2008–2009 honors physics review notestomstrong.org/physics/year0809h.pdf · honors physics review...

Honors Physics Review Notes

2008–2009

Tom StrongScience Department

Mt Lebanon High [email protected]

The most recent version of this can be found at http://www.tomstrong.org/physics/

Chapter 1 — The Science of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 2 — Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Chapter 3 — Two-Dimensional Motion and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Chapter 4 — Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Chapter 5 — Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Chapter 6 — Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Chapter 7 — Rotational Motion and the Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Chapter 8 — Rotational Equilibrium and Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Chapter 12 — Vibrations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Chapter 13 — Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Chapter 14 — Light and Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Chapter 15 — Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Chapter 16 — Interference and Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Chapter 17 — Electric Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Chapter 18 — Electrical Energy and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Chapter 19 — Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Chapter 20 — Circuits and Circuit Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Chapter 21 — Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Chapter 22 — Induction and Alternating Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Variables and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Mathematics Review for Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Physics Using Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Final Exam Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Final Exam Equation Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Mixed Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Mixed Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

ii

These notes are meant to be a summary of important points covered in the Honors Physics class at Mt. Lebanon HighSchool. They are not meant to be a replacement for your own notes that you take in class, nor are they a replacement foryour textbook. Much of the material in here is taken from the textbook without specifically acknowledging each case, inparticular the organization and overall structure exactly match the 2002 edition of Holt Physics by Serway and Faughnand many of the expressions of the ideas come from there as well.

The mixed review exercises were taken from the supplementary materials provided with the textbook. They are arepresentative sampling of the type of mathematical problems you may see on the final exam for the course. There willalso be conceptual questions on the exam that may not be covered by the exercises included here. These exercises areprovided to help you to review material that has not been seen in some time, they are not meant to be your only resourcefor studying. Exercises are included from chapters 1–8, 12–17, and 19–22.

The answers at the end of the review are taken from the textbook, often without verifying that they are correct. Usethem to help you to solve the problems but do not accept them as correct without verifying them yourself.

This is a work in progress and will be changing and expanding over time. I have attempted to verify the correctnessof the information presented here, but the final responsibility there is yours. Before relying on the information in thesenotes please verify it against other sources.

Honors Physics 2008–2009 Mr. Strong

1

Chapter 1 — The Science of Physics

1.1 What is Physics?Some major areas of Physics:Mechanics — motion and its causes — falling objects,

friction, weightThermodynamics — heat and temperature — melting

and freezing processes, engines, refrigeratorsVibrations and Waves — specific types of repeating

motions — springs, pendulums, soundOptics — light — mirrors, lenses, colorElectromagnetism — electricity, magnetism, and light

— electrical charge, circuitry, magnetsRelativity — particles moving at very high speeds — par-

ticle accelerators, particle collisions, nuclear energyQuantum Mechanics — behavior of sub-microscopic

particles — the atom and its parts

The steps of the Scientific Method1. Make observations and collect data that lead to a

question2. Formulate and objectively test hypotheses by experi-

ments (sometimes listed as 2 steps)3. Interpret results and revise the hypotheses if neces-

sary4. State conclusions in a form that can be evaluated by

others

1.2 Measurements in ExperimentsMeasurementsThere are 7 basic dimensions in SI (Systeme International),the 3 we will use most often are:

• Length — meter (m) — was 1/10,000,000 of the dis-tance from the equator to the North Pole — now thedistance traveled by light in 3.3× 10−9 s• Mass — kilogram (kg) — was the mass of 0.001 cubic

meters of water, now the mass of a specific platinum-iridium cylinder

• Time — second (s) — was a fraction of a mean so-lar day, now 9,162,631,700 times the period of a radiowave emitted by a Cesium-133 atom

Common SI Prefixes

Prefix Multiple Abbrev.

nano- 10−9 n deci- 10−1 dmicro- 10−6 µ kilo- 103 kmilli- 10−3 m mega- 106 Mcenti- 10−2 c giga- 109 G

Accuracy vs. Precision

• Accuracy describes how close a measured value is tothe true value of the quantity being measured

Problems with accuracy are due to error. To avoiderror:

– Take repeated measurements to be certain thatthey are consistent (avoid human error)

– Take each measurement in the same way (avoidmethod error)

– Be sure to use measuring equipment in goodworking order (avoid instrument error)

• Precision refers to the degree of exactness with whicha measurement is made and stated.

– 1.325 m is more precise than 1.3 m

– lack of precision is usually a result of the limi-tations of the measuring instrument, not humanerror or lack of calibration

– You can estimate where divisions would fall be-tween the marked divisions to increase the pre-cision of the measurement

1.3 The Language of Physics

There are many symbols that will be used in this class, someof the more common will be:

Symbol Meaning

∆x Change in xxi, xf Initial, final values of x∑

F Sum of all F

Dimensional analysis provides a way of checking tosee if an equation has been set up correctly. If the units re-sulting from the calculation are not those that are expectedthen it’s very unlikely that the numbers will be correct ei-ther. Order of magnitude estimates provide a quickway to evaluate the appropriateness of an answer — if theestimate doesn’t match the answer then there’s an errorsomewhere.


2

Counting Significant Figures in a Number

Rule Example

All counted numbers have an infinite numberof significant figures

10 items, 3 measurements

All mathematical constants have an infinitenumber of significant figures

1/2, π, e

All nonzero digits are significant 42 has two significant figures; 5.236 has four

Always count zeros between nonzero digits 20.08 has four significant figures; 0.00100409 has six

Never count leading zeros 042 and 0.042 both have two significant figures

Only count trailing zeros if the number con-tains a decimal point

4200 and 420000 both have two significant figures; 420.has three; 420.00 has five

For numbers in scientific notation apply theabove rules to the mantissa (ignore the expo-nent)

4.2010× 1028 has five significant figures

Counting Significant Figures in a Calculation

Rule Example

When adding or subtracting numbers, find the number which is knownto the fewest decimal places, then round the result to that decimalplace.

21.398 + 405 − 2.9 = 423 (3significant figures, roundedto the ones position)

When multiplying or dividing numbers, find the number with the fewestsignificant figures, then round the result to that many significant fig-ures.

0.049623 × 32.0/478.8 =0.00332 (3 significant figures)

When raising a number to some power count the number’s significantfigures, then round the result to that many significant figures.

5.82 = 34 (2 significantfigures)

Mathematical constants do not influence the precision of any compu-tation.

2× π × 4.00 = 25.1 (3 signif-icant figures)

In order to avoid introducing errors during multi-step calculations, keepextra significant figures for intermediate results then round properlywhen you reach the final result.

Rules for Rounding

Rule Example

If the hundredths digit is 0 through 4 drop it and all following digits. 1.334 becomes 1.3

If the hundredths digit is 6 though 9 round the tenths digit up to thenext higher value.

1.374 becomes 1.4

If the hundredths digit is a 5 followed by other non-zero digits thenround the tenths digit up to the next higher value.

1.351 becomes 1.4

If the hundredths digit is a 5 not followed by any non-zero digits thenif the tenths digit is even round down, if it is odd then round up.

1.350 becomes 1.4, 1.250 be-comes 1.2

(assume that the result was to be rounded to the nearest 0.1, for other precisions adjust accordingly)


3

Chapter 2 — Motion in One Dimension

2.1 Displacement and Velocity

The displacement of an object is the straight line (vector)drawn from the object’s initial position to its new position.Displacement is independent of the path taken and is notnecessarily the same as the distance traveled. Mathemati-cally, displacement is:

∆x = xf − xi

The average velocity, equal to the constant velocitynecessary to cover the given displacement in a certain timeinterval, is the displacement divided by the time intervalduring which the displacement occurred, measured in m

s

vavg =∆x∆t

=xf − xi

tf − ti

The instantaneous velocity of an object is equivalentto the slope of a tangent line to a graph of x vs. t at thetime of interest.

The area under a graph of instantaneous velocity vs.time (v vs. t) is the displacement of the object during thattime interval.

2.2 Acceleration

The average acceleration of an object is the rate ofchange of its velocity, measured in m

s2 . Mathematically, itis:

aavg =∆v∆t

=vf − vi

tf − ti

Like velocity, acceleration has both magnitude and di-rection. The speed of an object can increase or decreasewith either positive or negative acceleration, depending onthe direction of the velocity — negative acceleration doesnot always mean decrease in speed.

The instantaneous acceleration of an object is equiv-alent to the slope of a tangent line to the v vs. t graph atthe time of interest, while the area under a graph of instan-taneous acceleration vs. time (a vs. t) is the velocity ofthe object. In this class acceleration will almost always beconstant in any problem.

Displacement and velocity with constant uniform accel-eration can be expressed mathematically as any of:

vf = vi + a∆t

∆x = vi∆t+12a∆t2

∆x =12(vi + vf )∆t

v2f = v2

i + 2a∆x

2.3 Falling ObjectsIn the absence of air resistance all objects dropped near thesurface of a planet fall with effectively the same constantacceleration — called free fall. That acceleration is alwaysdirected downward, so in the customary frame of referenceit is negative, so:

ag = g = −9.81ms2


4

Chapter 3 — Two-Dimensional Motion and Vectors

3.1 Introduction to VectorsVectors can be added graphically.

Vectors can be added in any order:

~V1 + ~V2 = ~V2 + ~V1

To subtract a vector you add its opposite:

~V1 − ~V2 = ~V1 + (−~V2)

Multiplying a vector by a scalar results in a vector insame direction as the original vector with a magnitude equalto the original magnitude multiplied by the scalar.

3.2 Vector OperationsTo add two perpendicular vectors use the Pythagorean The-orem to find the resultant magnitude and the inverse of thetangent function to find the direction: Vr =

√V 2

x + V 2y and

the direction of Vr, θr, is tan−1 Vy

Vx

Just as 2 perpendicular vectors can be added, any vectorcan be broken into two perpendicular component vectors:~V = ~Vx + ~Vy where ~Vx = V cos θı and ~Vy = V sin θ;

Two vectors with the same direction can be added byadding their magnitudes, the resultant vector will have thesame direction as the vectors that were added.

Any two (or more) vectors can be added by first decom-posing them into component vectors, adding all of the x andy components together, and then adding the two remainingperpendicular vectors as described above.

3.3 Projectile MotionThe equations of motion introduced in chapter 2 are ac-tually vector equations. Once the new symbol for displace-ment, ~d = ∆xı+∆y has been introduced the most commonones can be rewritten as

~d = ~vi∆t+12~a∆t2

~vf = ~vi + ~a∆t

Since motion in each direction is independent of motion inthe other, objects moving in two dimensions are easier toanalyze if each dimension is considered separately.

~dx = ~vxi∆t+12~ax∆t2

~dy = ~vyi∆t+12~ay∆t2

In the specific case of projectile motion there is noacceleration in the horizontal (ı) direction (~ax = 0) andthe acceleration in the vertical () direction is constant(~ay = ~ag = ~g = −9.81 m

s2 ). While ~vy will change over time,~vx remains constant. Neglecting air resistance, the path fol-lowed by an object in projectile motion is a parabola. Thefollowing equations use these simplifications to describe themotion of a projectile launched with speed ~vi and directionθ up from horizontal:

~vx = vi cos θı = constant

~vyi = vi sin θ

~dx = vi cos θ∆tı

~dy = vi sin θ∆t+12~g∆t2

To solve a projectile motion problem use one part of theproblem to find ∆t, then once you know ∆t you can usethat to solve the rest of the problem. In almost every casethis will involve using the vertical part of the problem tofind ∆t which will then let you solve the horizontal partbut there may be some problems where the opposite ap-proach will be necessary. For the special case of a projectilelaunched on a level surface the range can be found with

dx =v2 sin 2θag


5

Chapter 4 — Forces and the Laws of Motion

4.1 Changes in MotionForce causes change in velocity. It can cause a station-ary object to move or a moving object to stop or otherwisechange its motion.

The unit of force is the newton (N), equivalent to kg·ms2 ,

which is defined as the amount of force that, when actingon a 1 kg mass, produces an acceleration of 1m

s2 .Contact forces act between any objects that are in

physical contact with each other, while field forces actover a distance.

A force diagram is a diagram showing all of the forcesacting on the objects in a system.

A free-body diagram is a diagram showing all of theforces acting on a single object isolated from its surround-ings.

4.2 Newton’s First LawNewton’s first law states that

An object at rest remains at rest, and an ob-ject in motion continues in motion with constantvelocity (that is, constant speed in a straightline) unless the object experiences a net exter-nal force.

This tendency of an object to not accelerate is calledinertia. Another way of stating the first law is that if thenet external force on an object is zero, then the accelerationof that object is also zero. Mathematically this is∑

~F = 0→ ~a = 0

An object experiencing no net external force is said tobe in equilibrium, if it is also at rest then it is in staticequilibrium

4.3 Newton’s Second and Third LawsNewton’s second law states that

The acceleration of an object is directly pro-portional to the net external force acting on theobject and inversely proportional to the object’smass.

Mathematically this can be stated as:∑~F = m~a

Which in the case of no net external force (∑ ~F = 0) also

illustrates the first law:∑~F = 0→ m~a = 0→ ~a = 0

Newton’s third law states that

If two objects interact, the magnitude of theforce exerted on object 1 by object 2 is equal tothe magnitude of the force simultaneously ex-erted on object 2 by object 1, and these twoforces are opposite in direction.

Mathematically, this can be stated as:

~F1,2 = −~F2,1

The two equal but opposite forces form an action-reactionpair.

4.4 Everyday Forces

The weight of an object (~Fg) is the gravitational force ex-erted on the object by the Earth. Mathematically:

~Fg = m~g where ~g = −9.81ms2j

The normal force (~Fn) is the force exerted on an objectby the surface upon which the object rests. This force isalways perpendicular to the surface at the point of contact.

The force of static friction (~Fs) is the force that op-poses motion before an object begins to move, it will pre-vent motion so long as it has a magnitude greater than theapplied force in the direction of motion. The maximummagnitude of the force of static friction is the product ofthe magnitude of the normal force times the coefficientof static friction (µs) and its direction is opposite thedirection of motion:

µs =Fs,max

Fn→ Fs,max = µsFn

The force of kinetic friction (~Fk) is the force that op-poses the motion of an object that is sliding against a sur-face. The magnitude of the force of kinetic friction is theproduct of the magnitude of the normal force times thecoefficient of kinetic friction (µk) and its direction isopposite the direction of motion:

µk =Fk

Fn→ Fk = µkFn

The force of friction between two solid objects dependsonly on the normal force and the coefficient of friction, itis independent of the the surface area in contact betweenthem.

For an object at rest with a force applied to it the fric-tional force will vary as the applied force is increased.


6

Fapp

Ffriction

µkFn

µsFn

µsFnstatic kinetic

Fs = Fapp

Fk = µkFn

Air resistance (~FR) is the force that opposes the mo-tion of an object though a fluid. For small speeds FR ∝ vwhile for large speeds FR ∝ v2. (Exactly what is small orlarge depends on things that are outside the scope of thisclass.) Air resistance is what will cause falling objects toeventually reach a terminal speed where FR = Fg.

Solving Friction Problems

When solving friction problems start by drawing a diagramof the system being sure to include the gravitational force(Fg), normal force (Fn), frictional force (Fk, Fs, or Fs,max)and any applied force(s).

Determine an appropriate frame of reference, rotatingthe x and y coordinate axes if that is convenient for theproblem (such as an object on an inclined plane). Resolveany vectors not lying along the coordinate axes into com-ponents.

Use Newton’s second law (∑ ~F = m~a) to find the rela-

tionship between the acceleration of the object (often zero)and the applied forces, setting up equations in both the xand y directions (

∑ ~Fx = m~ax and∑ ~Fy = m~ay) to solve

for any unknown quantities.


7

Chapter 5 — Work and Energy

5.1 WorkA force that causes a displacement of an object does work(W ) on the object. The work is equal to the product of thedistance that the object is displaced times the component ofthe force in the direction of the displacement, or if θ is theangle between the displacement vector and the (constant)net applied force vector, then:

W = Fnetd cos θ

The units of work are Joules (J) which are equivalentto N ·m or kg·m2

s2 . The sign of the work being done is sig-nificant, it is possible to do a negative amount of work.

5.2 EnergyWork done to change the speed of an object will accumu-late as the kinetic energy (KE , or sometimes K) of theobject. Kinetic energy is:

KE =12mv2

If work is being done on an object, the work-kineticenergy theorem shows that:

Wnet = ∆KE

In addition to kinetic energy, there is also potentialenergy (PE , or sometimes U) which is the energy storedin an object because of its position. If the energy is storedby lifting the object to some height h, then the equation forgravitational potential energy is:

PE g = mgh

Potential energy can also be stored in a compressedor stretched spring, if x is the distance that a spring isstretched from its rest position and k is the spring con-stant measuring the stiffness of the spring (in newtons permeter) then the elastic potential energy that is storedis:

PE elastic =12kx2

The units for all types of energy are the same as thosefor work, Joules.

5.3 Conservation of EnergyThe sum an objects kinetic energy and potential energy isthe object’s mechanical energy (ME ). In the absence offriction, the total mechanical energy of a system will remainthe same. Mathematically, this can be expressed as:

ME i = ME f

In the case of a single object in motion, this becomes:

12mv2

i +mghi =12mv2

f +mghf

5.4 PowerThe rate at which energy is transferred is called power (P ).The mathematical expression for power is:

P =W

∆t=Fd

∆t= F

d

∆t= Fv

The units for power are Watts (W) which are equivalentto J

s or kg·m2

s3


8

Chapter 6 — Momentum and Collisions

6.1 Momentum and ImpulseMomentum is a vector quantity described by the productof an object’s mass times its velocity:

~p = m~v

If an object’s momentum is known its kinetic energy canbe found as follows:

KE =p2

2m

The change in the momentum of an object is equal tothe impulse delivered to the object. The impulse is equalto the constant net external force acting on the object timesthe time over which the force acts:

∆~p = ~F∆t

Any force acting on on object will cause an impulse,including frictional and gravitational forces.

In one dimension the slope of a graph of the momen-tum of an object vs. time is the net external force actingon the object. The area under a graph of the net externalforce acting on an object vs. time is the total change inmomentum of that object.

Momentum and impulse are measured in kg·ms — there

is no special name for that unit.

6.2 Conservation of MomentumMomemtum is always conserved in any closed system:

Σ~pi = Σ~pf

For two objects, this becomes:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

Any time two objects interact, the change in momen-tum of one object is equal in magnitude and opposite indirection to the change in momentum of the other object:

∆~p1 = −∆~p2

6.3 Elastic and Inelastic CollisionsThere are three types of collisions:

• Elastic — momentum and kinetic energy are con-served. Both objects return to their original shapeand move away separately. Generally:

Σ~pi = Σ~pf

ΣKEi = ΣKEf

For two objects:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

12m1v

21i +

12m2v

22i =

12m1v

21f +

12m2v

22f

• Inelastic — momentum is conserved, kinetic energyis lost. One or more of the objects is deformed in thecollision. Generally:

Σ~pi = Σ~pf

ΣKEi > ΣKEf

For two objects:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

12m1v

21i +

12m2v

22i >

12m1v

21f +

12m2v

22f

• Perfectly inelastic — momentum is conserved, ki-netic energy is lost. One or more objects may bedeformed and the objects stick together after the col-lision. Generally:

Σ~pi = Σ~pf

ΣKEi > ΣKEf

For two objects:

m1~v1i +m2~v2i = (m1 +m2)~vf

12m1v

21i +

12m2v

22i >

12(m1 +m2)v2

f

True elastic or perfectly inelastic collisions are very rarein the real world. If we ignore friction and other small en-ergy losses many collisions may be modeled by them.

Newton’s Laws in terms of Momentum1. Inertia:

Σ~F = 0→ ~p = constant

2. ~F = m~a:∆~p = ~F∆t

3. Every action has an equal and opposite reaction:

∆~p1 = −∆~p2


9

Chapter 7 — Rotational Motion and the Law of Gravity

7.1 Measuring Ratational Motion

For an object moving in a circle with radius r through anarc length of s, the angle θ (in radians) swept by the objectis:

θ =s

r

The conversion between radians and degrees is:

θ(rad) =π

180oθ(deg)

The angular displacement (∆θ) through which anobject moves from θi to θf is, in rad:

∆θ = θf − θi =sf − si

r=

∆sr

The average angular speed (ω) of an object is, in rads ,

the ratio between the angular displacement and the timeinterval required for that displacement:

ωavg =θf − θi

∆t=

∆θ∆t

The average angular acceleration (α) is, in rads2 :

αavg =ωf − ωi

∆t=

∆ω∆t

For each quantity or relationship in linear motion thereis a corresponding quantity or relationship in angular mo-tion:

Linear Angular

x θv ωa α

vf = vi + a∆t ωf = ωi + α∆t

∆x = vi∆t+ 12a∆t

2 ∆θ = ωi∆t+ 12α∆t2

v2f = v2

i + 2a∆x ω2f = ω2

i + 2α∆θ

∆x = 12 (vi + vf )∆t ∆θ = 1

2 (ωi + ωf )∆t

7.2 Tangential and Centripetal AccelerationIn addition to the angular speed of an object moving withcircular motion it is also possible to measure the object’stangential speed (vt) or instantaneous linear speed whichis measured in m

s as follows:

vt = rω

An object’s tangential acceleration (at) can also bemeasured, in m

s2 , as follows:

at = rα

The centripetal acceleration (ac) of an object is di-rected toward the center of the object’s rotation and hasthe following magnitude:

ac =v2

t

r= rω2

7.3 Causes of Circular MotionThe centripetal force (Fc) causing a centripetal accel-eration is also directed toward the center of the object’srotation, and has the following magnitude:

Fc = mac =mv2

t

r= mrω2

The centripetal force keeping planets in orbit is a grav-itational force (Fg) and it is found with Newton’s Univer-sal Law of Gravitation:

Fg = Gm1m2

r2

where G is the constant of universal gravitation whichhas been determined experimentally to be

G = 6.673× 10−11 N ·m2

kg2

An object in a circular orbit around the Earth will sat-isfy the equation

v =

√GME

r

where ME is the mass of the Earth.


10

Chapter 8 — Rotational Equilibrium and Dynamics

8.1 Torque

Just as a net external force acting on an object causes linearacceleration, a net torque (τ) causes angular acceleration.The torque caused by a force acting on an object is:

τ = rF sin θ

where F is the force causing the torque, r is the distancefrom the center to the point where the force acts on theobject, and θ is the angle between the force and a radialline from the object’s center through the point where theforce is acting. Torque is measured in N ·m.

The convention used by the book is that torque in acounterclockwise direction is positive and torque in a clock-wise direction is negative (this corresponds to the right-hand rule). If more that one force is acting on an objectthe torques from each force can be added to find the nettorque:

τnet =∑

τ

8.2 Rotation and Inertia

The center of mass of an object is the point at which allthe mass of the object can be said to be concentrated. Ifthe object rotates freely it will rotate about the center ofmass.

The center of gravity of an object is the point throughwhich a gravitational force acts on the object. For mostobjects the center of mass and center of gravity will be thesame point.

The moment of inertia (I) of an object is the object’sresistance to changes in rotational motion about some axis.Moment of inertia in rotational motion is analogous to massin translational motion.

Some moments of inertia for various common shapes are:

Shape I

Point mass at a distance r from the axis mr2

Solid disk or cylinder of radius r aboutthe axis

12mr

2

Solid sphere of radius r about its diameter 25mr

2

Thin spherical shell of radius r about itsdiameter

23mr

2

Thin hoop of radius r about the axis mr2

Thin hoop of radius r about the diameter 12mr

2

Thin rod of length l about its center 112ml

2

Thin rod of length l about its end 13ml

2

An object is said to be in rotational equilibrium whenthere is no net torque acting on the object. If there is also

no net force acting on the object (translational equilib-rium) then the object is in equilibrium (without any qual-ifying terms).

