physics 12 notes
TRANSCRIPT
RRHS Physics 12 Course Notes
J. Burke
2009-2010
c©2001-2010
Contents
Textbook Correlations v
1 Dynamics Extension 11.1 Introduction to Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.2 Relative Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Force Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.1 Inclined Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.1 Translational Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.2 Rotational Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 2-D Motion 152.1 Projectiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.1.1 Objects Launched Horizontally . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.2 Objects Launched at an Angle . . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.1 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.2 Pendulum Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3 2D Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3.1 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3.2 Elastic and Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 Planetary Motion 253.1 Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.1.1 Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.1.2 Centripetal “Force” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.1.3 Centrifugal Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
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3.2 Universal Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.1 Newton’s Law of Universal Gravitation . . . . . . . . . . . . . . . . . . . . . 303.2.2 Acceleration Due to Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.3 Satellite Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.4 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4 Fields 354.1 Static Electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.1.1 Insulators and Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.1.2 Charging Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.1.3 Electroscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.1.4 Permanency of Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2 Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2.1 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2.2 Electric Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2.3 Lines of Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.2.4 Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.3 Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3.1 Electric Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3.2 Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3.3 Equipotential Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5 Electricity & Magnetism 455.1 Electric Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5.1.1 Electrical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.1.2 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.1.3 Electrical Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.2 *Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2.1 *Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2.2 *Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2.3 *Complex Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2.4 *Kirchhoff’s Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.2.5 *Safety Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.2.6 *Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.3 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3.1 Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3.2 Electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.3.3 Force on a Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.3.4 Force on a Charged Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.3.5 Electric Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
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5.3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.4 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5.4.1 Induced EMF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.4.2 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.4.3 Electric Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
6 Waves and Modern Physics 696.1 Quantum Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.1.1 Planck’s Quantum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . 696.1.2 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.1.3 Compton Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716.1.4 de Broglie Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.2 Wave-Particle Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.2.1 Historical Models of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.2.2 Modern Theory of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 756.2.3 Modern Theory of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.2.4 Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
6.3 Models of the Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.3.1 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.3.2 Bohr Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806.3.3 Quantum Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.3.4 Fluorescence and Phosphorescence . . . . . . . . . . . . . . . . . . . . . . . . 826.3.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
7 Nuclear Physics 837.1 The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
7.1.1 Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.1.2 Mass Defect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
7.2 Radioactive Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 867.2.1 Alpha Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 867.2.2 Beta Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 867.2.3 Gamma Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 877.2.4 Half-lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 877.2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
7.3 Artificial Radioactivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.3.1 Nuclear Fission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.3.2 Nuclear Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.3.3 Nuclear Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
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A Analysis of Data 93A.1 Experimental Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
A.1.1 Precision and Random Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.1.2 Accuracy and Systematic Errors . . . . . . . . . . . . . . . . . . . . . . . . . 94
A.2 Statistical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.2.1 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.2.2 Confidence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
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Section Pages in Textbook Problems in Textbook1.1 pgs 90-111,454-462 pg 93 #8,9; pg 463 #61.2 pgs 463-489 pg 475 #13; pg 489 #27,281.3 pgs 490-502 pg 495 #30; pg 501 #31,33,342.1 pgs 532-5502.2 pgs 598-621 pg 623 #18,24,25; pg 611 Conceptual Problems; pg 608 #3,42.3 pgs 503-508, 510-526 pg 509 #36,37; pg 515 #39,40; pg 526 #1,3; pg 529 #30; BLM #1,2,33.1 pgs 551-562 pg 567 #4,5,6; pg 571 #21,283.2 pgs 572-597 pg 594 #2; pg 595 #5,6,8; pg 596 #12,154.14.2 pgs 632-661 pg 641 #9,10; pg 655 #26,27,28; pg 661 #5; pg 685 #314.3 pgs 672-680,688-693 pg 681 #25.1 pgs 694-714, 734-7465.2 pgs 715-7335.3 pgs 752-780 pg 767 #1,2; pg 778 #1; pg 780 #2,3,45.4 pgs 781-796 pg 796 #1-4; pg 799 #266.1 pgs 840-860 pg 852 #1,3,4; pg 862 #6; pg 863 #8,196.2 pgs 8616.3 pgs 866-880 pg 876 #1-6; pg 886 #3,67.1 pgs 898-905 pg 905 #3,4,5,7,87.2 pgs 906-917 pg 917 #4,8; pg 918-919 #3,97.3 pgs 920-933 pg 925 #2,3,4; pg 933 #1,2,7; pg 934 #5,6,8,9,14; pg 936-937 #26,27
Appendix A pgs 938-939
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CHAPTER 0. TEXTBOOK CORRELATIONS
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Chapter 1
Dynamics Extension
1.1 Introduction to Vectors
In grade 11 physics, you probably discussedtwo kinds of quantities — vectors and scalars.A scalar is an ordinary quantity that has onlymagnitude (size); it does not have a direction.For example, temperature and mass have nodirection associated with them. A vector is aquantity that has both magnitude and direc-tion. For example, displacement, velocity, ac-celeration, force, and momentum are all quan-tities for which it is important to know thedirection. When writing, a vector is denotedby placing an arrow over it (−→v ); when typing,a vector is denoted using boldface (v).
Last year, you talked briefly about vectors inone dimension. This year, we will be extendingthat analysis to two dimensions. In university,the analysis will be extended again to threedimensions (this is a minor extension). Therest of this discussion will apply to vectors intwo dimensional space.
A vector is not just a single number, like ascalar is. In 2D space, it is actually two num-bers. You have used an x−y coordinate systemin math, and you know that two numbers areneeded to specify a position on one of thesegraphs. Likewise, two coordinates are neededto specify a vector in two-dimensional space.Consider the diagram below.
The vector d actually represents a step in spacefrom the origin to some point whose locationis given by (dx, dy). The symbol d repre-sents these components. It is often convenientto represent a vector by an arrow that indi-cates the direction of the vector. The vectorcan then be described using a magnitude (the“length” of the vector) and an angle θ (the di-rection of the vector).1
Vectors can be drawn using scale diagrams,where a protractor can be used to orient thevector correctly and an appropriate scale canbe used to represent the vector. The arrowrepresents the head of the vector and the tail isat the other end. For example, a scale of 1 cmfor every 5 m can be used; a 30 m displacement
1Note that if we know the magnitude d and the angleθ, we can use sin θ and cos θ identities to solve for dx
and dy in the above diagram.
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1.1. INTRODUCTION TO VECTORS CHAPTER 1. DYNAMICS EXTENSION
vector would then be drawn with an arrow thatis 6 cm long. Vectors can then be added in thescale diagram by drawing them head to tail.
Direction There are different conventionsfor describing the direction of a vector. Forthe examples that follow, assume that θ = 30o
in the previous diagram.
1. In math, you have probably described vec-tor directions as a counterclockwise rota-tion from the positive x-coordinate (eastusing compass directions). In the previ-ous diagram, the direction of the vectorwould then be 30o; north would be 90o,south would be 270o.
2. Bearings are another way of expressing di-rections. In this system, north is 0o andall directions are measured clockwise fromthis reference direction. In this system,the direction of the vector in our diagramwould be 60o.
3. The last convention I will discuss is theone that we are going to use. This con-vention describes a direction as a rotationfrom one of the four reference directions(north, east, south, west). The directionof the vector in our diagram would nowbe 30o north of east. This means that avector that was pointed east was rotated30o north. This convention is convenientbecause there is no ambiguity about whatthe reference direction (0o) is. (The di-rection in the diagram could also be ex-pressed as 60o east of north). A slightlydifferent way of expressing 30o north ofeast would be to say E30oN - this canbe interpreted as “go east and then ro-tate 30o toward the north” for the propervector direction. Your textbook uses thislast convention.
1.1.1 Vector Algebra
We must now look at rules to add and subtractvectors. Since vectors are not single numbers,our usual laws of algebra cannot be applied tothem; in other words, we cannot simply addthe magnitude of two vectors together to ob-tain a total magnitude.
Addition What does it mean to add twovectors? Consider two displacement vectors aand b which represent displacements of a per-son walking. The addition of these two dis-placements should tell us where the person isat the end of his journey relative to where hestarted. To help visualize this, we will draw avector diagram showing this (notice that thevectors are drawn head to tail when addingthem together)
The vector components have been drawn inhere as well (as dotted lines). The vector aactually represents the components (ax, ay);the other vector b represents the components(bx, by). If we add these two vectors, we areactually adding their components. So a + bwill give (ax + bx, ay + by), and the diagramwill look like this:
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The only difference between these two dia-grams is that the component vectors have beenmoved to show the x components together andthe y components together. Notice now thatwe have one large right angle, so we can againuse the pythagorean theorem and our trig func-tions to find the magnitude and direction.
When we add two scalars together, we geta sum. Similarly, when we add two vectorstogether we get a resultant vector. So we cansay that a + b = c. The resultant vector is asingle vector that goes from where we startedto where we ended.
Notice that the vector c represents the sum ofthe components (ax + bx, ay + by). Knowingthis, we can now find a magnitude for c usingthe pythagorean theorem and the appropriatetrigonometric identities.
Since we now have a single right angle triangle,we can use the pythagorean theorem
c =√
(Σx)2 + (Σy)2
to find the magnitude of c and the angle θ canbe found using
tan θ =Σy
Σx
Subtraction Just like subtraction of twoscalars is really the same as adding a nega-tive scalar (5− 3 is the same as 5 + (−3)), thesubtraction of two vectors a − b is the sameas a + (−b); but (−b) just means (−bx,−by);in other words, we are just changing the di-rection of the vector b and instead of addingthe components of the two vectors we subtractthem. Using the same vectors as our previousexample, a− b = c would look like
The resultant vector c can still be representedin component form
where, in this case, Σx = ax − bx and Σy =ay − by.
1.1.2 Relative Velocity
We saw in section 1.1 that an object’s positionis given by two coordinates (x, y). Remem-ber from grade 11 that velocity is the changein position, or displacement, over time; there-fore, velocity is also a vector which has twocomponents (vx, vy).
As was discussed in physics 11, there is noabsolute velocity; the velocity of an objectis always relative to some frame of reference.Consider the example of a dog on a boat. Theboat is moving north at 7 m/s relative to theshore. Now suppose that the dog is movingnorth at 2 m/s relative to the boat. In otherwords, the dog is moving 2 m/s faster than theboat. How fast is the dog actually moving? Itdepends on your point of view. To someone onthe boat, the dog is moving at 2 m/s; however,
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1.1. INTRODUCTION TO VECTORS CHAPTER 1. DYNAMICS EXTENSION
to somebody on the shore, the dog is movingits 2 m/s plus the boat’s 7 m/s (since theyare moving in the same direction), which is 9m/s.
The situation is similar in two dimensions.Suppose that a boat is crossing a body of waterat 5 m/s relative to the water (we will use thesymbol vbw to represent this speed).2 If thewater is not moving, a person on the shore seesthe boat moving at 5 m/s relative to the shoreas well. Now suppose that the body of wateris a river flowing perpendicular to the boat at3 m/s as measured by someone on the shore(vws).
The person on the shore now sees the rivercarrying the boat downstream at 3 m/s, butalso sees the boat moving across the river at 5m/s. Just like the dog on the boat, the personon the shore sees the addition of the two veloc-ities, so the velocity of the boat with respectto the shore is given by
vbs = vbw + vws (1.1)
Remember, however, that these quantities arevectors and must therefore be added as vec-tors! (as was described in section 1.1.1)
By using subscripts according to the conven-tion described above (Eq. 1.1), we see that theinner subscripts on the right-hand side of equa-tion 1.1 are the same and the outer subscriptson the right-hand side of equation 1.1 are thesame as the subscripts for the resultant vectoron the left vbs. This can be used as a check ifyou are not sure if you are adding the propervectors.
2Using this notation, the first subscript identifies theobject that is moving, the second subscript identifiesthe frame of reference with respect to which it is moving
Since they are vectors, however, these veloc-ities must be added as vectors (see section1.1.1).
The resultant vector (the velocity actually ob-served by someone on the shore) is the vectorvbs. This resultant velocity has two compo-nents (one across the river and one down theriver). Note that the component across theriver is the same as the original velocity of theboat that was directed across the river; there-fore, the boat will cross the river in the sameamount of time with the river flowing as with-out!
1.1.3 Problems
1. Slimy the slug crawled 34.0 cm E, then48.5 cm S. What is Slimy’s displacementfrom his starting point?
2. A delivery truck travels 18 blocks north,16 blocks east, and 10 blocks south. Whatis its final displacement from the origin?
3. A car is driven 30 km west and then 80km southwest. What is the displacementof the car from the point of origin (mag-nitude and direction)?
4. Break the following vectors into compo-nents: (a) 45 km in a direction 25o southof west; (b) 74 km, 35o E of N
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5. An explorer walks 22.0 km in a northerlydirection, and then walks in a direction60o south of east for 47.0 km.
(a) What distance has he travelled?
(b) What is his displacement from theorigin?
(c) What displacement vector must hefollow to return to his original loca-tion?
6. By breaking each of the following vectorsinto components, determine the resultantof the following vectors: 10.0 m, 30o northof east; 6.0 m, 37o east of north; and 12 m,30o west of south.
7. A man walks 3.0 km north, 4.5 km in adirection 40o east of north, and 6.0 km ina direction 60o south of east. What is hisdisplacement vector?
8. After the end of a long day of travelling,Slimy the Slug is 255 cm east of his home.If he started out the day by travelling90 cm in a direction 25o east of north inthe morning, how far did he travel in theafternoon (and in what direction) to getto his final location?
9. A dog walks at a speed of 1.8 m/s alongthe deck toward the front of a boat whichis travelling at 7.6 m/s with respect tothe water. What is the velocity of the dogwith respect to the water? What if thedog were walking toward the back of theboat?
10. An airplane is travelling 1000 km/h in adirection 37o east of north.
(a) Find the components of the velocityvector.
(b) How far north and how far east hasthe plane travelled after 2.0 hours?
11. An airplane whose airspeed is 200 km/hheads due north. But a 100 km/h windfrom the northeast suddenly begins toblow. What is the resulting velocity ofthe plane with respect to the ground?
12. A boat can travel 2.60 m/s in still water.
(a) If the boat heads directly across astream whose current is 0.90 m/s,what is the velocity (magnitude anddirection) of the boat relative to theshore?
(b) What will be the position of the boat,relative to its point of origin, after 4.0s?
13. An airplane is heading due north at aspeed of 300 km/h. If a wind begins blow-ing from the southwest at a speed of 50km/h, calculate
(a) the velocity of the plane with respectto the ground, and
(b) how far off course it will be after 30min if the pilot takes no correctiveaction.
(c) Assuming that the pilot has the sameairspeed of 300 km/h, what headingshould he use to maintain a coursedue north?
(d) What is his new groundspeed?
14. A swimmer is capable of swimming 1.80m/s in still water.
(a) If she aims her body directly across a200.0 m wide river whose current is0.80 m/s, how far downstream (froma point opposite her starting point)will she land?
(b) What is her velocity with respect tothe shore?
(c) At what upstream angle must theswimmer aim if she is to arrive at apoint directly across the stream?
RRHS Physics 5
1.1. INTRODUCTION TO VECTORS CHAPTER 1. DYNAMICS EXTENSION
15. A motorboat whose speed in still water is8.25 m/s must aim upstream at an angleof 25.5o (with respect to a line perpendicu-lar to the shore) in order to travel directlyacross the stream.
(a) What is the speed of the current?
(b) What is the resultant speed of theboat with respect to the shore?
16. A ferryboat, whose speed in still water is2.85 m/s, must cross a 260 m wide riverand arrive at a point 110 m upstream fromwhere it starts. To do so, the pilot musthead the boat at a 45o upstream angle.What is the speed of the river’s current?
17. Which of the following is a vector: veloc-ity, mass, wind speed?
18. The speed of a boat in still water is v.The boat is to make a round trip in a riverwhose current travels at speed u. Derivea formula for the time needed to make around trip of total distance D if the boatmakes the round trip by moving
(a) upstream and back downstream
(b) directly across the river and back.
We must assume u < v; why?
19. A plane’s velocity changes from 200 km/hN to 300 km/h 30o W of N. Find thechange in velocity.
20. A car travelling at 15 m/s N executes agradual turn, so that it then moves at 18m/s E. What is the car’s change in veloc-ity?
21. A plane is flying at 100 m/s E. The pilotchanges its velocity by 30 m/s in a direc-tion 30o N of E. What is the plane’s finalvelocity?
22. A football player is running at a constantspeed in a straight line up the field at an
angle of 15o to the sidelines. The coachnotices that it takes the player 4.0 s toget from the 25 m line to the goal line.How fast is the player running?
23. A pilot wishes to make a flight of 300 kmnortheast in 45 minutes. If there is tobe an 80 km/h wind from the north forthe entire trip, what heading and airspeedmust she use for the flight?
24. A ship leaves its home port expecting totravel to a port 500 km due south. Beforeit can move, a severe storm comes up andblows the ship 100 km due east. How faris the ship from its destination? In whatdirection must the ship travel to reach itsdestination?
25. A hiker leaves camp and, using a compass,walks 4 km E, 6 km S, 3 km E, 5 kmN, 10km W, 8 km N, and 3 km S. At the endof three days, the hiker is lost. Computehow far the hiker is from camp and whichdirection should be taken to get back tocamp.
26. Diane rows a boat at 8.0 m/s directlyacross a river that flows at 6.0 m/s.
(a) What is the resultant velocity of theboat?
(b) If the stream is 240 m wide, how longwill it take Diane to row across?
(c) How far downstream will Diane be?
27. Kyle wishes to fly to a point 450 km duesouth in 3.00 h. A wind is blowing fromthe west at 50 km/h. Compute the properheading and speed that Kyle must choosein order to reach his destination on time.
6 RRHS Physics
CHAPTER 1. DYNAMICS EXTENSION 1.2. FORCE VECTORS
1.2 Force Vectors
In Physics 11, you did many problems apply-ing Newton’s 2nd Law to different situationsusing free body diagrams. This will now beextended to situations where the forces are nolonger solely in the x or y directions. Remem-ber that Newton’s 2nd Law (Fnet = ma) isa vector equation, since it states a relation-ship between acceleration and net force, bothof which are vectors. This means that the ac-celeration and the net force will be in the samedirection; therefore, if we want to use scalaralgebra to solve a problem, we must use thisequation in only one dimension at a time (x ory).
In the diagram below, a man is pulling abox with a rope that makes an angle θ withthe ground.
Note that the expected acceleration (hori-zontal) for this box and the applied force areneither parallel nor perpendicular, so Newton’s2nd Law cannot be applied yet. A free body di-agram for this box would like like this.
Notice that although the normal, friction,and gravity forces are all solely in the x or ydirections, the force of the man pulling is not.This can be fixed if we break this force up intoits components.
As can be seen in the diagram above, all ofthe forces are now either in the x or y directionif we replace Fp with its components. We cannow analyze the forces in each dimension usingNewton’s 2nd Law.
First, the vertical forces. I will take up asthe positive direction; therefore, FN and Fpy
will both be positive and Fg will be negative.
may = ΣFy
may = FN + Fpy − Fg
and0 = FN + Fpy − Fg
since the vertical acceleration is zero. Noticethat FN 6= Fg. Because we often know Fg andFpy, we can solve for FN and use it in our calcu-lation of Ff (remember that Ff = µFN , whereµ is the coefficient of friction).
