand formulate concepts based on fundamental principles, such as conservation of energy,and reasonable conditions, such as lowest energies get filled first. These are rather difficultideas, but sometimes they help students grasp the difference between the very few funda-mental principles and the large body of laws that have been developed to explain variousphenomena.

4. Radiation, as one of the methods for heat transfer, allows you to get involved in a discussionof the interaction of radiation with matter. This could be delayed for later in the course, buta short introduction at this time will show the relevance of studies that lead to devices likedental X-ray machines, microwave ovens, MRI scanners, etc.

5. The last problem for this program was chosen to allow for some possible use of computercalculation and even a way to introduce the elements of calculus through a practical prob-lem that needs solving in a more precise manner.

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PROGRAM 10:THERMODYNAMICS

Key Terms and FormulasThe First Law of Thermodynamics states that the change in the internal energy of a systemequals the heat added minus the work done by the system, or ΔU = Q – W.

An isovolumetric process is when there is a gas and the volume is kept constant, but thepressure changes.

An isobaric process is when the pressure is kept constant, but the volume changes.

In an adiabatic process, no heat is added or taken from the system.

The Second Law of Thermodynamics states that heat flows naturally from a hot object to a colder one, and heat will not flow spontaneously from a cold object to a hot one. In otherwords, the efficiency of a heat engine can never equal one hundred percent. A third way tostate the second law of thermodynamics is to say that the entropy of a closed system can neverdecrease. It will always increase as heat is converted into work.

Heat engines turn heat energy into electrical or mechanical energy. The heat taken from thehot reservoir equals the work done plus the heat rejected to the cold reservoir, and efficiencyis work done divided by heat in.

Entropy is the measure of disorder of a system, and the change in entropy equals heat changeof a system divided by the absolute temperature, or ΔS = Q/T.

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Quiz

1. What is the fundamental difference, in the thermodynamic sense, between work and heat?

2. Suggest how the temperature of a gas is kept constant during an isothermal expansion.

3. What is meant by a heat engine? What is the best way of increasing the efficiency of a heatengine?

4. Why is it not possible to design an engine that has 100% heat efficiency?

5. When a system is taken from state A to state B along path ACB (see figure 1 below), thesystem absorbs 80 J of heat and does 30 J of work.

(a) Given that the work done is 10 J, how much heat is absorbed by the system along pathADB?

(b) The system is returned from state B to state A along the curved path. The work done on the system is 20 J. Does the system absorb or liberate heat, and how much?

(c) Given that UA = 20 J and UD = 60 J, determine the heat absorbed in the processes ADand AB.

6. A gas undergoes the cycle shown in Figure 2 below. The cycle is repeated 100 times perminute. Determine the power generated by this "heat engine."

7. Explain why it is rather unlikely for the entropy of an isolated system to decrease.

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C B

A D

V0

p p, kPa

30

20

10

Figure 1 Figure 2

B

A

C

0 0.2 0.4 0.6 0.8 1.0 V, m3

Solutions to Quiz

1. Work is a calculable quantity, by adding up all the pΔV terms; that is, the total amount ofwork done can be measured off a pV diagram. Because the value of p may be changing asyou go from one V to another, we need a plan to do that sum. On the other hand, heat isNOT calculable. In other words, both heat and work are forms of energy that can changethe energy of a system; work is "easy" to calculate while heat is not.

2. Isothermal means constant temperature. During an expansion, work is being done and, ifNO heat is added (Q = 0), then ΔU must decrease, which means the temperature must godown since the absolute temperature is directly proportional to the internal energy. So, dur-ing an isothermal expansion an amount of heat must be added exactly equal to the amountof work done. How you do this is your business—by conduction, convection, or radiation.

3. A heat engine is any device that converts heat [energy] into mechanical work [energy]. Thebest way to increase the efficiency of a heat engine is to throw away as little heat as possi-ble. This is done by making the ratio of the absolute temperature as great as possible.

4. Practically, trying to increase the temperature of the hot reservoir runs into containmentproblems (materials begin to melt at high temperatures and they undergo fatigue faster).Theoretically, it’s impossible to get to 0°K and it is very hard to maintain very cold tempera-tures at the cold reservoir for a long time—it’s expensive.

5. This problem is based on the first law of Thermodynamics: ΔU = Q – W, where ΔU is thechange in internal energy of, Q is the heat added to, and W is the work done by the system.No matter how you go from one set of conditions (called the state of the system), to theother—that is from A to B, the value of the change in internal energy is:

ΔU = 80 J – 30 J = 50 J.

(a) When we go from state A to state B by path ADB, ΔU = 50 J as we just found and W = 10 J. Therefore, we have

ΔU = Q – W = 50 J = Q – 10 J, or Q = 50 J + 10 J = 60 J.