Type Equation Meaning

TranslationalEquilibrium

∑ ~F = 0 The net force on the ob-ject is zero

Rotational Equi-librium

∑τ = 0 The net torque on the

object is zero

8.3 Rotational DynamicsNewton’s second law can be restated for angular motion as:

τnet = Iα

This is parallel to the equation for translational motionas follows:

Type of Motion Equation

Translational ~F = m~aRotational τ = Iα

Just as moment of inertia was analogous to mass, theangular momentum (L) in rotational motion is analogousto the momentum of an object in translational motion.

This is parallel to the equation for translational motionas follows:

Type of Motion Equation

Translational ~p = m~vRotational L = Iω

The angular momemtum of an object is conserved in theabsence of an external force or torque.

Rotating objects have rotational kinetic energy ac-cording to the following equation:

KE r =12Iω2 =

L2

2I

Just as other types of mechanical energy may be con-served, rotational kinetic energy is also conserved in theabsence of friction.

8.4 Simple MachinesThere are six fundamental types of machines, called sim-ple machines, with which any other machine can be con-structed. They are levers, inclined planes, wheels, wedges,pulleys, and screws.

The main purpose of a machine is to magnify the outputforce of the machine compared to the input force, the ratioof these forces is called the mechanical advantage (MA)


11

of the machine. It is a unitless number according to thefollowing equation:

MA =output forceinput force

=Fout

Fin

When frictional forces are accounted for, some of theoutput force is lost, causing less work to be done by themachine than by the original force. The ratio of work doneby the machine to work put in to the machine is called theefficiency (eff ) of the machine:

eff =Wout

Win

If a machine is perfectly efficient (eff = 1) then the idealmechanical advantage (IMA) can be found by comparingthe input and output distances:

IMA =din

dout

This leads to another way to find the efficiency of themachine as well:

eff =MAIMA


12

Chapter 12 — Vibrations and Waves

12.1 Simple Harmonic MotionHooke’s law states that for a given spring the force thespring exerts (~F ) is proportional to the negative of the dis-placement that the spring is stretched from rest (~x), or

~F = −k~x

Simple harmonic motion is the repetitive, back-and-forth motion of an object such as a pendulum or a massoscillating on the end of a spring. The motion is over thesame path each time and passes through the equilibriumposition twice each cycle under the influence of a restor-ing force (a force that pushes the object back toward theequilibrium position).

The displacement (x), velocity (v), and acceleration (a)of an object in simple harmonic motion when graphed withrespect to time are all sinusoids, if they are all plotted onthe same graph then their relationship can be seen

t

xva

12.2 Measuring Simple Harmonic MotionThe amplitude of harmonic motion measures how far theobject moves.

The period (T ) is the time that it takes for one com-plete cycle of motion.

The frequency (f) is the number of complete cycles ofmotion occurring in one second. Frequency is measured ininverse seconds (s−1) or Hertz (Hz). Frequency and periodare inverses of each other, so:

f =1T

T =1f

For a simple pendulum of length L, the period of smalloscillations is given by:

Tpendulum = 2π

√L

g

For a mass-spring system with mass m and spring constantk, the period of oscillation is:

Tspring = 2π√m

k

12.3 Properties of WavesMechanical waves require the presence of a mediumthrough which to move, such as a ripple traveling on thesurface of a pond, or a pulse traveling along a stretchedspring. If the vibrations of the wave are perpendicular to

the direction of motion of the wave then it is a transversewave, if the vibrations are parallel to the direction of mo-tion then it is a longitudinal wave.

The displacement of a wave moving with simple har-monic motion can be modeled by a sinusoidal wave.

rCrest

rTrough Wavelength (λ)

Amplitude

The wavelength (λ) of a wave is the shortest distancebetween corresponding parts of two waves, such as from onecrest or trough to the next one.

The wave speed (v) is the speed with which a wavepropagates, it is equal to the frequency of the wave timesits wavelength:

v = fλ =λ

T

The energy transferred by a wave is proportional to thesquare of its amplitude.

12.4 Wave Interactions

When two (or more) waves come together they pass througheach other, and through the principle of superposition thetotal displacement of the medium is equal to the sum of thedisplacements of the overlapping waves at each point.

When two waves overlap with displacements in the samedirection the resulting wave has an amplitude greater thaneither of the two overlapping waves, this is called construc-tive interference. In the other case, when two overlap-ping waves have displacements in opposite directions theresulting wave will have a displacement less than the dis-placement of the wave with larger amplitude, possibly evenno displacement at all. This is called destructive inter-ference.

→

←

→Destructive

→ ←→

Constructive

When a wave traveling through some medium reaches afixed boundary (a transition to a medium with a smaller,possibly zero, wave speed) the wave will reflect back in-verted with respect to the initial wave. If the wave insteadreflects off of a free boundary (a transition to a mediumwith a greater wave speed) then the wave will reflect off theboundary in the same orientation that it arrived.


13

If a wave is traveling in a confined space with just theright frequency then a standing wave may occur. A stand-ing wave results from a wave constructively and destruc-tively interfering with its own reflection. There will be somepoints (at least two, the ends) called nodes where completedestructive interference occurs, and between every pair ofnodes there will be an antinode where constructive inter-ference causes the oscillation to reach a relative maximumamplitude.s s

s s ss s s s

n = 1, λ = 2L

n = 2, λ = L

n = 3, λ = 23L

Node Node Node NodeAntinode Antinode Antinode

For a vibrating spring or string of length L the wave-

lengths of possible standing waves are:

λ = 2L, L,23L,

12L,

25L . . . λ =

2nL

where n is the number of the harmonic that correspondsto that standing wave. The nth harmonic will have n + 1nodes and n antinodes.

n Nodes Antinodes λ

1 2 1 λ = 2L2 3 2 λ = L3 4 3 λ = 2

3L4 5 4 λ = 1

2L5 6 5 λ = 2

5L· · ·

n n+ 1 n λ = 2nL


14

Chapter 13 — Sound

13.1 Sound WavesSound waves are longitudinal waves consisting of successivecompressions (areas of high pressure) and rarefactions(areas of low pressure) caused by and radiating outward inspherical wavefronts from a vibrating object.

The frequency a sound wave is perceived to have is calledthe pitch.

Sound waves travel outward in 3 dimensions along spher-ical wavefronts. If a ray is drawn from the source outthrough the wavefronts a graph of pressure vs. distancewill be a sinusoid. When r λ the wave fronts begin tolook like and can be approximated by plane waves.

Humans can hear sounds with frequencies from about20 Hz to about 20,000 Hz with the greatest sensitivity inthe range of 500 Hz to 5000 Hz. Frequencies outside thatrange are called either infrasonic (below 20 Hz) or ultra-sonic (above 20,000 Hz).

The speed at which a sound wave propagates dependson the medium. In air at sea level and room temperaturethe speed of sound is approximately 343 m

s .

Doppler EffectIf a sound source and the detector are moving relative toone another then the Doppler effect can be observed.

r rvs = 0 vs < v

r rvs = v vs > v

This is a change in the pitch (f ′) relative to the frequencyas emitted by the source (f). For a wave traveling at speedv, a detector moving at velocity vd (positive if toward thesource), and a source moving at velocity vs (positive if to-ward the detector), the pitch will be

f ′ = f

(v + vd

v − vs

)13.2 Sound Intensity and ResonanceIntensityAs a sound wave travels through a medium energy is trans-ferred from one molecule to the next. The rate at which

this energy is transferred is called the intensity and canbe described as power per unit area and measured in wattsper square meter.

In a spherical wave the energy is spread across the sur-face of a sphere, so the expression for intensity becomes

intensity =P

4πr2

The quietest sound audible to the human ear (thethreshold of hearing) is about 1.0×10−12 W

m2 . The loud-est sound a typical person can tolerate (the threshold ofpain) is about 1.0 W

m2 .

Perceived IntensityThe perceived loudness of a sound is dependent on the log-arithm of the intensity, so it is measured in decibels (dB).The sound level measured in decibels is:

dB = 10 log10

intensityintensityr

where intensity is the intensity of the sound being measuredand intensityr is the reference intensity. If nothing else isspecified the reference intensity is set equal to the thresholdof hearing or intensityr = 1.0× 10−12 W

m2 .An increase of 10 decibels will be perceived as being

twice as loud but will cause the intensity to increase by afactor of ten.

ResonanceJust as was observed with standing waves in a spring, cer-tain physical systems show a particular response at certainfrequencies called resonance. When a system is driven at(or very close to) its resonant frequency the amplitudeof its vibration will continue to increase.

13.3 HarmonicsStanding WavesStrings and air-filled pipes can exhibit standing waves atcertain frequencies. The lowest frequency able to form astanding wave is called the fundamental frequency (f1),those above it are harmonics of that frequency. The fre-quency of each harmonic frequency is equal to the funda-mental frequency multiplied by the harmonic number (n),this forms a harmonic series. The fundamental frequencyis also referred to as the first harmonic frequency.

Depending on the medium in which the standing wavesare formed the characteristics of the waves will vary. Apipe open at both ends has the same harmonic series asa vibrating string (producing all of the harmonics), whilea pipe closed at one end will only produce odd-numberedharmonics.

The harmonics that are present form a sound source willaffect the timbre or sound quality of the resulting sound.


15

Standing Waves on a String or Open Pipe of Length L

n Nodes Antinodes String Diagram Pipe Diagram Wavelength Frequency

1 2 1 λ1 = 2L f1 =v

2L

2 3 2 λ2 = L f2 =v

L= 2f1

3 4 3 λ3 =23L f3 =

3v2L

= 3f1

n n+ 1 n n = 1, 2, 3, . . . λn =2nL fn = n

v

2L= nf1

Standing Waves in a Pipe of Length L Closed at One End

n Nodes Antinodes Diagram Wavelength Frequency

1 1 1 λ1 = 4L f1 =v

4L

3 2 2 λ3 =43L f3 =

3v4L

= 3f1

5 3 3 λ5 =45L f5 =

5v4L

= 5f1

nn+ 1

2n+ 1

2n = 1, 3, 5, . . . λn =

4nL fn = n

v

4L= nf1

Beat Frequency

If two sources emit sound waves that are close to one an-other in frequency but not exactly the same then a beatwill be observed. When the amplitudes of the waves areadded together the resulting sound oscillates between loudand soft. The number of beats heard per second is equal tothe absolute value of the difference between the frequenciesof the two sounds. The diagram below shows one second of

the sum of a 34 Hz wave and a 40 Hz wave yielding a 6 Hzbeat frequency.

f = 40 Hz

f = 34 Hz

Sum


16

Chapter 14 — Light and Reflection

14.1 Characteristics of LightLight is an electromagnetic wave as opposed to a me-chanical wave, so it does not need a medium through whichto travel although it can pass through many media. Beingan electromagnetic wave it is composed of a vibrating elec-tric field and a vibrating magnetic field, the two fields areat right angles to each other.

The speed of light depends on the medium throughwhich it travels, it is fastest in a vacuum and slower indenser media. The speed of light in a vacuum is exactly299 792 458 m

s (the definition of a meter comes from thespeed of light so this is defined as the exact value). Formost purposes, a value of 3.00×108 m

s is sufficiently precise.The speed of light in a vacuum is a fundamental constantin many calculations and has been assigned the symbol c,thus for light and other electromagnetic waves the wavespeed equation v = fλ becomes:

c = fλ

Visible light is only one part of the electromagneticspectrum corresponding to wavelengths from 400 nm (vi-olet light) to 700 nm (red light).

Huygens’ principle states that all of the points on awave front can be treated as point sources and each of thosepoint sources will produce a circular or spherical wavelet.The wavelets will constructively interfere on a line tangentto the fronts of all of the wavelets, this will create the nextwave front. This principle will become more important laterwhen we discuss diffraction, for now we can use the ray ap-proximation which states that light can be considered tobe straight lines traveling outward from the source perpen-dicular to the wave fronts.

The brightness of a surface lit by a single light is in-versely proportional to the square of the distance betweenthe source and the surface, for a source twice as far awaythe surface is one-fourth as bright, etc.

14.2 Flat Mirrors

ReflectionLight traveling through a uniform medium will move in astraight line regardless of the medium.

When light encounters a boundary between media partof the light will pass through, part will be absorbed, andpart will be bounced off of the boundary, or reflected. Ifthe medium is one that is opaque to the light then noneof it will pass through, the light will either be absorbed orreflected instead.

If the surface is rough the result is diffuse reflection,if it’s smooth then it’s specular reflection. Smooth vs.rough depends on the relative sizes of the surface variationsand λ.

The light ray coming to the surface is the incident ray,the ray bouncing off of it is the reflected ray, and a lineperpendicular to the surface at the point where the incidentray strikes it is the normal line. The angle between theincident ray and the normal line is the angle of incidence(θi) and the angle between the normal line and the reflectedray is the angle of reflection (θ′). The angle of incidenceis always equal to the angle of reflection (θi = θ′).

@@

@R

Reflecting Surface

Incident Ray Reflected Ray

Normal Line

θi θ′

Flat MirrorsIf you put an object at some distance in front of a flat mirror(the object distance, p) light from the object will reflectoff of the mirror and appear to come from a location behindthe mirror’s surface. As a convention, the object’s image issaid to be at that location (the image distance, q) behindthe mirror. For a flat mirror the object is the same dis-tance in front of the mirror as the image is behind it andthe height of the image and object are equal (h′ = h).

Ray DiagramsRay diagrams are drawn to locate the image formed by amirror. To simplify matters we just consider a single pointon an object resting on the principal axis. Rays are drawnfrom the tip of the object to the mirror and allowed to reflectback. In the diagram below ray 1 is drawn perpendicularto the mirror, it reflects back along itself and away. Ray 2is drawn from the object to the mirror at the principal axiswhere it reflects back out at the same angle. The two raysas drawn do not meet to form an image so the need to be ex-tended backward behind the mirror to where the image willbe formed. The rays extended backward are not actuallythere but they seem to originate at the source of the image.Rays that are not there but seem to be are called virtualrays (real rays are the rays that are actually there) andthe image that they form is a virtual image as opposedto a real image that is formed by real rays. A real imagecould be projected onto a screen, there is no way to see avirtual image other than to observe it in the mirror.

HHHH

HHHHHj

-

h h′

p q1

2

Object Image

MirrorPrincipal Axis@

@@I

Ray diagram of a virtual image formed by a flat mirror


17

14.3 Curved Mirrors

Curved mirrors can be either concave (curving inward inthe middle) or convex (curving outward in the middle).They can be approximated as a section of a sphere withradius of curvature R, and will have a focal length (thedistance from the center of the mirror to the focal point,F ) of f = R/2. By convention f and R are positive for aconcave mirror and negative for a convex mirror. Similarly,p and q are considered to be positive if they are measuredto a point in front of the mirror and negative if the point isbehind the mirror. For a flat mirror f = R =∞.

For a given mirror of focal length f , the relationship be-tween focal length, object distance (p), and image distance(q) is:

1p

+1q

=1f

For an object of height h and an image of height h′ theirratio is the magnification, M and is given as:

M =h′

h= −q

p

Sign conventions for mirrors

Symbol Quantity Situation Sign

p Objectdistance

Object is in front of themirror

+

q Imagedistance

Image is in front of themirror (real image)

+

Image is behind the mirror(virtual image)

−

h Objectheight

Object is above the axis +

h′ Imageheight

Image is above the axis +

Image is below the axis −f Focal

lengthFocal point is in front ofthe mirror (concave mir-ror)

+

Focal point is behind themirror (convex mirror)

−

No focal point (flat mirror) ∞R Radius

of Cur-vature

Center of curvature is infront of the mirror (con-cave mirror)

+

Center of curvature is be-hind the mirror (convexmirror)

−

No center of curvature(flat mirror)

∞

Images can be located through ray diagrams as well.The general principle is the same as the ray diagram forthe flat mirror, the usual rays to be drawn are:

1. A ray that comes in parallel to the principal axis andgoes out through the focal point.

2. A ray that comes in through the focal point and goesout parallel to the principal axis.

3. A ray that goes in through the center of curvature, re-flects perpendicularly off of the mirror, and goes backout along the same line where it came in.

Each of these rays is labeled with its corresponding num-ber in the ray diagrams below. For more precise results therays are drawn to a line perpendicular to the principal axisrather than to the mirror itself.

sC

sFObject Image

1

2

1)

3

Concave mirror, p < f , virtual upright image, M > 1

sC

sFObject

:9

- 91

3

No Image

Concave mirror, p = f , no image


18

sC

sFObject

Image

HHHHH

HHHH

HHHH

HHj

1

19

-

22

*

3

3

Concave mirror, f < p < R, real inverted image, |M | > 1

sC

sFObject

Image

-

1

1

HHHHHH

HHHjH

HH

22

Concave mirror, f = R, real inverted image, |M | = 1

sC

sFObject

Image

-

1

1

XXXXXXXXXXXXXXXzXXX

2

2

HHHHHHH

HHHHHH

HHjHHH

HHY3

3

Concave mirror, f > R, real inverted image, |M | < 1

sC

sFObject Image

-HHHHHY1

XXXXXzXXXXXX2

-3 H

HHHHH

Convex mirror, virtual upright image, M < 1

Parabolic Mirrors

Spherical mirrors come close to focusing incoming parallelrays to a single point, particularly if all of the rays strikeclose to the center of the mirror. The farther the rays strikefrom the center the greater the error in focusing at a singlepoint, this is called spherical aberration. If a parabolicmirror is used instead of a spherical mirror the sphericalaberration is eliminated. High-quality optical instrumentsuse parabolic mirrors for their improved optical properties.

14.4 Color and Polarization

Color

The primary colors of light are red, green, and blue. Colorsof light mix through additive mixing, when all three primarycolors are mixed white light is produced. The primary col-ors of pigment are cyan, magenta, and yellow. Colors ofpigment mix through subtractive mixing, when all threeprimary colors are mixed black is produced. The primarycolors for additive mixing are the secondary colors for sub-tractive mixing and vice versa.

Pigments appear to be the color that they are becausethat color of light is reflected, all other colors are absorbed.When two pigments are mixed the resulting mix will absorbthe colors absorbed by both pigments that were mixed andonly reflect the colors that neither pigment will absorb.Primary Colors

In the diagrams below the circles represent the primary col-ors and the overlapping areas the result of mixing thosecolors.

Red

Green BlueCyan

White

Yellow Magenta

(Light)Additive

Yellow

Cyan MagentaBlue

Black

Green Red

(Pigment)Subtractive


19

Additive and subtractive primary colors

Color Additive (mixinglight)

Subtractive (mixingpigment)

red primary complimentary tocyan

green primary complimentary tomagenta

blue primary complimentary toyellow

cyan(bluegreen)

complimentary tored

primary

magenta(redblue)

complimentary togreen

primary

yellow complimentary toblue

primary

PolarizationLight can be polarized, or set to vibrate in the same planes(one for electric field, one for magnetic field), by passing itthrough certain materials.

If a polarized beam of light is passed through a secondpolarizing filter only the component of that light match-ing the polarization of the new filter will pass through, sothe overall intensity of the light will be proportional to thecosine of the angle between the polarizing filters.

Light can be polarized by reflection at certain angles(about 34 up from the surface of a glass plate works well,the best angle varies depending on the material). Polar-ization is perpendicular to the plane containing the rays orhorizontal for reflection off of a level surface. A pair of sun-glasses with vertical polarization will remove the polarizedglare from reflecting surfaces on a sunny day.


20

Chapter 15 — Refraction

15.1 RefractionAs light travels from one medium to another the speed oflight also changes depending upon the densities of the me-dia.

If a ray of light reaches a boundary between two me-dia part of the ray will be reflected and part may also berefracted or bent while passing across the boundary.

@@

@@@R

AAAAAU

AirGlass

θi θ′

θr

AAAAAU@

@@

@@R

GlassAir

θi θ′

θr

The index of refraction (n) of a medium is a measureof how much the speed of light changes in that mediumcompared to in a vacuum. For a medium with speed oflight v, the index of refraction is:

n =c

v

The amount that light will be refracted at a boundarydepends on the respective indices of refraction of the media.All angles are measured from a line normal to the boundaryat the point that the light intersects the boundary. For anangle of incidence θi in a medium with index of refractionni and an angle of refraction θr in a medium with indexof refraction nr the relationship is given by Snell’s law:

ni sin θi = nr sin θr

15.2 Thin LensesLenses operate similarly to curved mirrors with a few im-portant differences: lenses have a focal point on both sidesand some of the sign conventions are changed.

Sign conventions for lenses

Symbol Positive (+) Negative (−)

p Object is in front ofthe lens

Object is in back ofthe lens

q Image is in back ofthe lens

Image is in front ofthe lens

f Converging lens Diverging lens

With those sign conventions in place the following equa-tions still hold for focal length and magnification:

1p

+1q

=1f

M =h′

h= −q

p

Ray diagrams can also be drawn to locate and describethe image that will be formed by a lens. The usual rays tobe drawn are:

1. A ray that comes in parallel to the principal axis andgoes out through the focal point.

2. A ray that comes in through the focal point and goesout parallel to the principal axis.

3. A ray that goes straight through the center of the lens

Each of these rays is labeled with its corresponding numberin the ray diagrams below.

sF sF-

-

-

HHHH

HHHHHH

HHH

-*H

HHj

Object at ∞

Image

Converging lens, p =∞, point image, q = f

1

3

1

sF

sFhhhhhhhhhhhhhhhhhhhhhh

XXXXXXXXXXXXXXX

-XXXXXXXXz

HHHHHH

HH

HHHj

-

Object

Image

Converging lens, p > 2f , real image, f < q < 2f

1

2

3

sF

sF-XXXzXXXXXXXXXXXXXXXXXXXX

HHHjHHH

HHHHHHH

HHHHHH

HHHH

HHHj-

XXXXXXXXz

Object

ImageConverging lens, p = 2f , real image, q = 2f

1

2

3

sF

sFhhhhhhhhhhhhhhhhhhhhhh

HHHHH

HHH

-HHHHHj

-

XXXXXXXXXXXXXXX

XXXXXXXXz

Object

ImageConverging lens, f < p < 2f , real image, q > 2f

1

23


21

sF

sF-H

HHjH

HHHHHH

HHHHHHj

HHHHHHH

HHHHHH

HH

HHHHHH

HHHHHHH

HHHj

Object

No image

Converging lens, p = f , no image

1

3

sF

sF

HHHHH

HHHHHH

HH

HHHHH

HHHHHj

XXXXXXXXXXXXX

XXXXXXXXz

-HHHXXXXX

Object

Image

Converging lens, p < f , virtual image

1

23

sF sF

HHHHHH

HHHHj

PPPPP -

Image

Object

Diverging lens, virtual image

1

2

3

Lenses can be used for the correction of defects in hu-man vision. If a person has hyperopia (farsightedness)light rays tend to focus behind the retina, a converging lenswill cause them to focus on the retina. Similarly, if a personhas myopia (nearsightedness) a diverging lens will causethe light rays to focus on the retina instead of in front of it.

Lenses can be used in combination, the image formedby the first lens is treated as the object for the second lens.The overall magnification of the system is the product ofthe magnification of the individual lenses.

15.3 Optical Phenomena

*

HH

HHHHj

-

7

12

3 3

2

1

refra

cted

ray

reflected ray

θr

θi

AirGlass

When light passes from a medium with a higher indexof refraction to one with a lower index of refraction an ef-fect called total internal reflection can sometimes beobserved. If the angle of incidence is greater than somecritical angle θc then there will be no refracted ray. Theexpression for that critical angle is:

sin θc =nr

nifor ni > nr

The diagram above shows incident rays at less than the crit-ical angle (1), equal to the critical angle (2), and greaterthan the critical angle (3). All three incident rays wouldproduce reflected rays but they have been omitted from 1and 2 for clarity.

Refraction can occur even if there is not an abruptboundary between the two media, if for example warm (lessdense) air is trapped below cooler (more dense) air thenlight rays passing through the air will be refracted slightly.It is very difficult to predict the exact amount of refractionthat will be observed.