Now for the horizontal forces:
max = ΣFx
max = Fpx − Ff
This can then be used with the horizontalacceleration. These are not equations tobe memorized and applied to all prob-lems!!! This is a sample analysis of a typicalfree body diagram involving forces at an angle.Analysis should always start with a free bodydiagram.
1.2.1 Inclined Planes
We are now going to apply force vectors andNewton’s second law to an inclined plane (aramp). If we place a box on a ramp (ignoring
RRHS Physics 7
1.2. FORCE VECTORS CHAPTER 1. DYNAMICS EXTENSION
friction for now), as in the following diagram, itcan be observed that there are only two forcesacting on the box - the normal force FN (whichis perpendicular to the surface) and the forceof gravity Fg. Drawing a free body diagram,we get
Notice that these vectors exist in two dimen-sions and are not in component form (they arenot either parallel or perpendicular to one an-other). In order to apply Newton’s second law,we want to analyze the forces one dimensionat a time. Instead of using our usual coordi-nate system containing horizontal and verticalaxes, it makes more sense in this situation torotate our axes so that they are perpendicu-lar and parallel to the surface of the inclinedplane (the same direction as the acceleration).In other words, our x direction will be parallelto the plane and the y direction will by per-pendicular to the plane.
Since the normal force is already perpendic-ular to the plane, only the force of gravity mustbe broken up into components. This can bedone as shown in the following diagram (wherethe Fg from the previous diagram has been en-larged).
The angle θ in the top of the triangle isthe same angle as the slope of the inclined
plane (try showing this using geometry). Us-ing trigonometry, it can be found that the twocomponents are
Fgx = mg sin θ (1.2)
andFgy = mg cos θ (1.3)
We see now by analyzing the perpendicu-lar forces
may = ΣFy
may = FN − Fgy
m(0) = FN − Fgy
since there is no acceleration perpendicular tothe plane, and
FN = Fgy
where Fgy can be found using equation 1.3. Iffriction is present, the normal force can thenbe used in this calculation. Again notice thatFN 6= Fg.
Similarly, the parallel forces can be usedto obtain an expression for the parallel accel-eration on the inclined plane
max = ΣFx
max = Fgx
where Fgx can be found using equation 1.2.Notice that this is just a simple analysis wherefriction and other external forces have not beenincluded; if present, these would have to beconsidered in the force analysis. Again, it isextremely important to draw a free bodydiagram at the start of the problem!
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CHAPTER 1. DYNAMICS EXTENSION 1.2. FORCE VECTORS
1.2.2 Problems
1. A 15.0 kg sled is being pulled along a hor-izontal surface by a rope that is held at a20.0o angle with the horizontal. The ten-sion in the rope is 110.0 N . If the coeffi-cient of friction is 0.30, what is the accel-eration of the sled?
2. A 25.0 kg sled is accelerating at 2.3 m/s2.A force of 300.0 N is pulling the sled alonga rope that is being held at an angle of35o with the horizontal. What is the co-efficient of friction?
3. A 55.0 kg rock is being pulled at a con-stant speed. The coefficient of friction is0.76. If the rope pulling the rock is at a40.0o angle with the horizontal, with whatforce is the rock being pulled?
4. A man pushes a 15 kg lawnmower at con-stant speed with a force of 90 N directedalong the handle, which is at an angle of30o to the horizontal. What is the coeffi-cient of friction?
5. An 18.0 kg box is released on a 33.0o in-cline and accelerates at 0.300 m/s2. Whatis the coefficient of friction?
6. A dead slug (mass is 455 g)is lying on ahill which has an inclination of 15o.
(a) Ignoring friction, what is the acceler-ation of the slug down the hill?
(b) If there is a coefficient of friction of0.20, will the slug slide down the hill?If so, at what acceleration?
(c) How much force is required to pushthe slug up the ramp at a constantspeed?
7. A 165 kg piano is on a 25o ramp. Thecoefficient of friction is 0.30. Jack is re-sponsible for seeing that nobody is killedby a runaway piano.
(a) How much force (and in what direc-tion) must Jack exert so that the pi-ano descends at a constant speed?
(b) How much force (and in what direc-tion) must Jack exert so that the pi-ano ascends at a constant speed?
8. A car can decelerate at -5.5 m/s2 whencoming to rest on a level road. Whatwould the deceleration be if the road in-clines 15o uphill?
9. If a bicyclist (75 kg) can coast down a 5.6o
hill at a steady speed of 7.0 km/h, howmuch force must be applied to climb thehill at the same speed?
10. A 5.0 kg mass is on a ramp that is in-clined at 30o with the horizontal. A ropeattached to the 5.0 kg block goes up theramp and over a pulley, where it is at-tached to a 4.2 kg block that is hangingin mid air. The coefficient of friction be-tween the 5.0 kg block and the ramp is0.10. What is the acceleration of this sys-tem?
11. A physics student is skiing down Ben EoinSki Hill. He wipes out 225 m from thebottom. It takes 13.5 s for him to reachthe bottom. His speed when he wiped outwas approximately 6.0 m/s. If the slopeof the ski hill is 30o, what is the coeffi-cient of friction between the ski hill andthe person’s rear end?
12. A bicyclist can coast down a 4.0o hill at6.0 km/h. The force of friction is propor-tional to the speed v so that Ffr = cv.The total mass is 80 kg.
(a) Find the average force that that mustbe applied in order to descend the hillat 20 km/h.
(b) Using the same power as in (a), atwhat speed can the cyclist climb thesame hill? (Hint: P = Fv)
RRHS Physics 9
1.3. EQUILIBRIUM CHAPTER 1. DYNAMICS EXTENSION
1.3 Equilibrium
You saw in Physics 11 that if two equal butopposite forces are applied to an object, thenet force is zero and the object is said to bein equilibrium.3 This is a somewhat simplifiedview of equilibrium; we will now extend ourdiscussion of equilibrium to two dimensions.
1.3.1 Translational Equilibrium
This is the type of equilibrium discussed ingrade 11. Consider a mass being supported inmidair by two ropes. The mass is stationary;therefore, it is obviously not accelerating. Thenet force must therefore be zero and the objectis said to be in translational equilibrium.
As can be seen by the free-body diagram, thereare three forces acting on the mass. As we said,the net force acting on the mass must be zero;therefore, F1 +F2 +Fg = 0. Remember, theseare vectors so they must add as vectors to bezero, as shown in the following vector diagram:
3A body in equilibrium at rest in a particular refer-ence frame is said to be in static equilibrium; a bodymoving uniformly at constant velocity is in dynamicequilibrium. We will be dealing with mainly static equi-librium, although the net force is zero in both cases.
Note that our vector diagram starts and endsat the same point; therefore, the resultant vec-tor (the net force) is zero.
Since force is a vector, the components ofthe net force on a body in equilibrium musteach be zero, so
ΣFx = 0
andΣFy = 0
Looking at the components in the x and ydirection separately, this tells us that in the xdirection
F2x − F1x = 0
and in the y direction
F1y + F2y − Fg = 0
The requirement that the net force be zerois only the first condition for equilibrium. Thesecond condition will be discussed in the nextsection.
Equilibrant Force If the vector sum of allof the forces acting on an object is not zero,there will be a net force in some direction.There is a single additional force that can beapplied to balance this net force. This addi-tional force is called the equilibrant force.The equilibrant force is equal in magnitude tothe sum of all of the forces acting on the object,but opposite in direction.
1.3.2 Rotational Equilibrium
Even if all of the forces acting on an objectbalance, it is possible for the object not to be
10 RRHS Physics
CHAPTER 1. DYNAMICS EXTENSION 1.3. EQUILIBRIUM
in total equilibrium. Consider a board whereequal forces are applied at opposite ends of theboard, but one up and one down.
Obviously, even though the forces are equaland opposite, the board will begin to spin.It is not in rotational equilibrium. Rota-tional equilibrium refers to the situationwhere there is no rotary motion. To exam-ine this more, we must introduce the notion ofa torque.
A torque has the same relationship to rota-tion as force does to linear movement. It canbe thought of as a twisting force. To measurethe rotating effect of a torque, it is necessaryto choose a stationary reference point for themeasurements (the pivot point). This pivotpoint can be chosen arbitrarily, since the pointof rotation is often not known until the rota-tion begins.4 The further away from this pivot,the greater the torque. A line drawn from thepivot to the force that is providing the torqueis known as the torque arm.
A torque τ is the product of a force multi-plied by a distance from the pivot.
τ = F⊥d (1.4)
where it is only the component of the forcethat is perpendicular to the torque arm thatcontributes to the torque (try opening a doorby pushing parallel to the door). This conceptof multiplying only the perpendicular compo-nents of two vectors is called a cross product.When you calculated work, you multiplied only
4If there is a natural pivot point (for example, on asee-saw) then it usually makes sense to choose this asthe pivot point.
the parallel components of two vectors. Thisis called a dot product. You will learn moreabout these in university.
While forces were described using up, down,left, right, etc., torques are described usingthe terms clockwise and counterclockwise. Aclockwise torque added to an equal (in mag-nitude) counterclockwise torque will be zero.Rotational equilibrium is attained if the sumof all of the torques is zero.
Στ = 0
This is the second condition for equilibrium.As can be seen from equation 1.4, the units
for torque are usually N ·m (this is not calleda Joule, as it was when discussing work; whencalculating the work, the force and the dis-placement used had to be parallel).
As we have seen, there are two conditionsfor equilibrium: that the sum of the forces iszero (translational equilibrium), and that thesum of the torques is zero (rotational equi-librium). An equilibrant force should provideboth translational and rotational equilibrium.When finding an equilibrant force to satisfyboth of these conditions, it is necessary to findboth the force itself (magnitude and direction)and the location of application.
Centre of Gravity One of the forces ofteninvolved in calculating the torques on an ob-ject is the force of gravity. Before dealing withtorques, we were not usually concerned withthe location of the force on a body, but forcalculating torques, this is important. Wheredoes gravity act on a body? Of course, it actson every particle in the body, but there is apoint called the centre of gravity (cg) wherethe entire force of gravity can be considered tobe acting. The center of gravity is the pointat which we could apply a single upward forceto balance the object. For a mass with a uni-form distribution of mass (such as a ruler), thecenter of gravity would be in the center of themass (the middle of the ruler).
RRHS Physics 11
1.3. EQUILIBRIUM CHAPTER 1. DYNAMICS EXTENSION
1.3.3 Problems
1. A 20.0 kg sack of potatoes is suspendedby a rope. A man pushes sideways witha force of 50.0 N . What is the tension inthe rope?
2. A high wire is 25.0 m long and sags 1.0m when a 50.0 kg tightrope walker standsin the middle. What is the tension in thewire? Is it possible to apply enough ten-sion in the wire to eliminate the sag com-pletely? Explain.
3. Joe wishes to hang a sign weighing 750N so that cable A attached to the storemakes a 30o angle as shown in the picturebelow. Cable B is attached to an adjoiningbuilding. Calculate the necessary tensionin cable B.
4. Find the unknown mass in the diagrambelow:
5. A sign with a mass of 1653.7 kg is sup-ported by a boom and a cable. The ca-ble makes an angle of 36o with the boom.Find the tension in the boom and the ca-ble.
6. Find the tensions T1 and T2 in the twostrings indicated:
7. Two tow trucks attach ropes to a strandedvehicle. The first tow truck pulls with aforce of 25000 N , while the second truckpulls with a force of 15000 N . The tworopes make an angle of 15.5o with eachother. Find the resultant force on the ve-hicle.
8. You mother asks you to hang a heavypainting. The frame has a wire across theback, and you plan to hook this wire overa nail in the wall. The wire will break ifthe force pulling on it is too great, and youdon’t want it to break. If the wire mustbe fastened at the edges of the painting,should you use a short wire or a long wire?Explain.
9. When lifting a barbell, which grip will ex-ert less force on the lifter’s arms: one inwhich the arms are extended straight up-ward from the body so that are at rightangles to the bars, or on in which the armsa re spread apart so that the bar is grippedcloser to the weights? Explain.
10. A 40 kg iceboat is gliding across a frozenlake with a constant velocity of 14 m/s E,when a gust of wind from the southwestexerts a constant force of 100 N on its sailsfor 3.0 s. With what velocity will the sledbe moving after the wind has subsided?Ignore any frictional forces.
12 RRHS Physics
CHAPTER 1. DYNAMICS EXTENSION 1.3. EQUILIBRIUM
11. Three students are pulling ropes that areattached to a car. Barney is pulling northwith a force of 235 N ; Wilma is pullingwith a force of 175 N in a direction 23o
E of N; Betty is pulling with 205 N east.What equilibrant force must a fourth stu-dent, Fred, apply to prevent acceleration?
12. A force of 500.0 N applied to a rope heldat 30.0o above the surface of a ramp is re-quired to pull a wagon weighing 1000.0 Nat a constant velocity up the plane. Theplane has a base of 14.0 m and a length of15.0 m. What is the coefficient of friction?
13. If there is a spring on the door 5.0 cm fromthe hinges which exerts a force of 60.0 N,how much force must be used to open thedoor if the force is applied at the outeredge of the door? How much force mustbe used if the force is applied 15 cm fromthe hinges? Assume that the door is 90.0cm wide.
14. A 60.0 kg person is sitting 1.2 m from thepivot on a see-saw. A 50.0 kg person issitting 0.90 m away from the pivot on theother side. Where must a 22.0 kg child sitto balance the see-saw?
15. A long platform is holding your physicsteacher in the air above some hungry al-ligators. Your physics teacher has a massof 75 kg and is located 2 m from one end.The 10.0 m platform has a mass of 10.0kg, and its center of gravity is located 4.0m from the same end. The platform isbeing held up by two students, one at ei-ther end. What force is required by eachstudent to hold the platform up?
16. Find the size and correct location for thesingle force which will stabilize the follow-ing beam:
17. Find the equilibrant force:
18. In the following diagram, determine themagnitude, direction, and point of appli-cation of the necessary equilibrant force.
19. Calculate the forces F1 and F2 that thesupports exert on the diving board whena 50.0 kg person stands at its tip.
(a) ignoring the mass of the board(b) If the board has a mass of 40.0 kg
(uniformly distributed)
RRHS Physics 13
1.3. EQUILIBRIUM CHAPTER 1. DYNAMICS EXTENSION
14 RRHS Physics
Chapter 2
2-D Motion
2.1 Projectiles
An object that is launched in the air followsa trajectory and is called a projectile. Themotion of a projectile is described in terms ofits position, velocity, and acceleration. Theseare all vector quantities, and we are going toapply our knowledge of vectors to analyze thismotion.
2.1.1 Objects Launched Horizon-tally
Consider a train that drives horizontally off theedge of a cliff, as seen in the picture below:
Notice that the train follows a parabolic tra-jectory.1 We have already discussed this yearthat horizontal and vertical motion are inde-pendent of one another; only a horizontal forcecan contribute to horizontal motion and only avertical force can contribute to vertical motion.
Ignoring air resistance, a free body diagramof the train (after it has left the ground) wouldlook like this
1We can show this later on.
Horizontal Motion Notice that there areNO horizontal forces acting on the train!There is no force either speeding up or slow-ing down the train horizontally (as long as weare ignoring air resistance); therefore, sincemax = ΣFx, there is no horizontal accelera-tion. The horizontal speed does not change.This makes the horizontal analysis very easy— all analysis of the motion can be performedusing the equation
dx = vxt (2.1)
where dx is the horizontal distance travelled,vx is the horizontal speed, and t is the time inthe air.
Vertical Motion Looking at the verticalforces in our free body diagram, we see thatthere is only one - gravity. This also makesthings somewhat simple, since we now knowthat the vertical acceleration is going to be 9.8m/s2 (assuming that we are at the surface ofthe earth and we are ignoring air resistance).Since we know our vertical acceleration, all ofour motion equations for acceleration can be
15
2.1. PROJECTILES CHAPTER 2. 2-D MOTION
used.dy = vyit +
12at2 (2.2)
dy =v2yf − v2
yi
2a(2.3)
dy =vyi + vyf
2t (2.4)
where dy is the vertical displacement, vyi is theinitial vertical velocity, vyf is the final verticalvelocity, t is the time in the air, and a is theacceleration due to gravity. Since in this sec-tion we are dealing with horizontally launchedprojectiles, vyi will be zero in equations 2.2,2.3, and 2.4.
Remember from grade 11 that you must usethe appropriate sign conventions for up anddown for each quantity. Notice that the onequantity that the horizontal and vertical mo-tion have in common is t, the time in the air.For this reason, you will find yourself most of-ten using equations 2.1 and 2.2 as both of theseequations make use of this quantity.
2.1.2 Objects Launched at an Angle
We are now going to analyze an object that islaunched at an angle, instead of horizontally.The analysis is essentially the same as that forthe horizontally launched projectile. In thiscase, we are not usually given a horizontal andvertical speed; The object does, however, havea velocity that can be resolved into horizontaland vertical components.
Consider a soccer ball that is kicked in theair as shown below:
Extremely Important!! The arrow in thediagram above represents the velocity vector
for the soccer ball, not the ball’s actual path!The direction of the arrow indicates the ball’sinitial direction, and the length of the vector(if drawn to scale) indicates its magnitude. Re-member, the ball follows a parabolic path; itdoes not follow a straight line!!!
Your first step in any problem with an objectlaunched at an angle should be to resolve theobject’s velocity into its components, as shownin the diagram below.
This is done using trigonometry as shown backin section 1.1. Once this is done, the analysiscan be done as it was for the horizontal projec-tiles, namely using equations 2.1 to 2.4. Again,remember to keep your horizontal and verticalmotion separate from one another and to becareful with your sign conventions.
The horizontal speed vx is constant, sincethere are no horizontal forces. The verticalspeed vy is initially upward in this example,but gravity will act to slow it down. As theball rises, the vertical speed gets smaller andsmaller, until it reaches zero at its highestpoint. The ball then begins speeding up ver-tically downward and continues speeding upuntil it returns to the ground.
If a projectile such as the ball above leavesthe ground and returns to the same height (theground), then the vertical displacement dy iszero (why?). The horizontal distance travelleddx is called the range in this situation.
Notice that equation 2.2 is a quadratic equa-tion if t is an unknown; therefore, you mayhave to use the quadratic formula from timeto time
t =−b±
√b2 − 4ac
2a(2.5)
16 RRHS Physics
CHAPTER 2. 2-D MOTION 2.1. PROJECTILES
2.1.3 Problems
1. A diver running 3.6 m/s dives out hori-zontally from the edge of a vertical cliffand reaches the water below 2.0 s later.How high was the cliff and how far fromits base did the diver hit the water?
2. A hunter aims directly at a target (onthe same level) 220 m away. If the bul-let leaves the gun at a speed of 550 m/s,by how much will it miss the target?
3. An athlete throws the shotput with an ini-tial speed of 14 m/s at a 40o angle to thehorizontal. The shot leaves the shotput-ter’s hand at a height of 2.2 m above theground. Calculate the horizontal displace-ment travelled.
4. An Olympic longjumper is capable ofjumping 8.0 m. Assuming his horizontalspeed is 9.0 m/s as he leaves the ground,how long was he in the air and how highdid he go?
5. A sniper on a building is trying to hit atarget on the ground. The building is 13.0m high. The sniper aims his rifle at apoint 19.5 m away from the building inorder to hit the target. If the bullet travelsat 135 m/s, how far from the building isthe target?