(b) Same thought process: ΔU = Q – W = -50 J = Q – (-20 J). Q = -50 J – 20 J = -70 J. Thismeans that the system liberated [gave off] 70 J of heat. We used -50 J for ΔU because thepath was from B to A.

(c) Given that UA = 20 J and UD = 60 J. Therefore, we have ΔUAD = UD – UA = 60 J – 20 J= 40 J. When you go from state D to state B no work is done (because ΔV = 0 alongpath DB). We were given that the total work on the path ADB is 10 J, and so all thiswork is done along AD. So we write ΔUAD = QAD – WAD = 40 J = QAD – 10 J

Therefore QAD = 40 J + 10 J = 50 J.

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6. The total work done by this heat engine each cycle is the area contained within the triangle.Therefore:

Wcycle = 1/2 [0.8 m3 – 0.2 m3][3 x 104 Pa – 1 x 104 Pa] = 1/2 (0.6 m3)(2 x 104 Pa) = 6000 J

Since power is work per unit time, we have:

P = Wcycle[cycles/min][1 min/60 s] = [6 x 103 J][100 cycles/min][1 min/60 s] = 10 kW.

7. ΔS = Q/T, where ΔS is change in entropy, Q is the heat absorbed during this change, and Tis the (constant) absolute temperature during this change. Consider an isolated system com-posed of a totally sealed room containing "everything" to maintain the well-being of a fewhumans, including trees, animals, an air-recirculating system, a waste recycling facility (allkinds of waste), etc. Initially, this isolated system would be highly organized into a numberof very specific entities, such as plant life. But, to cook food it would be necessary to cutdown a tree and burn the wood. As time went on, the ultra-high organization of plants, min-erals, and animals would become less organized—or more dispersed throughout the room.Every time some action took place within the isolated system, there would be a ΔS thatwould be positive, even in those cases where some parts would have a decrease in S. Thatis, overall even though the energy within the system would remain conserved (unchanged),the entropy would increase. As the system became less and less organized—more uniform innature (every fire reduces a highly organized group of molecules that made up the fiber of atree into a uniform pile of carbonized molecules that are all pretty much the same)—theentropy increases. The end result is a system that can no longer increase its entropy and atthat point everything is pretty uniform—or in a state of chaos, if you prefer that term.

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Suggestions for Instructors

1. The laws of thermodynamics are hardly understandable without the general equation of statefor gases, namely pV = kNT or pV = nRT, where k is the Boltzmann constant, N is the num-ber of molecules of the material, n is the number of moles, R is the gas constant, and T is the temperature of the material in kelvin. Recall that this equation of state producesBoyle’s law when the absolute temperature is held constant (isothermal case); that is, pV =a constant. When the pressure or the volume is held constant we get Charles’ law and Gay-Lussac’s law, respectively. These empirical laws, along with the relation for the internal ener-gy of a system, U = (3/2)kNT or U = (3/2)nRT, are sufficient to do most heat problems untilentropy must be defined.

2. Entropy is a rather difficult concept to explain. I suggest you use examples such as placing adrop of ink into a glass of water. Initially the system is "structured" with all the ink in oneplace. The entropy of the ink drop and water system is low. When the ink dispersesthroughout the water and the ink-water system is uniform, the entropy is higher. Somewould say the second state is more chaotic than the first, while some insist the second stateis more uniform. So exactly what is meant by chaos must also be clearly defined. The natu-ral tendency of an enclosed system is to go toward uniformity, or the highest state ofentropy (or the most probable state—that’s where statistics comes into play). If your studentswould like to investigate an interesting experiment in an isolated system, have them look atthe Web site of the Biosphere II, located in Tucson, Arizona: www.bio2.com.

3. Drawing pV diagrams and placing on them various processes is an excellent exercise.Isothermal pV lines are parabolas [because pV = constant], constant pressure lines are hori-zontal, and constant volume lines are vertical. Isentropic [i.e., Q = 0] lines are steeper thanisotherms and you’ll have to get into exponentials to get the right shape. Nevertheless, alllines are interesting to draw. And don’t forget—the area enclosed within a closed path on apV diagram represents the total work done during the cycle.

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ContributorsThe Core Curriculum Series: Physics was designed by the Standard Deviants Academic Team,including:

Edward Finn, Ph.D., Georgetown UniversityBarry Berman, Ph.D., George Washington University

Headed by:

David Sturdevant, Director of Writing, MFA, University of Louisville

with assistance from Margaret Packard, Films for the Humanities & Sciences.

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1 0 5 4 6for use with

11745, 10277

PO Box 2053, Princeton, NJ 08543-2053

CALL 800-257-5126 • FAX 609-671-0266

A Wealth of Information. A World of Ideas.Films for the Humanities & Sciences