When light travels through the atmosphere it is scat-tered with the most scattering happening to the shorterwavelengths of light, this causes the sky to appear to beblue and sunsets, after the sunlight has had most of theblue scattered when traveling through the atmosphere, toappear orange or red. The name for this phenomenon isRayliegh scattering.

The index of refraction in a medium is dependent onthe wavelength of light involved, because of this a boundarybetween media may refract different wavelengths by differ-ent amounts, a phenomenon known as dispersion. Thiscauses the display of a spectrum by a prism, rainbows, andchromatic aberration or the focusing of different colorsof light by a lens at slightly different distances behind thelens.


22

Chapter 16 — Interference and Diffraction

16.1 InterferenceInterference can only happen between waves of the samewavelength. If the crests of two waves arrive at the samepoint then they have a phase difference of 0 and they aresaid to be in phase, if the crest of one wave matches atrough of the other then they have a phase difference of180 and they are said to be out of phase.

If monochromatic light from a single source is passedthrough two narrow parallel slits they will serve as coher-ent sources and will produce an interference pattern witha bright fringe where constructive interference occurs anddark ones in between as shown below. s

sd

1st bright fringe

0th dark fringe

The fringes are numbered with an order number, thecenter one being assigned the number zero (the zeroth-order maximum or central maximum) and others num-bering outward from there. The dark fringes are also num-bered from zero although there are two zeroth-order darkfringes and only one bright one. The diagram below showshow the fringes produced by two slits a distance d apartare numbered (one side of the diagram is omitted to savespace), the angle θ shown below corresponds to the 2ndbright fringe.

dθ

m = 2 (bright)

m = 1 (bright)

m = 0 (bright)m = 0 (dark)

m = 0 (dark)

m = 1 (dark)

m = 2 (dark)

For two slits separated by a distance d the mth ordermaximum is located at some angle θ from a normal linedrawn between the slits, that angle also depends on thewavelength of the light and can be found from the equation

d sin θ = mλ m = 0,±1,±2,±3 . . .

with the corresponding dark fringes being found from thesimilar equation

d sin θ =(m+

12

)λ m = 0,±1,±2,±3 . . .

If the difference in distances is an integral multiple ofthe wavelength then a bright fringe will be seen, if the dif-ference is half a wavelength from an integral multiple of thewavelength then it will be a dark fringe.

When a white light source is used, instead of simplefringes a very different pattern is observed since each colorof light refracts differently and the resulting fringes mix toform other colors.16.2 Diffraction

Since each point on a wave front is a source of Huygenswavelets interference patterns can be observed from a sin-gle slit. Light deviates from a straight-line path and entersthe area that would normally be in shadow (diffraction)and produces a diffraction pattern when the wavelets in-terfere with each other. This can occur when light passesthrough a slit or when it passes by the edge of an object.

When light passes through a diffraction grating each ad-jacent pair of grooves in the grating will behave like the twoslits described previously, the same equations will work todescribe the location of the bright and dark fringes.

Diffraction plays a mayor part in determining the re-solving power of an optical instrument. The resolvingpower is the instrument’s ability to distinguish two objectsthat appear to be close to each other as separate imagesand it is measured as the minimum angle between the twoobjects (with the instrument at the vertex of the angle) thatcan be discerned. This angle is proportional to the wave-length of the light and inversely proportional to the size ofthe aperture through which the light must pass.16.3 Lasers

In a typical light source the individual rays of light areemitted haphazardly, the waves do not share a commonwavelength or phase.

-

6 AAAU

-

Ground energy level

Metastable energy level

Excited energy level

Pumping energy

Spontaneous energy decay

Stimulated emission

Q0

Qe

Qm

As shown above, lasers take advantage of properties ofthe energy levels of an electron in certain types of atoms.They do so by having a set of energy levels where some


23

initial energy (the pumping energy) will raise an electronto an unstable, excited state. The electron will then spon-taneously decay to a lower, metastable state (a state theelectron can hold for some time) where it will remain untilthe electron is struck by a photon from another decayingelectron, this will trigger the electron to drop back to theground state and release an additional photon. Since all ofthe stimulated emission photons come from the same en-ergy transition in the atoms they all have the same energyand thus frequency and wavelength. The waves are alsoproduced in phase with each other, yielding what is calledcoherent light. The beam of a laser appears much moreintense than a similar light source because of the coherenceand the tendency of lasers to emit a nearly parallel beamof light (it does not spread very much from its source untilit strikes an object).

Lasers are often used in information storage and re-trieval (reading and writing CDs and DVDs), in measure-ment (finding the time for a laser to reflect from a distantobject) and in medicine (precisely delivering energy to spe-cific targets in the body that might not otherwise be reach-able).

Thin-Film InterferenceWhen light bounces off of a soap bubble it is reflected offof both the outside and inside surfaces. Depending on theexact thickness of the soap bubble the light reflecting off ofone of the two surfaces will be either in or out of phase withthe light reflecting from the other one. If they are in phasethen a bright band will be observed, if not then it will bedark. To see this sort of interference the bubble must be atleast 1

4λ thick, any thinner and the waves will be less than180 out of phase.

R@@

@

AAsn = 1

n = 1.33n = 1

airsoap filmair

The black dot on the diagram above indicates that thephase of the reflected light at that point is shifted by 180

because the reflecting surface has a higher index of refrac-tion. Since all other reflections are from a reflecting surfacewith a lower index of refraction they will not change thephase of the incident rays. If the thickness of the soap filmis 1

4λ then there will be constructive interference, if it is 12λ

then there will be descructive interference, and the patternwill repeat with each 1

4λ added to the thickness.Since the bubbles vary in thickness from top to bottom

the interference pattern observed will also vary. The pat-tern will generally consist of horizontal bands with someirregularities as the fluid in the bubble moves around. Ifthe bubble is lit with only one color of light then the bandswill be light and dark, if the bubble is lit with white lightthen a rainbow pattern will be observed.

Thin films of oil floating on water (as you may see onyour driveway after it rains) exhibit similar properties, in-stead of seeing horizontal bands around a bubble you seeconcentric irregular rings of color as the thickness of the oildecreases toward the edges.

R@@

@

AAsn = 1

n = 1.4n = 1.33

airoilwater

A similar principle controls the function of anti-reflection coatings on lenses, a very thin coating of mag-nesium fluoride is deposited on the surface of a glass lensto provide destructive interference for most wavelengths ofvisible light.

R@@

@

AAs sn = 1

n = 1.38n = 1.72

airMgF2

glass

The main difference between the anti-reflection coatingsand the soap or oil films is that there is an extra phase in-version at the coating-glass boundary. Because of this theinterference at 1

4λ will be destructive and at 12λ it will be

constructive.


24

Chapter 17 — Electric Forces and Fields

17.1 Electric Charge

Electric charge comes in two varieties, negative (an excessof electrons) and positive (a lack of electrons). Positivecharges attract negative charges and vice versa, two likecharges will repel each other.

As shown by Robert Millikan in 1909 charge must comein multiples of a fundamental unit. This fundamental unitof charge, e, is the charge on a single electron or proton. Theunit by which charge is usually measured is the coulomb(C), one coulomb is 6.2× 1018 e and the charge on a singleelectron is −1.60× 10−19 C.

Electric charge is conserved. It can me transferred fromone object to another by moving electrons but it can notbe created or destroyed.

Materials can be either conductors (materials throughwhich electrons flow freely), insulators (materials that donot permit the free flow of electrons) and semiconductors(materials that will either conduct or not depending on thecircumstances).

Both conductors and insulators can be charged by con-tact (by placing two dissimilar objects together, the effectcan be intensified by rubbing them together) although aconductor must not be connected to a ground for the effectto accumulate.

A conductor can also be charged by induction by bring-ing another charged object such as a rubber rod near it (1,below), allowing the displaced charge to drain off on theother side (2) then removing the drain while the charged ob-ject remains (3) then finally removing the original chargedobject leaving the induced charge (4).

h− h− h− h+ h+h+h+ h+h+h−h−h− h−h−h−

1

h− h− h− h+ h+h+h+ h+h+2

h− h− h− h+ h+h+h+ h+h+3

h+h+h+ h+h+h+

4A surface charge can be induced on an insulator by po-

larization (by bringing a charged object close to the sur-face, this will cause an opposite charge on the near surfaceand a charge the same as the charged object on the far side).

h+h+h+ h+h+h+ h−h−

h−h−h−

h+h+h+h+h+

17.2 Electric ForceAccording to Coulomb’s law, the electric force between twocharges is proportional to the product of the charges andinversely proportional to the square of the distance betweenthem.

Felectric = kCq1q2r2

This is dependent on the Coulomb constant (kC):

kC = 8.987 551 788× 109 N ·m2

C2≈ 8.99× 109 N ·m2

C2

This is very similar to the equation for universal gravita-tion, substituting charge for mass and making correspond-ing changes in the constant.

Like gravity, the electric force is a field force so it doesnot require contact to act. The resultant electric force onany charge is the vector sum of the individual vector forceson that charge (the superposition principle). An timemultiple forces act on a single charged object and the sumof the forces is zero the object is said to be in equilibrium.

17.3 The Electric Field

Electric Field StrengthA charged object sets up an electric field around it, whena second charge is present in this field an electric force willresult. The strength of the electric field (E) is defined to bethe force that would be experienced by a small positive testcharge (q0) divided by the magnitude of the test charge.Since the field strength is a force divided by a charge theunits are newtons per coulomb (N

C ). Mathematically, thiscan be expressed as

E =Felectric

q0

Combining the above equation with Coulomb’s law gives anequation that will give the electric field strength around acharge

E =kCq

r2

Since the test charge is always positive a positive charge willproduce a field directed away from the charge, a negativecharge will produce a field directed inward. The directionof the electric field vector, ~E, is the direction in which anelectric force would act on a positive test charge. The equa-tion will provide the magnitude of the vector, the directionis found from the sign of the charge.

Electric Field LinesA convenient aid for visualizing electric field patterns is todraw electric field lines. They consist of lines drawn tan-gent to the electric field vector at any point, the numberof lines drawn being proportional to the magnitude of the


25

field strength. It is important to remember that electricfield lines do not actually exist as drawn since the field iscontinuous at every point outside of a conductor; the fieldlines are only drawn as a representation of the field.Rules for drawing field lines

• Field lines must begin on positive charges or at in-finity and must terminate on negative charges or atinfinity.

• The number of lines drawn leaving a positive chargeor approaching a negative charge is proportional tothe magnitude of the charge.

• No two field lines from the same field can cross eachother.

6

6

66

6

66

6

66

6

6

A single positive charge

????

??????

??

A single negative charge

Equal positive and negative charges

Equal positive charges

+2 −1

A postive charge twice the size of a negative charge

Conductors in Electrostatic EquilibriumAn electrical conductor such as a metallic object containsexcess electrons (or a lack of them) that are free to movearound the conductor. When no net motion of charge isoccurring within the conductor it is said to be in electro-static equilibrium. Any conductor in electrostatic equi-librium has the following properties:

• The electric field is zero everywhere inside the con-ductor.

• Any excess charge on an isolated conductor is entirelyon the surface of the conductor.

• The electric field immediately outside of a chargedconductor is perpendicular to the surface of the con-ductor.

• On an irregularly shaped conductor the charge willtend to accumulate where the radius of outward cur-vature is smallest. (This will produce the strongestelectric fields just outside of those points.)

For example, consider this irregular conductor with an ex-cess of electrons:

−−

−

−

−

−−

−

−

−

−−

−

−−−

−


26

Chapter 18 — Electrical Energy and Capacitance

18.1 Electrical Potential EnergyA uniform electric field is an electric field that has all ofthe field lines parallel to and equidistant from each other.An electric field caused by a charge that is very far away iseffectively uniform over a small area since the field lines arenearly parallel and the difference in strength over a smalldistance is negligible. Such a field could also be caused bya pair of oppositely charged plates, the field will be uniformin the region between the plates.

A uniform electric field will always exert a constant forceon a charged object, if that object is displaced in the fieldthen a force will be required causing work to be done andenergy to be accumulated.

-

-

-

-

-

-E

-q F = qE

q q

∆d

Just as gravitational potential energy was determinedby taking the product of the weight of an object and itsheight from some reference height the same can be donefor electrical potential energy (PEelectric) by taking theproduct of the force generated by the field and its positionin the field. Mathematically, in a uniform electric field ofstrength E with a displacement d measured in the directionof E

PEelectric = −qEd

The negative sign is needed because the displacementof a positive charge to increase the potential energy shouldbe in the direction opposite the direction of the field. Ifthe initial and final positions are taken before and after adisplacement the change in potential energy can be foundthusly

∆PEelectric = −qE∆d

18.2 Potential Difference

Electric potential (V ) is defined as the electrical potentialenergy per unit charge, so

V =PEelectric

q=−qEd

q→ V = −Ed

The relative difference in electric potential between twopoints is the potential difference (∆V ) which, like elec-tric potential, is measured in J

C which are given the namevolts (V).

∆V =∆PEelectric

q= −E∆d

Note that the electric potential at a given point is inde-pendent of the charge at that point and that it’s only theelectric potential difference between two points that is use-ful, the electric potential at a single point is based on anarbitrary reference similar to position.

18.3 Capacitance

Because of time constraints the topic of capacitance wasomitted this year.


27

Chapter 19 — Current and Resistance

19.1 Electric CurrentCurrentElectric current (I) is the rate at which positive chargesmove through a conductor past a fixed point. For a charge∆Q moving during a time ∆t the current is

I =∆Q∆t

Current is measured in amperes (A) which are a fun-damental SI unit from which coulombs are derived.(1 A = 1 C

s or more properly 1 C = 1 A · s)

Charge CarriersElectric charge can be either positive or negative, so eitherpositive or negative charges could be carried by the parti-cles moving to make up the current. The convention is toregard current as being the movement of positive charges,but in common conductors such as metals it is actually thenegatively charged electrons that are moving. In such a casethe current is considered to be positive charges moving inthe opposite direction as shown below.

i+ -

-

i+ - i+ -

-i++i+ i− i -

i−

EquivalentConventional

Current

Motion of ChargeCarriers

When studying current on a macroscopic scale there isno discernible difference between the actual motion of thecharge carriers and the conventional current, it is whenstudying the charge carriers themselves that a distinctionmust be made.

Sources and Types of CurrentThe charge carriers (electrons, protons) are always present,all source of current does is move them to points with apotential difference between them and then allow them toflow back to a lower energy state through some device.

Batteries do this by converting chemical energy to elec-trical energy, generators do this by converting mechanicalenergy to electrical energy. Batteries always produce a fixedpotential difference between their terminals (direct cur-rent, limited of course as the battery runs out of storedchemical energy), generators can either produce the sametype of potential difference or one that reverses itself manytimes a second (alternating current). Alternating cur-rent has several practical advantages for energy transmis-sion, we will address them later.

19.2 ResistanceThe current through some circuit depends on the potentialdifference in the current source, a larger potential differ-

ence produces a larger current. If the circuit is replacedwith a different one that can also change the current thatwill flow, some materials allow current to flow more easilythan others. The opposition to the flow of current througha conductor is called resistance (R) which is defined to be

R =∆VI

Resistance is measured in VA which are called ohms (Ω).

Ohm’s LawOhm’s law states that for many materials the resistance isconstant over a wide range of applied potential differences.Mathematically

∆VI

= constant

Materials that follow this relationship are said to be ohmic,others (such as semiconductors) are non-ohmic. Materi-als may be ohmic at one temperature and non-ohmic atanother.

An easy way to remember Ohm’s law as well as toquickly derive it for any of the values is to show it thisway:

∆V

I R

If you put our finger on top of any one of the three quan-tities (∆V , I, or R) the remaining two will be in the rightposition to tell you what to do with them to find the oneyou have covered. For example, to find current cover the Iwith your finger, what you then see is ∆V

R which is equal tothe current.

R depends on length, cross-section, material, tempera-ture. Longer conductors, smaller cross-sections, and highertemperatures all increase the resistance of a conductor andvice versa, replacing the material making up the conductorwith another material will also change the resistance (forexample, copper has lower resistance than aluminum).

Human Body ResistanceThe resistance of the human body is about 500 000 Ω if theskin is dry, if the skin is soaked with salt water (or sweat)the resistance can drop to 100 Ω or so. A current throughyour body less than about 0.01 A is either imperceptibleor felt as a slight tingling, greater currents are painful, anda current of about 0.15 A through the chest cavity can befatal.

SuperconductorsSome materials have zero resistance below a certain crit-ical temperature, these materials are known as super-


28

conductors. The critical temperatures are generally verylow (less than 10 K for most materials that superconduct)although some have been found that superconduct up toabout 125 K.

19.3 Electric PowerJust as power was studied previously as the rate of con-version of mechanical energy it is also the rate at whichelectrical energy is transferred. Since

P =W

∆t=

∆PE∆t

=q∆V∆t

and I =q

∆t

it follows that power is then

P = I∆V

or that power is the product of the current and the potentialdifference. Combining the equation for power with Ohm’s

law (∆V = IR) yields the relationships

P = I2R and P =∆V 2

R

When electric utilities sell energy they use the derivedunit kilowatt-hours (kW · h), one of which is defined tobe the energy delivered at a constant rate of 1 kW in1 h. A brief exercise in algebra will show that 1 kW · h =3.6× 106 J.

Since power converted by a resistive load is P = I2R,the power lost to resistance in a wire is proportional to thesquare of the current. Reducing the current provides a con-siderable reduction in lost power so electrical transmissionlines operate at very high potential differences to keep thecurrent low, up to 765 000 V in the highest potential differ-ence lines. Transformers are used to reduce that potentialdifference to 120 V close to the point where it will be used.


29

Chapter 20 — Circuits and Circuit Elements

20.1 Schematic Diagrams and Circuits

Electric circuits are generally drawn as schematic dia-grams that show the electrical components and their con-nections but leave out unnecessary detail. There are severalversions of the symbols used, the ones most commonly usedby your textbook are:

Component Symbol Comments

Wire Wires are representedby drawing lines be-tween the elements to beconnected

Resistor orload

All resistive loads sharethis symbol. If a distinc-tion needs to be madeit is generally made byadding a label.

Lamp A lamp (or bulb) isshown as a resistor witha circle around it to rep-resent the glass

Battery The longer line rep-resents the positiveterminal.

Switch An open switch is shown.A closed switch wouldbring the arm down tomake contact on bothsides.

A schematic diagram of a simple flashlight could look likethis

−∆V+

The battery is shown on the right, the lamp on the left, anda switch is at the top to allow the circuit to be opened orclosed as desired. No details such as the flashlight housing,lens, or reflector are shown, they do not affect the electricalfunction of the flashlight so they are omitted.

The battery provides a potential difference at two pointsin the circuit, when the switch is open nothing will happen.When the switch is closed then the circuit is completed andthe current can flow through the lamp, the amount of cur-rent flowing will depend on the potential difference providedby the battery and the resistance of the lamp.

20.2 Resistors In Series or In ParallelFinding the current in a circuit that only contains a singleresistor is simple, but finding the current in circuits withmore than one resistor becomes more complex.

Resistors In Series

−∆V+

R1

R2

In the case where multiple resistors are connected in se-ries the fact that the same current must flow through all ofthem leads to the equations

Req = R1 +R2 +R3 + . . .

∆Vt = ∆V1 + ∆V2 + ∆V3 + . . .

It = I1 = I2 = I3 . . .

which indicates that in a series circuit the equivalent re-sistance is equal to the sum of the individual resistances.All of the resistances can be replaced with a single resistorequal to their sum and the current will be the same

−∆V+

Req

Once the current has been found through the equivalent re-sistance the potential difference across each resistor (∆Vi)can be found by once again taking advantage of the fact thethe current is the same through resistors in series and thesum of the individual potential differences is equal to thetotal potential difference, so

∆Vi = ItRi

If any element in a series circuit stops conducting then theflow of current through the entire circuit will stop as if aswitch had been opened at that point.

Resistors In Parallel

−∆V+

R1 R2

For the case where multiple resistors are connected in seriesthe fact that the potential difference across all of the resis-tors is the same and the total current must be the sum ofthe individual currents through each resistor leads to

1Req

=1R1

+1R2

+1R3

. . .

∆Vt = ∆V1 = ∆V2 = ∆V3 . . .


30

It = I1 + I2 + I3 . . .

The current through each resistor (Ii) can be found with

Ii =∆Vt

Ri

Because of the reciprocal relationship used in findingthe equivalent resistance for a parallel circuit the equiva-lent resistance will always be smaller than the smallest ofthe individual resistances.

If any element in a parallel circuit stops conducting thenthe flow of electricity will continue through the remainingparallel elements.

20.3 Complex Resistor Combinations

−∆V+

R1

R2 R3

In a circuit like the one above with both parallel and serieselements the methods to find each type of equivalent resis-tance must be applied repetitively until only one equivalentresistance is left. In this circuit R1 is in series with the com-bination of R2 and R3 in parallel, so first the two parallelresistors R2 and R3 combine in parallel to give the equiva-lent resistance R2,3, then R2,3 and R1 combine in series togive the overall equivalent resistance for the entire circuit.

Once the equivalent resistance has been found you workbackward from there to find the current in and potentialdifference across each resistor.


31

Chapter 21 — Magnetism

21.1 Magnets and Magnetic Fields

Every magnet has two magnetic poles, one designated asnorth, one as south. A single magnet can be cut or brokeninto two pieces, each piece of a magnet will also be a magnetwith two poles. (It is impossible to have a single magneticpole without the corresponding opposite pole.) Like mag-netic poles repel each other, unlike poles attract.

Just as an electric charge creates an electric field, amagnetic dipole (a pair of magnetic poles) creates a mag-netic field around it and interacting with other magneticdipoles in the vicinity.

The magnetic field strength (B) is measured in tesla(T) which are N

A·m (this is related to the force exerted on amoving charge).

The direction of a magnetic field is defined as the direc-tion that the north pole of a magnet would point if placedin the field. The magnetic field of a magnet points from thenorth pole of the magnet to the south pole. By convention,when magnetic field lines are drawn they are representedas arrows when laying in the plane of the page, if they arecoming up out of the page they are represented by a solidcircle, and if they are going down into the page they arerepresented by an ‘x’. It may help to think of an arrow likewould be shot from a bow, if it is coming toward you thenyou will see the point, if it is moving away then you will seethe feathers on the end.

The magnetic north pole of the earth corresponds to thegeographic south pole and vice versa.

21.2 Electromagnetism and Magnetic Do-mains

A magnetic field exists around any current-carrying wire,the direction of the field around the wire follows a circularpath around the wire according to the right-hand rule — ifyou point the thumb of your right hand in the direction ofthe current in the wire and curl your fingers the magneticfield will follow your fingers.

The magnetic field created by a solenoid or coil is similarto the magnetic field of a permanent magnet, the directionis once again determined by the right-hand rule — this timeif you wrap the fingers of your right hand in the directionthat the current flows in the coil then the magnetic fieldinside the coil will be in the direction of your thumb andwill be nearly uniform within the coil.

A magnetic domain is a group of atoms whose mag-netic fields are aligned. In some materials (wood, plastic,etc.) the magnetic fields of the individual atoms are ar-ranged randomly and can not be aligned, others such asiron, cobalt, and nickel the magnetic fields do not com-

pletely cancel. These are known as ferromagnetic materi-als. When a substance becomes magnetized by an externalmagnetic field the individual domains are oriented to pointin the same direction. If a material is easily magnetized thenthe domains will return to a random organization when theexternal field is removed, if it is not easily magnetized thenthe organization of the domains will remain when the fieldis removed and the substance will remain magnetized.

21.3 Magnetic Force

Charged Particles in a Magnetic FieldWhen a charge moves through a magnetic field it will expe-rience a force caused by the magnetic field. This force hasa maximum when the charge moves perpendicularly to themagnetic field, decreases at other angles, and becomes zerowhen the charge moves along the field lines.