6. A football is kicked with a speed of 21.0m/s at an angle of 37o to the horizontal.How much later does it hit the ground?
7. A basketball player tries to make a half-court jump-shot, releasing the ball at theheight of the basket. Assuming the ballis launched at 51.0o, 14.0 m from the bas-ket, what velocity must the player give theball?
8. A hunter is trying to shoot a monkeyhanging from a tree. As soon as the hunterfires, the monkey is going to let go of the
tree. Should the hunter aim directly at,above, or below the monkey in order tohit him?
9. Trailing by two points, and with only 2.0s remaining in a basketball game, Patmakes a jump-shot at an angle of 60o withthe horizontal, giving the ball a velocityof 10 m/s. The ball is released at theheight of the basket, 3.05 m above thefloor. YES! It’s a score.
(a) How much time is left in the gamewhen the basket is made?
(b) The three-point line is a distance of6.02 m from the basket. Did the Pattie the game or put his team ahead?
10. A baseball is hit at 30.0 m/s at an an-gle of 53.0o with the horizontal. Immedi-ately, an outfielder runs 4.00 m/s towardthe infield and catches the ball at the sameheight it was hit. What was the originaldistance between the batter and the out-fielder?
11. A football is kicked at an angle of 37o withthe horizontal with a velocity of 20.0 m/s.The field goal poles are 31.0 m away andare 3.5 m high. Is the field goal good?
12. A person is in a moving elevator. Hethrows a rotten egg horizontally out ofthe moving elevator with a velocity of 5.0m/s. At the time of the throw, the ele-vator was 8.7 m above the ground. Therotten egg landed 4.2 m away from theelevator. What was the velocity of the el-evator? Was the elevator moving up ordown?
13. An airplane is in level flight at a velocity of500 km/h and an altitude of 1500 m whena wheel falls off. What horizontal distancewill the wheel travel before it strikes theground and what will the wheel’s velocitybe when it strikes the ground?
RRHS Physics 17
2.1. PROJECTILES CHAPTER 2. 2-D MOTION
14. (a) Show that the range R of a projec-tile, which is defined as the horizon-tal distance travelled when the finalpoint is at the same level as the ini-tial point, is given by the equation
R =v2 sin 2θ
g
where v is the initial velocity of theprojectile and θ is the angle with thehorizontal. (Hint: use the trigono-metric identity sin 2θ = 2 sin θ cos θ)
(b) Assuming that the initial velocity isv, what angle will provide the maxi-mum range?
15. What minimum initial velocity must aprojectile have to reach a target 90.0 maway?
16. Police agents flying a constant 200.0 km/hhorizontally in a low-flying airplane wishto drop an explosive onto a master crimi-nal’s car travelling 130 km/h (in the samedirection) on a level highway 78.0 m be-low. At what angle (with the horizontal)should the car be in their sights when thebomb is released?
17. A basketball leaves a player’s hands at aheight of 2.1 m above the floor. The bas-ket is 2.6 m above the floor. The playerlikes to shoot the ball at a 35o angle. If theshot is made from a horizontal distance of12.0 m and must be accurate to ±0.22 m(horizontally). what is the range of initialspeeds allowed to make the basket?
18. A ball is thrown horizontally from the topof a cliff with initial speed vo. At anymoment, its direction of motion makes anangle of θ with the horizontal. Derive aformula for θ as a function of time.
19. Two baseballs are pitched horizontallyfrom the same height but at different
speeds. The fatser ball crosses home platewithin the strike zone, but the slower oneis below the batter’s knees. Why does thefaster ball not fall as far as the slower one?After all, they travel the same distanceand accelerate down at the same rate.
20. A teflon hockey puck slides without fric-tion across a table at constant velocity.When it reaches the end of the table, itflies of and lands on the ground. For eachof the following questions, draw all vectorsto scale.
(a) Draw the situation above, drawingvectors showing the force on the puckat two positions while it is on the ta-ble and at two more while it is in theair.
(b) Draw vectors showing the horizon-tal and vertical components of thepuck’s velocity at the four points.
(c) Draw the total velocity vector at thefour points.
21. Suppose an object is thrown with thesame initial velocity on the moon, whereg is one-sixth as large as on Earth. Willthe following quantities change? If so, willthey become larger or smaller?
(a) vxi and vyi
(b) time of flight
(c) maximum height
(d) range
18 RRHS Physics
CHAPTER 2. 2-D MOTION 2.2. SIMPLE HARMONIC MOTION
2.2 Simple Harmonic Motion
When a mass is hung on a spring, a force equalto the weight of the mass is exerted on thespring, which causes the spring to stretch; itwill often be found that this is a linear rela-tionship. If you double the mass hanging onthe spring, you will double the distance thespring stretches. The spring exerts an equaland opposite force on the mass. This force canbe given by the relationship
F = kx (2.6)
where k is what is known as the spring constantand x is the displacement of the spring in me-tres (how far it stretched from the equilibriumposition). The relationship is sometimes givenas F = −kx, where F is the restoring force ofthe spring and the negative sign indicates thatthis force is in the opposite direction of thedisplacement x. This relationship is known asHooke’s Law.
The spring constant k is constant for anygiven spring, but is dependent on the spring;different springs will have different spring con-stants. The units for the spring constant areN/m, meaning that a spring constant of 45N/m indicates that it would take 45 N tostretch this spring 1 m (assuming that thislength was within the limits of the spring; ifyou exceed the limits of the spring, this for-mula no longer holds).
Consider a spring that is allowed to hangvertically with no mass attached. (See Fig2.1a). Notice that the spring has a naturallength to which it always wants to return ifyou stretch or compress it. This is the equilib-rium position.
Suppose that you place a mass on the spring(see Fig 2.1b)). The mass will cause the springto stretch a certain distance, depending on itsspring constant. This is now its new equilib-rium position - at this point, the force exertedby the spring upwards is equal to the force ex-erted by gravity downwards. Suppose that you
Figure 2.1: Simple Harmonic Motion
now pull this mass down a bit (Fig 2.1c)andlet it go. What happens? You should noticethat it bobs up and down repeatedly. Whenthe mass is below its equilibrium position, thespring exerts a greater force than the forceof gravity and provides an upward accelera-tion. When the spring is above the equilib-rium point, the spring exerts a smaller forcethan gravity, which results in a downward ac-celeration.2 This type of oscillation (when therestoring force follows Hooke’s Law) is referredto as simple harmonic motion. The period(the time for one complete vibration, or oscil-lation) of this motion in seconds is given by
T = 2π
√m
k(2.7)
where m is the mass in kg and k is the springconstant again. Also, remember from grade 11that frequency is the inverse of period (f =1/T ). Simple harmonic motion can be appliedto many real world situations : a raft bobbingup and down in the water, the suspension of acar, a mattress, suspension bridges, etc.
2Of course, we can also have simple harmonic mo-tion with a horizontal spring; in this case, the springitself exerts a force towards equilibrium as it is com-pressed or stretched.
RRHS Physics 19
2.2. SIMPLE HARMONIC MOTION CHAPTER 2. 2-D MOTION
2.2.1 Conservation of Energy
When we stretch or compress a spring, workis done on the spring; therefore, a compressedor stretched spring will have potential energy.Remember that
∆E = W
so∆E = Fd
But F is not constant; it increases linearly aswe move away from equilibrium (Eq 2.6). Sothe average force exerted will be F = 1
2kx and
∆E = (12kx)(x)
or, since the increase in energy becomes thepotential energy of the spring,
Ep =12kx2 (2.8)
where k is the spring constant of the spring(in N/m)and x is the displacement from equi-librium (in m).
Consider a spring supporting a mass wherethe mass is pulled a distance x from its restposition and then released.
Since the total mechanical energy of a sys-tem is the sum of the kinetic and potential en-ergies of that system, the total energy of anoscillating system can be given by3
3If we are dealing with a vertically held spring that
Et =12mv2 +
12kx2 (2.9)
If no energy is being introduced to, or re-moved from, the system, the total energy re-mains the same. At equilibrium, x = 0 andall of the energy is kinetic; at the maximumdisplacement (the amplitude A), v = 0 andall of the energy is potential. The total en-ergy of the system can therefore be expressedas Et = 1
2kA2.
2.2.2 Pendulum Motion
For small displacements (θ less than ≈ 15o), itcan be shown that a pendulum exhibits simpleharmonic motion with a spring constant of
k =mg
L
where L is the length of the pendulum. Sub-stituting this into Eq 2.7 we get
T = 2π
√l
g(2.10)
Notice that the period of a pendulum doesnot depend on its mass!
is supporting a mass, then there is also gravitationalpotential energy involved in the system; however, thiscan be ignored if all displacements (x) are measuredfrom the new equilibrium position (b) shown in Fig 2.1instead of the original equilibrium position (a).
20 RRHS Physics
CHAPTER 2. 2-D MOTION 2.2. SIMPLE HARMONIC MOTION
2.2.3 Problems
1. A piece of rubber is 45 cm long when aweight of 8.0 N hangs from it and is 58cm long when a weight of 12.5 N hangsfrom it. What is the spring constant ofthis piece of rubber?
2. If a particle undergoes SHM with an am-plitude A, what is the total distance ittravels in one period?
3. When an 80.0 kg person climbs into an1100 kg car, the car’s springs compressvertically by 1.2 cm. What will be thefrequency of vibration when the car hits abump?
4. A spring vibrates with a frequency of 2.4Hz when a weight of 0.60 kg is hung fromit. What will its frequency be if only 0.30kg hangs from it?
5. A mass m at the end of a spring vibrateswith a frequency of 0.62 Hz; when an ad-ditional 700 g mass is added to m, thefrequency is 0.48 Hz. What is the value ofm?
6. A 300 kg wooden raft floats on a lake.When a 75 kg man stands on the raft,it sinks deeper into the water by 5.0 cm.When the man steps off, the raft vibratesbriefly. What is the frequency of vibra-tion?
7. A small cockroach of mass 0.30 g is caughtin a spider’s web. The web vibrates at afrequency of 15 Hz. At what frequencywould you expect the web to vibrate if aninsect of mass 0.10 g were trapped?
8. A mass of 2.70 kg stretches a verticalspring 0.325 m. If the spring is stretchedan additional 0.110 m and released, howlong does it take to reach the (new) equi-librium position again?
9. It takes a force of 60 N to compress thespring of a popgun 0.10 m to load a 0.200kg ball. With what speed will the ballleave the gun?
10. How much would a spring scale with k =120 N/m stretch, if it had 3.75 J of workdone on it?
11. A block of mass 0.50 kg is placed on alevel, frictionless surface, in contact witha spring bumper, with a spring constant of100 N/m that has been compressed by anamount 0.30 m. The spring, whose otherend is fixed, is then released. What is thespeed of the block at the instant when thespring is still compressed by 0.10 m?
12. A spring stretches 0.150 m when a 0.30kg mass is hung from it. The spring isthen stretched an additional 0.100 m fromthis equilibrium point and released. De-termine:
(a) The maximum velocity(b) The velocity when the mass is 0.050
m from equilibrium(c) The maximum acceleration.
13. A geologist’s simple pendulum, whoselength is 37.10 cm, has a frequency of0.8190 Hz at a particular location. Whatis the acceleration of gravity?
14. How long must a pendulum be to make ex-actly one complete vibration per second?
15. Given the following position-time graphfor a simple harmonic oscillator, drawthe appropriate velocity-time graph andacceleration-time graph for the oscillator.
RRHS Physics 21
2.3. 2D COLLISIONS CHAPTER 2. 2-D MOTION
2.3 2D Collisions
As with many of our topics so far in this course,we are now going to look at one of our grade11 topics (collisions), and extend our analysisto two dimensions.
2.3.1 Conservation of Momentum
You learned in grade 11 that the total mo-mentum of an isolated system remains con-stant. Also, if you remember from grade 11,momentum is a product of mass and velocity(p = mv). Since velocity is a vector, so is mo-mentum. This vector nature of momentum be-comes extremely important in two dimensionalcollisions.
When you analyzed one dimensional colli-sions, you could show that in an isolated sys-tem the momentum of each object before thecollision added up to equal the total momen-tum after the collision. This still applies intwo dimensional collisions, but remember thatmomentum is a vector so it must be added asa vector!! For a collision involving two objectsin one dimension, you would write
pa + pb = p′a + p′b (2.11)
or, since p = mv,
mava + mbvb = mav′a + mbv′b (2.12)
where primed quantities (′) mean after the col-lision and unprimed mean before the collision.The vector nature of the momentum could beaddressed in this one dimensional situation us-ing positive or negative values for the veloci-ties.
In two dimensions, the vector nature of mo-mentum does not allow simple algebraic op-erations using equation 2.12. Although youcan still express the conservation of momen-tum using equations 2.11 and 2.12, the specialattention must be paid to the vector nature ofmomentum. To add momentum vectors in twodimensions, a vector diagram must be drawn.
Equation 2.12 could only be used algebraicallyif you first break the vectors into componentsand then apply the equation in each dimension.
Consider the example of a ball moving to theright that collides with another ball at rest.
If the collision is not head on, the two ballswill go in different directions after the collision.
Just as with one dimensional collisions, thesum of all of the momentum vectors after thecollision (p′a and p′b) is equal to the total of themomentum vectors before the collision (pa).
pa = p′a + p′b (2.13)
Since momentum is a product of mass (ascalar) and velocity (a vector), the momen-tum vector for an object will be in the samedirection as the velocity vector of the object;however, remember that it is momentum thatis conserved, not velocity. Do not draw a ve-locity vector diagram when solving theseproblems! The momentum vector diagramfor equation 2.13 would look like this:
where pt is really just pa, since there is onlyone momentum vector before the collision.
The individual momentum vectors can befound using the formula p = mv. We can now
22 RRHS Physics
CHAPTER 2. 2-D MOTION 2.3. 2D COLLISIONS
use our usual methods of component analysisfor solving vector problems.
If we draw our components into the momen-tum vector diagram, we see that the momen-tum is conserved in each dimension.
In other words, the sum of the x components ofmomentum before the collision are equal to thesum of the x components after the collision.
pa = p′ax + p′bx
where the momentum components can befound using the appropriate velocity compo-nents (p′ax = mav
′ax and p′bx = mbv
′bx).
Similarly the sum of the y components ofmomentum before the collision are equal to thesum of the y components after the collision.Since the original y momentum is zero in thisexample, the y momentum after the collisionis still zero
0 = p′ay − p′by
2.3.2 Elastic and Inelastic Collisions
Elastic Collisions As you learned in grade11, an elastic collision is one in which no kineticenergy is lost; the total kinetic energy of theparticles before the collision is the same as thetotal kinetic energy of the particles after thecollision. For a two body collision, this wouldbe expressed as
12mav
2a +
12mbv
2b =
12mav
′2a +
12mbv
′2b (2.14)
Remember that energy is not a vector; there-fore, it is only the magnitude of the velocitythat is used in Eq 2.14.
Consider the special case where particle b isinitially at rest. We now have
12mav
2a =
12mav
′2a +
12mbv
′2b
If the mass of each particle is the same, thenafter cancelling the mass and the factor of onehalf, our conservation of energy equation (2.14)reduces to
v2a = v′2a + v′2b (2.15)
which is really an expression of thepythagorean theorem. Since the massesare equal, the velocity vectors are propor-tional to the momentum vectors. A velocityvector diagram in this situation4 wouldtherefore show that the vectors v′a and v′bwould add to give the vector va. Since themagnitudes of these vectors are related bythe pythagorean theorem, the vector diagrammust be a right angle triangle.
In other words, v′a and v′b (and p′a and p′b)are perpendicular to one another; after thiscollision, the two particles move off at rightangles to one another. Remember, though,that this is only true for the special casewhere the two objects have the samemass, the collision is elastic, and one ofthe particles is initially at rest.
Inelastic Collisions An inelastic collision isone in which the kinetic energy is not con-served; some of the energy is transformed intoother types of energy, such as thermal energy.A completely inelastic collision is one in whichthe objects stick together; some energy is lost,but a completely inelastic collision does notmean that all of the energy is lost. In thistype of collision, it may be possible to calcu-late the amount of energy lost by comparingthe total initial kinetic energy with the totalfinal kinetic energy.
4A velocity vector diagram can be applied here onlybecause the masses are all the same; therefore, everyvelocity vector is multiplied by the same factor to ob-tain the corresponding momentum vector.
RRHS Physics 23
2.3. 2D COLLISIONS CHAPTER 2. 2-D MOTION
2.3.3 Problems
1. A collision between two vehicles occurs ata right angled intersection. Vehicle A is acar of mass 1800 kg travelling at 60 km/hnorth. Vehicle B is a delivery truck ofmass 3500 kg initially travelling east at 45km/h. If the two vehicles remain stuck to-gether after the impact, what will be theirvelocity after the impact? How much ki-netic energy was lost in the collision?
2. A radioactive nucleus at rest decays intoa second nucleus, an electron, and a neu-trino. The electron and neutrino are emit-ted at right angles and have momenta of8.6×10−23 kg·m/s and 6.2×10−23 kg·m/s.What is the magnitude and direction ofthe momentum of the recoiling nucleus?
3. A collision investigator is called to an ac-cident scene where two vehicles collidedat a right-angled intersection. From skidmarks, the investigator determined thatcar A, mass 1400 kg was travelling 50km/h west before impact. The two vehi-cles remained stuck together after impactand the velocity of the cars after impactwas 10 km/h in a direction 30o W of N.
(a) What was the mass of car B?
(b) How fast was car B travelling beforethe accident?
4. Two streets intersect at a 40o angle. CarA has a mass of 1500 kg and is travellingat 50 km/h. Car B has a mass of 1250 kgand is travelling 60 km/h. If they collideand remain stuck together, what will bethe velocity of the combined mass imme-diately after impact?
5. A billiard ball of mass 0.400 kg movingwith a speed of 2.00 m/s strikes a secondball, initially at rest, of mass 0.400 kg.The first ball is deflected off at an angleof 30o with a speed of 1.20 m/s. Find
the speed and direction of the second ballafter the collision.
6. A proton travelling with speed 8.2 × 105
m/s collides elastically with a stationaryproton. One of the protons is observedto be scattered at a 60o angle. At whatangle will the second proton be observed,and what will be the velocities of the twoprotons after the collision?
7. A particle of mass m travelling with aspeed v collides elastically with a targetparticle of mass 2m (initially at rest) andis scattered at 90o.
(a) At what angle does the target parti-cle move after the collision?
(b) What are the particles’ final speeds?(c) What fraction of the initial kinetic
energy is transferred to the targetparticle?
8. A billiard ball is moving North at 3.00m/s, and another is moving East with aspeed of 4.80 m/s. After the collision (as-sumed elastic), the second ball is movingNorth. What is the final direction of thefirst ball, and what are their final speeds?
9. A billiard ball of mass ma = 0.40 kgstrikes a second ball, initially at rest, ofmass mb = 0.60 kg. As a result of thiselastic collision, ball A is deflected at anangle of 30o and ball B at 53o. What is theratio of their speeds after the collision?
10. Two cars collide at an intersection. Thefirst car has a mass of 925 kg and was trav-elling North. The second car has a massof 1075 kg and was travelling West. Im-mediately after impact, the first car hada velocity of 52.0 km/h, 40.0o North ofWest, and the second car had a velocityof 40.0 km/h, 50.0o North of West. Whatwas the speed of each car prior to the col-lision?