The magnitude of the magnetic field (B) is given by

B =Fmagnetic

qv

where q is the moving charge and v is its velocity.The direction of the force on a positive charge moving

through a magnetic field can be found using the right-handrule. If you point the fingers in the direction of the mag-netic field and your thumb in the direction of the motion ofthe charge and then bend your fingers at an angle the forcewill then be in the direction of your fingers. If the chargeis negative then find the direction for a positive charge andthen reverse it.

A charged particle moving perpendicularly to a uniformmagnetic field will move in a circular path. If the particle isnot moving perpendicularly to the magnetic field then thecomponent of the particle’s velocity in the direction of thefield will remain unchanged and the particle will move in ahelical path.

Magnetic Force on a Current-Carrying ConductorA length of wire, l, within an external magnetic field, B,carrying a current, I, undergoes a magnetic force of

Fmagnetic = BIl

Two parallel current-carrying wires exert on one anotherforces that are equal in magnitude and opposite in direction.If the currents are in the same direction the two wires at-tract one another, if they are in opposite directions thenthey repel each other.

The magnetic fields developed by conductors in an elec-tric field are used to produce sound with loudspeakers andare also used to cause the needle to move in a galvanometer.


32

Chapter 22 — Induction and Alternating Current

22.1 Induced CurrentMoving a conductor in an electric field or changing the mag-netic field around a conductor will induce an emf (alsoknown as an electromotive force, but it is actually a po-tential difference and not a force) in the conductor throughelectromagnetic induction. The charges already presentin the conductor will be caused to move in opposite di-rections by the motion of the conductor through the field,thus a current can be induced without a battery or othersource present. This effect is greatest when the motion ofthe conductor is perpendicular to both the conductor andthe magnetic field.

If a loop (or coil) of wire is moved linearly through amagnetic field the orientation of the loop will affect the cur-rent induced, the current will be greatest when the planeof the loop is perpendicular to the magnetic field and zerowhen it is parallel to the field.

Induced emf for a moving loop can be calculated usingFaraday’s law of magnetic induction

emf = −N∆(AB cos θ)∆t

where A is the area of the loop, B is the strength of themagnetic field, θ is the angle the field lines make with anormal vector to the plane of the loop, and N is the num-ber of turns in the loop. The effect of this equation is thatif more magnetic field lines are cut by the loop of wire theemf will be increased, whether by moving the loop or bychanging the field itself.

Lenz’s law states that the magnetic field of the inducedcurrent opposes the change in the applied magnetic field.

22.2 Alternating Current, Generators, andMotorsGenerators use induction to convert mechanical energy toelectrical energy. The produce a continuously changing emfby rotating a coil of wire inside a uniform magnetic field.When the loop is perpendicular to the magnetic field everyportion of the loop is moving parallel to the magnetic fieldso there is no current induced, when the loop is parallelto the magnetic field then the sides of the loop are cross-ing the largest number of magnetic field lines so the largest

emf is then produced. The emf will vary from zero to themaximum twice every rotation of the coil and will follow asinusoidal path, this is called alternating current.

The maximum potential difference is only reached dur-ing a small part of each cycle, the effective potential dif-ference is the square root of the average of the square ofthe potential difference over time, known as the root-mean-square potential difference (∆Vrms). This and the similarexpression for current are

∆Vrms =∆Vmax√

2= 0.707∆Vmax

Irms =Imax√

2= 0.707Imax

The rms potential difference can be used as the potentialdifference term and the rms current can be used as the cur-rent term in all of the electrical equations from previouschapters. Electrical meters also usually read rms values foralternating current.

Motors use an arrangement similar to that of generatorsto convert electrical energy into mechanical energy, the dif-ference is that in a motor the electrical energy is convertedto mechanical energy, a generator does the opposite.

22.3 Inductance

Mutual inductance involves the induction of a current inone circuit by means of a changing current in another cir-cuit. The example seen most often is a transformer, twocoils of wire wound on the same iron core, this will changethe potential difference of an alternating current source. Fora transformer with primary and secondary windings of N1

and N2 turns, an input potential difference of ∆V1 will bechanged into an output potential difference ∆V2

∆V2 =N2

N1∆V1

The power produced by the output of a transformer isthe same as the power consumed by the transformer, soP1 = P2 and combining the two equations eventually yields∆V1I1 = ∆V2I2.


33

Variables and Notation

SI Prefixes

Prefix Mult. Abb. Prefix Mult. Abb.

yocto- 10−24 y yotta- 1024 Yzepto- 10−21 z zetta- 1021 Zatto- 10−18 a exa- 1018 Efemto- 10−15 f peta- 1015 Ppico- 10−12 p tera- 1012 Tnano- 10−9 n giga- 109 Gmicro- 10−6 µ mega- 106 Mmilli- 10−3 m kilo- 103 kcenti- 10−2 c hecto- 102 hdeci- 10−1 d deka- 101 da

Units

Symbol Unit Quantity Composition

kg kilogram Mass SI base unitm meter Length SI base units second Time SI base unitA ampere Electric current SI base unitcd candela Luminous intensity SI base unitK kelvin Temperature SI base unit

mol mole Amount SI base unit

Ω ohm Resistance VA or m2·kg

s3·A2

C coulomb Charge A · sF farad Capacitance C

V or s4·A2

m2·kg

H henry Inductance V·sA or m2·kg

A2·s2

Hz hertz Frequency s−1

J joule Energy N ·m or kg·m2

s2

N newton Force kg·ms2

rad radian Angle mm or 1

T tesla Magnetic field NA·m

V volt Electric potential JC or m2·kg

s3·A

W watt Power Js or kg·m2

s3

Wb weber Magnetic flux V · s or kg·ms2·A

Notation

Notation Description

~x Vectorx Scalar, or the magnitude of ~x|~x| The absolute value or magnitude of ~x∆x Change in x∑x Sum of all x

Πx Product of all xxi Initial value of xxf Final value of xx Unit vector in the direction of x

A→ B A implies BA ∝ B A is proportional to BA B A is much larger than B

Greek Alphabet

Name Maj. Min. Name Maj. Min.

Alpha A α Nu N νBeta B β Xi Ξ ξ

Gamma Γ γ Omicron O oDelta ∆ δ Pi Π π or $

Epsilon E ε or ε Rho P ρ or %Zeta Z ζ Sigma Σ σ or ςEta H η Tau T τ

Theta Θ θ or ϑ Upsilon Υ υIota I ι Phi Φ φ or ϕ

Kappa K κ Chi X χLambda Λ λ Psi Ψ ψ

Mu M µ Omega Ω ω


34

Variables

Variable Description Units

α Angular acceleration rads2

θ Angular position radθc Critical angle o (degrees)θi Incident angle o (degrees)θr Refracted angle o (degrees)θ′ Reflected angle o (degrees)∆θ Angular displacement radτ Torque N ·mω Angular speed rad

sµ Coefficient of friction (unitless)µk Coefficient of kinetic friction (unitless)µs Coefficient of static friction (unitless)~a Acceleration m

s2

~ac Centripetal acceleration ms2

~ag Gravitational acceleration ms2

~at Tangential acceleration ms2

~ax Acceleration in the x direction ms2

~ay Acceleration in the y direction ms2

A Area m2

B Magnetic field strength TC Capacitance F~d Displacement m

d sin θ lever arm m~dx or ∆x Displacement in the x direction m~dy or ∆y Displacement in the y direction m

E Electric field strength NC

f Focal length m~F Force N~Fc Centripetal force N

~Felectric Electrical force N~Fg Gravitational force N~Fk Kinetic frictional force N

~Fmagnetic Magnetic force N~Fn Normal force N~Fs Static frictional force N~F∆t Impulse N · s or kg·m

s~g Gravitational acceleration m

s2

Variable Description Units

h Object height mh′ Image height mI Current AI Moment of inertia kg ·m2

KE or K Kinetic energy JKE rot Rotational kinetic energy JL Angular Momentum kg·m2

sL Self-inductance Hm Mass kgM Magnification (unitless)M Mutual inductance HMA Mechanical Advantage (unitless)ME Mechanical Energy Jn Index of refraction (unitless)p Object distance m~p Momentum kg·m

s

P Power WPE or U Potential Energy JPE elastic Elastic potential energy JPE electric Electrical potential energy J

PE g Gravitational potential energy Jq Image distance m

q or Q Charge CQ Heat, Entropy JR Radius of curvature mR Resistance Ωs Arc length mt Time s

∆t Time interval s~v Velocity m

s

vt Tangential speed ms

~vx Velocity in the x direction ms

~vy Velocity in the y direction ms

V Electric potential V∆V Electric potential difference VV Volume m3

W Work Jx or y Position m


35

Constants

Symbol Name Established Value Value Used

ε0 Permittivity of a vacuum 8.854 187 817× 10−12 C2

N·m2 8.85× 10−12 C2

N·m2

φ Golden ratio 1.618 033 988 749 894 848 20

π Archimedes’ constant 3.141 592 653 589 793 238 46

g, ag Gravitational acceleration constant 9.79171 ms2 (varies by location) 9.81 m

s2

c Speed of light in a vacuum 2.997 924 58× 108 ms (exact) 3.00× 108 m

s

e Natural logarithmic base 2.718 281 828 459 045 235 36

e− Elementary charge 1.602 177 33× 1019 C 1.60× 1019 C

G Gravitational constant 6.672 59× 10−11 N·m2

kg2 6.67× 10−11 N·m2

kg2

kC Coulomb’s constant 8.987 551 788× 109 N·m2

C2 8.99× 109 N·m2

C2

NA Avogadro’s constant 6.022 141 5× 1023 mol−1

Astronomical Data

Symbol Object Mean Radius Mass Mean Orbit Radius Orbital Period

Á Moon 1.74× 106 m 7.36× 1022 kg 3.84× 108 m 2.36× 106 s

À Sun 6.96× 108 m 1.99× 1030 kg — —

Â Mercury 2.43× 106 m 3.18× 1023 kg 5.79× 1010 m 7.60× 106 s

Ã Venus 6.06× 106 m 4.88× 1024 kg 1.08× 1011 m 1.94× 107 s

Ê Earth 6.37× 106 m 5.98× 1024 kg 1.496× 1011 m 3.156× 107 s

Ä Mars 3.37× 106 m 6.42× 1023 kg 2.28× 1011 m 5.94× 107 s

Ceres1 4.71× 105 m 9.5× 1020 kg 4.14× 1011 m 1.45× 108 s

Å Jupiter 6.99× 107 m 1.90× 1027 kg 7.78× 1011 m 3.74× 108 s

Æ Saturn 5.85× 107 m 5.68× 1026 kg 1.43× 1012 m 9.35× 108 s

Ç Uranus 2.33× 107 m 8.68× 1025 kg 2.87× 1012 m 2.64× 109 s

È Neptune 2.21× 107 m 1.03× 1026 kg 4.50× 1012 m 5.22× 109 s

É Pluto1 1.15× 106 m 1.31× 1022 kg 5.91× 1012 m 7.82× 109 s

Eris21 2.4× 106 m 1.5× 1022 kg 1.01× 1013 m 1.75× 1010 s

1Ceres, Pluto, and Eris are classified as “Dwarf Planets” by the IAU2Eris was formerly known as 2003 UB313


36

Mathematics Review for Physics

This is a summary of the most important parts of math-ematics as we will use them in a physics class. There arenumerous parts that are completely omitted, others aregreatly abridged. Do not assume that this is a completecoverage of any of these topics.

AlgebraFundamental properties of algebra

a+ b = b+ a Commutative law for ad-dition

(a+ b) + c = a+ (b+ c) Associative law for addi-tion

a+ 0 = 0 + a = a Identity law for additiona+ (−a) = (−a) + a = 0 Inverse law for addition

ab = ba Commutative law formultiplication

(ab)c = a(bc) Associative law for mul-tiplication

(a)(1) = (1)(a) = a Identity law for multipli-cation

a 1a = 1

aa = 1 Inverse law for multipli-cation

a(b+ c) = ab+ ac Distributive law

Exponents(ab)n = anbn (a/b)n = an/bn

anam = an+m 0n = 0an/am = an−m a0 = 1(an)m = a(mn) 00 = 1 (by definition)

Logarithms

x = ay → y = loga x

loga(xy) = loga x+ loga y

loga

(x

y

)= loga x− loga y

loga(xn) = n loga x

loga

(1x

)= − loga x

loga x =logb x

logb a= (logb x)(loga b)

Binomial Expansions

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)n =n∑

i=0

n!i!(n− i)!

aibn−i

Quadratic formulaFor equations of the form ax2 +bx+c = 0 the solutions are:

x =−b±

√b2 − 4ac

2a

Geometry

Shape Area Volume

Triangle A = 12bh —

Rectangle A = lw —

Circle A = πr2 —

Rectangularprism

A = 2(lw + lh+ hw) V = lwh

Sphere A = 4πr2 V = 43πr

3

Cylinder A = 2πrh+ 2πr2 V = πr2h

Cone A = πr√r2 + h2 + πr2 V = 1

3πr2h

TrigonometryIn physics only a small subset of what is covered in atrigonometry class is likely to be used, in particular sine, co-sine, and tangent are useful, as are their inverse functions.As a reminder, the relationships between those functionsand the sides of a right triangle are summarized as follows:

θ

opp

adj

hypsin θ =

opp

hypcos θ =

adj

hyp

tan θ =opp

adj=

sin θcos θ

The inverse functions are only defined over a limited range.The tan−1 x function will yield a value in the range −90 <θ < 90, sin−1 x will be in −90 ≤ θ ≤ 90, and cos−1 xwill yield one in 0 ≤ θ ≤ 180. Care must be taken toensure that the result given by a calculator is in the correctquadrant, if it is not then an appropriate correction mustbe made.

Degrees 0o 30o 45o 60o 90o 120o 135o 150o 180o

Radians 0 π6

π4

π3

π2

2π3

3π4

5π6 π

sin 0 12

√2

2

√3

2 1√

32

√2

212 0

cos 1√

32

√2

212 0 − 1

2 −√

22 −

√3

2 −1

tan 0√

33 1

√3 ∞ −

√3 −1 −

√3

3 0

sin2 θ + cos2 θ = 1 2 sin θ cos θ = sin(2θ)


37

Trigonometric functions in terms of each other

sin θ = sin θ√

1− cos2 θ tan θ√1+tan2 θ

cos θ =√

1− sin2 θ cos θ 1√1+tan2 θ

tan θ = sin θ√1−sin2 θ

√1−cos2 θcos θ tan θ

csc θ = 1sin θ

1√1−cos2 θ

√1+tan2 θtan θ

sec θ = 1√1−sin2 θ

1cos θ

√1 + tan2 θ

cot θ =√

1−sin2 θ

sin θcos θ√

1−cos2 θ1

tan θ

sin θ = 1csc θ

√sec2 θ−1sec θ

1√1+cot2 θ

cos θ =√

csc2 θ−1csc θ

1sec θ

cot θ√1+cot2 θ

tan θ = 1√csc2 θ−1

√sec2 θ − 1 1

cot θ

csc θ = csc θ sec θ√sec2 θ−1

√1 + cot2 θ

sec θ = csc θ√csc2 θ−1

sec θ√

1+cot2 θcot θ

cot θ =√

csc2 θ − 1 1√sec2 θ−1

cot θ

Law of sines, law of cosines, area of a triangle

HH

HHHH

A

B

C

a

b

c

a2 = b2 + c2 − 2bc cosAb2 = a2 + c2 − 2ac cosBc2 = a2 + b2 − 2ab cosC

a

sinA=

b

sinB=

c

sinC

let s = 12 (a+ b+ c)

Area=√s(s− a)(s− b)(s− c)

Area= 12bc sinA = 1

2ac sinB = 12ab sinC

VectorsA vector is a quantity with both magnitude and direction,such as displacement or velocity. Your textbook indicatesa vector in bold-face type as V and in class we have beenusing ~V . Both notations are equivalent.

A scalar is a quantity with only magnitude. This caneither be a quantity that is directionless such as time ormass, or it can be the magnitude of a vector quantity suchas speed or distance traveled. Your textbook indicates ascalar in italic type as V , in class we have not done any-thing to distinguish a scalar quantity. The magnitude of ~Vis written as V or |~V |.

A unit vector is a vector with magnitude 1 (a dimen-sionless constant) pointing in some significant direction. Aunit vector pointing in the direction of the vector ~V is in-dicated as V and would commonly be called V -hat. Anyvector can be normalized into a unit vector by dividing it byits magnitude, giving V = ~V

V . Three special unit vectors, ı,, and k are introduced with chapter 3. They point in thedirections of the positive x, y, and z axes, respectively (asshown below).

6

?

-

y+

−

x+−

z+

−6

-

ı

k

Vectors can be added to other vectors of the same di-mension (i.e. a velocity vector can be added to anothervelocity vector, but not to a force vector). The sum of allvectors to be added is called the resultant and is equivalentto all of the vectors combined.

Multiplying VectorsAny vector can be multiplied by any scalar, this has theeffect of changing the magnitude of the vector but not itsdirection (with the exception that multiplying a vector bya negative scalar will reverse the direction of the vector).As an example, multiplying a vector ~V by several scalarswould give:

- -

-

~V 2~V

.5~V −1~V

In addition to scalar multiplication there are also twoways to multiply vectors by other vectors. They will not bedirectly used in class but being familiar with them may helpto understand how some physics equations are derived. Thefirst, the dot product of vectors ~V1 and ~V2, represented as~V1 · ~V2 measures the tendency of the two vectors to point inthe same direction. If the angle between the two vectors isθ the dot product yields a scalar value as

~V1 · ~V2 = V1V2 cos θ

The second method of multiplying two vectors, thecross product, (represented as ~V1× ~V2) measures the ten-dency of vectors to be perpendicular to each other. It yieldsa third vector perpendicular to the two original vectors withmagnitude

|~V1 × ~V2| = V1V2 sin θ

The direction of the cross product is perpendicular to thetwo vectors being crossed and is found with the right-handrule — point the fingers of your right hand in the directionof the first vector, curl them toward the second vector, andthe cross product will be in the direction of your thumb.

Adding Vectors GraphicallyThe sum of any number of vectors can be found by drawingthem head-to-tail to scale and in proper orientation thendrawing the resultant vector from the tail of the first vector


38

to the point of the last one. If the vectors were drawn ac-curately then the magnitude and direction of the resultantcan be measured with a ruler and protractor. In the ex-ample below the vectors ~V1, ~V2, and ~V3 are added to yield~Vr

-

AAAAAAU

~V1

~V2 ~V3

~V1, ~V2, ~V3

-AAAAAAU

~V1

~V2

~V3

~V1 + ~V2 + ~V3

-AAAAAAU

~V1

~V2

~V3

-~Vr

~Vr = ~V1 + ~V2 + ~V3

Adding Parallel VectorsAny number of parallel vectors can be directly added byadding their magnitudes if one direction is chosen as pos-itive and vectors in the opposite direction are assigned anegative magnitude for the purposes of adding them. Thesum of the magnitudes will be the magnitude of the resul-tant vector in the positive direction, if the sum is negativethen the resultant will point in the negative direction.

Adding Perpendicular VectorsPerpendicular vectors can be added by drawing them as aright triangle and then finding the magnitude and directionof the hypotenuse (the resultant) through trigonometry andthe Pythagorean theorem. If ~Vr = ~Vx + ~Vy and ~Vx ⊥ ~Vy

then it works as follows:

-~Vx

6

~Vy

-~Vx

6

~Vy

~Vr

θ

Since the two vectors to be added and the resultant forma right triangle with the resultant as the hypotenuse thePythagorean theorem applies giving

Vr = |~Vr| =√V 2

x + V 2y

The angle θ can be found by taking the inverse tangent ofthe ratio between the magnitudes of the vertical and hori-zontal vectors, thus

θ = tan−1 Vy

Vx

As was mentioned above, care must be taken to ensure thatthe angle given by the calculator is in the appropriate quad-rant for the problem, this can be checked by looking at the

diagram drawn to solve the problem and verifying that theanswer points in the direction expected, if not then makean appropriate correction.

Resolving a Vector Into ComponentsJust two perpendicular vectors can be added to find asingle resultant, any single vector ~V can be resolved intotwo perpendicular component vectors ~Vx and ~Vy so that~V = ~Vx + ~Vy.

~V

θ ~Vx

~Vy

~V

θ

-

6

As the vector and its components can be drawn as aright triangle the ratios of the sides can be found withtrigonometry. Since sin θ = Vy

V and cos θ = Vx

V it followsthat Vx = V cos θ and Vy = V sin θ or in a vector form,~Vx = V cos θı and ~Vy = V sin θ. (This is actually an appli-cation of the dot product, ~Vx = (~V · ı)ı and ~Vy = (~V · ),but it is not necessary to know that for this class)

Adding Any Two Vectors AlgebraicallyOnly vectors with the same direction can be directly added,so if vectors pointing in multiple directions must be addedthey must first be broken down into their components, thenthe components are added and resolved into a single resul-tant vector — if in two dimensions ~Vr = ~V1 + ~V2 then

*

~V1

~V2

~Vr

~Vr = ~V1 + ~V2

*

-

-6

6

~V1

~V2

~Vr

~V1x

~V2x

~V1y

~V2y

~V1 = ~V1x + ~V1y

~V2 = ~V2x + ~V2y

~Vr = (~V1x + ~V1y) + (~V2x + ~V2y)

*

- -

6

6

~V1

~V2

~Vr

~V1x~V2x

~V1y

~V2y

~Vr = (~V1x + ~V2x) + (~V1y + ~V2y)

Once the sums of the component vectors in each direc-tion have been found the resultant can be found from themjust as an other perpendicular vectors may be added. Since


39

from the last figure ~Vr = (~V1x+~V2x)+(~V1y +~V2y) and it waspreviously established that ~Vx = V cos θı and ~Vy = V sin θit follows that

~Vr = (V1 cos θ1 + V2 cos θ2) ı+ (V1 sin θ1 + V2 sin θ2)

and

Vr =√

(V1x + V2x)2 + (V1y + V2y)2

=√

(V1 cos θ1 + V2 cos θ2)2 + (V1 sin θ1 + V2 sin θ2)

2

with the direction of the resultant vector ~Vr, θr, being foundwith

θr = tan−1 V1y + V2y

V1x + V2x= tan−1 V1 sin θ1 + V2 sin θ2

V1 cos θ1 + V2 cos θ2

Adding Any Number of Vectors Algebraically

For a total of n vectors ~Vi being added with magnitudes Vi

and directions θi the magnitude and direction are:

*

- - -

6

6

6

~V1

~V2

~V3

~Vr

~V1x~V2x

~V3x

~V1y

~V2y

~V3y

~Vr =

(n∑

i=1

Vi cos θi

)ı+

(n∑

i=1

Vi sin θi

)

Vr =

√√√√( n∑i=1

Vi cos θi

)2

+

(n∑

i=1

Vi sin θi

)2

θr = tan−1

n∑i=1

Vi sin θi

n∑i=1

Vi cos θi

(The figure shows only three vectors but this method willwork with any number of them so long as proper care istaken to ensure that all angles are measured the same wayand that the resultant direction is in the proper quadrant.)

CalculusAlthough this course is based on algebra and not calculus,it is sometimes useful to know some of the properties ofderivatives and integrals. If you have not yet learned cal-culus it is safe to skip this section.

DerivativesThe derivative of a function f(t) with respect to t is afunction equal to the slope of a graph of f(t) vs. t at everypoint, assuming that slope exists. There are several waysto indicate that derivative, including:

ddtf(t)

f ′(t)

f(t)

The first one from that list is unambiguous as to the in-dependent variable, the others assume that there is only onevariable, or in the case of the third one that the derivativeis taken with respect to time. Some common derivativesare:

ddttn = ntn−1

ddt

sin t = cos t

ddt

cos t = − sin t

ddtet = et

If the function is a compound function then there are afew useful rules to find its derivative:

ddtcf(t) = c

ddtf(t) c constant for all t

ddt

[f(t) + g(t)] =ddtf(t) +

ddtg(t)

ddtf(u) =

ddtu

dduf(u)

IntegralsThe integral, or antiderivative of a function f(t) withrespect to t is a function equal to the the area under agraph of f(t) vs. t at every point plus a constant, assumingthat f(t) is continuous. Integrals can be either indefinite ordefinite. An indefinite integral is indicated as:∫

f(t) dt

while a definite integral would be indicated as∫ b

a

f(t) dt

to show that it is evaluated from t = a to t = b, as in:∫ b

a

f(t) dt =∫f(t) dt|t=b

t=a =∫f(t) dt|t=b −

∫f(t) dt|t=a

Some common integrals are:∫tn dt =

tn+1

n+ 1+ C n 6= −1


40

∫t−1 dt = ln t+ C∫

sin tdt = − cos t+ C∫cos tdt = sin t+ C∫et dt = et + C

There are rules to reduce integrals of some compoundfunctions to simpler forms (there is no general rule to reducethe integral of the product of two functions):∫

cf(t) dt = c

∫f(t) dt c constant for all t∫

[f(t) + g(t)] dt =∫f(t) dt+

∫g(t) dt

Series Expansions

Some functions are difficult to work with in their normalforms, but once converted to their series expansion can bemanipulated easily. Some common series expansions are:

sinx = x− x3

3!+x5

5!− x7

7!+ · · · =

∞∑i=0

(−1)i x2i+1

(2i+ 1)!

cosx = 1− x2

2!+x4

4!− x6

6!+ · · · =

∞∑i=0

(−1)i x2i

(2i)!

ex = 1 + x+x2

2!+x3

3!+x4

4!+ · · · =

∞∑i=0

xi

(i)!


41

Physics Using Calculus

In our treatment of mechanics the majority of the equa-tions that have been covered in the class have been eitherapproximations or special cases where the acceleration orforce applied have been held constant. This is requiredbecause the class is based on algebra and to do otherwisewould require the use of calculus.

None of what is presented here is a required part of theclass, it is here to show how to handle cases outside theusual approximations and simplifications. Feel free to skipthis section, none of it will appear as a required part of anyassignment or test in this class.

Notation

Because the letter ‘d’ is used as a part of the notation ofcalculus the variable ‘r’ is often used to represent the posi-tion vector of the object being studied. (Other authors use‘s’ for the spatial position or generalize ‘x’ to two or moredimensions.) In a vector form that would become ~r to giveboth the magnitude and direction of the position. It is as-sumed that the position, velocity, acceleration, etc. can beexpressed as a function of time as ~r(t), ~v(t), and ~a(t) butthe dependency on time is usually implied and not shownexplicitly.

Translational Motion

Many situations will require an acceleration that is not con-stant. Everything learned about translational motion sofar used the simplification that the acceleration remainedconstant, with calculus we can let the acceleration be anyfunction of time.

Since velocity is the slope of a graph of position vs. timeat any point, and acceleration is the slope of velocity vs.time these can be expressed mathematically as derivatives:

~v =ddt~r

~a =ddt~v =

d2

dt2~r

The reverse of this relationship is that the displacementof an object is equal to the area under a velocity vs. timegraph and velocity is equal to the area under an accelera-tion vs. time graph. Mathematically this can be expressedas integrals:

~v =∫~adt

~r =∫~v dt =

∫∫~adt2

Using these relationships we can derive the main equa-tions of motion that were introduced by starting with the

integral of a constant velocity ~a(t) = ~a as∫ tf

ti

~adt = ~a∆t+ C

The constant of integration, C, can be shown to be the ini-tial velocity, ~vi, so the entire expression for the final velocitybecomes

~vf = ~vi + ~a∆t

or~vf (t) = ~vi + ~at

Similarly, integrating that expression with respect to timegives an expression for position:∫ tf

ti

(~vi + ~at) dt = ~vi∆t+12~a∆t2 + C

Once again the constant of integration, C, can be shown tobe the initial position, ~ri, yielding an expression for positionvs. time

~rf (t) = ~ri + ~vit+12~at2

While calculus can be used to derive the algebraic formsof the equations the reverse is not true. Algebra can han-dle a subset of physics and only at the expense of needingto remember separate equations for various special cases.With calculus the few definitions shown above will sufficeto predict any linear motion.

Work and PowerThe work done on an object is the integral of dot product ofthe force applied with the path the object takes, integratedas a contour integral over the path.

W =∫

C

~F · d~r

The power developed is the time rate of change of thework done, or the derivative of the work with respect totime.

P =ddtW

MomentumNewton’s second law of motion changes from the familiar~F = m~a to the statement that the force applied to an objectis the time derivative of the object’s momentum.

~F =ddt~p

Similarly, the impulse delivered to an object is the in-tegral of the force applied in the direction parallel to themotion with respect to time.

∆~p =∫

~F dt


42

Rotational MotionThe equations for rotational motion are very similar tothose for translational motion with appropriate variablesubstitutions. First, the angular velocity is the time deriva-tive of the angular position.

~ω =d~θdt

The angular acceleration is the time derivative of the an-gular velocity or the second time derivative of the angularposition.

~α =d~ωdt

=d2~θ

dt2

The angular velocity is also the integral of the angularacceleration with respect to time.

~ω =∫~α dt

The angular position is the integral of the angular veloc-ity with respect to time or the second integral of the angularacceleration with respect to time.

~θ =∫~ω dt =

∫∫~α dt2

The torque exerted on an object is the cross product ofthe radius vector from the axis of rotation to the point ofaction with the force applied.

~τ = ~r × ~F

The moment of inertia of any object is the integral ofthe square of the distance from the axis of rotation to eachelement of the mass over the entire mass of the object.

I =∫r2 dm

The torque applied to an object is the time derivativeof the object’s angular momentum.

~τ =ddt~L

The integral of the torque with respect to time is theangular momentum of the object.

~L =∫~τ dt


43

Data Analysis

Comparative MeasuresPercent ErrorWhen experiments are conducted where there is a calcu-lated result (or other known value) against which the ob-served values will be compared the percent error betweenthe observed and calculated values can be found with

percent error =∣∣∣∣observed value − accepted value

accepted value

∣∣∣∣× 100%

Percent DifferenceWhen experiments involve a comparison between two ex-perimentally determined values where neither is regardedas correct then rather than the percent error the percentdifference can be calculated. Instead of dividing by the ac-cepted value the difference is divided by the mean of thevalues being compared. For any two values, a and b, thepercent difference is

percent difference =∣∣∣∣ a− b

12 (a+ b)

∣∣∣∣× 100%

Dimensional AnalysisUnit ConversionsWhen converting units you need to be careful to ensure thatconversions are done properly and not accidentally reversed(or worse). The easiest way to do this is to be careful toalways keep the units associated with each number and ateach step always multiply by something equal to 1.

As an example, if you have a distance of 78.25 km andwant to convert it to meters the immediate reaction is tosay that you can multiply by 1000, or is that divide by1000. Knowing that 1 km = 1000 m it is obvious then that1 km

1000 m = 1 = 1000 m1 km . Since both of them are equal to 1

they can be multiplied by what you wish to convert with-out changing the quantity itself, it will only change the unitsin which it is shown. So, to convert 78.25 km to meters youwould multiply by 1000 m

1 km :

(78.25 km)(

1000 m1 km

)= 78 250 m

As you can see, the km in the original quantity and in thedenominator cancel leaving meters. If you were to reversethe conversion accidentally you would get this:

(78.25 km)(

1 km1000 m

)= 0.07825

km2

m

Instead of canceling the km and leaving the measurementin meters the result is in km2

m , a meaningless quantity. Thenonsensical result is an indication that the conversion wasreversed and should be redone. If the conversion was donewithout the units then it could be reversed without realizingthat it happened.

Units on ConstantsConsider a mass vibrating on the end of a spring. The equa-tion relating the mass to the frequency on graph might looksomething like

y =3.658x2− 1.62

This isn’t what we’re looking for, a first pass would be tochange the x and y variables into f and m for frequencyand mass, this would then give

m =3.658f2

− 1.62

This is progress but there’s still a long way to go. Massis measured in kg and frequency is measured in s−1, sincethese variables represent the entire measurement (includ-ing the units) and not just the numeric portion they do notneed to be labeled, but the constants in the equation doneed to have the correct units applied.

To find the units it helps to first re-write the equationusing just the units, letting some variable such as u (or u1,u2, u3, etc.) stand for the unknown units. The exampleequation would then become

kg =u1

(s−1)2+ u2

Since any quantities being added or subtracted must havethe same units the equation can be split into two equations

kg =u1

(s−1)2and kg = u2

The next step is to solve for u1 and u2, in this case u2 = kgis immediately obvious, u1 will take a bit more effort. Afirst simplification yields

kg =u1

s−2

then multiplying both sides by s−2 gives

(s−2)(kg) =( u1

s−2

)(s−2)

which finally simplifies and solves to

u1 = kg · s−2

Inserting the units into the original equation finally yields

m =3.658 kg · s−2

f2− 1.62 kg

which is the final equation with all of the units in place.Checking by choosing a frequency and carrying out the cal-culations in the equation will show that it is dimensionallyconsistent and yields a result in the units for mass (kg) asexpected.


44

Final Exam Description

• 95 Multiple Choice Items (22 problems, 73 conceptual questions)

• Calculators will not be permitted

• Exam items are from the following categories:

– Chapter 1 — The Science of Physicsunits of measurementsignificant figures

– Chapter 2 — Motion in One Dimensiongraphsequations of motionacceleration

– Chapter 3 — Two-Dimensional Motion andVectorsvectors and scalarsvector additioncomponentsprojectilesrelative motion

– Chapter 4 — Forces and the Laws of Motioncauses of motioninertiabalanced forcesaccelerated motionfriction

– Chapter 5 — Work and Energyworktypes of energyconservation of energypower

– Chapter 6 — Momentum and Collisionsmomentum and impulseconservation of momentumtypes of collisions

– Chapter 7 — Rotational Motion and the Lawof Gravityrotational motion equationscentripetal force and accelerationcauses of centripetal forcegravitational force equation

– Chapter 8 — Rotational Equilibrium andDynamicstorquecenter of gravityequilibrium

– Chapter 12 — Vibrations and Wavessimle harmonic motionwave propertiesinterferencestanding waves

– Chapter 13 — Soundtypes of waveswave speedinterferenceDoppler shiftresonance

– Chapter 14 — Light and Reflectioncharacteristics of lightflat mirrorscurved mirrorspolarization

– Chapter 15 — Refractionindex of refractionSnell’s lawlenses

– Chapter 16 — Interference and Diffractioninterferencediffraction

– Chapter 17 — Electric Forces and Fieldselectric chargemethods of producing net chargeconductors and insulatorselectric forces

– Chapter 18 — Electrical Energy and Capac-itanceelectric potential energypotential difference

– Chapter 19 — Current and ResistancecurrentresistanceOhm’s lawpower

– Chapter 20 — Circuits and Circuit Elementsseries and parallel resistorsresistor combinations

– Chapter 21 — Magnetismcauses of magnetismmagnetic fieldscurrent and magnetismmagnetic forces

– Chapter 22 — Induction and AlternatingCurrentinduced currentLenz’s lawmotors and generatorstransformers


45

Final Exam Equation Sheet

This is a slightly reduced copy of the front and back of the equation sheet that you will receive with the final exam.

Honors Physics – 2nd Semester Final Constants and Equations

KinematicsLinear

avd v t=

( )12 i fd v v t= + Δ

( ) 21

2id v t a t= Δ + Δ

f iv v a t= + Δ

2 2 2f iv v ad= +

x xd v t= Δ

( ) 21

2y y yd v t a t= Δ + Δ

sinyv v θ=

cosxv v θ=

KinematicsRotational

dr

θ =

vr

ω =

a

rα =

av tθ ω= Δ

( ) 21

2i t tθ ω α= Δ + Δ

( )12 i f tθ ω ω= + Δ

f i tω ω α= + Δ

2 2f i 2ω ω αθ= +

DynamicsLinear

F ma=∑gF mg=

cosF w θ =

sinF w θ=

Nf Fμ=

tanμ θ=

springF kx= −

1 22grav

m mF G

r=

DynamicsCircular

2 rv

Tπ

=

2 2

2

4c

v ra

r Tπ

= =

2 2

2

4c

v rF m m

r Tπ

= =

1T f=

DynamicsRotational

Iτ α=∑2

point massI mr=

counterclockwise clockwiseτ τ=∑ ∑

Work, Power, EnergyLinear

cosW Fd θ=

gPE mgh=

212elasticPE kx=

212KE mv=

0PE KE WΔ + Δ + =

WP Fv

t= =

Work, Power, EnergyRotational

W τθ=21

2KE Iω=

P τω=

MomentumLinear

P mv=

1 1 2 2 1 1 2 2i f f fm v m v m v m v+ = +

F t m vΔ = Δ

Rotational

L Iω=

Constants

210. msg =

2

2

11 N mkg

6.67 10G − = ×

8 ms3.00 10c = ×

2

2

9 N mC

9.00 10K = ×


46

Simple Harmonic Motion

2m

Tk

π=

2pendulumTg

π=

( )cosx A tω=

( )sinv A tω ω= −

( )2 cosa A tω ω= −

ka x

m= −

( )2 2kv A x

m= ± −

22 f

Tπ

ω π= =

1T

f=

Waves and Sound

v f λ=

20.1 331 1273

cmsk

Tv T= = +

331 .6ms cv T= +

os

s

v vf f

v v

±=

1 2# Beats f f= −

Light

i r =

in vacuum

in medium

light

light

vn

v

sinsin

c in

v r

= =

sin sini rn i n r =

sinsin

i

r

v iv r

=

1sin ci n =

1 1 1

o if d d= +

i i

o o

s dM

s d= =

Electrostatics

1 22

q qF k

r=

2o

F qE k

q r= =

eB A

PEV V V

q

ΔΔ − =

qV k

r=

( )B AW q V q V V= − Δ = − −

ePE qEd=

Currents and Circuits

qI

t

ΔΔ

V IR=2

2 VP VI I R

R= = =

qC

V

Series Circuits

1 2 3 ...tI I I I= = = =

1 2 3 ...tV V V V= + + +

1 2 3 ...eqR R R R= + + +

1 2 3

1 1 1 1...

eqC C C C= + + +

Parallel Circuits

1 2 3 ...tI I I I= + + +

1 2 3 ...tV V V V= = = =

1 2 3

1 1 1 1...

eqR R R R= + + +

1 2 3 ...eqC C C C= + + +

Magnetism

sinF qvB θ=

sinF I B θ=

V El Blv= =

Transformers

p s

p s

V V

N N=

p p s sV I V I=

Note that some equations may be slightly different than the ones we have used in class, the sheet that you receivewill have these forms of the equations, it will be up to you to know how to use them or to know other forms that youare able to use.


47

Mix

edRev

iew

Exe

rcises

HRW material copyrighted under notice appearing earlier in this book.

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48HRW material copyrighted under notice appearing earlier in this book.

Mot

ion

in O

ne D

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sion

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

2

Chap

ter 2

9

1.D

uri

ng

a re

lay

race

alo

ng

a st

raig

ht

road

,th

e fi

rst

run

ner

on

a t

hre

e-

pers

on te

am r

un

s d 1

wit

h a

con

stan

t ve

loci

ty v

1.T

he

run

ner

th

en h

ands

off

the

bato

n to

th

e se

con

d ru

nn

er,w

ho

run

s d 2

wit

h a

con

stan

t ve

loci

ty

v 2.T

he

bato

n is

th

en p

asse

d to

th

e th

ird

run

ner

,wh

o co

mpl

etes

th

e ra

ce

by t

rave

ling

d 3w

ith

a c

onst

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velo

city

v3.

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term

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d th

e ti

me

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akes

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plet

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a se

gmen

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the

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.

3 ren

nu

R2 re

nn

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1 ren

nu

R

b.W

hat

is t

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tan

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fth

e ra

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ours

e?

c.W

hat

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e it

tak

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he

team

to c

ompl

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the

race

?

2.T

he e

quat

ion

s be

low

incl

ude

the

equ

atio

ns

for

stra

ight

-lin

e m

otio

n.

For

each

of

the

follo

win

g pr

oble

ms,

indi

cate

whi

ch e

quat

ion

or

equ

atio

ns

you

wou

ld u

se to

sol

ve th

e pr

oble

m,b

ut d

o n

ot a

ctu

ally

per

form

the

calc

ula

tion

s.

a.D

uri

ng

take

off,

a pl

ane

acce

lera

tes

at 4

m/s

2an

d ta

kes

40 s

to r

each

take

off

spee

d.W

hat

is t

he

velo

city

of

the

plan

e at

tak

eoff

?

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car

wit

h a

n in

itia

l spe

ed o

f31

.4 k

m/h

acc

eler

ates

at

a u

nif

orm

rat

e

of1.

2 m

/s2

for

1.3

s.W

hat

is t

he

fin

al s

peed

an

d di

spla

cem

ent

ofth

e

car

duri

ng

this

tim

e?

∆x=

1 2 (v i

+v f

)∆t

∆x=

1 2 (v f

)∆t

∆x=v i

(∆t)

+1 2 a

(∆t)

2∆x

=1 2 a

(∆t)

2

v f=v i

+a(

∆t)

v f=

a(∆t

)

v f2

=v i

2+

2a∆x

v f2

=2a

∆x


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s103.

Bel

ow is

th

e ve

loci

ty-t

ime

grap

h o

fan

obj

ect

mov

ing

alon

g a

stra

igh

t

path

.Use

th

e in

form

atio

n in

th

e gr

aph

to fi

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th

e ta

ble

belo

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HO

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Mix

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Ch

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Tim

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ocit

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9.8

m/s

from

th

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build

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a.Fi

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belo

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how

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the

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at

the

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4 s

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ion

Vel

ocit

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(s)

(m)

(m/s

)(m

/s2 )

1 2 3 4

b.In

wh

ich

sec

ond

does

th

e ba

ll re

ach

th

e to

p of

its

flig

ht?

c.In

wh

ich

sec

ond

does

th

e ba

ll re

ach

th

e le

vel o

fth

e ro

of,o

n t

he

way

dow

n?

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each

oft

he le

tter

ed in

terv

als

belo

w,i

ndi

cate

the

mot

ion

oft

he o

bjec

t

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wn

,or

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est)

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city

(+,

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the

dire

ctio

n o

fthe

acc

eler

atio

n (

+,−,

or 0

).

Velocity (m/s)

time

(s)

15 10 5 070

100

AB

CD

E30

5020

6040


49

Two-

Dim

ensi

onal

Mot

ion

and

Vect

ors

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

3

Chap

ter 3

15

1.T

he

diag

ram

bel

ow in

dica

tes

thre

e po

siti

ons

to

whi

ch a

wom

an tr

avel

s.Sh

e st

arts

at p

osit

ion

A,

trav

els

3.0

km to

the

wes

t to

poin

t B,t

hen

6.0

km

to

the

nor

th to

poi

nt C

.She

then

bac

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trav

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2.0

km to

the

sou

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poi

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.

a.In

th

e sp

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prov

ided

,dia

gram

th

e di

spla

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each

seg

men

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the

wom

an’s

tri

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hat i

s th

e to

tal d

ispl

acem

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fth

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oman

from

her

init

ial p

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,A,t

o he

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nal

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c.W

hat i

s th

e to

tal d

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trav

eled

by

the

wom

an

from

her

init

ial p

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ion,

A,t

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r fin

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D?

2.Tw

o pr

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e la

un

ched

from

th

e gr

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each

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the

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as p

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ow la

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le A

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c.Su

ppos

e pr

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A is

lau

nch

ed a

t an

an

gle

of45

°to

th

e h

oriz

onta

l.

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at is

th

e ra

tio,v B

/vA

,of

the

spee

d of

proj

ecti

le B

,vB

,com

pare

d

wit

h t

he

spee

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proj

ecti

le A

,vA

?


C D BA


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s163.

A p

asse

nge

r at

an

air

port

ste

ps o

nto

a m

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that

is 1

00.0

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th

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elat

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to t

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follo

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th

e m

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g si

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per

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sta

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e m

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per

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the

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k w

ith

a sp

eed

of5.

0 m

/s

and

in th

e sa

me

dire

ctio

n in

whi

ch th

e si

dew

alk

is m

ovin

g.

e.A

per

son

rid

ing

in a

car

t w

ith

a s

peed

of

4.0

m/s

an

d in

a d

irec

tion

perp

endi

cula

r to

th

e di

rect

ion

in w

hic

h t

he

side

wal

k is

mov

ing.

4.U

se t

he

info

rmat

ion

giv

en in

item

3 to

an

swer

th

e fo

llow

ing

ques

tion

s:

a.H

ow lo

ng

does

it t

ake

for

the

pass

enge

r w

alki

ng

on t

he

side

wal

k to

get

from

on

e en

d of

the

side

wal

k to

th

e ot

her

en

d?

b.H

ow m

uch

tim

e do

es th

e pa

ssen

ger

save

by

taki

ng

the

mov

ing

side

-

wal

k in

stea

d of

wal

kin

g al

ongs

ide

it?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

3


50

Forc

es a

nd th

e La

ws

ofM

otio

n

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

4

Chap

ter 4

21

1.A

cra

te r

ests

on

the

hori

zont

al b

ed o

fa p

icku

p tr

uck.

For

each

sit

uati

on d

e-

scri

bed

belo

w,i

ndic

ate

the

mot

ion

ofth

e cr

ate

rela

tive

to th

e gr

ound

,the

mot

ion

ofth

e cr

ate

rela

tive

to th

e tr

uck,

and

whe

ther

the

crat

e w

ill h

it th

e

fron

t wal

l oft

he tr

uck

bed,

the

back

wal

l,or

nei

ther

.Dis

rega

rd fr

icti

on.

a.St

arti

ng

at r

est,

the

tru

ck a

ccel

erat

es to

th

e ri

ght.

b.T

he

crat

e is

at

rest

rel

ativ

e to

th

e tr

uck

wh

ile t

he

tru

ck m

oves

to t

he

righ

t w

ith

a c

onst

ant

velo

city

.

c.T

he

tru

ck in

item

b s

low

s do

wn

.

2.A

bal

l wit

h a

mas

s of

mis

th

row

n t

hro

ugh

th

e ai

r,as

sh

own

in t

he

figu

re.


a.W

hat

is t

he

grav

itat

ion

al fo

rce

exer

ted

on t

he

ball

by E

arth

?

b.W

hat

is t

he

forc

e ex

erte

d on

Ear

th b

y th

e ba

ll?

c.If

the

surr

oun

din

g ai

r ex

erts

a fo

rce

on t

he

ball

that

res

ists

its

mot

ion

,

is t

he

tota

lfor

ce o

n t

he

ball

the

sam

e as

th

e fo

rce

calc

ula

ted

in p

art

a?

d.If

the

surr

oun

din

g ai

r ex

erts

a fo

rce

on t

he

ball

that

res

ists

its

mot

ion

,

is t

he

grav

itat

iona

l for

ce o

n t

he

ball

the

sam

e as

th

e fo

rce

calc

ula

ted

in

part

a?


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s223.

Two

bloc

ks o

fm

asse

s m

1an

dm

2,r

espe

ctiv

ely,

are

plac

ed in

con

tact

wit

h

each

oth

er o

n a

sm

ooth

,hor

izon

tal s

urf

ace.

A c

onst

ant

hor

izon

tal f

orce

F

to t

he

righ

t is

app

lied

to m

1.A

nsw

er t

he

follo

win

g qu

esti

ons

in te

rms

of

F,m

1,an

dm

2.

a.W

hat

is t

he

acce

lera

tion

of

the

two

bloc

ks?

b.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

2?

c.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

1?

d.W

hat

is t

he

mag

nit

ude

of

the

con

tact

forc

e be

twee

n t

he

two

bloc

ks?