24 RRHS Physics
Chapter 3
Planetary Motion
3.1 Uniform Circular Motion
We know from Newton’s First Law of Motionthat an object with no net force acting on itwill continue to move in a straight line at aconstant speed. If a force acts on the objectparallel to the direction of motion, the objectwill speed up or slow down. We also saw withprojectiles that if a force acts perpendicularto the motion, the object moves in a curve.With projectile motion, however, the force act-ing (gravity) was always perpendicular to theoriginal direction of motion. We will now lookat the situation where the force acts so that itchanges direction and is always perpendicularto the motion.
If we consider a force that is always perpen-dicular to the motion, we realize that the speedof the object should not change.1 An objectthat moves in a circle at constant speed is saidto undergo uniform circular motion.
3.1.1 Centripetal Acceleration
Since the force is never in the same directionas the motion, there will be no acceleration inthe direction of motion; in other words, theobject will not speed up or slow down. Thereis, however, an acceleration present. Remem-ber from grade 11 that acceleration was de-fined as the change of velocity with time, not
1Since the force is never in the direction of the mo-tion, the acceleration is never in the direction of themotion.
the change of speed. So even though the speedis not changing, there is still an acceleration.
Consider an object revolving at the end ofa string in a circle. Note that the velocity isalways tangential to the circular motion (it isalways perpendicular to the string).
To calculate the speed of the object, we cansimply use
v =d
t(3.1)
and since the distance travelled in one periodT is the circumference (2πr), we get
v =2πr
T(3.2)
The only force acting on the object is thestring, which is pulling inward. Since this isthe only force, the acceleration must also beinward. This inward acceleration is what iscalled the centripetal acceleration. Know-ing that the acceleration is always perpendic-ular to the velocity, and if we rearrange thevelocity vectors so that they all start from thesame point in our diagram, we see
25
3.1. UNIFORM CIRCULAR MOTION CHAPTER 3. PLANETARY MOTION
You can see that this diagram is very similarto our first one, but where r in the first one hasbeen replaced with v, and v in the first onehas been replaced by a. Looking at equation3.2, the corresponding equation for the seconddiagram would be
a =2πv
T(3.3)
Combining equations 3.2 and 3.3, we get theequation for the magnitude of the centripetalacceleration
ac =v2
r(3.4)
This centripetal acceleration is, by defini-tion, always inward toward the center of thecircle. Also note that the units for this accel-eration are still m/s2.
3.1.2 Centripetal “Force”
The word “Force” in this heading is in quotesbecause it should not be confused with an ac-tual force on an object. It is in reality anotherterm for the net force acting on an object thatis exhibiting a centripetal acceleration. In fact,when solving centripetal force problems, we aredoing nothing more than applying Newton’sSecond Law
Fnet = ma (3.5)
If the acceleration is a centripetal accelera-tion, then equation 3.5 becomes
Fc = mac (3.6)
where you can see that the centripetal force Fc
is just the net force required for a particular
centripetal acceleration. Centripetal force isnot, however, an actual force and should notbe included in any free body diagram. This isa common misconception of students. In ourexample of an object being swung in a circle ona string, the only force acting on the object isthe force exerted by the string; this providesthe required centripetal force for circular mo-tion.
To summarize the directions of each of thevectors that have been discussed (see figure 3.1below), consider an object being swung by astring at constant speed on a frictionless, hor-izontal surface.
1. the centripetal force (which is a combi-nation of all of the actual forces acting onthe object) is always directed toward thecenter of the circle;
2. the centripetal acceleration is also al-ways directed toward the center of the cir-cle;
3. the velocity is perpendicular to the ra-dius of the circle (tangential)
Figure 3.1: This is not a free body diagram; itjust shows the direction of the three quantities.
Vertical Circles Consider the case of an ob-ject being swung in a vertical circle; in particu-lar, we will look first at the object at its lowestpoint in the circle. There are only two forcesacting on the object — The force of gravity Fg
26 RRHS Physics
CHAPTER 3. PLANETARY MOTION 3.1. UNIFORM CIRCULAR MOTION
and the tension of the string T . Drawing a freebody diagram of this situation would look likethis:
Notice that there is no centripetal force inthis diagram! The acceleration (centripetal) inthis case is upward; we will also choose the up-ward direction to be upward. Applying New-ton’s Second Law to this situation, we get
mac = Fc
mac = T − Fg
where we have made T positive because it isupward and Fg negative because it is down-ward. Remember, also, that ac can be foundusing ac = v2/r.
3.1.3 Centrifugal Force
The term centrifugal force (“center-fleeing”) isprobably one that you have heard before. It isa common misconception that circular motionintroduces a force on an object that is directedaway from the center of the circle. When youare spinning a ball around in a circle, you knowthat you feel a force pulling outward on yourhand. This is wrongly interpreted as an out-ward force on the ball which is transmittedalong the string to your hand. We have already
seen that the force required to move in a cir-cle is inward (since the acceleration is inward),not outward. Your hand is actually exertingan inward force on the ball; because of New-ton’s Third Law, the ball exerts an equal butopposite force on your hand. The term cen-trifugal force is used to explain this apparentsensation of being pulled outward. Centrifugalforce is what is called a pseudoforce — it is nota real force.
In this situation, the ball is not being pushedoutward; it is, in fact, being pulled inward bythe string. Newton’s First Law states that ob-jects in motion continue in motion at a con-stant velocity. If you break the string, the ballwill fly off in the direction of the velocity2 thatit had when the string broke. If there were, infact, some centrifugal force pushing outwardon the ball, the ball would fly outward awayfrom the center of the circle.
Centrifugal force is simply a term used toexplain the apparent force that a rotating ob-ject experiences. Pretend you are the ball inour example; because of inertia, you wouldnaturally want to travel in a straight line.You are moving in a circle (away from thisstraight line path); from your point of view(a rotating reference frame), it would appearthat some force is trying to push you backto this straight line path (your natural ten-dency). This “fake” force has been called thecentrifugal force. Someone watching from anon-rotating reference frame (for example, afixed position above the rotating ball) wouldobviously see that there is only a force actinginward on the ball and that you simply wantto keep going straight because of your inertia.
2tangent to the circle
RRHS Physics 27
3.1. UNIFORM CIRCULAR MOTION CHAPTER 3. PLANETARY MOTION
3.1.4 Problems
1. A 150 g ball at the end of a string is swing-ing in a horizontal circle of radius 1.15 m.The ball makes exactly 2.00 revolutions ina second. What is its centripetal acceler-ation?
2. The moon’s nearly circular orbit aboutthe earth has a radius of about 385,000km and a period of 27.3 days. Determinethe acceleration of the moon towards theearth.
3. Sue whirls a yo-yo in a horizontal circle.The yo-yo has a mass of 0.20 kg and isattached to a string 0.80 m long.
(a) If the yo-yo makes 1.0 complete rev-olution each second, what force doesthe string exert on it?
(b) If Sue increases the speed of theyo-yo to 2.0 revolutions per second,what force does the string now exert?
4. How large must the coefficient of frictionbe between the tires and the road if a 1600kg car is to round a level curve of radius62 m at a speed of 55 km/h?
5. A 1000 kg car rounds a curve on a flat roadof radius 50 m at a speed of 50 km/h. Willthe car make the turn if (a) the pavementis dry and the coefficient of static frictionis 0.60, (b) the pavement is icy and µ =0.20?
6. What is the maximum speed at whicha car can safely travel around a circulartrack of radius 80.0 m if the coefficient offriction between the tire and the road is0.30?
7. A gravitron circus ride has a 2.0 m radiusand rotates 1.1 times per second.
(a) Draw a free body diagram indicatingall of the forces involved.
(b) What coefficient of friction is nec-essary to prevent the people fromfalling?
8. What minimum speed must a rollercoaster be travelling when upside down atthe top of a circle if the passengers are notto fall out. Assume a radius of curvatureof 8.0 m.
9. A cat is stuck in a washing machine whileit is in spin mode. The diameter of thewashing machine is 65 cm. If the coeffi-cient of friction between the cat and thevertical wall of the washing machine is0.42, how fast must the washing machinespin (rotations per minute) if the cat isnot to slide down the side?
10. A coin is placed 18.0 cm from the axisof a rotating turntable of variable speed.When the speed of the turntable is slowlyincreased, the coin remains fixed on theturntable until a rate of 58 rpm is reached.What is the coefficient of static frictionbetween the coin and the turntable?
11. A ball on a string is revolving at a uni-form rate in a vertical circle of radius 96.5cm. If its speed is 3.15 m/s and its massis 0.335 kg, calculate the tension in thestring
(a) at the top of its path
(b) at the bottom of its path
(c) at the middle of its path (halfway be-tween top and bottom)
12. A 5.0 kg mass is being swung in a verticalcircle on a 3.0 m rope. What is the criticalspeed (i.e. the minimum speed at whichthe ball will maintain a circular path) forthis mass?
28 RRHS Physics
CHAPTER 3. PLANETARY MOTION 3.1. UNIFORM CIRCULAR MOTION
13. For the previous question, assuming thatthe ball is travelling at its critical speed atthe top of the circle, calculate the tensionin the rope at the ball’s lowest point. As-sume no change in energy for the system.
14. Tarzan plans to cross a gorge by swingingin an arc from a hanging vine. If his armsare capable of exerting a force of 1500 Non the vine, what is the maximum speedhe can tolerate at the lowest point of hisswing? His mass is 85 kg; the vine is 4.0m long.
15. A projected space station consists of a cir-cular tube which is set rotating about itscenter (like a tubular bicycle tire). Thecircle formed by the tube has a diameterof 1.6 km.
(a) On which part of the inside of thetube will people be able to walk?
(b) What must be the rotation speed(revolutions per day) if an effectequal to gravity at the surface of theearth (1 g) is to be felt?
16. A person has a mass of 75.0 kg. If theperson is standing on the equator, by howmuch is the person’s weight changed be-cause of the earth’s rotation? The radiusof the earth is 6370 km.
17. When you drive rapidly on a hilly road orride in a roller coaster, you feel lighter asyou go over the top of a hill and heavierwhen you go through a valley. Sketch thesituation, including the relevant forces,and explain this sensation.
18. For a car travelling with speed v arounda curve of radius r, determine a formulafor the angle at which a road should bebanked so that no friction is required.
19. If a curve with a radius of 60 m is properlybanked for a car travelling 60 km/h, what
must be the coefficient of friction for a carnot to skid when travelling at 90 km/h?
20. A 1200 kg car rounds a curve of radius 65m banked at an angle of 14o. If the car istravelling at 80 km/h, will a friction forcebe required? If so, how much and in whatdirection?
RRHS Physics 29
3.2. UNIVERSAL GRAVITATION CHAPTER 3. PLANETARY MOTION
3.2 Universal Gravitation
For readings on this unit, you should also referto chapter 12 in your textbook. Any planetarydata needed for the problems can be obtainedfrom the table on page 955 of your textbook.
Everyone has experienced gravity on earth,and many people are aware that there is a forceof gravity on other planets; however, gravity ismuch more common than this. In fact, a forceof gravity exists between any two masses.
3.2.1 Newton’s Law of UniversalGravitation
In the 1600’s, Newton discovered that thisforce depends on the two masses involved andthe distance separating them.
Fg ∝m1m2
r2
where m1 and m2 are the masses of the twoobjects and r is the distance between them;specifically, Newton realized that there is aninverse square relationship between the dis-tance and the force of gravity. This type ofrelationship appears often in physics, and hasled scientists to believe that there may be someunifying theory for apparently unrelated phe-nomena.
Newton, however, could not determine theconstant needed to form an equation out of thisproportionality. It was not for another hun-dred years before Henry Cavendish devised anexperiment to determine this proportionalityconstant, given by G in the equation below.
Newton’s Law of Universal Gravitation canbe expressed as
Fg =Gm1m2
r2(3.7)
where G is the proportionality constant and isequal to 6.67 × 10−11 Nm2/kg2. It should benoted that this law allows us to accurately pre-dict results, but not to understand why theyare so. We don’t understand exactly whatgravity is.
3.2.2 Acceleration Due to Gravity
In grade 11, you used the equation Fg = mgto calculate the force of gravity, where g wasthe acceleration due to gravity (9.8 m/s2 onthe surface of the earth). Equation 3.7 is amore general expression for the force of gravitybetween any two objects. Consider a mass mon a planet of mass M with a radius of R;equating the two expressions, we get
mg =GMm
R2
or
g =GM
R2(3.8)
We now have a general expression which canbe used to calculate the acceleration due togravity on any planet (or, if the accelerationdue to gravity is known then the mass of theplanet can be calculated; this is how the massof the earth was found.)
3.2.3 Satellite Motion
If a projectile is thrown horizontally, it fallsin a parabolic trajectory toward the ground.If the object is given a higher speed, it trav-els a further distance. On a completelysmooth earth (with no atmosphere to slowthings down) one can imagine an object thatis thrown fast enough so that when it falls to-ward the earth, it has actually travelled farenough that the earth’s curvature matches thecurvature of the falling object. In this way,a satellite can be launched so that it actually“falls” around the earth.
People often ask what keeps a satellite up.Nothing is actually keeping a satellite up; itis falling toward the earth. It is just that itsspeed and the curvature of the earth prevent itfrom actually hitting the earth. To determinethis necessary speed, we must consider the or-bit. Assuming a circular orbit, the accelerationof the satellite is a centripetal acceleration; us-ing Newton’s Second Law we get
30 RRHS Physics
CHAPTER 3. PLANETARY MOTION 3.2. UNIVERSAL GRAVITATION
F = mac (3.9)
What is providing the centripetal force forthis satellite? The force of gravity between theearth and the satellite, as given in equation 3.7.Substituting this (as well as equation 3.4)intoequation 3.9, we get
GMm
r2=
mv2
r(3.10)
where M is the mass of the earth (or otherplanet), m is the mass of the satellite, and ris the radius of the orbit which is the same asthe distance between the objects. Solving thisequation for v, one can obtain the necessaryspeed for the satellite to obtain a circular orbit.
v =
√GM
r(3.11)
Notice that the mass of the satellite is notimportant. If the satellite goes slower than thisspeed, its orbit will decay and the satellite willspiral towards the earth; faster than this speed,and the satellite will enter an elliptical orbit(unless the satellite attains the required escapevelocity to escape the earth’s gravity).
Since the satellite is in free fall around theearth, it can be understood why astronauts inthe space shuttle experience apparent weight-lessness. It is the same situation as a person ina freely falling elevator. Gravity is still quitesignificant at the height of most satellites, andif there were no gravity at this location thesatellite would not be able to maintain its or-bit.
3.2.4 Kepler’s Laws
More than half a century before Newton pro-posed his law of gravitation, Johannes Ke-pler published astronomical works examiningthe motion of the planets around the sun.Among these works were Kepler’s laws of plan-etary motion, which were determined experi-mentally:
1. The path of each planet around the sun isan ellipse with the sun at one focus.
2. Each planet moves so that an imaginaryline drawn from the sun to the planetsweeps out equal areas in equal times.
3. The ratio of the squares of the periods (T )of any two planets is the same as the ratioof the cubes of their average distances (r)from the sun.
T 21
T 22
=r31
r32
Newton’s Law of Universal Gravitation canin fact be used to derive Kepler’s third law (seeproblem 15).
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3.2. UNIVERSAL GRAVITATION CHAPTER 3. PLANETARY MOTION
3.2.5 Problems
1. Calculate the force of gravity on a space-craft 12800 km above the earth’s surfaceif its mass is 700 kg.
2. The force of gravity between two similarbowling balls is 1.71×10−8 N . If the bowl-ing balls are 0.50 m apart, what is themass of each bowling ball?
3. A hypothetical planet has a radius 1.6times that of the earth, but has the samemass. What is the acceleration due togravity near its surface?
4. Another hypothetical planet (there’s a lotof these planets out there!) has a radius20.0 times that of earth and a mass 100times that of earth. What is g near thesurface?
5. What is the effective value of g at a heightof 1000.0 km above the earth’s surface?That is, what is the acceleration due togravity of objects allowed to fall freelyat this altitude? Just for fun, sketch avelocity-time graph of the object as it fallstoward the earth.
6. Determine the net force on the moon(mm = 7.36× 1022kg) due to the gravita-tional attraction of both the earth (me =5.98 × 1024kg) and the sun (ms = 1.99 ×1030kg), assuming that they are pulling inopposite directions on the moon. The dis-tance between the moon and the earth is3.85 × 105km, and the distance betweenthe moon and the sun is 1.50 × 108km.All distances are center to center.
7. Do the previous question again, this timeassuming that the earth and the sun arepulling at right angles to one another.
8. Frank is really concerned about hisweight. But Frank is lazy, and doesn’treally want to exercise in order to lose
weight. How far above the surface of theearth will Frank have to go so that hisweight will be only half of what it is onthe surface of the earth? How will thisaffect Frank’s mass?
9. Dick and Jane are on a joyride from theearth to the moon. See Dick and Janefly. At what distance from the earthwill they experience zero net force be-cause the earth and the moon pull withequal and opposite forces? (See Dick andJane float.) The distance (center to cen-ter) between the earth and the moon is3.85× 105km.
10. A force of 40.0 N is required to pull a10.0 kg wooden block at a constant ve-locity across a smooth glass surface onearth. A physics class is planning a classtrip to Jupiter (m = 1.90 × 1027kg, r =6.98× 107m), and would like to figure outbeforehand what force would be necessaryto pull the same wooden block across thesame glass surface on Jupiter. Can youhelp them out? Try anyway!!
11. Four 8.0 kg spheres are located at the cor-ners of a square of sides 0.50 m. Calculatethe magnitude and direction of the grav-itational force on one sphere due to theother three.
12. Calculate the speed of a satellite movingin a stable circular orbit about the earthat a height of 3200 km.
13. One of the moons of Jupiter discovered byGalileo has a rotational period of 1.44 ×106 s and it is 1.9× 109 m (center to cen-ter) from Jupiter. From this data, deter-mine the mass of Jupiter.
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CHAPTER 3. PLANETARY MOTION 3.2. UNIVERSAL GRAVITATION
14. On July 19, 1969, Apollo 11’s orbit aroundthe moon was adjusted to an average orbitof 111 km. The radius of the moon is 1785km and the mass of the moon is 7.3×1022
kg.
(a) At what velocity did it orbit themoon?
(b) How many minutes did it take to or-bit once?
15. Using Newton’s Law of Universal Grav-itation, show that for any satellite in acircular orbit around the earth, the ratioR3/T 2 is a constant. Find the value ofthis constant.
16. A geosynchronous satellite is one whichstays above the same part of the earth allof the time(in other words, it’s period isthe same as that of the earth). How highabove the surface of the earth is this satel-lite?
17. What happens to the gravitational forcebetween 2 masses when the distance be-tween the masses is doubled?
18. What happens to the gravitational forcebetween two objects if the distance be-tween the objects is tripled and one of themasses is doubled?
19. What happens to the gravitational forcebetween two objects if the distance be-tween the objects is halved and each ofthe masses is tripled?
20. What is the apparent weight of a 65 kgastronaut 4200 km from the center of theearth’s moon in a space vehicle
(a) moving at constant velocity?
(b) accelerating toward the moon at 3.6m/s2?
(c) in orbit around the moon?
State “direction” in each case.
21. Does a satellite with a large or small or-bital radius have a greater velocity?
22. If a space shuttle goes into a higher orbit,what happens to the shuttle’s period?
23. As an astronaut in an orbiting space shut-tle, how would you go about “dropping”an object down to earth?
24. How long would a day be if the earth wererotating so fast that objects at the equatorwere weightless?
25. The asteroid Icarus, though only a fewhundred meters across, orbits the sun likeother planets. Its period is 410 days.What is its average distance from the sun?
26. Use Kepler’s third law and the period ofthe moon (27.4 days) to do problem 16.
27. The mass of Pluto was not known untila satellite of the planet was discovered.Why?
28. A satellite is going around Earth. Onwhich of the following does the speed de-pend?
(a) mass of the satellite(b) distance from Earth(c) mass of Earth
29. If Earth were twice as massive but re-mained the same size, what would happento the value of G?
30. Jupiter is 5.2 times farther than Earth isfrom the sun. Find Jupiter’s orbital pe-riod in Earth years.
31. Uranus requires 84 years to circle the sun.Find Uranus’ orbit as a multiple of Earth’sorbital radius.
32. A satellite is placed in an orbit with a ra-dius that is half the radius of the moon’sorbit. Find its period in units of the pe-riod of the moon.
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3.2. UNIVERSAL GRAVITATION CHAPTER 3. PLANETARY MOTION
34 RRHS Physics
Chapter 4
Fields
The electric force plays a very important rolein our lives, even more important than manypeople think. According to atomic theory, theforces that holds atoms and molecules togetherto form liquids and solids are electrical forces;electric forces are responsible for the metabolicprocesses that occur in our body; even ordi-nary pushes and pulls are the result of the elec-tric force between the molecules of your handand those of the object being pushed or pulled.
4.1 Static Electricity
Everyone has experienced static electricity intheir lives. If you rub a balloon in your hair,you notice that it will stick to the wall; youmay have felt a shock when you touched ametal door knob after walking across a car-pet; a plastic ruler rubbed with a cloth will beable to pick up small pieces of paper. In eachcase, two objects are being rubbed togetherand each obtains a charge.
The two types of charge were referred to aspositive and negative by Benjamin Franklin;the choice of what was negative and what waspositive was arbitrary, and was chosen long be-fore our present knowledge of the atom andthe charges present in it. Objects that havelike charges (either both negative or both pos-itive) are found to repel one another; objectsthat have unlike charges (one negative and onepositive) are found to attract. During any ofthe processes described above, the net change
in the amount of charge is zero; for example,when a plastic ruler is rubbed with a papertowel the plastic acquires a negative charge andthe towel acquires an equal amount of positivecharge. This is the law of conservation ofelectric charge.
You have learned in chemistry that the ba-sic structure of the atom consists of a posi-tively charged nucleus (which has its chargedue to the positively charged protons in it)that is surrounded by one or more negativelycharged electrons. In a normal state, the posi-tive charges and negative charges in the atomare equal and the atom is electrically neutral.Sometimes (as in the examples involving fric-tion earlier) an atom may gain or lose one ormore electrons, giving it a net negative or pos-itive charge. This kind of atom is called anion. Remember, it is the negative electronsthat are free to move from atom to atom(or object to object), not the protons.
4.1.1 Insulators and Conductors
A conductor is a material in which many of theelectrons are bound very loosely to the nucleiand can move about freely within the material.When a conductor is given a negative charge,the excess electrons will spread themselves overthe whole conductor (since they are trying toget away from one another). Likewise, a posi-tively charged conductor will have a deficiencyof electrons over the whole conductor. Metalsare generally very good conductors.
35
4.1. STATIC ELECTRICITY CHAPTER 4. FIELDS
An insulator is a material in which there arealmost no loosely bound electrons. An insula-tor can be charged (such as the plastic ruleris when rubbed with a cloth), but the chargeremains only on the particular part of the ma-terial that was charged. For example, if it ischarged negatively, the excess electrons do notdistribute themselves over the entire material.
Nearly all materials fall into one of these twocategories; there are ,however, some materialsknown as semiconductors (such as silicon, ger-manium, and carbon) which generally have afew free electrons. These semiconductors of-ten have interesting properties, such as onlyconducting electrons in one direction or onlyconducting when illuminated by light.
4.1.2 Charging Objects
We have already seen that an object can becharged using friction (in which case thecharge is actually separated, with each objectgaining an equal and opposite charge). Anobject can also be charged by conduction.Consider the case where you have a negativelycharged rod, i.e. there are more electrons thanprotons on the rod. This rod is touched to aneutral sphere. Since the extra electrons onthe rod all repel one another, they are tryingto get as far away from one another as possi-ble. As soon as you touch the neutral sphere,these electrons now have somewhere to go toget away from one another, so the sphere nowbecomes negatively charged.
The other way of charging an object is calledinduction. With induction, the charged ob-ject does not actually touch the neutral one,but is just brought near it. Consider our exam-ple of the negative rod and the neutral sphere.When the negative rod is brought near theneutral sphere, some of the free electrons inthe sphere will be repelled from the rod; inthis way, the side of the sphere near the rodwill be left with a positive charge and the sideof the sphere furthest away from the rod will
gain a negative charge (see diagram below).No charge has been created, it has merelybeen separated; however, you could break thesphere in two and have two oppositely chargedobjects. If you ground the sphere, it wouldalso be possible to make the charge permanent(think about how this would work).
Induction and conduction can also work to-gether. If you take a charged plastic ruler andput it near a pile of little pieces of paper, thepieces of paper will actually jump through theair to the ruler. When the ruler is placed nearthe pieces of paper, a charge is induced in thepapers just as in the diagram above. The pos-itive side of the paper is then attracted to thenegatively charged ruler, and they touch. Assoon as they touch, conduction occurs. Someof the excess electrons on the ruler can nowmove into the paper, giving it excess electrons.The ruler and paper are now both charged neg-atively, and you will observe the tiny pieces ofpaper flying off (being repelled) of the rulerafter a few seconds.
4.1.3 Electroscopes
An electroscope is a device that detects thepresence of an electric charge. One of themore common types of electroscope is calleda thin-leaf electroscope. This type of electro-scope consists of two metal leaves that are ona hinge and are therefore free to swing. The
36 RRHS Physics
CHAPTER 4. FIELDS 4.1. STATIC ELECTRICITY
two leaves are connected by a conductor whichextends outside of the case. If the electroscopeis neutral, the two leaves just hang vertically,as shown here. They have been shown here tobe slightly separated for illustration purposes.Notice the equal number of positive and nega-tive charges, particularly on each leaf.
Suppose, now that a negatively charged rodis brought near the electroscope. Some of theelectrons will be repelled down into the leaves;the leaves, now negatively charged, will repeleach other and will spread out.
If we then touch the electroscope with thecharged rod, some of the excess electrons inthe rod will be transferred to the electroscope,giving it a permanent charge. The leaves willthen stay spread apart, even after we removethe charged rod.
Note that an electroscope does not tell youwhat kind of charge is present; a positivecharge will also cause the leaves to repel. Youcan, however, use an electroscope to determinethe sign of the charge if you first use conduc-tion to charge the electroscope with a knowncharge (positive or negative). Think abouthow you may do this.
4.1.4 Permanency of Charge
A charged object can sometimes be observedto lose its charge, even when nothing is ap-parently done to them. In some cases, objectscan be neutralized by charged ions in the air;more often, the charge is neutralized by watermolecules in the air. Water molecules are whatare known as polar molecules - even thoughthey are neutral, each end of the molecule isoppositely charged. Suppose you have a neg-atively charged plastic ruler. The excess elec-trons on the ruler can be attracted to the pos-itive end of the polar water molecule and car-ried away. The more water molecules in theair, the faster the charge will be carried away.
Air can also become a conductor under cer-tain circumstances. If charges become largeenough, they will exert a large enough force torip electrons off of molecules in the air; theseions are free to move and form a conductorthrough the air called a plasma. Sparks andlightning are examples of this.
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4.1. STATIC ELECTRICITY CHAPTER 4. FIELDS
4.1.5 Problems
1. Using a charged rod and an electroscope,how can you find if an object is a conduc-tor?
2. A charged rod is brought near a pile oftiny plastic spheres. Some of the spheresare attracted to the rod, but as soon asthey touch the rod, they fly away in dif-ferent directions. Explain.
3. Explain what happens to the leaves of apositively charged electroscope when rodswith the following charges are nearby butnot touching the electroscope:
(a) positive
(b) negative
4. Explain how to charge a conductor nega-tively if you only have a positively chargedrod.
5. You find that object A repels object B, Aattracts C, and C repels D. If you knowthat D is positively charged, what kind ofcharge does B have?
6. Can you charge a metal rod by holding itin your hand? Why or why not?
7. If you move a charged rod toward a pos-itively charged electroscope, the leaves atfirst collapse and then diverge. Whatcharge is on the rod?
8. Three metal blocks in contact are restingon a plastic tabletop. You place two ob-jects with strong positive charges, one ateach end of the line of blocks, close to butnot touching the blocks. You then pokethe blocks apart with an uncharged insu-lating rod, while the objects with strongpositive charges are nearby. Finally, youremove the two positively charges objects.
(a) What charge is now on each block?
(b) Explain how the blocks acquiredthese charges by describing the mo-tion of the negative particles.
9. When an electroscope is charged, theleaves rise to a certain angle and remain atthat angle. Why don’t they rise farther?
10. Why would trucks carrying flammable flu-ids drag a metal strip along the ground?
11. Will an object hold its charge longer on adry day or a humid day? Explain.
12. If you wipe a stereo record with a cleancloth, why does the record now attractdust?
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CHAPTER 4. FIELDS 4.2. FORCES AND FIELDS
4.2 Forces and Fields
4.2.1 Coulomb’s Law
The French physicist Charles Coulomb inves-tigated electric forces in the 1780’s using atorsion balance similar to that used by HenryCavendish for his studies of the universal grav-itation constant. By varying the charges on avariety of spheres, he was able to deduce thatthe electric force between two charged spheresis directly proportional to the magnitude ofeach charge and inversely proportional to thedistance between the spheres.
Coulomb’s Law is given by the equation
F =kq1q2
r2(4.1)
where q1 and q2 represent the magnitude ofeach charge in Coulombs, r is the distance be-tween the charges in meters, and k is a pro-portionality constant whose value is 9.0 × 109
Nm2/C2.The inverse square relation is one of the re-
curring mathematical patterns in nature. Ein-stein once said “The most incomprehensiblething about the universe is its utter compre-hensibility.” Scientists often discover that atheory which is very complex is often wrong.The search for simple, comprehensive explana-tions is one of the driving forces in physics.The current search for a unified theory thatrelates the four forces of nature (gravitational,electromagnetic, strong nuclear forces, andweak nuclear forces) continues.
Charges produced by rubbing ordinary ob-jects (such as a comb) are typically 1 µC orless. The smallest known charge is that of anelectron (or a proton, which has an equal butopposite charge); this is known as the elemen-tary charge
e = 1.60× 10−19C
It should be noted that equation 4.1 onlyapplies to objects whose size is much smaller
than the distance between them; in fact, it isprecise for only point charges. If the two ob-jects are spheres, then the r in equation 4.1 isthe distance between the centers.
4.2.2 Electric Fields
Forces like gravity and electric force behavevery differently than the forces that peopleare used to in everyday life. When peoplethink of forces, they think of pushing or pullingan object. This may require pushing withyour hand, tying a rope to something, or someother type of contact. Forces between electriccharges and masses are different in that theyappear to act over empty space. This worriedpeople, since it appeared to behave like magic.To help explain this idea, Michael Faraday firstsuggested the concept of an electric field inthe 1800’s.
Faraday suggested that any charged objecthas an electric field surrounding it. When an-other charged object is placed in this electricfield, it is the field that interacts with the sec-ond object and applies the force. The electricfield is not a kind of matter - it is a concept.1
Since the electric field is something associ-ated with only the source charge, it should beindependent of any test charge being used tomap the electric field; however, without usingsome test charge, we can’t measure the electricfield. Using some test charge q, we can mea-sure the force exerted on q by the electric field.The electric field E can then be defined as theforce exerted per unit charge at any locationaround a source charge.
E =Fq
(4.2)
Notice that E is a vector and therefore has adirection. The direction of the electric field atany point is defined as the direction of the forceon a positive test charge at that point.
1It is in fact an invention of the human mind thatis very useful.
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4.2. FORCES AND FIELDS CHAPTER 4. FIELDS
For a point source Q, we know that theforce on any test charge q can be found us-ing Coulomb’s Law, equation 4.1. Substitutingequation 4.1 into equation 4.2, we obtain
E =kQ
r2(4.3)
for the magnitude of the electric field.Notice that the test charge q is absent in
this equation, showing that the electric field Eis independent of the test charge q - it dependsonly on the source charge Q and the distancefrom this charge r. If there is more than onesource charge, then equation 4.3 can be appliedto each source to obtain the electric field; thesefields can then be added vectorially.
4.2.3 Lines of Force
In order to visualize an electric field, we drawa series of lines to indicate the direction of theelectric field at various points in space. Theseelectric field lines, or lines of force, are drawnso they indicate the direction of the force on apositive test charge.
For example, consider a positive sourcecharge. If a positive test charge is placed any-where in the vicinity of the source, the force onthe test charge will be away from the source.Drawing these lines of force around the posi-tive test charge, the representation of the elec-tric field will then look like this:
To draw an electric field around two or morepoint sources, consider what direction the forceon the positive test charge would be at various
points around the sources. For example, con-sider a positive and a negative source (of equalstrength). The electric field would look likethis:
The lines of force in the previous two dia-gram do a number of things:
1. They indicate the direction of the electricfield;
2. They are drawn so that the magnitudeof the electric field is proportional to thenumber of field lines in a unit area. Thecloser together the field lines, the strongerthe electric field. Note in our diagramsabove that the lines are closer togethernear the charges than they are furtheraway from the charges.
The electric field lines are sometimes visual-ized as the path that would be followed by atiny test charge placed on it; however, this isonly true if the test charge has no iner-tia or moves extremely slowly. In reality,as the test charge is accelerated by the force,it would gain momentum and would not followthe field lines.
4.2.4 Gravitational Fields
In the same way that electric fields can be usedto explain electric forces acting over a distance,gravitational fields can be used to explain grav-ity acting over a distance. The earth can besaid to possess a gravitational field, which in-teracts with all objects near the earth. In thesame way that the electric field was defined as
40 RRHS Physics
CHAPTER 4. FIELDS 4.2. FORCES AND FIELDS
the force per unit charge (equation 4.2), thegravitational field is defined as the force perunit mass. We have already seen that this ra-tio is equal to g (F/m = g). In other words,the acceleration due to gravity g can also bethought of as the gravitational field intensity.
4.2.5 Problems
1. How many excess electrons are on a ballwith a charge of −4.00× 10−17 C?
2. A strong lightning bolt transfers about 25C to Earth.
(a) How many electrons are transferred?
(b) If each water molecule donates oneelectron, what mass of water lost anelectron to the lightning? One moleof water has a mass of 18 g.
3. How far apart are two electrons if theyexert a force of repulsion of 1.0 N on eachother?
4. Two electrons in an atom are separatedby 1.5 × 10−10 m, the typical size of anatom. What is the force between them?
5. The hydrogen atom contains a proton,mass 1.67 × 10−27 kg, and an electron,mass 9.11×10−31 kg. What is the ratio ofthe magnitude of the average electrostaticforce of attraction between them to thegravitational force of attraction betweenthem?
6. Two electrons are arranged so that oneis above the other. The bottom electronis resting on a table. How high will thesecond electron “float” above this bot-tom electron? In other words, at whatheight will the electrical force of repulsionbe equal and opposite to the gravitationalforce of attraction of the earth?
7. Three particles are placed in a line. Theleft particle has a charge of -67 µC, the
middle +45 µC, and the right -83 µC.The middle particle is 72 cm from eachof the others.
(a) Find the net force on the middle par-ticle.
(b) Find the net force on the right par-ticle.
8. A positive charge of 3.0 µC is pulled onby two negative charges. One, -2.0 µC is0.050 m to the north and the other, -4.0µC, is 0.030 m to the east. What totalforce is exerted on the positive charge?
9. A charged ball has a charge of +16 µC.A second ball, located 16 cm to the right,has a charge of -20 µC. A third ball, lo-cated 25 cm above the second ball, has acharge of +25 µC. What is the total force(magnitude and direction) which acts onthe first ball?
10. You are given two similar spheres, A andB. You want to charge the spheres so thatB has exactly half the charge on A. Whatshould you do?
11. Two charged bodies exert a force of 0.145N on each other. If they are moved sothat they are one fourth as far apart, whatforce is exerted?
12. Two charges, q1 and q2, are separated bya distance d and exert a force F . Whatnew force will exist if
(a) q1 is doubled?
(b) q1 and q2 are cut in half?
(c) d is tripled?
(d) d is cut in half?
(e) q1 is tripled and d is doubled?
13. In one model of the hydrogen atom, theelectron revolves in a circular orbit aroundthe proton with a speed of 1.1× 106 m/s.What is the radius of the electron’s orbit?
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4.2. FORCES AND FIELDS CHAPTER 4. FIELDS
14. Two charges, −Qo and −3Qo, are a dis-tance l apart. These two charges are freeto move but do not because there is a thirdcharge nearby. What must be the chargeand placement of the third charge for thefirst two to be in equilibrium?
15. Two nonconducting spheres have a totalcharge of 850 µC. When placed 1.30 mapart, the force each exerts on the otheris 28.5 N and is repulsive. What is thecharge on each? What if the force wereattractive?
16. Draw the electric field lines for the follow-ing situations. Assume all of the chargesare of the same magnitude.
(a) two positively charged point sources;
(b) one positively charged point sourceand two negatively charged pointsources, one at each corner of anequilateral triangle;
(c) one positive plate and one negativeplate (across from and parallel to oneanother).
17. A negative charge of 2.0× 10−8 C experi-ences a force of 0.060 N to the right in anelectric field. What is the field magnitudeand direction?
18. You are probing the field of a charge ofunknown magnitude and sign. You firstmap the field with a 1.0 × 10−6 C testcharge, then repeat your work with a 2.0×10−6 C charge.
(a) Would you measure the same forceswith the two test charges? Explain.
(b) Would you find the same fields? Ex-plain.
19. Electrons are accelerated by the electricfield in a television, which is about 1×105
N/C. Find the force on an electron.
20. What is the electric field 2.0 cm away froma 1.0 µC charged particle?
21. A lead nucleus has the charge of 82 pro-tons.
(a) What is the direction and magnitudeof the electric field at 1.0× 10−10 mfrom the nucleus?
(b) What is the direction and magnitudeof the force exerted on an electron atthis distance?
22. What is the magnitude and direction ofthe electric field at a point midway be-tween -20.0 µC and a +60.0 µC charge40.0 cm apart?
23. A proton (m = 1.67 × 10−27 kg) is sus-pended at rest in a uniform field E. Takeinto account gravity and determine E.
24. What is the acceleration of an electron ina 2200 N/C electric field?
25. Measurements indicate that there is anelectric field surrounding the earth. Itsmagnitude is about 150 N/C at theearth’s surface and points inward towardsthe centre. What is the electric charge onthe earth?