4.A

ssu

me

you

hav

e th

e sa

me

situ

atio

n a

s de

scri

bed

in it

em 3

,on

ly t

his

tim

e th

ere

is a

fric

tion

al fo

rce,

F k,b

etw

een

th

e bl

ocks

an

d th

e su

rfac

e.

An

swer

th

e fo

llow

ing

ques

tion

s in

term

s of

F,F k

,m1,

and

m2.

a.W

hat

is t

he

acce

lera

tion

of

the

two

bloc

ks?

b.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

2?

c.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

1?

d.W

hat

is t

he

mag

nit

ude

of

the

con

tact

forc

e be

twee

n t

he

two

bloc

ks?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

4


51

Wor

k an

d En

ergy

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

5

Chap

ter 5

27

1.A

bal

l has

a m

ass

of3

kg.W

hat

is t

he

wor

k do

ne

on t

his

bal

l by

the

grav

i-

tati

onal

forc

e ex

erte

d by

Ear

th if

the

ball

mov

es 2

m a

lon

g ea

ch o

fth

e

follo

win

g di

rect

ion

s?

a.do

wnw

ard

(alo

ng

the

forc

e)

b.u

pwar

d (o

ppos

ite

the

forc

e)

2.A

sto

ne

wit

h a

mas

s of

mis

th

row

n o

ffa

build

ing.

As

the

ston

e pa

sses

poin

tA

,it

has

a s

peed

ofv A

at a

n a

ngl

e of

qto

th

e h

oriz

onta

l.T

he

ston

e

then

tra

vels

a v

erti

cal d

ista

nce

hto

poi

nt

B,w

her

e it

has

a s

peed

vB

.


a.W

hat

is t

he

wor

k do

ne

on t

he

ston

e by

th

e gr

avit

atio

nal

forc

e du

e to

Ear

th w

hile

th

e st

one

mov

es fr

om A

to B

?

b.W

hat

is t

he

chan

ge in

th

e ki

net

ic e

ner

gy o

fth

e st

one

as it

mov

es fr

om

Ato

B?

c.W

hat

is t

he

spee

d v B

ofth

e st

one

in te

rms

ofv A

,g,a

nd

h?

d.D

oes

the

chan

ge in

th

e st

one’

s sp

eed

betw

een

Aan

dB

depe

nd

on t

he

mas

s of

the

ston

e?

e.D

oes

the

chan

ge in

th

e st

one’

s sp

eed

betw

een

Aan

dB

depe

nd

on t

he

angl

e q

?

h

A

Bθ


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s283.

An

em

pty

coff

ee m

ug

wit

h a

mas

s of

0.40

kg

gets

kn

ocke

d of

fa

tabl

etop

0.75

m a

bove

th

e fl

oor

onto

th

e se

at o

fa

chai

r 0.

45 m

abo

ve t

he

floo

r.

Ass

um

e th

at t

he

grav

itat

ion

al p

oten

tial

en

ergy

,PE

g,is

mea

sure

d u

sin

g

the

floo

r as

th

e ze

ro e

ner

gy le

vel.

a.W

hat

is t

he

init

ial g

ravi

tati

onal

pot

enti

al e

ner

gy a

ssoc

iate

d w

ith

th

e

mu

g’s

posi

tion

on

th

e ta

ble?

b.W

hat

is t

he

fin

al g

ravi

tati

onal

pot

enti

al e

ner

gy a

ssoc

iate

d w

ith

th

e

mu

g’s

posi

tion

on

th

e ch

air

seat

?

c.W

hat

was

th

e w

ork

don

e by

th

e gr

avit

atio

nal

forc

e as

it fe

ll fr

om t

he

tabl

e to

th

e ch

air?

d.Su

ppos

e th

at z

ero

leve

l for

th

e en

ergy

was

tak

en to

be

the

ceili

ng

of

the

room

rat

her

th

an t

he

floo

r .W

ould

th

e an

swer

s to

item

s a

to c

be

the

sam

e or

dif

fere

nt?

4.A

car

ton

of

shoe

s w

ith

a m

ass

ofm

slid

es w

ith

an

init

ial s

peed

ofv i

m/s

dow

n a

ram

p in

clin

ed a

t an

an

gle

of23

°to

the

hor

izon

tal.

Th

e ca

rton

’s

init

ial h

eigh

t is

hi,

and

its

fin

al h

eigh

t is

hf,

and

it t

rave

ls a

dis

tan

ce o

f

ddo

wn

th

e ra

mp.

Th

ere

is a

fric

tion

al fo

rce,

F k,b

etw

een

th

e ra

mp

and

the

cart

on.

a.W

hat i

s th

e in

itia

l mec

han

ical

en

ergy

,ME

i,of

the

cart

on?

(Hin

t:

App

ly th

e la

w o

fco

nse

rvat

ion

of

ener

gy.)

b.Ifm

is t

he

coef

fici

ent

offr

icti

on b

etw

een

th

e ra

mp

and

the

cart

on,

wh

at is

Fk?

c.E

xpre

ss t

he

fin

al s

peed

,vf,

ofth

e ca

rton

in te

rms

ofv i

,g,d

,an

dm.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

5


52

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s32

Mom

entu

m a

nd C

ollis

ions

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

6

1.A

pit

cher

th

row

s a

soft

ball

tow

ard

hom

e pl

ate.

Th

e ba

ll m

ay b

e h

it,s

end-

ing

it b

ack

tow

ard

the

pitc

her

,or

it m

ay b

e ca

ugh

t,br

ingi

ng

it to

a s

top

in

the

catc

her

’s m

itt.

a.C

ompa

re t

he

chan

ge in

mom

entu

m o

fth

e ba

ll in

th

ese

two

case

s.

b.D

iscu

ss t

he

mag

nit

ude

of

the

impu

lse

on t

he

ball

in t

hes

e tw

o ca

ses.

c.In

th

e sp

ace

belo

w,d

raw

a v

ecto

r di

agra

m fo

r ea

ch c

ase,

show

ing

the

init

ial m

omen

tum

of

the

ball,

the

impu

lse

exer

ted

on t

he

ball,

and

the

resu

ltin

g fi

nal

mom

entu

m o

fth

e ba

ll.

2.a.

Usi

ng

New

ton’

s th

ird

law

,exp

lain

why

th

e im

puls

e on

on

e ob

ject

in a

colli

sion

is e

qual

in m

agn

itu

de b

ut

oppo

site

in d

irec

tion

to t

he

im-

puls

e on

th

e se

con

d ob

ject

.

b.E

xten

d yo

ur

disc

uss

ion

of

impu

lse

and

New

ton’

s th

ird

law

to t

he

case

ofa

bow

ling

ball

stri

kin

g a

set

of10

bow

ling

pin

s.



Chap

ter 6

33

3.St

arti

ng

wit

h t

he

con

serv

atio

n o

fto

tal m

omen

tum

,pf

=p

i,sh

ow

that

th

e fi

nal

vel

ocit

y fo

r tw

o ob

ject

s in

an

inel

asti

c co

llisi

on is

v f=

m1m +1 m

2

v 1,i

+ m

1m +2 m2

v 2

,i.

4.Tw

o m

ovin

g bi

lliar

d ba

lls,e

ach

wit

h a

mas

s of

M,u

nde

rgo

an e

last

ic

colli

sion

.Im

med

iate

ly b

efor

e th

e co

llisi

on,b

all A

is m

ovin

g ea

st a

t

2 m

/s a

nd

ball

B is

mov

ing

east

at

4 m

/s.

a.In

term

s of

M,w

hat i

s th

e to

tal m

omen

tum

(m

agn

itu

de a

nd

dire

c-

tion

) im

med

iate

ly b

efor

e th

e co

llisi

on?

b.T

he

fin

al m

omen

tum

,M(v

A,f

+v B

,f),

mu

st e

qual

th

e in

itia

l mom

en-

tum

.If

the

fin

al v

eloc

ity

ofba

ll A

incr

ease

s to

4 m

/s e

ast

beca

use

of

the

colli

sion

,wh

at is

th

e fi

nal

mom

entu

m o

fba

ll B

?

c.Fo

r ea

ch b

all,

com

pare

th

e fi

nal

mom

entu

m o

fth

e ba

ll to

th

e in

itia

l

mom

entu

m o

fth

e ot

her

ball.

Th

ese

resu

lts

are

typi

cal o

fh

ead-

on

elas

tic

colli

sion

s.W

hat

gen

eral

izat

ion

abo

ut

hea

d-on

ela

stic

col

lisio

ns

can

you

mak

e?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

6



Rota

tiona

l Mot

ion

and

the

Law

ofG

ravi

ty

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

7

Chap

ter 7

37

1.C

ompl

ete

the

follo

win

g ta

ble.

s(m

)r(

m)

q(r

ad)

t(s)

w(r

ad/s

)v t

(m/s

)a c

(m/s

2 )

a.4.

51.

50.

50

b.5.8

5.805.0

c.85

02.02.3

d.

1250

2.0

17

e.68

0570573

2.D

escr

ibe

the

forc

e th

at m

ain

tain

s ci

rcu

lar

mot

ion

in t

he

follo

win

g ca

ses.

a.A

car

exi

ts a

free

way

an

d m

oves

aro

un

d a

circ

ula

r ra

mp

to r

each

th

e

stre

et b

elow

.

b.T

he

moo

n o

rbit

s E

arth

.

c.D

uri

ng

gym

cla

ss,a

stu

den

t h

its

a te

ther

bal

l on

a s

trin

g.

3.D

eter

min

e th

e ch

ange

in g

ravi

tati

onal

forc

e u

nde

r th

e fo

llow

ing

chan

ges.

a.on

e of

the

mas

ses

is d

oubl

ed

b.bo

th m

asse

s ar

e do

ubl

ed

c.th

e di

stan

ce b

etw

een

mas

ses

is d

oubl

ed

d.th

e di

stan

ce b

etw

een

mas

ses

is h

alve

d

e.th

e di

stan

ce b

etw

een

mas

ses

is t

ripl

ed


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s384.

Som

e pl

ans

for

a fu

ture

spa

ce s

tati

on m

ake

use

ofro

tati

onal

forc

e to

sim

u-

late

gra

vity

.In

ord

er to

be

effe

ctiv

e,th

e ce

ntr

ipet

al a

ccel

erat

ion

at t

he o

uter

rim

oft

he s

tati

on s

houl

d eq

ual a

bout

1 g

,or

9.81

m/s

2 .How

ever

,hum

ans

can

wit

hsta

nd

a di

ffer

ence

ofo

nly

1/1

00 g

bet

wee

n th

eir

head

an

d fe

et

befo

re th

ey b

ecom

e di

sori

ente

d.A

ssum

e th

e av

erag

e hu

man

hei

ght i

s

2.0

m,a

nd

calc

ulat

e th

e m

inim

um r

adiu

s fo

r a

safe

,eff

ecti

ve s

tati

on.

(Hin

t:T

he r

atio

oft

he c

entr

ipet

al a

ccel

erat

ion

ofa

stro

nau

t’s fe

et to

the

cen

trip

etal

acc

eler

atio

n o

fthe

ast

ron

aut’s

hea

d m

ust b

e at

leas

t 99/

100.

)

5.A

s an

ele

vato

r be

gin

s to

des

cen

d,yo

u fe

el m

omen

tari

ly li

ghte

r.A

s th

e

elev

ator

sto

ps,y

ou fe

el m

omen

tari

ly h

eavi

er.S

ketc

h t

he

situ

atio

n,a

nd

expl

ain

th

e se

nsa

tion

s u

sin

g th

e fo

rces

in y

our

sket

ch.

6.Tw

o ca

rs s

tart

on

opp

osit

e si

des

ofa

circ

ula

r tr

ack.

On

e ca

r h

as a

spe

ed

of0.

015

rad/

s;th

e ot

her

car

has

a s

peed

of

0.01

2 ra

d/s.

Ifth

e ca

rs s

tart

pra

dian

s ap

art,

calc

ula

te t

he

tim

e it

tak

es fo

r th

e fa

ster

car

to c

atch

up

wit

h t

he

slow

er c

ar.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

7


54

Rota

tiona

l Equ

ilibr

ium

and

Dyn

amic

s

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

8

Chap

ter 8

43

1.a.

On

som

e do

ors,

the

door

knob

is in

th

e ce

nte

r of

the

door

.Wh

at

wou

ld a

phy

sici

st s

ay a

bou

t th

e pr

acti

calit

y of

this

arr

ange

men

t? W

hy

wou

ld p

hysi

cist

s de

sign

doo

rs w

ith

kn

obs

fart

her

from

th

e h

inge

?

b.H

ow m

uch

mor

e fo

rce

wou

ld b

e re

quir

ed to

ope

n t

he

door

from

th

e

cen

ter

rath

er t

han

from

th

e ed

ge?

2.Fi

gure

ska

ters

com

mon

ly c

han

ge t

he

shap

e of

thei

r bo

dy in

ord

er to

ach

ieve

spi

ns

on t

he

ice.

Exp

lain

th

e ef

fect

s on

eac

h o

fth

e fo

llow

ing

quan

titi

es w

hen

a fi

gure

ska

ter

pulls

in h

is o

r h

er a

rms.

a.m

omen

t of

iner

tia

b.an

gula

r m

omen

tum

c.an

gula

r sp

eed

3.Fo

r th

e fo

llow

ing

item

s,as

sum

e th

e ob

ject

s sh

own

are

in r

otat

ion

al e

quili

briu

m.

a.W

hat

is t

he

mas

s of

the

sph

ere

to t

he

righ

t?

b.W

hat

is t

he

mas

s of

the

port

ion

of

the

met

er-

stic

k to

th

e le

ft o

fth

e pi

vot?

(H

int:

20%

of

the

mas

s of

the

met

erst

ick

is o

n t

he

left

.How

mu

ch m

ust

be

on t

he

righ

t?)


1.0

kg

0 cm

50 c

m25

cm

75 c

m10

0 cm

1.0

kg0 cm

50 c

m25

cm

75 c

m10

0 cm


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s444.

A fo

rce

of25

N is

app

lied

to t

he

end

ofa

un

ifor

m r

od t

hat

is 0

.50

m lo

ng

and

has

a m

ass

of0.

75 k

g.

a.Fi

nd

the

torq

ue,

mom

ent o

fin

erti

a,an

d an

gula

r ac

cele

rati

on if

the

rod

is a

llow

ed to

piv

ot a

rou

nd

its

cen

ter

ofm

ass.

b.Fi

nd

the

torq

ue,

mom

ent o

fin

erti

a,an

d an

gula

r ac

cele

rati

on if

the

rod

is a

llow

ed to

piv

ot a

rou

nd

the

end,

away

from

the

appl

ied

forc

e.

5.A

sat

ellit

e in

orb

it a

rou

nd

Ear

th is

init

ially

at

a co

nst

ant

angu

lar

spee

d of

7.27

×10

–5ra

d/s.

Th

e m

ass

ofth

e sa

telli

te is

45

kg,a

nd

it h

as a

n o

rbit

al

radi

us

of4.

23 ×

107

m.

a.Fi

nd

the

mom

ent

ofin

erti

a of

the

sate

llite

in o

rbit

aro

un

d E

arth

.

b.Fi

nd

the

angu

lar

mom

entu

m o

fth

e sa

telli

te.

c.Fi

nd

the

rota

tion

al k

inet

ic e

ner

gy o

fth

e sa

telli

te a

rou

nd

Ear

th.

d.Fi

nd

the

tan

gen

tial

spe

ed o

fth

e sa

telli

te.

e.Fi

nd

the

tran

slat

ion

al k

inet

ic e

ner

gy o

fth

e sa

telli

te.

6.A

ser

ies

oftw

o si

mpl

e m

ach

ines

is u

sed

to li

ft a

133

00 N

car

to a

hei

ght

of3.

0 m

.Bot

h m

ach

ines

hav

e an

eff

icie

ncy

of

0.90

(90

per

cen

t).M

ach

ine

A m

oves

th

e ca

r,an

d th

e ou

tpu

t of

mac

hin

e B

is t

he

inpu

t to

mac

hin

e A

.

a.H

ow m

uch

wor

k is

don

e on

th

e ca

r?

b.H

ow m

uch

wor

k m

ust

be

don

e on

mac

hin

e A

in o

rder

to a

chie

ve t

he

amou

nt

ofw

ork

don

e on

th

e ca

r?

c.H

ow m

uch

wor

k m

ust

be

don

e on

mac

hin

e B

in o

rder

to a

chie

ve t

he

amou

nt

ofw

ork

from

mac

hin

e A

?

d.W

hat

is t

he

over

all e

ffic

ien

cy o

fth

is p

roce

ss?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

8


55

Vibr

atio

ns a

nd W

aves

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Chap

ter 1

267

1.A

pen

dulu

m w

ith

a m

ass

of0.

100

kg w

as r

elea

sed.

Th

e st

rin

g m

ade

a

7.0°

angl

e w

ith

th

e ve

rtic

al.T

he

bob

ofth

e pe

ndu

lum

ret

urn

s to

its

low

est

poin

t ev

ery

0.10

s.

a.W

hat

is it

s pe

riod

? W

hat

is it

s fr

equ

ency

?

b.T

he

pen

dulu

m is

rep

lace

d by

on

e w

ith

a m

ass

of0.

300

kg a

nd

set

to

swin

g w

ith

a 1

5°an

gle.

Do

the

follo

win

g qu

anti

ties

incr

ease

,dec

reas

e,

or r

emai

n t

he

sam

e?

per

iod

freq

uen

cy

tota

l en

ergy

spee

d at

th

e lo

wes

t po

int

2.A

nar

row

,fla

t st

eel r

od is

an

chor

ed a

t it

s lo

wer

en

d,w

ith

a 0

.500

kg

ball

wel

ded

to t

he

top

end.

A fo

rce

of6.

00 N

is r

equ

ired

to h

old

the

ball

10.0

cm

aw

ay fr

om it

s ce

ntr

al p

osit

ion

.

Ifth

is a

rran

gem

ent

is m

odel

ed a

s an

osc

illat

ing

hor

izon

tal m

ass-

spri

ng

syst

em,v

ibra

tin

g w

ith

a s

impl

e h

arm

onic

mot

ion

,fin

d

a.th

e fo

rce

con

stan

t,k,

ofth

e sp

rin

g.

b.th

e pe

riod

an

d fr

equ

ency

of

the

osci

llati

ons.

3.Fi

nd

the

acce

lera

tion

du

e to

gra

vity

at

a pl

ace

wh

ere

a si

mpl

e pe

ndu

lum

0.15

0 m

lon

g co

mpl

etes

1.0

0 ×

102

osci

llati

ons

in 3

.00

×10

2se

con

ds.

Cou

ld t

his

pla

ce b

e on

Ear

th?

Ch

apte

r


AC

B

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s684.

Con

side

r th

e fi

rst

two

cycl

es o

fa

pen

dulu

m s

win

gin

g fr

om p

osit

ion

A

wit

h a

per

iod

of2.

00 s

.

a.A

t w

hic

h t

imes

is t

he

bob

fou

nd

at p

osit

ion

s A

,B,a

nd

Cdu

rin

g th

e

firs

t tw

o cy

cles

?

b.A

t w

hic

h t

imes

an

d lo

cati

ons

is g

ravi

tati

onal

pot

enti

al e

ner

gy a

t a

max

imu

m?

At

wh

ich

tim

es is

kin

etic

en

ergy

at

a m

axim

um

?

c.A

t w

hic

h t

imes

an

d lo

cati

ons

is t

he

velo

city

at

a m

axim

um

? th

e

rest

orin

g fo

rce?

th

e ac

cele

rati

on?

5.T

he fr

eque

ncy

ofa

pre

ssur

e w

ave

is 1

.00

×10

2H

z.It

s w

avel

engt

h is

3.0

0 m

.

Fin

d th

e sp

eed

ofw

ave

prop

agat

ion

.

6.A

pre

ssu

re w

ave

of0.

50 m

wav

elen

gth

pro

paga

tes

thro

ugh

a 3

.00

m lo

ng

coil

spri

ng

at a

spe

ed o

f2.

00 m

/s.H

ow lo

ng

does

it t

ake

for

the

wav

e to

trav

el fr

om o

ne

end

ofth

e co

il to

th

e ot

her

? H

ow m

any

wav

elen

gth

s fi

t

in t

he

coil?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

12


AC

B


56

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s72

Soun

d

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

13

1.T

he s

peed

ofs

ound

incr

ease

s w

ith

tem

pera

ture

.It i

s 33

1 m

/s in

air

at 0

°Can

d 34

3 m

/s in

air

at 2

0°C

.A g

lass

pip

e vi

brat

es w

ith

a fr

eque

ncy

of15

1 H

z.

a.W

hat

is t

he

wav

elen

gth

of

the

sou

nd

prod

uce

d by

th

e co

lum

n o

fai

r

in t

he

pipe

on

a c

old

day

(0°C

) an

d on

a w

arm

er d

ay (

20°C

)?

b.H

ow d

oes

air

tem

pera

ture

aff

ect t

he w

avel

engt

h of

the

soun

d pr

oduc

ed

by th

e pi

pe?

2.T

he

driv

er o

fan

am

bula

nce

tu

rns

on it

s si

ren

as

the

ambu

lan

ce h

eads

east

at

30 m

ph.A

pol

ice

car

is fo

llow

ing

the

ambu

lan

ce a

t 30

mph

.A

tru

ck b

ehin

d th

e po

lice

car

is m

ovin

g at

20

mph

.A v

an is

tra

velin

g w

est

in t

he

oppo

site

lan

e at

20

mph

.A s

mal

l car

is s

topp

ed a

t th

e si

de o

fth

e

road

.Th

e ve

hic

les

are

posi

tion

ed a

s sh

own

.

a.O

n t

he

diag

ram

,ske

tch

an

d la

bel a

rrow

s to

indi

cate

th

e ve

loci

ty o

f

each

veh

icle

.


b.R

ank

the

sou

nds

per

ceiv

ed b

y th

e pa

ssen

gers

in e

ach

of

the

veh

icle

s

in o

rder

of

decr

easi

ng

freq

uen

cy.

Tru

ckPo

lice

car

Am

bula

nce

Van

Smal

lca

r

Chap

ter 1

373

3.A

330

Hz

tun

ing

fork

is v

ibra

tin

g af

ter

bein

g st

ruck

.It i

s pl

aced

on

a ta

ble

nea

r bu

t not

dir

ectl

y to

uchi

ng

othe

r ob

ject

s,in

clud

ing

othe

r tu

nin

g fo

rks.

Even

tual

ly o

ne

glas

s an

d on

e ot

her

tun

ing

fork

sta

rt v

ibra

tin

g.E

xpla

in w

hy

this

hap

pen

s.

4.T

he

firs

t h

arm

onic

in a

pip

e cl

osed

at

one

end

is 4

87 H

z.

a.Fi

nd

the

nex

t tw

o h

arm

onic

freq

uen

cies

th

at w

ill o

ccu

r in

th

is p

ipe.

b.W

hat

are

th

e co

rres

pon

din

g w

avel

engt

hs

ofth

e fi

rst

thre

e h

arm

onic

s?

(Hin

t:as

sum

e th

e sp

eed

ofso

un

d is

345

m/s

.)

c.W

hat

is t

he

len

gth

of

this

pip

e?

d.R

epea

t th

is e

xerc

ise

for

a pi

pe o

pen

at

both

en

ds.