26. Two positive charges, one 33.0 µC andthe other 68.0 µC are 8.2 cm apart. Atwhat location between them will the elec-tric field be zero?
27. A water droplet of radius 0.020 mm re-mains stationary in the air. If the electricfield of the earth is 150 N/C, how manyexcess electrons must the water droplethave?
28. Explain why it is not possible for two elec-tric field lines to cross.
42 RRHS Physics
CHAPTER 4. FIELDS 4.3. ELECTRIC POTENTIAL
4.3 Electric Potential
We have seen that energy can be extremelyuseful in dealing with mechanical systems – itis a conserved quantity and is an importantaspect of nature. We are now going to extendthis concept to include electrical phenomena.
4.3.1 Electric Potential Energy
As was true when dealing with gravitationalpotential energy, a change in electric potentialenergy is equal to the work required to move acharge2 from one location to another. Remem-ber that W = ∆E.
If positive work is required to move thecharge, then you will increase the potential en-ergy of the system; for example, if you have apositive charge that you want to move closer toanother positive charge, you have to do workto move it (you have to overcome the force ofrepulsion between the two positive charges).This will add energy to the system, namelypotential energy. This is similar to doing workto lift an object from one level to a higher level.
Suppose you want a negative particle tomove closer to a positive charge. In this case,you don’t have to do anything; because of theforce of attraction between the two charges,the negative charge will move on its own to-ward the positive charge. The potential en-ergy here will decrease; as the negative parti-cle accelerates toward the positive charge, theelectrical potential energy will actually be con-verted into kinetic energy.
4.3.2 Electric Potential
Just as the electric field was defined as theforce per unit charge, it is useful to define anelectric potential as the potential energy perunit charge. Note that the electric potential isnot the same thing as the electric potentialenergy. The symbol for electric potential is V .
2without accelerating it
The potential at some point a can be expressedas
Va =Epa
q
where Epa is the potential energy of a chargeq placed at point a. Just as with gravitationalpotential energy, electric potential energy canonly be measured relative to some referencepoint; therefore, only differences in electricalpotential energy (and thus electric potential)are measurable. The difference in potential be-tween two points is called the potential differ-ence. The potential difference between pointsa and b would be Vab = Va − Vb, which is just
Vab =Epa − Epb
q
but the change in potential energy is just thework done in moving the charge, so
Vab =Wab
q(4.4)
The unit of electric potential (and potentialdifference) is joule/coulomb, which is calledthe volt. Potential difference is often referredto as voltage.
Sharing Charge All systems come to equi-librium when the energy of the system is at aminimum. For example, a ball on a hill willcome to rest in the valley below where the po-tential energy is zero. Suppose you have twospheres, one negatively charged (A) and oneneutral (B). Since the excess electrons are be-ing held close together on sphere A, we saythat it is at a high potential; sphere B is saidto be neutral.
If the two spheres are touched together, elec-trons will go from sphere A into sphere B, sincethey are trying to get away from one another.It can be seen that the potential of A is de-creasing while that of B is increasing. This willcontinue until the work done adding charge to
RRHS Physics 43
4.3. ELECTRIC POTENTIAL CHAPTER 4. FIELDS
sphere B is equal to the work gained in remov-ing a charge from sphere A; at this point, thetwo spheres will be at the same potential. Ifthe two spheres are different sizes, than a largersphere would be able to hold more charge thana smaller sphere and still be at the same po-tential (since it has more space for the chargeto spread itself over).
4.3.3 Equipotential Lines
The electric potential can be represented inour electric field diagrams by drawing equipo-tential lines3. An equipotential line is one inwhich all of the points are at the same poten-tial; that is, the potential difference betweenany two points on the line is zero and no workis done moving from one point to another onthe line.
Equipotential lines are perpendicular to theelectric field at any point; if they were not,there would be some component of the electricfield parallel to the equipotential line and workwould be required to move the charge along thesurface against this electric field. We usuallyuse dashed lines to represent the equipotentiallines, as shown below.
4.3.4 Problems
1. A 12 V battery does 1200 J of work trans-ferring charge. How much charge is trans-ferred?
3or equipotential surfaces in three dimensions
2. What work is done when 5.0 C is raisedin potential by 1.5 V ?
3. How much kinetic energy will an electrongain if it falls through a potential differ-ence of 800 V ?
4. A force of 0.053 N is needed to move acharge of 37 µC a distance of 25 cm in anelectric field. What is the size of the po-tential difference between the two points?
5. A -30.0 µC charge is moved towards a+45.0 µC charge. The change in energywhile doing this is 4.5× 10−4 J .
(a) Is the potential energy increased ordecreased?
(b) What is the potential difference?
6. If a large charged sphere is touched bya smaller uncharged sphere, what can besaid about
(a) the potentials of the two spheres?(b) the charges on the two spheres?
7. An electron in a picture tube of a TV set isaccelerated from rest through a potentialdifference of 5000 V . What is the speedof the electron as a result of this acceler-ation?
8. A lightning flash transfers 30 C of chargeto earth through a potential difference of3.5× 107 V . How much water at 0oC canbe brought to boiling temperature?
9. Draw the electric field lines and theequipotential lines for the following situa-tions:
(a) two positively charged point sources;(b) two equally but oppositely charged
point sources;(c) one positive plate and one negative
plate (across from and parallel to oneanother).
44 RRHS Physics
Chapter 5
Electricity & Magnetism
5.1 Electric Current
Until 1800, the idea of electricity was restrictedto producing a static charge by friction onsmall scales; it was only in 1752 that BenjaminFranklin showed that lightning was an electricdischarge, indicating that electricity can trans-fer large amounts of energy. In 1800, Alessan-dro Volta produced the first steady flow of elec-tric charge when he invented the electric bat-tery1. A battery produces electricity by trans-forming chemical energy into electrical energy;you will study this in more detail in chemistry.In short, a chemical reaction inside the batteryresults in an excess of electrons on one termi-nal of the battery (negative terminal) and adeficit of electrons on the other terminal of thebattery (positive terminal).
5.1.1 Electrical Quantities
Current When a conductor such as a wire isconnected to the terminals of a battery, chargecan flow from one terminal of the battery to theother through the wire. This flow of chargeis referred to as an electric current. Theelectric current (I) is defined as the net amountof charge that passes a given point per unittime.
1The small devices that we commonly refer to asbatteries are really cells; a battery is several cells con-nected together.
I =Q
t(5.1)
where Q is the charge that passes a given pointin coulombs and t is the time interval in sec-onds. Electric current is therefore measured inC/s; this is given a special name, an ampere(A), also referred to as an amp.
Remember that in solids, it is the electronsthat are free to move and not the protons;therefore, the current must actually be a flowof electrons through the wire. A wire is a con-ductor, so its electrons are held very loosely.When a wire is connected to the two terminalsof a battery, free electrons in the end of thewire attached to the positive terminal immedi-ately are attracted to this positive terminal; atthe same time, electrons on the negative termi-nal enter the end of the wire attached to thisterminal. It can be seen that there is a chainreaction of moving electrons through the wirefrom the negative terminal to the positive ter-minal. Contrary to a common belief, electronsdo not move through a wire at the speed oflight.
As was discussed in the previous chapter,when the conventions for positive and negativewere established two centuries ago, little wasknown about the structure of the atom. Whenpeople discussed current, it was assumed thatit was positive charge that flowed in the wire.Even though we now know that it is the nega-tive electrons that actually flow in the wire, westill refer to a positive flow of charge in a wire
45
5.1. ELECTRIC CURRENT CHAPTER 5. ELECTRICITY & MAGNETISM
as conventional current. The actual flow ofnegative charge in a wire is referred to as elec-tron flow. For practical purposes, the flow ofpositive charge in one direction is nearly iden-tical (mathematically and conceptually) to theflow of negative charge in the opposite direc-tion so it really doesn’t make a difference whichconvention we are using.
In liquids and gases, positive and negativeions are both free to flow so a current couldreally be the movement of either positive ornegative charges.
Potential Difference (Voltage) A differ-ence in potential is required for an electriccurrent to flow. When discussing sharing ofcharge in the last chapter, it was observedthat when two spheres at different potentialstouched, charges flowed from the object at ahigher potential to the one at a lower poten-tial. A difference in potential was required forthe flow of charge.
Remember that when a charged particle un-dergoes a change in potential, it gains or losesenergy. We used the gravitational analogy be-fore to discuss electric potential; we can useit here as well. Consider a pipe carrying wa-ter that is perfectly horizontal; since each endof the pipe is at the same height, the waterat each end has the same potential energy andthere is no flow of water. If one end of the pipeis raised, however, the water at one end has ahigher potential energy than the other end andthe water will begin to flow. The higher thepipe is raised (or the greater the difference inpotential energy), the greater the flow of wa-ter.
Comparing this to electricity, when we in-crease the potential difference (or voltage) be-tween two points more current will flow. Witha battery, there is a potential difference be-tween the two terminals because of their oppo-site charges. Potential difference is measuredin volts (V ).
Resistance The amount of current that ac-tually flows depends not only on the volt-age (potential difference) but on the resistancepresent. In our gravity/water analogy above,the walls of the pipe offer resistance. If we in-serted a series of screens or grates in the pipe,this would offer more resistance as it wouldinterfere with the flow of water by slowing itdown. In the same way, electrons in a wire areslowed down because of their interaction withatoms of the wire. Resistance is measured inohms, and the symbol for an ohm is Ω (theGreek letter Omega).
When charges are moved through a resis-tance, they lose potential; therefore, there isa loss of potential across any resistor (and again in potential across a battery). Rememberthat we can only measure a potential differ-ence between two points. If we consider a wireto be an ideal conductor (no resistance), thenthe potential difference between any two pointson this wire is zero (no voltage is lost in thewire).
Resistance of a wire can depend on a numberof things:
1. Type of material : Different materials, be-cause of their atomic structure, offer dif-ferent levels of resistance to the move-ment of electrons. Silver is one of the bet-ter conductors (low resistance); insulatorshave a very high resistance.
2. Temperature: In general, the resistanceof most materials increases with temper-ature. This makes sense, since at highertemperatures atoms move faster and areless orderly, thereby interfering with themoving electrons more. At very low tem-peratures (within a few degrees of abso-lute zero), the resistance of certain mate-rials becomes essentially zero. These ma-terials are than said to be superconduct-ing.
3. Thickness: A thicker wire has more cross-
46 RRHS Physics
CHAPTER 5. ELECTRICITY & MAGNETISM 5.1. ELECTRIC CURRENT
sectional area for the electrons to passthrough, so it will have a lower resistance.
4. Length: A longer wire has more obsta-cles in total for the electrons to pass by,thereby increasing the resistance.
5.1.2 Ohm’s Law
Ohm’s “Law” is really a misnomer, since it isnot really a law that applies in all situations.Ohm’s Law was discovered experimentally byGeorg Ohm to apply to many materials. Inmaterials that follow Ohm’s Law, the currentis proportional to the voltage. That is, if youdouble the voltage, the current also doubles.
I ∝ V
In order for this proportionality to be true,the voltage must be the only variable changingthat affects the current; the resistance mustbe constant. Most (but not all) metals obeyOhm’s Law. A resistor that follows Ohm’s lawis said to be ohmic.
Since we know that current is directly pro-portional to the voltage, and inversely propor-tional to the resistance (from our discussionsin the last section), current can be expressedas
I =V
R(5.2)
where the unit of resistance is defined so that1 Ω = 1 V/A. Note that equation 5.2 itself isnot Ohm’s Law. Ohm’s Law refers to the factthat the resistance for most conductors doesnot depend on the potential difference acrossthe conductor (in other words, the current isproportional to voltage). A device that has aconstant resistance that is independent of thepotential difference is said to obey Ohm’s law.
5.1.3 Electrical Power
In most electric circuits, we want to transformelectrical energy into some other form of en-ergy (such as heat, light, or mechanical). We
are often interested in how much energy is be-ing transformed per unit time; from physics 11you may remember that this quantity is power:
P =∆E
t(5.3)
Since ∆E = qV (from equation 4.4) we have
P =qV
t
but I = q/t (equation 5.1) so
P = IV (5.4)
This gives us the power transformed by anydevice, as long as we know the current flowingthrough the device and the potential differenceacross the device. The unit for electrical poweris the same as any other kind of power, thewatt (W ). Remember that one watt is equalto one joule per second.
If we are specifically talking about the powerdissipated in a resistor, we can replace the po-tential difference V in equation 5.4 with equa-tion 5.2 to obtain
P = I2R (5.5)
This equation is often useful since we may notknow how much voltage is lost in the resis-tor, but we probably know the current flowingthrough it and the resistance of the resistor.
Consider a wire that is carrying a current.Since wires have a resistance in the real world,power will be dissipated in the form of heat en-ergy according to equation 5.5. Looking at thisequation, we can see that the power dissipatedin the wire depends on both the current in thewire and the resistance in the wire. By keepingboth of these quantities as small as possible, wecan minimize the amount of power lost in thewire.
Cost of Electricity Although we often referto paying for power, it is really energy that wepay for; power is just the amount of energy
RRHS Physics 47
5.1. ELECTRIC CURRENT CHAPTER 5. ELECTRICITY & MAGNETISM
used per unit time. Because the joule2 is afairly small unit of energy, electrical companiesusually measure energy usage in units calledkilowatt hours. Remember that energy is givenby the equation
E = Pt (5.6)
The energy E can be found in kilowatt hours(kWh) if the power P is measured in kilowattsand the time t is measured in hours. Thecost of electricity is usually expressed as a costper kilowatt hour (our cost in Nova Scotia isroughly $0.085/kWh).
5.1.4 Problems
1. A service station charges a battery usinga current of 5.5 A for 6.0 h. How muchcharge passes through the battery?
2. A current of 1.10 A flows in a wire. Howmany electrons are flowing past any pointin the wire per second?
3. What voltage will produce 12.0 A of cur-rent through a 150 Ω resistor?
4. What is the resistance of a toaster if 110V produces a current of 4.0 A?
5. A resistance of 60 Ω has a current of 400mA through it when it is connected to theterminals of a battery. What is the volt-age of the battery?
6. A 12 V battery is connected to a deviceand 24 mA of current flows through it. Ifthe device obeys Ohm’s law, how muchcurrent will flow when a 24 V battery isused?
7. Sue finds a device that looks like a resistor.When she connects it to a 1.5 V battery,only 45× 10−6 A flows, but when a 3.0 Vbattery is used, 25 × 10−3 A flows. Doesthe device obey Ohm’s law?
2the standard SI unit of energy
8. What is the effect on the current in a cir-cuit if both the resistance and voltage aredoubled?
9. If the voltage across a circuit is kept con-stant and the resistance is doubled, whateffect does this have on the circuit’s cur-rent?
10. Joe argues that, since R = V/I, if he in-creases the voltage the resistance will in-crease. Is Joe correct? Explain.
11. A 1.5 V battery is connected to a bulbwhose resistance is 10 Ω. How many elec-trons leave the battery each minute?
12. A bird stands on an uninsulated transmis-sion line carrying 1200 A. The line has aresistance of 1.0 × 10−5 Ω per meter andthe bird’s feet are 3.0 cm apart. Whatvoltage does the bird feel?
13. The resistance of the human body whenthe skin is perfectly dry is about 105 Ω. Itdrops to about 1500 Ω for wet skin. Thedamage caused by electric shock dependson the current flowing through the body– 1 mA can be felt; 5 mA can be painful;10-20 mA can cause muscular effects; at20 mA, a person may not be able to let goof a conducting wire; respiratory paralysisoccurs between 20 and 100 mA; above 100mA can be fatal. Calculate the amount ofcurrent flowing through a person’s body(for dry skin and for wet skin) if they sticktheir finger in a household socket (120 V ).
14. Assuming the same values of resistance fora bird, calculate the amount of currentflowing through the bird in question 12.What effect does this have on the bird?(Does tweety fry?)
15. What is the current through a 6.0 W lightbulb if it is connected to its proper sourcevoltage of 12 V ?
48 RRHS Physics
CHAPTER 5. ELECTRICITY & MAGNETISM 5.1. ELECTRIC CURRENT
16. How many kWh does a 1300 W frying panuse in 15 minutes?
17. Calculate the resistance of a 40 W auto-mobile headlight designed for 12 V .
18. An electric heater draws 15 A on a 120 Vline. How much power does it use and howmuch does it cost per month (30 days) if itoperates 3.0 hours per day and the electriccompany charges $0.06 per kWh?
19. How many 100 W light bulbs, operated at120 V , can be used without blowing a 10A fuse?
20. A power station delivers 360 kW of powerto a factory through 3.2 Ω lines. Howmuch less power is wasted if the electric-ity is delivered at 40,000 V rather than12,000 V ?
21. A transistor radio operates by means of a9.0 V battery that supplies it with a 50mA current.
(a) If the cost of the battery is $0.90 andit lasts for 300 hours, what is the costper kWh to operate the radio in thismanner?
(b) The same radio, by means of a con-verter, is plugged into a householdcircuit by a homeowner who pays$0.08 per kWh. What does it nowcost to operate the radio for 300hours?
22. A modern television set draws 2.0 A whenoperated on 120 V . At $0.11 per kWh,what is the cost of operating the set permonth (at an average of 7.0 hours per dayfor 30 days)?
23. A small immersion water heater can beused in a car to heat a cup of water forcoffee. If the heater can heat 200 ml ofwater from 5oC to 95oC in 5.0 minutes,
how much current does it draw from the12 V battery?
24. The resistance of an electric stove elementat operating temperature is 11 Ω.
(a) 220 V are applied across it. Whatis the current through the stove ele-ment?
(b) How much energy does the elementconvert to thermal energy in 30.0 s?
(c) The element is used to heat a kettlecontaining 1.20 kg of water. Assumethat 70 % of the heat is absorbed bythe water. What is its increase intemperature during the 30.0 s?
25. A stove element operating on 220 V isbeing used to heat 2.5 kg of water. Itis observed that it takes 12.0 minutes forthe temperature of the water to go from21.0oC to 55.0oC. If the resistance of theelement is 75 Ω, what is the efficiency ofthe burner?
26. What is the efficiency of a 0.50 hp (1horsepower = 750 W ) electric motor thatdraws 4.4 A from a 120 V line?
27. The current in an electromagnet con-nected to a 240 V line is 60 A. At whatrate (in kg/s) must cooling water passover the coils if the water temperature isto rise by no more than 10oC?
28. An electric heater is used to heat a roomof volume 36 m3. Air is brought into theroom at 5oC and is changed completelytwice an hour. Heat loss through the wallsamounts to approximately 2090 kJ/h. Ifthe air is to be maintained at 20oC, whatminimum wattage must the heater have?(The specific heat of air is 0.71 kJ/kgoCand the density of air is 1.29 kg/m3.)
RRHS Physics 49
5.2. *CIRCUITS CHAPTER 5. ELECTRICITY & MAGNETISM
5.2 *Circuits
In this section we will be looking at directcurrent (dc) circuits, applying equation 5.2 toanalyze the resistances, currents, and voltagesthroughout the circuit. In our circuit diagramswe will be using some of these symbols:
5.2.1 *Series Circuits
A series circuit is one in which two or moreresistors are connected end to end so thatthe same current passes through each resistor;there cannot be any junction points betweenthe resistors that would allow the current tochange while going from one resistor to theother.