5.A

pia

no

tun

er u

ses

a 44

0 H

z tu

nin

g fo

rk to

tu

ne

a st

rin

g th

at is

cu

rren

tly

vibr

atin

g at

445

Hz.

a.H

ow m

any

beat

s pe

r se

con

d do

es h

e h

ear?

b.W

hat

oth

er fr

equ

ency

cou

ld p

rodu

ce t

he

sam

e so

un

d ef

fect

? E

xpla

in w

hy.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

13



57

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s78

Ligh

t and

Re

ectio

n

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

14

1.P

roxi

ma

Cen

tau

ri,t

he

nea

rest

sta

r in

ou

r ga

laxy

,is

4.30

ligh

t-ye

ars

away

.

Wh

at is

its

dist

ance

in m

eter

s?

2.R

adio

sig

nal

s em

itte

d fr

om a

nd

rece

ived

by

an a

irpl

ane

hav

e a

freq

uen

cy

of3.

00 ×

1012

Hz

and

trav

el a

t th

e sp

eed

oflig

ht.

a.H

ow lo

ng

is t

he

dela

y in

eac

h m

essa

ge g

oin

g fr

om t

he

con

trol

tow

er

to a

jet

flyi

ng

at 1

.00

×10

4m

of

alti

tude

?

b.W

hat

is t

he

wav

elen

gth

of

thes

e si

gnal

s?

3.A

lase

r be

am is

sen

t to

the

moo

n fr

om E

arth

.The

ref

lect

ed b

eam

is r

ecei

ved

on E

arth

aft

er 2

.56

seco

nds

.Wha

t is

the

dist

ance

from

Ear

th to

the

moo

n?

4.T

he

back

grou

nd

radi

atio

n in

th

e u

niv

erse

(be

lieve

d to

com

e fr

om t

he

Big

Ban

g) in

clu

des

mic

row

aves

wit

h w

avel

engt

hs

of0.

100

cm.W

hat

is t

he

freq

uen

cy o

fth

is r

adia

tion

?

5.Li

st fi

ve o

bjec

ts t

hat

ref

lect

ligh

t di

ffu

sely

.Lis

t th

ree

obje

cts

that

ref

lect

ligh

t sp

ecu

larl

y fo

r th

e m

ost

part

.

Dif

fuse

ref

lect

ion

Spec

ula

r re

flec

tion


Chap

ter 1

479

6.A

mir

ror

door

is lo

cate

d n

ext

to a

larg

e w

all m

irro

r.

Th

e do

or is

clo

sed

to c

reat

e a

90°a

ngl

e w

ith

th

e

wal

l.Yo

u s

tan

d 2.

00 m

from

th

e do

or a

nd

1.00

m

from

th

e w

all.

a.O

n t

he

diag

ram

at

righ

t,sk

etch

a to

p-vi

ew d

ia-

gram

of

the

situ

atio

n a

t sc

ale.

Labe

l th

e ob

ject

(you

rsel

f) A

.

b.Lo

cate

you

r fi

rst

imag

e in

th

e m

irro

r on

th

e

door

.Lab

el it

B.L

ocat

e B

’s im

age

in t

he

mir

ror

on t

he

wal

l.La

bel i

t C

.

c.Lo

cate

you

r fi

rst

imag

e in

th

e m

irro

r on

th

e w

all a

nd

its

imag

e in

th

e

mir

ror

on t

he

door

.Lab

el t

hem

Dan

dE

.

d.W

her

e w

ill t

he

nex

t im

ages

of

the

imag

es b

e lo

cate

d?

7.A

n o

bjec

t lo

cate

d 36

.0 c

m fr

om a

con

cave

mir

ror

prod

uce

s a

real

imag

e

loca

ted

12.0

cm

from

th

e m

irro

r.

a.Fi

nd

the

foca

l len

gth

of

this

mir

ror

b.Fi

nd

the

loca

tion

,typ

e,an

d si

ze o

fth

e im

age

form

ed b

y a

6.00

cm

tal

l

obje

ct lo

cate

d 30

.0 c

m,2

4.0

cm,1

8.0

cm,1

2.0

cm,a

nd

6.00

cm

in

fron

t of

the

mir

ror.

8.T

he

con

cave

mir

ror

in t

he

prob

lem

abo

ve is

rep

lace

d by

a c

onve

x on

e

wit

h t

he

sam

e cu

rvat

ure

.Fin

d th

e lo

cati

on o

fth

e im

ages

pro

duce

d w

hen

the

obje

ct is

loca

ted

30.0

cm

,24.

0 cm

,18.

0 cm

,12.

0 cm

,an

d 6.

00 c

m in

fron

t of

the

mir

ror.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

14


wal

l

door


58

Refractio

n

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Chap

ter 1

583

1.Tw

o pa

ralle

l ray

s en

ter

an a

quar

ium

as

show

n.

Ray

1 fo

rms

a 70

.0°a

ngl

e w

ith

th

e n

orm

al to

the

surf

ace.

Ray

2 fo

rms

a 20

.0°a

ngl

e w

ith

the

nor

mal

to t

he

wal

l.(H

int:

the

inde

x of

refr

acti

on fo

r w

ater

is 1

.33.

)

a.C

alcu

late

th

e an

gle

ofre

frac

tion

of

each

ray

.

b.Tr

ace

the

path

of

each

ligh

t ra

y in

side

th

e w

ater

.

c.A

re t

he

refr

acte

d ra

ys in

side

th

e w

ater

sti

ll pa

ralle

l? W

ill t

hey

inte

rsec

t

in t

he

wat

er?

2.A

larg

e be

aker

con

tain

s la

yers

of

wat

er o

fin

crea

sin

g sa

linit

y,se

para

ted

by

a th

in p

last

ic p

late

.Th

e lo

wes

t la

yer

has

th

e h

igh

est

salin

ity

and

refr

acti

ve

inde

x,as

sh

own

in t

he

diag

ram

.A r

ay o

flig

ht

stri

kes

the

surf

ace

offr

esh

wat

er a

t th

e to

p,at

a 7

0.0°

angl

e fr

om t

he

nor

mal

.

Ch

apte

r


1

2

70.0

°

20.0

°

air

fres

h w

ater

salt

wat

er

hig

h s

alin

ity

n =

1.0

0

n =

1.3

3

n =

1.4

5

n =

1.5

7

70.0

°

a.Fi

nd

the

angl

es o

fre

frac

tion

an

d th

e an

gles

of

inci

den

ce a

t ea

ch

bou

nda

ry.

b.T

here

is a

flat

mir

ror

at th

e bo

ttom

oft

he c

onta

iner

.Tra

ce th

e pa

th o

f

one

light

ray

com

ing

from

the

air

to th

e bo

ttom

oft

he b

eake

r an

d ba

ck.

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s843.

An

obj

ect

loca

ted

36.0

cm

from

a t

hin

con

verg

ing

len

s h

as a

rea

l im

age

loca

ted

12.0

cm

from

th

e le

ns.

a.Fi

nd

the

foca

l poi

nt

ofth

is le

ns.

b.Fi

nd

the

loca

tion

,typ

e,an

d si

ze o

fth

e im

age

form

ed b

y a

6.00

cm

tall

obje

ct lo

cate

d 30

.0 c

m,2

4.0

cm,1

8.0

cm,1

2.0

cm,a

nd

6.00

cm

in fr

ont

ofth

e le

ns.

4.T

he

conv

ergi

ng

len

s in

item

3 is

rep

lace

d by

a d

iver

gin

g le

ns.

Now

th

e

imag

e of

the

firs

t ob

ject

is lo

cate

d 12

.0 c

m in

fron

t of

the

len

s.Fi

nd

the

foca

l dis

tan

ce o

fth

e di

verg

ing

len

s an

d th

e lo

cati

on o

fth

e im

ages

pro

-

duce

d w

hen

th

e ob

ject

is p

lace

d at

th

e di

stan

ces

desc

ribe

d in

item

3b.

5.A

bu

g pl

aced

1.0

0 cm

un

der

a m

agn

ifyi

ng

glas

s ap

pear

s ex

actl

y si

x

tim

es la

rger

.

a.W

her

e is

th

e bu

g’s

imag

e lo

cate

d?

b.W

hat

is t

he

foca

l poi

nt

ofth

e le

ns

in t

he

mag

nif

yin

g gl

ass?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

15



59

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s88

Inte

rfer

ence

and

Di

ract

ion

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

16

1.T

he

seco

nd-

orde

r br

igh

t fr

inge

s of

inte

rfer

ence

are

obs

erve

d at

an

8.5

3°an

gle

in a

dou

ble-

slit

exp

erim

ent

wit

h li

ght

of5.

00 ×

102

nm

wav

elen

gth

.

a.D

eter

min

e th

e sl

its’

sepa

rati

on.

b.Fi

nd

the

angl

e of

the

ten

th-o

rder

bri

ght

frin

ge.

c.In

th

is e

xper

imen

t,th

e sc

reen

is 2

.00

m w

ide.

Its

dist

ance

from

th

e

sou

rce

is 1

.00

m.C

ould

th

e te

nth

-ord

er fr

inge

be

obse

rved

? W

hy o

r

why

not

?

2.D

iffr

acti

on o

fw

hit

e lig

ht

wit

h a

sin

gle

slit

pro

duce

s br

igh

t lin

es o

f

diff

eren

t co

lors

.

a.W

hic

h w

avel

engt

hs

are

mor

e di

ffra

cted

by

the

sam

e sl

it s

ize?

b.In

the

spac

e be

low

,ske

tch

a di

agra

m s

how

ing

the

loca

tion

ofr

ed,g

reen

and

blue

lin

es o

fthe

firs

t an

d se

con

d or

der.

Des

crib

e th

e se

quen

ce in

whi

ch th

e co

lors

app

ear,

begi

nn

ing

wit

h th

e co

lor

clos

est t

o th

e ce

nte

r.

c.W

hat

is t

he

colo

r of

the

cen

tral

imag

e?


Chap

ter 1

689

3.Yo

u h

ave

thre

e di

ffra

ctio

n g

rati

ngs

.Gra

tin

g A

has

2.0

×10

5lin

es p

er m

eter

.

Gra

tin

g B

has

9.0

×10

6lin

es p

er m

eter

.Gra

tin

g C

has

3.0

×10

7lin

es

per

met

er.

a.W

hat

is t

he

slit

dis

tan

ce o

fea

ch g

rati

ng?

b.W

hic

h g

rati

ngs

can

dif

frac

t th

e fo

llow

ing:

•vi

sibl

e lig

ht

of50

0 n

m w

avel

engt

h

•X

ray

s of

5.00

nm

wav

elen

gth

•in

frar

ed li

ght

of50

00 n

m w

avel

engt

h

4.Yo

u d

rop

pebb

les

into

th

e w

ater

on

a r

ocky

bea

ch.W

hen

th

e w

aves

you

mad

e re

ach

th

e ro

cks,

new

wav

es a

ppea

r to

sta

rt in

th

e sp

aces

bet

wee

n

the

rock

s.

a.A

re t

hes

e w

aves

coh

eren

t?

b.H

ow is

th

is li

ke a

dou

ble

slit

illu

min

ated

by

a si

ngl

e lig

ht

sou

rce?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

16



60

Elec

tric

For

ces

and

Fiel

ds

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Chap

ter 1

793

Use

kC

=8.

99×

109

N•m

2 /C2 .

1.Tw

o sp

her

es,A

and

B,a

re p

lace

d 0.

60 m

apa

rt,a

s sh

own

.Sph

ere

Ah

as

+3.0

0mC

exc

ess

char

ge.S

pher

e B

has

+5.

00 m

C e

xces

s ch

arge

.

Ch

apte

r


AB

a.H

ow m

any

elec

tron

s ar

e m

issi

ng

on s

pher

e A

? on

sph

ere

B?

b.H

ow d

o th

e fo

rces

of

Bon

Aan

dA

onB

com

pare

? D

oes

the

grea

ter

char

ge e

xert

a g

reat

er fo

rce?

2.A

th

ird

sph

eric

al c

har

ge,C

,of

+2.

00 m

C,i

s pl

aced

on

th

e lin

e co

nn

ecti

ng

sph

eres

Aan

dB

.Fin

d th

e re

sult

ant

forc

e ex

erte

d by

Aan

dB

onC

wh

en

Cis

pla

ced

in t

he

follo

win

g lo

cati

ons.

a.0.

20 m

to t

he

left

of

A

b.0.

20 m

to t

he

righ

t of

Abe

twee

nA

and

B

c.ex

actl

y in

th

e m

iddl

e be

twee

n A

and

B

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s943.

Alp

ha

part

icle

s ar

e m

ade

oftw

o pr

oton

s an

d tw

o n

eutr

ons.

mp

=1.

673

×10

–27

kg;m

n=

1.67

5×

10–2

7kg

;qe

=1.

60×

10–1

9C

a.Fi

nd

the

elec

tric

forc

e ac

tin

g on

an

alp

ha

part

icle

in a

hor

izon

tal

elec

tric

fiel

d of

6.00

×10

2N

/C.

b.W

hat

is t

he

acce

lera

tion

of

this

alp

ha

part

icle

?

c.H

ow d

oes

this

acc

eler

atio

n c

ompa

re w

ith

gra

vity

? D

escr

ibe

the

part

i-

cle’

s tr

ajec

tory

.Will

it b

e cl

ose

to h

oriz

onta

l? to

ver

tica

l fre

e fa

ll?

4.A

2.0

0 mC

poi

nt

char

ge o

fm

ass

5.00

g is

su

spen

ded

on a

str

ing

and

plac

ed in

a h

oriz

onta

l ele

ctri

c fi

eld.

Th

e m

ass

is in

equ

ilibr

ium

wh

en t

he

stri

ng

form

s a

17.3

°an

gle

wit

h t

he

vert

ical

.

a.In

the

spac

e be

low

,ske

tch

a fr

ee-b

ody

diag

ram

oft

he p

robl

em.S

how

the

vert

ical

an

d ho

rizo

nta

l com

pon

ents

oft

he te

nsi

on fo

rce

in th

e st

rin

g.

b.Fi

nd

the

elec

tric

forc

e on

th

e ch

arge

in t

his

fiel

d.

c.Fi

nd

the

stre

ngt

h o

fth

e el

ectr

ic fi

eld.

5.H

ow m

any

elec

tron

s ar

e th

ere

in 1

.00

C?

How

man

y el

ectr

ons

are

ther

e

in 1

.00 mC

?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

17



61

Curr

ent a

nd R

esis

tanc

e

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Chap

ter 1

910

3

1.A

60.

0 cm

met

al w

ire

draw

s 0.

185

A fr

om a

36.

0 V

bat

tery

.Will

th

e

curr

ent

incr

ease

or

decr

ease

wh

en t

he

follo

win

g ch

ange

s ar

e pe

rfor

med

?

Exp

lain

wh

eth

er t

he

chan

ge is

du

e to

a c

han

ge in

res

ista

nce

,a c

han

ge in

pote

nti

al d

iffe

ren

ce,o

r ot

her

rea

son

s.

a.T

he

wir

e is

cu

t in

to fo

ur

piec

es,a

nd

only

on

e se

gmen

t is

use

d.

b.T

he

wir

e is

ben

t to

form

an

Msh

ape.

c.T

he

wir

e is

hea

ted

to 5

00°C

.

d.T

he

36.0

V b

atte

ry is

rep

lace

d by

a 2

4.0

V b

atte

ry.

2.A

25

Ωre

sist

ance

hea

ter

is c

onn

ecte

d to

a p

oten

tial

dif

fere

nce

of

120

V

for

5.00

h.

a.H

ow m

uch

cu

rren

t do

es t

he

hea

ter

draw

?

b.H

ow m

uch

ele

ctri

c ch

arge

tra

vels

th

rou

gh t

he

hea

tin

g el

emen

t du

rin

g

this

tim

e?

c.W

hat

is t

he

pow

er c

onsu

mpt

ion

of

the

hea

ter?

d.U

se th

e po

wer

an

d ti

me

to c

alcu

late

how

mu

ch e

ner

gy w

as c

onsu

med

.

Ch

apte

r



Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s10

4

3.T

he

labe

l on

a t

hre

e-w

ay li

ght

bulb

pac

kage

spe

cifi

es 1

00 W

,150

W,

250

W,1

20 V

.

a.H

ow m

uch

cu

rren

t doe

s th

e lig

ht b

ulb

dra

w in

eac

h of

the

thre

e w

ays?

(Ass

um

e th

ree

sign

ific

ant f

igu

res

in e

ach

ofth

ese

mea

sure

men

ts.)

b.W

hat

is t

he

bulb

’s r

esis

tan

ce in

eac

h w

ay?

c.C

ompa

re t

he

cost

of

usi

ng

the

ligh

t bu

lb fo

r 10

0.0

h in

eac

h w

ay.

(Ass

um

e th

at t

he

pric

e is

7.0

0 ¢/

kWh

.)

4.A

n e

lect

ric

hot p

late

dra

ws

6.00

A o

fcu

rren

t whe

n it

s re

sist

ance

is 2

4.0

Ω.

a.W

hat

is t

he

volt

age

acro

ss t

he

hot

pla

te’s

hea

tin

g el

emen

t?

b.H

ow m

uch

pow

er d

oes

it c

onsu

me?

c.Fo

r w

hat

len

gth

of

tim

e sh

ould

it b

e ke

pt o

n in

ord

er to

su

pply

9×

104

J to

a c

offe

epot

? (A

ssu

me

that

all

elec

tric

al e

ner

gy is

tra

ns-

ferr

ed to

th

e co

ffee

pot

by h

eat.

)

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

19


62

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s10

8

Circ

uits

and

Circ

uit E

lem

ents

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

20

1.C

onsi

der

the

circ

uit

sh

own

bel

ow.


A

1

2

34

5

BC

D

a.D

o an

y of

the

bulb

s h

ave

a co

mpl

ete

circ

uit

wh

en a

ll th

e sw

itch

es a

re

open

? W

hic

h o

ne(

s)?

b.D

o an

y of

the

swit

ches

cau

se a

sho

rt c

ircu

it w

hen

clos

ed? W

hich

one

(s)?

c.W

hic

h s

wit

ches

sh

ould

be

kept

ope

n,a

nd

wh

ich

sh

ould

be

clos

ed fo

r

the

follo

win

g to

occ

ur?

•on

ly b

ulb

s A

and

Bar

e of

f

•on

ly b

ulb

s A

and

Car

e of

f

•on

ly b

ulb

s B

and

Car

e of

f

2.A

ligh

t bu

lb o

fu

nkn

own

res

ista

nce

is c

onn

ecte

d in

ser

ies

wit

h a

9.0

Ωre

sist

or to

a 1

2.0

V b

atte

ry.T

he

curr

ent

in t

he

bulb

is 0

.80

A.

a.In

th

e sp

ace

belo

w,s

ketc

h a

sch

emat

ic d

iagr

am o

fth

e ci

rcu

it.

b.Fi

nd

the

equ

ival

ent

resi

stan

ce o

fth

e ci

rcu

it.

c.Fi

nd

the

resi

stan

ce o

fth

e lig

ht

bulb

.


Chap

ter 2

010

9

3.A

ligh

t bu

lb o

fu

nkn

own

res

ista

nce

is c

onn

ecte

d in

par

alle

l to

a 48

.0 Ω

resi

stor

an

d to

a 1

2.0

V b

atte

ry.T

he

curr

ent

thro

ugh

th

e ba

tter

y is

2.5

0 A

.

a.In

th

e sp

ace

belo

w,s

ketc

h a

sch

emat

ic d

iagr

am o

fth

e ci

rcu

it.

b.Fi

nd

the

pote

nti

al d

iffe

ren

ce a

cros

s th

e re

sist

or a

nd

acro

ss t

he

bulb

.

c.Fi

nd

the

curr

ent

in t

he

resi

stor

an

d in

th

e bu

lb.

d.Fi

nd

the

resi

stan

ce o

fth

e lig

ht

bulb

.

4.In

th

e ci

rcu

it b

elow

,fin

d th

e eq

uiv

alen

t re

sist

ance

for

the

follo

win

g

situ

atio

ns.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

20

a.R

a=

Rb

=R

c=

Rd

=R

e=

Rf

=10

.0Ω

b.R

a=

10.0

Ω;R

b=

20.

0 Ω

;Rc

=30

.0Ω

;Rd

=40

.0Ω

;Re

=50

.0Ω

;Rf

=60

.0Ω

Ra

Rb

Rd

Rc

Re

Rf


63

Mag

netism

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Chap

ter 2

111

3

1.A

wir

e fr

ame

carr

ies

an e

lect

ric

curr

ent i

n th

e di

rect

ion

sho

wn

.

Con

side

r th

e m

agn

etic

fiel

d co

ntr

ibut

ed b

y ea

ch s

egm

ent o

fthe

fram

e at

poi

nts

A,B

,C,D

,an

dE

.

a.U

se th

e co

nven

tion

sym

bols

(×,

•,an

d →

) to

rep

rese

nt th

e

dire

ctio

n of

mag

net

ic fi

elds

cre

ated

at p

oin

t Aby

the

vert

ical

segm

ents

oft

he fr

ame.

Do

they

hav

e th

e sa

me

dire

ctio

n?

the

sam

e st

ren

gth?

b.R

epea

t fo

r th

e h

oriz

onta

l seg

men

ts.

c.A

nsw

er it

ems

a an

d b

for

poin

ts B

,C,D

,an

dE

,an

d fi

ll in

th

e ta

ble

belo

w.

Ch

apte

r


EA

CB

D

d.D

o th

e co

ntr

ibu

tion

s of

each

seg

men

t to

th

e m

agn

etic

fiel

d ca

nce

l

out

at t

he

cen

ter?

Exp

lain

.

e.Is

th

e m

agn

etic

fiel

d re

sult

ing

from

th

e co

mbi

ned

eff

ects

of

the

fou

r

side

s of

the

fram

e st

ron

ger

insi

de o

r ou

tsid

e th

e fr

ame?

left

mos

tri

ghtm

ost

upp

erlo

wer

B C D E


114

2.A

2.0

m lo

ng

con

duct

ing

wir

e h

as a

cu

rren

t of

5.0

in a

un

ifor

m m

agn

etic

fiel

d of

0.43

T.T

he

fiel

d is

par

alle

l to

the

x-ax

is.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

21 IB

(a)

I

B

(b)

a.W

hat

is t

he

forc

e on

th

e w

ire

wh

en it

is v

erti

cal,

para

llel t

o th

e y-

axis

as s

how

n a

?

b.W

hat i

s th

e fo

rce

on th

e w

ire

whe

n it

is h

oriz

onta

l,pa

ralle

l to

the

x-ax

is

as s

how

n in

b?

3.T

he

wir

e in

item

2 is

ben

t to

form

a 0

.50

m ×

0.50

m s

quar

e ca

rryi

ng

the

sam

e 5.

0 A

cu

rren

t,w

ith

th

e po

siti

ve c

har

ges

mov

ing

cloc

kwis

e in

th

e

fram

e.T

he

fram

e is

in t

he

sam

e m

agn

etic

fiel

d (B

=0.

43 T

).

a.Sk

etch

a d

iagr

am o

fth

e si

tuat

ion

.Use

arr

ows

to in

dica

te t

he

dire

c-

tion

of

the

curr

ent

in e

ach

seg

men

t of

the

fram

e.

b.Fi

nd

the

forc

es a

ctin

g on

eac

h s

ide

ofth

e fr

ame.

Spec

ify

thei

r m

agn

i-

tude

an

d di

rect

ion

.

c.D

o th

e fo

rces

on

th

e fr

ame

can

cel e

ach

oth

er?

Will

th

e fr

ame

be a

ble

to m

ove?

Will

it b

e ab

le to

rot

ate?

Exp

lain

.