Consider three resistors in series as shownbelow:
The same current must pass through each re-sistor; since there is only one path, the charge(and therefore the current) cannot leave or en-ter the circuit between resistors. If V1, V2, V3
are the potential differences across R1, R2, R3
respectively, then by applying equation 5.2 weknow that V1 = IR1, V2 = IR2, and V3 = IR3.By conservation of energy, we know that thetotal voltage provided by the battery is equal
to the sum of the voltage drops across eachresistor
V = V1 + V2 + V3 (5.7)
orIRt = IR1 + IR2 + IR3
Rt = R1 + R2 + R3 (5.8)
which makes sense; when we put several resis-tance in series, the total resistance (also calledthe equivalent resistance) is just the sum of theseparate resistances. When you add more re-sistances, you increase the total resistance; thisdecreases the current going through each re-sistor and therefore decreases the voltage dropacross each resistor. The sum of the voltagedrops would then still be the same as the volt-age of the battery. Of course, in this examplewe used just three resistors but equation 5.8could be applied to any number of resistors inparallel.
Knowing the equivalent resistance, the equa-tion I = V/R can then be used to find thecurrent flowing from the battery.
5.2.2 *Parallel Circuits
A parallel circuit is one in which the currentsplits up; each resistor has its own path. Con-sider the parallel circuit shown below:
If I is the total current that leaves the bat-tery, I1, I2, and I3 will be the currents througheach of the resistors R1, R2, and R3. Becausecharge must be conserved, the total currentmust equal the sum of the individual currentsin each branch.
I = I1 + I2 + I3 (5.9)
50 RRHS Physics
CHAPTER 5. ELECTRICITY & MAGNETISM 5.2. *CIRCUITS
In the parallel circuit, the voltage of the bat-tery is applied to each resistor3, so
V
Rt=
V
R1+
V
R2+
V
R3
and dividing out the V from each term gives
1Rt
=1
R1+
1R2
+1
R3(5.10)
so we now have a way of finding the total (orequivalent) resistance of a parallel circuit.
For example, if three 30 Ω resistors areplaced in parallel, the net resistance is
1Rt
=130
+130
+130
so Rt = 10Ω. Notice that the total resistance isless than any of the individual resistances! Butremember, every time you add a resistance inparallel, you are also adding another path forthe current to follow.
Again, equation 5.10 can be applied to anynumber of resistors that are connected in par-allel.
5.2.3 *Complex Circuits
Circuits are often not simply either series orparallel circuits, but are often some combina-tion of the two. In this case it is necessary toanalyze the circuit in steps:
1. If any resistors are in series, calculate anew equivalent resistance that can replacethem. Draw the circuit again (an equiv-alent circuit), replacing the original re-sistors with the new equivalent resistancethat was calculated. Remember, resistorsare in series if there is one and only onecurrent path between them; if there is ajunction between the resistors, then theyare not in series.
3Since the loss of potential must be the same regard-less of the path that the charge follows.
2. If any resistors are in parallel, calculate anew equivalent resistance that can replacethem. Draw the circuit again (an equiv-alent circuit), replacing the original re-sistors with the new equivalent resistancethat was calculated. Remember, resistorsare in parallel only if each resistor has aseparate current path.
3. Repeat steps 1 and 2 until the circuit hasbeen reduced to a simple series or paral-lel circuit. You can then work backwardsthrough your equivalent circuits to findthe required information about each in-dividual resistor.
Consider the following example.
In this example, R1 and R2 are not in series,since there is a junction in between the two.Also, R2, R3, and R4 are not in parallel sinceR3 and R4 share the same path (all of the cur-rent that goes through R3 also goes throughR4).
R3 is in series with R4, however, so these canbe added together to give Req1 (see diagram1 below). This equivalent resistance is thenin parallel with R2, so they can be combinedusing equation 5.10 to give Req2 (see diagram2 below). This combination is then in serieswith R1, so they can then be added to findthe total resistance. The equivalent circuitsfor each step are shown below.
RRHS Physics 51
5.2. *CIRCUITS CHAPTER 5. ELECTRICITY & MAGNETISM
5.2.4 *Kirchhoff’s Rules
Most of the circuits that you will see this yearcan be solved by finding equivalent resistancesand applying the equation I = V/R. Somecircuits4 are, however, too complicated for thisanalysis; for example, circuits that have mul-tiple batteries in different paths. To deal withthese circuits, we use Kirchhoff’s rules. Theserules actually apply to all circuits, and we havein fact already discussed them although theyhave not yet been formally stated.
Kirchhoff’s two rules are:
1. At any junction point, the sum of all of thecurrents entering the junction must equalthe sum of all of the currents leaving thejunction. Note that this is just an expres-sion of equation 5.9.
2. The algebraic sum of the changes in po-tential around any closed path of the cir-cuit must be zero. This is just an expres-sion of equation 5.7.
By applying these rules to the junctionpoints (rule #1) and closed paths (rule #2)of a circuit, a system of equations can then befound and solved.
5.2.5 *Safety Devices
Houses commonly have either fuses or circuitbreakers to ensure against too much currentflowing. If too much current flows, a lot ofpower will be dissipated in the wires (since P =I2R). The wires may overheat and start a fire.There are two reasons that too much currentmay be flowing.
1. Houses are wired in parallel; each objectreceives the full voltage across the cir-cuit. This means that as more devices areplugged into a circuit, the total resistanceof the circuit decreases and more current
4such as many of the ones found in a first year uni-versity physics course
will flow. If a fuse (or circuit breaker)blows, then it may be an indication thattoo many things were being operated onthe circuit.
2. The second reason is potentially evenmore dangerous. There could be a shortcircuit somewhere in the house. A shortcircuit exists when a current finds a wayto avoid the resistance in the circuit. Forexample, consider a lamp cord which hastwo insulated wires leading to the lightbulb. If the insulation were to becomedamaged and the wires allowed to touch,the current could bypass the light bulb al-together. This resistance is then taken outof the circuit, dramatically increasing thecurrent flowing.
Most newer houses have circuit breakers,which serve the same purpose as the fusesfound in older homes. A fuse is simply a thinstrip of metal that is designed to melt if a cur-rent higher than desired tries to flow throughit. If this strip melts, the current can no longerflow and the fuse must be replaced. A cir-cuit breaker consists of a bimetallic strip whichmakes contact to complete the circuit; whenthis strip heats up because of too much cur-rent flowing, the two metals expand at differ-ent rates. This causes the bimetallic strip tobend, breaking the circuit. It then cools downand can be pushed back in place by a springmechanism.
A third type of safety device is slightly dif-ferent. It is called a ground fault interrupter(GFI) and is usually required in bathroomsand kitchens. Instead of being designed to shutoff when the current exceeds a certain level, itis designed to detect small changes in the cur-rent. For example, if you are using a hair dryerin the bathroom and it fell in the sink, the wa-ter would provide another path for the currentand the total current flowing would increase.The GFI would sense this change and wouldturn itself off.
52 RRHS Physics
CHAPTER 5. ELECTRICITY & MAGNETISM 5.2. *CIRCUITS
5.2.6 *Problems
1. Find the potential difference across eachresistor.
2. Find the current in each branch.
3. Find the potential difference across eachresistor.
4. Find the voltage drop across each resistorand the current in each branch.
5. Eight lights are connected in series acrossa 120 V line.
(a) What is the voltage across eachbulb?
(b) If the current is 0.50 A, what is theresistance of each bulb and the powerdissipated in each?
6. Find V.
7. Find the unknown currents and voltages.
8. Find each resistance.
9. Three 100 Ω resistors can be connected tomake four different equivalent resistances.What is the resistance in each case?
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5.2. *CIRCUITS CHAPTER 5. ELECTRICITY & MAGNETISM
10. If each resistor is 10 Ω, find the currentleaving the battery.
11. Find It,V2,I3, and P1.
12. If each resistor is 10 Ω, find the currentleaving the battery.
13. Find the current in each branch.
14. Suppose that you have a 6.0 V batteryand you wish to apply a voltage of only1.0 V . Given an unlimited supply of 1.0
Ω resistors, how could you connect themso as to produce a 1.0 V output for a 6.0V input?
15. Find R3,I2,I3, and I4.
16. Find the potential difference across eachresistor and the current going througheach resistor.
17. Eight lights are connected in parallel to a120 V source by two leads of total resis-tance 2.0 Ω. If 100 mA flows through eachbulb, what is the resistance of each andwhat percent of the total power is wastedin the leads?
18. A three-way light bulb can produce 50 W ,100 W , or 150 W at 120 V . Such a bulbcontains two filaments that can be con-nected to the 120 V individually or in par-allel. Describe how the connections to thetwo filaments are made to give each of the
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CHAPTER 5. ELECTRICITY & MAGNETISM 5.2. *CIRCUITS
three wattages; what must be the resis-tance of each filament?
19. Consider the circuit below.
(a) Compare the brightness of the threebulbs.
(b) What happens to the brightness ofeach bulb when bulb 1 is unscrewedfrom its socket? What happens tothe three currents?
(c) Bulb 1 is screwed in again and bulb 3is unscrewed. What happens to thebrightness of each bulb? What hap-pens to the three currents?
(d) What happens to the brightness ofeach bulb if a wire is connected be-tween points B and C?
(e) A fourth bulb is connected in parallelwith bulb 3 alone. What happens tothe brightness of each bulb?
(f) The wire at point C is broken anda small resistor is inserted in serieswith bulbs 2 and 3. What happensto the brightness of the two bulbs?
20. Two lamps have different resistances, onelarger than the other.
(a) If they are connected in parallel,which is brighter (dissipates morepower)?
(b) When connected in series, which isbrighter?
21. Find the value of the resistors in the fol-lowing circuit.
22. Using Kirchhoff’s rules, determine thecurrents I1, I2, and I3 in the following cir-cuit.
23. Lamp dimmers often consist of rheostats(variable resistors).
(a) Would a dimmer be hooked in seriesor parallel with the lamp to be con-trolled. Why?
(b) Should the resistance of the dimmerbe increased or decreased to dim thelamp?
(c) Can the dimmer be used to savemoney?
24. Two resistors when connected in series toa 120 V source use one-fourth the powerthat is used when they are connected inparallel. If one resistor is 2.8 kΩ, what isthe resistance of the other?
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5.3. MAGNETISM CHAPTER 5. ELECTRICITY & MAGNETISM
5.3 Magnetism
As was the case with electric and gravita-tional forces, magnetic forces act over dis-tances. Since these forces do behave similarly,the concept of fields and lines of force will alsobe used to explain magnetic forces.
5.3.1 Magnetic Fields
Whereas electric fields were the result of pos-itive and negative charges, magnetic fields arethe result of north and south poles. Also simi-lar to electric field is the fact that like magneticpoles repel and unlike poles attract. Magneticpoles are not, however, the same as electriccharges.
Materials that are strongly magnetic (theycan be turned into magnets and are attractedby magnets) are called ferromagnetic materi-als. Some examples of ferromagnetic materialsare iron, nickel, and cobalt. Materials that arenot ferromagnetic show slight magnetic effects,but these effects are very small and not usuallynoticeable.
A compass needle is really a small magnet;the north pole of the compass points towardsthe earth’s north magnetic pole.5 The northpole of the compass is also observed to pointaway from the north pole of another magnet,however. This means that the north magneticpole of the earth is really a south pole!
When drawing the magnetic field linesaround a magnet, we follow the same conven-tions as for electric field lines — namely, thatthe direction of the magnetic field is tangentto the field line at any point and the numberof lines per unit area is proportional to thestrength of the magnetic field. The directionof the magnetic field is defined as the directionthat the north pole of a compass needle wouldpoint when placed at that point in the field —
5The earth’s north magnetic pole is actually about1500 km away from the north geographic pole. The an-gular difference between magnetic north and true (ge-ographic) north is called the magnetic declination.
away from the north pole of the magnet andtowards the south pole.
Domain Theory One of the major differ-ence between magnets and electric charges isthat electric charges can be isolated while mag-netic poles cannot. A positive or negativecharge can be isolated, but north and southpoles always appear in pairs. If you cut a mag-net in two, the result is two magnets, each witha north and south pole.
On a small scale, ferromagnetic materialsare actually made up of tiny regions knownas domains. Each domain behaves like a tinymagnet with a north and south pole. In anunmagnetized piece of iron, for example, thesedomains are arranged randomly pointing in alldirections. The magnetic effects of the do-mains end up cancelling each other out. In amagnetized piece of iron, however, the domainsare more lined up in one direction. Whenevera ferromagnetic material is placed in a mag-netic field, the domains attempt to line up andthe material (at least temporarily) becomes amagnet. This is how ferromagnetic materialsare attracted to other magnets.
The explanation of the domain theory has itsroots at the atomic level. Electrons in atomscan be visualized as orbiting a nucleus. Theelectrons produce a magnetic field, almost asif they were spinning on their axis. In mostmaterials, these spins cancel each other outand there is no net magnetic field; in ferro-magnetic materials, the electrons in a domainseem to cooperate and “spin” in the same di-rection. As a result, the magnetic fields due toeach electron add together so that the domainbehaves as a tiny magnet.
The idea that all magnetic fields are a re-sult of electric currents supports the idea thatnorth and south poles must always exist inpairs, since an electric current will always pro-duce both. This will be seen in the next sec-tion.
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5.3.2 Electromagnetism
The first person to uncover a connection be-tween electricity and magnetism was HansOersted, around 1820. He first tried deflectinga compass needle with a static charge, but thiswas found to have no effect. It was only witha moving charge, or a current, that he foundhe was able to deflect the compass needle.
We will be required in this section to rep-resent three dimensional diagrams. Since wedraw on two dimensional paper, we will be us-ing a sign convention to represent the thirddimension. Anything directed into the page(away from us) will be identified with an ‘×’;anything pointing out of the page (toward us)will be identified with a ‘·’.
Straight Wire It is observed that a compassneedle placed near a straight current carryingwire will align itself so that it is perpendicularto the wire, tangent to a circle drawn aroundthe wire. In other words, the magnetic fieldlines are actually circles around the wire. Thedirection of this magnetic field can be foundusing the first right hand rule.6
The first right hand rule is used to deter-mine the direction of the magnetic field arounda straight conductor. To use this hand rule,point your thumb in the direction of the con-ventional current (positive flow); if you thencurl your fingers (as if making a fist), your fin-gers point in the direction of the magnetic field.
Coil of Wire If you take a straight wire andform a single loop, the first right hand rule canbe applied to show that the field inside the loopis in the same direction everywhere (and in theopposite direction outside the loop). Since thefield lines are more concentrated inside of theloop, the field will be stronger here. By usingmore than one loop, this increases the strength
6Some people use left hand rules instead; when usingleft hand rules, electron flow is used instead of conven-tional current.
of the field even more. A coil of wire containingmany loops is called a solenoid. The strengthof the solenoid can also be increased by increas-ing the current.
This solenoid actually behaves as a magnet,with a north pole at one end and a south poleat the other end. This is an electromagnet.In addition to adding loops and increasing thecurrent, the strength of the electromagnet canbe increased by using a ferromagnetic core in-side the coil; the domains in the core will bealigned by the magnetic field of the current,turning the ferromagnetic material into a mag-net as well. To determine the direction of themagnetic field in a solenoid, the second righthand rule can be used.
This hand rule is used to determine the di-rection of the magnetic field inside of a solenoid(a coil). To use this hand rule, curl you fin-gers around the coil in the direction of the con-ventional current (positive flow); your thumbpoints in the direction of the magnetic field in-side the coil. Another way of thinking aboutthis is that your thumb will point to the northpole of the electromagnet created by the coil.
5.3.3 Force on a Wire
We have already seen in section 5.3.2 that awire carrying a current exhibits a magneticfield; it makes sense, then, that the wire’s mag-netic field will interact with another externalmagnetic field. In fact, when a wire is placedin another magnetic field, it often7 experiencesa force.
The force on a wire can be calculated withthe following formula:
F = IlB sin θ (5.11)
where I is the current in the wire in amperes,l is the length of the wire (in metres) in themagnetic field, B is the strength of the mag-netic field in Tesla, and θ is the angle between
7depending on its orientation
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5.3. MAGNETISM CHAPTER 5. ELECTRICITY & MAGNETISM
the wire and the magnetic field. It can be seenthat if the wire is parallel to the magnetic field(θ = 0o or θ = 180o) then there is no force onthe wire.
The third right hand rule is used to predictthe force exerted on a current carrying wire inan external magnetic field. To use this rule,hold your hand flat with your four fingers to-gether and your thumb perpendicular to yourfingers; point your thumb in the direction ofthe conventional current and extend your fin-gers straight out in the direction of the exter-nal magnetic field; your palm will then pointin the direction of the force on the wire.
5.3.4 Force on a Charged Particle
We saw in the last section that a current carry-ing wire in a magnetic field experiences a force.The current in the wire is the result of movingcharges. The charges do not, however, have tobe moving through a wire. A charged particlemoving on its own can experience a force dueto a magnetic field.
The magnitude of the force on a charged par-ticle can be found in a way similar to the forceon a wire. Remember that I = q/t; substitut-ing this into equation 5.11 we get
F =qlB sin θ
t
but l/t is just the speed of the particle, so
F = qvB sin θ (5.12)
where q is the charge of the particle incoulombs and v is the speed in m/s.
The third right hand rule can also be ap-plied to a moving charged particle in a mag-netic field; instead of the thumb pointing inthe direction of the conventional current, thethumb points in the direction of a moving posi-tive particle. If the moving particle is negative,you must point your thumb in the direction op-posite the motion of the particle.8
8Remember, you may also use the left hand rule; to
Notice when using the third right hand rulethat the force on the particle (direction of yourpalm) is always perpendicular to the directionof the motion of the particle (direction of yourthumb). It will therefore not change the speedof the particle. Even when this force causesthe particle to change direction, the force con-tinues to be perpendicular to the motion. Aswe learned before, as long as this force remainsthe same magnitude, a force perpendicular tothe velocity of the particle will produce circu-lar motion.
5.3.5 Electric Motor
An electric motor is an extremely useful devicethat changes electric energy into mechanicalenergy. To do this, it makes use of the factthat a current carrying wire experiences a forcein a magnetic field. The simplest design of anelectric motor consists of a loop of wire (thearmature) suspended on an axis in a magneticfield, as shown below.
If we examine the part of the wire betweena and b, we find by applying the third righthand rule that there will be a force on the wireinto the page. There will be no force between band c, since the wire is parallel to the magneticfield. Between c and d, the force will be out ofthe page. There will therefore be a torque onthe loop of wire. This loop of wire will rotate,as shown in the side view below.
use left hand rules, your thumb points in the directionof a moving negative charge.
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CHAPTER 5. ELECTRICITY & MAGNETISM 5.3. MAGNETISM
If this analysis is repeated after the loop hasmade a quarter turn (a′ and d′ in the abovepicture), it will be seen that the forces on theloop are no longer perpendicular to the planeof the loop so there will be no torque effect.Also, if the loop goes past this point, the forceswill try to bring the loop back to this verticalposition.
To make efficient use of a motor, we want itto turn continuously. In order to make the loopcontinue turning, it is necessary to change thedirection of the current at the point where theloop is vertical. This is done in a direct current(DC) motor using a split ring commutator andbrushes, as shown below.