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s


64

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s11

8

Indu

ctio

n an

d Al

tern

atin

g Cu

rren

t

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

22

1.W

hic

h o

fth

e fo

llow

ing

acti

ons

will

indu

ce a

n e

mf

in a

con

duct

or?

a.M

ove

a m

agn

et n

ear

the

con

duct

or.

b.M

ove

the

con

duct

or n

ear

a m

agn

et.

c.R

otat

e th

e co

ndu

ctor

in a

mag

net

ic fi

eld.

d.C

han

ge t

he

mag

net

ic fi

eld

stre

ngt

h.

e.al

l of

the

abov

e

2.A

cir

cula

r lo

op (

10 t

urn

s) w

ith

a r

adiu

s of

29 c

m is

in a

mag

net

ic fi

eld

that

osc

illat

es u

nif

orm

ly b

etw

een

0.9

5 T

an

d 0.

45 T

wit

h a

per

iod

of1.

00 s

.

a.H

ow m

uch

tim

e is

req

uir

ed fo

r th

e fi

eld

to c

han

ge fr

om 0

.95

T

to 0

.45

T?

b.W

hat

is t

he

cros

s-se

ctio

nal

are

a of

one

turn

of

the

loop

?

c.A

ssu

min

g th

at t

he

loop

is p

erp

endi

cula

r to

th

e m

agn

etic

fiel

d,w

hat

is

the

indu

ced

emf

in t

he

loop

?

3.E

lect

ric

gen

erat

ors

conv

ert

mec

han

ical

en

ergy

into

ele

ctri

cal e

ner

gy.

a.W

hat

are

th

e re

quir

emen

ts fo

r ge

ner

atin

g em

f?

b.T

he

mec

han

ical

en

ergy

inpu

t is

usu

ally

rot

atio

nal

mot

ion

.Wh

at a

re

two

poss

ible

sou

rces

of

rota

tion

al m

otio

n?


Chap

ter 2

211

9

4.A

250

-tu

rn g

ener

ator

wit

h c

ircu

lar

loop

s of

radi

us

15 c

m r

otat

es a

t

60.0

rpm

in a

mag

net

ic fi

eld

wit

h a

str

engt

h o

f1.

00 T

.

a.W

hat

is t

he

angu

lar

spee

d of

the

loop

s?

b.W

hat

is t

he

area

of

one

loop

?

c.W

hat

is t

he

max

imu

m e

mf?

d.W

hat

is t

he

rms

emf?

5.A

n e

lect

ric

mot

or is

som

etim

es c

alle

d a

gen

erat

or in

rev

erse

.Exp

lain

you

r u

nde

rsta

ndi

ng

ofth

is s

tate

men

t.

6.C

onsi

der

a tw

o-co

il tr

ansf

orm

er jo

ined

by

a co

mm

on ir

on c

ore.

a.If

the

curr

ent

in t

he

prim

ary

side

is in

crea

sed,

wh

at h

appe

ns

to t

he

mag

net

ic fi

eld

in t

he

core

?

b.W

hat

eff

ect

does

th

e an

swer

to it

em 6

a h

ave

on t

he

seco

nda

ry c

oil?

c.Fu

lly e

xpla

in t

he

effe

ct o

fre

duci

ng

the

curr

ent

to t

he

prim

ary

side

of

a tr

ansf

orm

er.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

22



65

Mix

edRev

iew

Answ

ers

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–1

1.a.

2.2

×10

5s

b.3.

5×

107

mm

c.4.

3×

10−4

km

d.2.

2×

10−5

kg

e.6.

71×

1011m

g

f.8.

76×

10−5

GW

g.1.

753

×10

−1ps

2.a.

3

b.4

c.10

d.3

e.2

f.4

3.a.

4

b.5

c.3

4.a.

1.00

54;−

0.99

52;5

.080

×10

−3;

5.07

6×

10−3

b.4.

597

×10

7 ;3.8

66 ×

107 ;

1.54

6×

1014

;11.

58

5.15

.9 m

2

6.T

he g

raph

sho

uld

be a

str

aigh

t lin

e.

Chap

ter 1

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–2

III

2.a.

v f=

a(∆t

)

b.v f

=v i

+a(

∆t);

∆x=

1 2 (v i

+v f

)∆t

or∆x

=v i

(∆t)

+1 2 a

(∆t)

2

1.a.

t 1=

d 1/v

1;t 2

=d 2

/v2;t

3=

d 3/v

3

b.to

tal d

ista

nce

=d 1

+d 2

+d 3

c.to

tal t

ime

=t 1

+t 2

+t 3

3.

Tim

e in

terv

alTy

pe o

fmot

ion

v(m

/s)

a(m

/s2 )

Asp

eedi

ng

up

++

Bsp

eedi

ng

up

++

Cco

nst

ant

velo

city

+0

Dsl

owin

g do

wn

+−

Esl

owin

g do

wn

+−

.b

.a

.4

1 s

c.2

sTi

me

(s)

Posi

tion

(m)

v(m

/s)

a(m

/s2 )

14.

90

−9.8

1

20

−9.8

−9.8

1

3−1

4.7

−19.

6−9

.81

4−3

9.2

−29.

4−9

.81

Chap

ter 2

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–4

1.a.

Th

e di

agra

m s

hou

ld in

dica

te t

he

rela

tive

dis

tan

ces

and

dire

ctio

ns

for

each

seg

men

t of

the

path

.

b.5.

0 km

,slig

htl

y n

orth

of

nor

thw

est

c.11

.0 k

m

2.a.

Th

e sa

me

b.Tw

ice

as la

rge

c.1.

58

3.a.

2.5

m/s

,in

th

e di

rect

ion

of

the

side

wal

k’s

mot

ion

b.1.

0 m

/s,i

n t

he

dire

ctio

n o

fth

e si

dew

alk’

s m

otio

n

c.4.

5 m

/s,i

n t

he

dire

ctio

n o

fth

e si

dew

alk’

s m

otio

n

d.2.

5 m

/s,i

n t

he

dire

ctio

n o

ppos

ite

to t

he

side

wal

k’s

mot

ion

e.4.

7 m

/s,q

=32

°

4.a.

4.0

×10

1se

con

ds

b.6.

0×

101

seco

nds

Chap

ter 3

Mix

ed R

evie

w

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–5

1.a.

at r

est,

mov

es to

th

e le

ft,h

its

back

wal

l

b.m

oves

to t

he

righ

t (w

ith

vel

ocit

y v)

,at

rest

,nei

ther

c.m

oves

to t

he

righ

t,m

oves

to t

he

righ

t,h

its

fron

t w

all

2.a.

mg,

dow

n

b.m

g,u

p

c.n

o

d.ye

s

3.a.

a=

m1

+F

m2

b.m

2a

c.F

−m

2a

=m

1a

d. m

1m +1 m2

F

4.a.

a= mF 1− +F mk

2

b.m

2a

−F k

c.F

−m

2a

−F k

=m

1a−

F k

d. m

1m +1 m2

(F

−F k

)

Chap

ter 4

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–6

1.a.

60 J

b.−6

0 J

2.a.

mgh

b.m

gh

c.v B

=v A

2+

2gh

d.n

o

e.n

o

3.a.

2.9

J

b.1.

8 J

c.1.

2 J

d.a,

b:di

ffer

ent;

c:sa

me

4.a.

1 2 mv i

2+

mgh

i=

1 2 mv f

2+

mgh

f+

F kd

b.F k

=mm

g(co

s 23

°)

c.v f

= mv i

2+

2g(d

sin

23°−

mco

s23

°)

Chap

ter 5

Mix

ed R

evie

w

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–7

III

Copyright ©by Holt, Rinehart and Winston.All rights reserved.

1.a.

Th

e ch

ange

du

e to

th

e ba

t is

gre

ater

th

an t

he

chan

gedu

e to

th

e m

itt.

b.T

he

impu

lse

due

to t

he

bat

is g

reat

er t

han

th

e im

-pu

lse

due

to t

he

mit

t.

c.C

hec

k st

ude

nt

diag

ram

s.B

at:v

ecto

r sh

owin

g in

itia

lm

omen

tum

an

d a

larg

er v

ecto

r in

th

e op

posi

te d

i-re

ctio

n s

how

ing

impu

lse

ofba

t,re

sult

is t

he

sum

of

the

vect

ors.

Mit

t:ve

ctor

sh

owin

g in

itia

l mom

entu

man

d an

equ

al le

ngt

h v

ecto

r sh

owin

g im

puls

e of

mit

t,re

sult

is t

he

sum

,wh

ich

is e

qual

to z

ero.

2.a.

Th

e im

puls

es a

re e

qual

,bu

t op

posi

te fo

rces

,occ

ur-

rin

g du

rin

g th

e sa

me

tim

e in

terv

al.

b.T

he

tota

l for

ce o

n t

he

bow

ling

ball

is t

he

sum

of

forc

es o

n p

ins.

Th

e fo

rce

on t

he

pin

s is

equ

al b

ut

op-

posi

te o

fto

tal f

orce

on

bal

l.

3.m

1v1i

+m

2v2i

=(m

1+

m2)v f

;

m1v

1i/(

m1

+m

2)+

m2v

2i/(

m1

+m

2)=v f

4.a.

M(6

m/s

)

b.2

m/s

c.ob

ject

s tr

ade

mom

entu

m;i

fm

asse

s ar

e eq

ual

,ob-

ject

s tr

ade

velo

citi

es

Chap

ter 6

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–8

1.a.

3.0,

3.0,

9.0,

27

b.4.

3,1.

0,4.

3,37

c.16

,0.2

8,11

,6.0

×10

2

d.63

0,0.

11,7

4,8.

7

e.5.

0,44

,0.1

1,9.

9

2.a.

fric

tion

b.gr

avit

atio

nal

forc

e

c.te

nsi

on in

str

ing

3.a.

dou

bled

b.qu

adru

pled

c.re

duce

d to

1 4

d.qu

adru

pled

e.re

duce

d to

1 9

4.19

0 m

5.St

uden

t dia

gram

s sh

ould

sho

w v

ecto

rs fo

r w

eigh

t an

dn

orm

al fo

rce

from

ele

vato

r;de

scen

t sho

uld

show

nor

mal

forc

e le

ss th

an w

eigh

t;st

oppi

ng

shou

ld s

how

nor

mal

forc

e gr

eate

r th

an w

eigh

t;“w

eigh

tles

snes

s”fe

elin

g is

due

to a

ccel

erat

ion

.

6.10

50 s

(17

.5 m

in)

Chap

ter 7

Mix

ed R

evie

w


66

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–9

1.a.

Ifth

e kn

ob is

fart

her

from

th

eh

inge

,tor

que

is in

crea

sed

torq

ue

for

a gi

ven

forc

e.

b.tw

ice

as m

uch

2.a.

Rot

atio

nal

iner

tia

is r

edu

ced.

b.A

ngu

lar

mom

entu

m r

emai

ns

the

sam

e.

c.A

ngu

lar

spee

d in

crea

ses.

3.a.

2.0

kg

b.0.

67 k

g

4.a.

6.2

N• m

,0.0

16 k

g• m

2 ,39

0 ra

d/s2

b.12

N• m

,0.0

62 k

g•m

2 ,190

rad

/s2

5.a.

8.1

×10

16kg

• m2

b.5.

9×

1012

kg• m

2 /s

c.2.

1×

108

J

d.3.

1×

103

m/s

e.2.

2×

108

J

6.a.

4.0

×10

4J

b.4.

4×

104

J

c.4.

9×

104

J

d.0.

81

Chap

ter 8

Mix

ed R

evie

w

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–1

3

1.a.

0.20

s;5

.0 H

z

b.sa

me,

sam

e,in

crea

se,i

ncr

ease

2.a.

60.0

N/m

b.0.

574

seco

nds

;1.7

4 H

z

3.6.

58 m

/s2 ;n

o

4.a.

A:0

s,2

s,4

s;B

:0.5

s,1

.5 s

,2.5

s,3

.5 s

;C:1

,3 s

b.P

E:0

s a

t A,1

s a

t C

,2 s

at A

,3 s

at

C,4

s a

t A;K

E:

0.5

s,1.

5 s,

2.5

s,3.

5 s

at B

c.0.

5 s,

2.5

s at

B to

th

e ri

ght

1.5

s,3.

5 s

at B

to t

he

left

;0

s,2

s,4

s at

A to

th

e ri

ght,

1 s,

3 s

at C

to t

he

left

5.3.

00×

102

m/s

6.3.

0 s;

6.0

Chap

ter 1

2 M

ixed

Rev

iew

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–14

III

Copyright ©by Holt, Rinehart and Winston.All rights reserved.

1.a.

2.19

m;2

.27

m

b.w

avel

engt

h in

crea

ses

wh

en te

mpe

ratu

re in

crea

ses

2.a.

arro

ws

poin

tin

g E

ast

on a

mbu

lan

ce,p

olic

e,an

dtr

uck

,Wes

t on

van

.

b.po

lice

and

ambu

lan

ce (

equ

al),

tru

ck,s

mal

l car

,van

3.T

hes

e ob

ject

s h

ad t

he

sam

e n

atu

ral f

requ

ency

of

330

Hz,

so r

eson

ance

occ

urr

ed.

4.a.

1460

Hz,

2440

Hz

b.70

.8 c

m,2

3.6

cm,1

4.1

cm

c.0.

177

m

d.97

4 H

z,14

60 H

z;70

.8 c

m,3

5.4

cm,2

3.6

cm;0

.354

m

5.a.

5

b.43

5 H

z,be

cau

se it

will

als

o pr

ovid

e a

diff

eren

ce

of5

Hz.

Chap

ter 1

3 M

ixed

Rev

iew

7.a.

9.00

cm

b.p

=30

.0 c

m;q

=12

.9 c

m;r

eal;

inve

rted

;2.5

8 cm

tal

l

p=

24.0

cm

;q=

14.4

cm

;rea

l,in

vert

ed;3

.60

cm t

all

p=

18.0

cm

;q=

18.0

cm

;rea

l;in

vert

ed;6

.00

cm t

all

p=

12.0

cm

;q=

36.0

cm

;rea

l;in

vert

ed;2

.00

cm t

all

p=

6.0

cm;q

=−1

8 cm

;vir

tual

;upr

igh

t;18

cm

tal

l

8.p

=30

.0 c

m;q

=−6

.92

cm;v

irtu

al;u

prig

ht;

1.38

cm

tal

l

p=

24.0

cm

;q=

−6.5

5 cm

;vir

tual

,upr

igh

t;1.

64 c

m t

all

p=

18.0

cm

;q=

−6.0

0 cm

;vir

tual

;upr

igh

t;2.

00 c

m t

all

p=

12.0

cm

;q=

−5.1

4 cm

;vir

tual

;upr

igh

t;2.

57 c

m t

all

p=

6.0

cm;q

=−3

.6 c

m;v

irtu

al;u

prig

ht;

3.6

cm t

all

1.4.

07×

1016

m

2.a.

3.33

×10

−5s

b.1.

00×

10−4

m

3.3.

84×

108

m

4.3.

00×

1011

Hz

5.D

iffu

se r

efle

ctio

n:(

non

shin

y su

rfac

es)

tabl

e to

p,fl

oor,

wal

ls,c

ar p

ain

t,po

ster

s (a

nsw

ers

will

var

y)

Spec

ula

r re

flec

tion

:met

allic

su

rfac

es,w

ater

,mir

rors

(an

swer

s w

ill v

ary)

6.a.

Ch

eck

stu

den

t dr

awin

gs fo

r ac

cura

cy.

b.B

is 4

m fr

om A

hor

izon

tally

,Cis

2 m

bel

ow B

vert

ical

ly

c.D

is 2

m b

elow

Ave

rtic

ally

,Eco

inci

des

wit

h C

d.th

ey w

ill o

verl

ap t

he

exis

tin

g im

ages

or

obje

cts

Chap

ter 1

4 M

ixed

Rev

iew

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–1

7

1.a.

Ray

1 a

t 45

°;R

ay 2

at

14.9

°

b.R

ays

shou

ld in

ters

ect

insi

de t

he

aqu

ariu

m.

c.B

ecau

se t

he

rays

are

no

lon

ger

para

llel,

they

will

in-

ters

ect

in t

he

wat

er.

2.a.

Firs

t bo

un

dary

:70.

0°,4

5.0°

Seco

nd

bou

nda

ry:4

5.0°

,40.

4°

Th

ird

bou

nda

ry:4

0.3°

,36.

8°

b.In

com

ing

rays

get

clo

ser

and

clos

er to

th

e n

orm

al.

Ref

lect

ed r

ays

get

fart

her

aw

ay fr

om t

he

nor

mal

wit

hth

e sa

me

angl

es.

3.a.

9.00

cm

b.12

.9 c

m,1

4.4

cm,1

8.0

cm,3

6.0

cm,−

18.0

cm

2.58

cm

,3.6

cm

,6.0

0 cm

,18.

0 cm

,−18

.0 c

m

real

,rea

l,re

al,r

eal,

virt

ual

4.18

.0 c

m,w

ith

all i

mag

es v

irtu

al a

nd o

n th

e le

ft o

fthe

len

s

−11.

2,−1

0.3,

−9.0

0,−7

.20,

−4.5

0

5.a.

6.00

cm

in fr

ont

ofth

e le

ns

b.0.

857

cm

Chap

ter 1

5 M

ixed

Rev

iew

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–18

III

1.a.

6.74

×10

−6m

b.47

.9°

c.T

he

max

imu

m a

ngl

e fo

r lig

ht

to r

each

th

e sc

reen

inth

is a

rran

gem

ent

is 4

5°.

2.a.

Lon

ger

wav

elen

gths

are

dif

frac

ted

wit

h a

grea

ter

angl

e.

b.Fi

rst

orde

r gr

oup

oflin

es:b

lue,

gree

n,r

ed;s

econ

dor

der:

the

sam

e

c.W

hit

e

3.a.

A=

5.0

×10

−6m

,B =

1.1

×10

−7m

,C =

3.3

×10

−8m

b.vi

sibl

e:A

;x-r

ay:A

,B,o

r C

;IR

:non

e

4.a.

Nei

ther

wou

ld w

ork

beca

use

th

ey w

ould

act

as

dif-

fere

nt

sou

rces

,so

even

wit

h t

he

sam

e fr

equ

ency

,th

eysh

ould

not

be

in p

has

e.

b.In

terf

eren

ce is

occ

urr

ing.

Chap

ter 1

6 M

ixed

Rev

iew

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–1

9

1.a.

A;1

.87

×10

13el

ectr

ons;

B:3

.12

×10

13el

ectr

ons

b.th

e fo

rces

are

equ

al a

nd

oppo

site

,no

2.a.

Res

ult

ant =

1.49

N,l

eft;

F(A

-C)

=1.

35 N

,lef

t;F

(B-C

)=

0.14

0 N

,lef

t

b.R

esu

ltan

t =0.

788

N,r

igh

t;F(

A-C

) =

1.35

N,r

igh

t;F

(B-C

)=

0.56

2 N

,lef

t

c.R

esu

ltan

t =0.

400

N,l

eft;

F(A

-C)

=0.

599

N,r

igh

t;F

(B-C

)=

0.99

9 N

,lef

t

3.a.

1.92

×10

16N

b.2.

87×

1010

m/s

2

c.9.

81 m

/s2 ;t

his

is n

eglig

ible

in c

ompa

riso

n w

ith

th

eac

cele

rati

on a

;alp

ha

part

icle

s w

ill m

ove

hor

izon

tally

4.a.

Ch

eck

stu

den

ts d

iagr

ams

for

accu

racy

.

b.1.

53×

10−2

N

c.7.

65×

103

N/C

5.1

C =

6.25

×10

18;1

mC

=6.

25×

1012

Chap

ter 1

7 M

ixed

Rev

iew

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–2

1

1.a.

Iin

crea

ses

beca

use

Rde

crea

ses

(sh

orte

r)

b.n

o ch

ange

c.I

decr

ease

s be

cau

se R

incr

ease

sw

ith

tem

pera

ture

d.I

decr

ease

s

2.a.

4.8

A

b.8.

64×

104

J

c.58

0 W

d.1.

0×

107

J

3.a.

0.83

3 A

;1.2

5 A

;2.0

8 A

b.14

4Ω

;96.

0 Ω

;57.

6 Ω

c.70

.0 ¢

;$1.

05;$

1.75

4.a.

144

V

b.86

4 W

c.10

4 se

con

ds

Chap

ter 1

9 M

ixed

Rev

iew

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–22

1.a.

D

b.sw

itch

5

c.•

swit

ches

1 a

nd

3 op

en,s

wit

ches

2,4

,an

d 5

clos

ed

• sw

itch

es 1

an

d 4

open

,sw

itch

es 2

,3,a

nd

5 cl

osed

• sw

itch

2 o

pen

,sw

itch

es 1

,3,4

,an

d 5

clos

ed;o

rsw

itch

es 3

an

d 4

open

,sw

itch

es 1

,2,a

nd

5 cl

osed

;or

sw

itch

es 2

,3,a

nd

4 op

en,s

wit

ches

1 a

nd

5 cl

osed

2.a.

Ch

eck

stu

den

ts’d

iagr

ams,

wh

ich

sh

ould

sh

ow a

bulb

an

d a

resi

stor

in s

erie

s w

ith

a b

atte

ry.

b.15

Ω

c.6

Ω

3.a.

Ch

eck

stu

den

ts d

iagr

ams.

b.12

.0 V

,12.

0 V

c.0.

25 A

,2.2

5 A

d.5.

33Ω

4.a.

R=

6.15

Ω

b.R

=30

.4Ω

Chap

ter 2

0 M

ixed

Rev

iew


67

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–2

3

1.a.

Th

e m

agn

etic

fiel

d fr

om t

he

left

mos

t se

gmen

t is

•an

d st

ron

ger.

Th

e m

agn

etic

fiel

d fr

om t

he

righ

tmos

tse

gmen

t is

×an

d w

eake

r.

b.A

t A,b

oth

hor

izon

tal s

egm

ents

con

trib

ute

a ×

mag

net

ic fi

eld

ofeq

ual

str

engt

h

c.B

;×;×

wea

ker;

×;×

sam

e

C;×

;×sa

me;

×;×

sam

e

D;×

;×st

ron

ger;

×;×

sam

e

E;×

;•st

ron

ger;

×;×

sam

e

d.N

o.T

hey

rei

nfo

rce

each

oth

er in

th

e sa

me

dire

ctio

n.

e.in

side

2.a.

F=

4.3

N in

to t

he

page

b.F

=0

3.a.

Dia

gram

s sh

ould

sh

ow c

lock

wis

e cu

rren

t.

b.St

arti

ng

from

th

e le

ft s

ide:F

=1.

1 N

into

th

e pa

ge;

F=

0;F

=1.

1 N

ou

t of

the

page

;F=

0

c.Fo

rces

are

equ

al a

nd

oppo

site

,so

no

tran

slat

ion

alm

otio

n w

ill o

ccu

r,bu

t it

cou

ld r

otat

e ar

oun

d a

vert

i-ca

l axi

s.

Chap

ter 2

1 M

ixed

Rev

iew

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–24

1.e

2.a.

0.50

s

b.0.

26 m

2

c.2.

6 V

3.a.

mag

net

ic fi

eld,

con

duct

or,r

elat

ive

mot

ion

b.an

swer

s m

ay v

ary,

but

cou

ld in

clu

de t

he

follo

win

g:w

ater

wh

eel,

win

dmill

,ele

ctri

c m

otor

,com

bust

ion

engi

ne

4.a.

6.28

rad

/s

b.7.

1×

10−2

m2

c.11

0 V

d.78

V

5.A

mot

or c

onve

rts

elec

tric

en

ergy

to r

otat

ion

al e

ner

gy;

gen

erat

ors

conv

erts

rot

atio

nal

en

ergy

to e

lect

ric

ener

gy.

6.a.

incr

ease

s

b.in

duce

s cu

rren

t w

hile

ch

ange

occ

urs

c.It

dec

reas

es m

agn

etic

fiel

d w

hic

h w

ill in

duce

a c

ur-

ren

t w

hile

th

e ch

ange

occ

urs

.

Chap

ter 2

2 M

ixed

Rev

iew


2008–2009 honors physics review notestomstrong.org/physics/year0809h.pdf · honors physics review...

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