Notice that the split ring commutator andthe brushes are not attached to each other, butjust touch one another. As the armature turns,the split ring commutator turns with it whilethe brushes remain fixed in place. The brushesare contact points which allow the current toflow into the split ring commutator. As a re-sult, every half turn (when the loop is vertical)the commutator changes its connection to the
other brush. This allows the current to changedirection in the loop; the direction of the forceon each side of the loop is reversed and theloop continues to rotate.
In reality, motors do not consist of a singleloop of wire as described above. Many loopsof wire are usually used, as well as a ferromag-netic core, both of which increase the size ofthe force on the armature. The speed of themotor can also be increased by increasing thecurrent or the strength of the external magnets(since F = IlB).
5.3.6 Problems
1. Sketch the magnetic field in the followingsituations:
(a) A bar magnet.(b) Two opposite poles.(c) A wire carrying a current towards
you (out of the paper)
2. Locate the North pole for the followingelectromagnets.
(a)
(b)
3. A strong current is suddenly switched onin a wire, but no force acts on the wire.Can you conclude that there is no mag-netic field at the location of the wire?
4. A wire is carrying a current to the eastin the earth’s magnetic field. What is thedirection of the force on the wire?
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5.3. MAGNETISM CHAPTER 5. ELECTRICITY & MAGNETISM
5. Find the direction of the force on the wirein each of the following magnetic fields.
(a)
(b)
(c)
6. A wire carrying a 30 A current has alength of 12 cm between the pole facesof a magnet at an angle of 60o. The uni-form magnetic field is approximately 0.90T . What is the force on the wire?
7. A copper wire 40 cm long carries a current0f 6.0 A and weighs 0.35 N . A certainmagnetic field is strong enough to balancethe force of gravity on the wire. What isthe strength of the magnetic field?
8. Electrons in a vertical wire are moving up-ward. The wire is placed in a magnetic
field directed from east to west. What isthe direction of the force on the wire?
9. If the force on the wire below is into thepage, identify the poles of the magnets.
10. A straight 2.0 mm diameter copper wirecan just “float” horizontally in air becauseof the force of the earth’s magnetic fieldB which is horizontal and of magnitude5.0× 10−5 T . What current does the wirecarry? The density of copper is 8.9× 103
kg/m3.
11. A current carrying wire is pointing to theEast. An external magnetic field is di-rected vertically upward. What is the di-rection of the force on the wire?
12. An electron is moving alongside a wirecarrying a current in the opposite direc-tion. What is the direction of the force onthe electron?
13. A beam of protons is moving from theback to the front of the room. It is de-flected upward by a magnetic field. Whatis the direction of the field?
14. A proton having a speed of 5.0 × 106
m/s in a magnetic field feels a force of8.0 × 10−14 N toward the west when itmoves vertically upward. When movinghorizontally in a northerly direction, itfeels zero force. What is the magnitudeand direction of the magnetic field?
15. Describe the path (quantitatively) of aproton (m = 1.67 × 10−27kg) that movesperpendicular to a 0.120 T magnetic field
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with a speed of 9.25× 106 m/s. The fieldpoints directly toward the observer.
16. A proton moves in a circular path perpen-dicular to a 1.10 T magnetic field. Theradius of its path is 4.5 cm. Calculate theenergy of the proton.
17. If a long straight wire carrying a currentwere placed flat on a paper and iron filingswere sprinkled on the paper, what wouldyou expect the iron filings to do?
18. An electron experiences the greatest forceas it travels 2.1 × 105 m/s in a mag-netic field when it is moving southward.The force is upward and of magnitude5.6 × 10−13 N . What is the magnitudeand direction of the magnetic field?
19. A charged particle moves in a straightline through a particular region of space.Could there be a nonzero magnetic fieldin this region? Why or why not?
20. Charged cosmic ray particles from outsidethe earth tend to strike the earth morefrequently at the poles than at lower lati-tudes. Explain.
21. A force of 5.78 × 10−16 N acts on an un-known particle travelling at a 90o anglethrough a magnetic field. If the velocity ofthe particle is 5.65×104 m/s and the fieldis 0.032 T , how many elementary chargesdoes the particle carry?
22. An electron (m = 9.11×10−31 kg) is accel-erated from rest through a potential dif-ference of 20,000 V , which exists betweenthe two parallel plates below. The elec-tron then passes through a small open-ing into a magnetic field of uniform fieldstrength 0.25 T .
(a) What is the speed of the electron asit leaves the second plate?
(b) Describe the motion (radius and di-rection) of the electron.
23. An electron is accelerated through a po-tential difference of 5000 V before enteringa magnetic field. What is the strength ofthe magnetic field if the radius of its pathin the field is 3.4 mm?
24. A doubly charged helium atom whosemass is 6.7 × 10−27 kg is accelerated bya voltage of 2800 V . What is its period ofrevolution if it encounters a 0.240 T uni-form magnetic field?
25. A beam of singly charged ions move in aregion of space where there is a uniformelectric field, E=1000 N/C, and a uniformmagnetic field, B=0.02 T. The electric andmagnetic fields are at right angles to eachother and both are perpendicular to theion beam so that the electric and magneticforces on an ion oppose each other. If anion is to pass through these fields withoutbeing deflected, what must be the speedof the ion?
26. Protons move in a circle of radius 8.10cm in a 0.385 T magnetic field. Whatvalue of electric field could make theirpath straight? In what direction must itpoint?
27. A particle with a charge of 2.0× 10−18 Cis accelerated by 400 V . It then entersa magnetic field (B=0.4 T) and follows apath with a radius of 0.08 m. Calculatethe mass of the particle.
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5.4. INDUCTION CHAPTER 5. ELECTRICITY & MAGNETISM
5.4 Induction
We have already discovered two ways in whichelectricity and magnetism are related: (1) anelectric current produces a magnetic field, and(2) a magnetic field exerts a force on an elec-tric current or moving electric charge. Scien-tists then began to wonder: if electric currentsproduce magnetic fields, could magnetic fieldsproduce electric current?
5.4.1 Induced EMF
Around 1831, Michael Faraday found that achanging magnetic field can produce a cur-rent as if there were a source of emf9 in thecircuit. We call this an induced emf. Such acurrent is called an induced current. For ex-ample, if a magnet is moved quickly into a coilof wire, a current will flow in the wire whilethe magnet is moving; when the magnet is re-moved, a current will flow in the opposite di-rection. No current flows while the magnet isstationary.
Magnetic flux (φ, measured in webers Wb)refers to the total magnetic field in a certainarea (or the number of field lines) and is givenby φ = B⊥A (where B⊥ is the component ofB that is perpendicular to the area surroundedby the conductor). Faraday found that the in-duced emf is not simply related to the changein the magnetic field strength B; it turns outthat it is actually the rate of change of the fluxthat induces a current.
Faraday’s law of induction states all ofthis in mathematical terms. Suppose we havea coil of wire which is perpendicular to a mag-netic field, and we move this wire so that theflux changes. The induced emf V (or the volt-age) which is observed in the wire is given by
V = −N∆φ
∆t(5.13)
9EMF stands for electromotive force; it is a histor-ical term and was in use before we actually knew thatemf was a potential difference, and not a force
where N is the number of loops (if there aremore than one). The minus sign is part of theequation to remind us that the induced emf al-ways opposes the change in magnetic flux (seeLenz’s Law below).
The rule for determining the direction of theinduced emf is called Lenz’s Law and it statesthat an induced emf always gives rise to a cur-rent whose magnetic field opposes the originalchange in flux. In other words, whatever theexternal magnetic field is doing, the current isinduced in such a way to create a magnetic fieldwhich opposes this external magnetic field.
For example, suppose the bar magnet belowis brought towards the coil. The current willbe induced in the coil in a direction so that thecoil becomes an electromagnet which will tryto push the bar magnet away.
The current must flow in such a way that theleft end of the electromagnet will become asouth pole, opposing the motion of the barmagnet.
If the bar magnet is pulled away from thecoil, the current will be induced so that the coilbecomes an electromagnet which tries to pullthe bar magnet back towards the coil. Fill inthe direction of the current in this example.
Now we will look at a straight wire (of lengthl) going through a magnetic field. The inducedemf in this situation is given by
V = Blv (5.14)
where B, v, and the conductor itself are allperpendicular to one another. The direction of
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CHAPTER 5. ELECTRICITY & MAGNETISM 5.4. INDUCTION
the induced current in this wire can be foundusing the same hand rule as we had for theforce on a wire before (3rd right hand rule).Just like before, our fingers go straight out inthe direction of the external magnetic field andthe thumb gives the direction of the current.But remember, the current is always inducedso that force opposes the motion. So the forcethat the magnetic field exerts on the wire hasto be opposite the direction of motion. Justthink about it — if the magnetic field startedpushing the wire in the same direction that itwas moving originally (the applied force), thiswould create more current which would createa stronger force which would cause the wireto move faster. But this would mean the wireis moving on its own and creating an electriccurrent. This is called perpetual motion, andit would mean that we are getting somethingfor nothing!!!
Remember that the motion of the wire andthe wire itself must be perpendicular to themagnetic field B; therefore, it is only when thewire cuts through the lines of flux that a po-tential is induced in the conductor.
5.4.2 Transformers
When we discussed transmission of power, webrought up the idea of increasing or decreas-ing the voltage while keeping the power thesame. This is accomplished through what iscalled a transformer. A transformer consistsof two coils of wire called the primary and thesecondary. The primary coil has the incomingcurrent; it is this coil that would be connectedto the source of the power. The secondary coilwould be considered to be the output current.
In the example shown below, the two coilsare wrapped around a common soft iron core.There is, however, no current passedthrough the iron core from coil to coil;the two wires are insulated from one another.
When a current flows in the primary coil,we know that a magnetic field will be createdaround this coil. This magnetic field will alsopass through the secondary coil. From equa-tion 5.13, we know that the induced voltage inthe secondary coil is given by
Vs = Ns∆φ
∆t
where Ns is the number of turns in the sec-ondary coil and ∆φ
∆t is the rate at which themagnetic flux changes. The input primaryvoltage Vp is also related to the change in fluxby
Vp = Np∆φ
∆t
where Np is the number of turns in the primarycoil. Combining these two equations, we get
Vs
Vp=
Ns
Np(5.15)
This is called the transformer equation. No-tice that if Ns > Np, the secondary voltagewill be larger than the primary voltage; this iscalled a step-up transformer. If Ns < Np, thesecondary voltage will be smaller than the pri-mary voltage; this is a step-down transformer.
Remember, however, that it is only a changein flux that will induce a voltage; therefore,to maintain a current in the secondary coil,there must be a constantly changing magneticfield from the primary coil. This is achieved byusing an alternating current in the primary coil(which also means there will be an alternatingcurrent in the secondary coil.)
Even though the voltage is being changedin a transformer, conservation of energy tells
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5.4. INDUCTION CHAPTER 5. ELECTRICITY & MAGNETISM
us that the power output can be no greaterthan the power input. Since P = V I, thismeans that if the voltage goes up, then thecurrent must be lowered. If we assume thatthe transformer is 100% efficient (no power islost), then
VpIp = VsIs
or
Vs
Vp=
Ip
Is(5.16)
5.4.3 Electric Generators
A generator transforms mechanical energy intoelectrical energy, and is in effect a motor inreverse. Consider the picture below, which isalmost the same as the one used to explain theelectric motor.
If we begin turning the loop with our hand sothat ab comes out of the page and cd goes intothe page, we can apply Lenz’s law to each wire.The wire ab is moving out, so the current mustflow in a direction so that there will be a forceinto the page; applying our third right handrule we see that the induced current must flowfrom a to b. Similarly, the force on cd must beout of the page, so the current must flow fromc to d.
Now if we look at a side view and only followthe line ab in a complete rotation, we see whathappens to the current.
At position 1, the wire is moving perpendic-ular to the magnetic field and the maximumcurrent is induced (in this case, into the page);at positions 2 and 4, the wire is moving parallelto the magnetic field so no current is induced;at position 3, the wire is again moving perpen-dicular to the magnetic field and the currentinduced is a maximum (in this case, out of thepage).
Shown below is a graph of the potential dif-ference (the graph for the current would lookthe same) for one complete rotation, with thenumbers on the graph corresponding to the ex-planation above. Notice the sinusoidal natureof the graph.
Unlike the DC motor described earlier, anAC generator does not need to change the di-rection of the current every half turn; there-fore, the split ring commutator does not haveto be used. An AC generator uses two sliprings as shown below.
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To make a DC generator, the slip rings canbe replaced with split rings, as were used withthe DC motor. The result of this is a rectifiedcurrent (the current always flows in the samedirection).
This current can be smoothed out by usingmany sets of armatures and commutators.
Back EMF As was previously stated, a mo-tor and a generator are constructed similarly.When a motor is operating, the armature is be-ing turned by the force exerted on the currentcarrying wire; however, we have just seen thatan armature moving through a magnetic fieldalso generates an emf. This emf will opposethe emf connected to the motor. The greaterthe speed of the motor, the greater the back(or counter) emf.
In a generator, the situation is the reverse.As we turn the generator, current is inducedthrough the armature so there is a force onthe armature that opposes the motion. This is
called a counter torque. The more current thatis drawn, the greater this counter torque andthe greater the applied torque must be to keepthe generator turning.
Alternating Current As we have seen,generators can produce alternating currentand this is also what is required for trans-formers. Alternating current is just what itsname suggests – the current changes direc-tion.10 The current is actually sinusoidal, aswas seen in a previous graph. Since the cur-rent is not constant, we want to come up withsome way to refer to the average, or effective,value;11 we cannot just average the currentover time, since this result would be zero (cur-rent would cancel out since it changes direc-tion). Instead, we take a root mean squareaverage (rms). This simply means that wesquare the values before averaging them, andthen take the square root of the average whenwe are finished.
If we square an AC electric current graph, weget a sin2 θ graph. The average of the squaresof the currents can be shown to be
I2 = 0.5I2max
Taking the square root of each side, we getthe rms (or effective) current in terms of themaximum (or peak) current
Irms = 0.707Imax (5.17)
Similarly, the rms (or effective) voltage canbe found to be
Vrms = 0.707Vmax (5.18)
Since power is P = V I, the average powercan be found by multiplying the rms voltageby the rms current, giving
Pavg = 0.5VmaxImax
10In North America, the frequency of this alternatingcurrent is 60 Hz.
11the equivalent direct current that would producethe same power
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5.4. INDUCTION CHAPTER 5. ELECTRICITY & MAGNETISM
or
Pavg = 0.5Pmax (5.19)
Also note that since P = V I, and voltageand current are both sinusoidal, a power vstime graph would be a sin2 θ graph so the av-erage power should be half the maximum (orpeak) power.
A direct current whose values of I and Vequal the rms values of I and V for an alter-nating current will produce the same power.Hence, it is usually the rms value of a currentor voltage that is specified.
5.4.4 Problems
1. A 10 cm diameter circular loop of wire isin a 0.50 T magnetic field. It is removedfrom the field in 0.10 s. What is the aver-age induced emf?
2. The rectangular loop below is being pulledto the right, out of the magnetic fieldwhich points inward as shown. In whatdirection is the induced current?
3. The magnetic flux through a coil of wirecontaining 2 loops changes from -20 Wbto +15 Wb in 1.4 s. What is the inducedemf ?
4. If the solenoid below is being pulled awayfrom the loop shown, in what direction isthe induced current in the part of the loopclosest to the viewer?
5. A rod is moving perpendicular to a mag-netic field with a speed of 15.0 cm/s. Ifthe rod is 12.0 cm long and the magneticfield is 0.800 T , calculate the emf devel-oped.
6. A square coil of sides 5.0 cm contains100 loops and is positioned perpendicu-lar to a uniform 0.60 T magnetic field. Itis quickly and uniformly pulled from thefield (moving perpendicularly to B) to aregion where B drops abruptly to zero. Ittakes 0.10 s for the whole coil to reach thefield free region. How much energy is dis-sipated in the coil if its resistance is 100.0Ω? How much work was done in pullingthe coil out of the field?
7. An airplane travels 1000 km/h in a regionwhere the earth’s magnetic field is 5.0 ×10−5 T and is nearly vertical.What is thepotential difference induced between thewing tips that are 70 m apart? What partof the earth would this be?
8. A 12.0 cm diameter circular loop of wirehas a resistance of 8.15 Ω. It is initiallyin a 0.405 T magnetic field, with its planeperpendicular to B, but is removed fromthe field in 100 ms. Calculate the electricenergy dissipated in the process.
9. The magnetic field perpendicular to a sin-gle 12.0 cm diameter circular loop of cop-per wire decreases uniformly from 0.350 Tto zero. If the wire has a resistance of 0.5Ω, how much charge moves through thecoil during this operation?
10. A step-down transformer has 7500 turnson its primary and 125 turns on its sec-ondary. The voltage across the primary is7200 V .
(a) What voltage is across the sec-ondary?
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(b) The current in the secondary is 36 A.What current flows in the primary?
11. A hair dryer uses 10 A at 120 V . It is usedwith a transformer in England, where theline voltage is 240 V . What should be theratio of turns in the transformer? Whatcurrent will it draw from the 240 V line?
12. A transformer for a transistor radio re-duces 120 V AC to 9.0 V AC. The sec-ondary contains 30 turns and the radiodraws 400 mA. Calculate:
(a) the number of turns in the primary,
(b) the current in the primary, and
(c) the power transformed
13. A 150 W transformer has an input voltageof 9.0 V and an output current of 5.0 A.
(a) Is this a step-up or step-down trans-former?
(b) What is the ratio of output voltageto input voltage?
14. A transformer has input voltage and cur-rent of 12 V and 3.0 A respectively, andan output current of 0.75 A. If there are1200 turns on the secondary side of thetransformer, how many turns are on theprimary side?
15. Scott connects a transformer to a 24.0 Vsource and measures 8.0 V at the sec-ondary. If the primary and secondarywere reversed, what would the new out-put voltage be?
16. The output voltage of a 180 W trans-former is 16.0 V and the input current is11.0 A.
(a) Is this a step-up or step-down trans-former?
(b) By what factor is the voltage multi-plied?
17. Frequently, transformer windings thathave only a few turns are made of verythick (low-resistance) wire, while thosewith many turns are made of thin wire.Why is this true?
18. Would permanent magnets make goodtransformer cores? Explain.
19. You hang a coil of wire with its ends joinedso it can swing easily. If you now plungea magnet into the coil, the coil will swing.Which way will it swing with respect tothe magnet and why?
20. If you unplug a running vacuum cleanerfrom the wall outlet, you are much morelikely to see a spark than if you unplug alighted lamp from the wall. Why?
21. Thomas Edison proposed distributingelectrical energy using constant voltages(DC). Georger Westinghouse proposed us-ing the present AC system. What arethe reasons the Westinghouse system wasadopted?
22. Why is a generator more difficult to ro-tate when it is connected to a circuit andsupplying current that when it is standingalone?
23. An ac voltage, whose peak value is 90 V , isacross a 35 Ω resistor. What is the value ofthe rms and peak currents in the resistor?
24. What is the resistance of an ordinary 60W, 120 V light bulb when it is on?
25. Calculate the peak current in a 2.2 kΩ re-sistor connected to a 240 V ac source.
26. The peak value of an alternating currentpassing through a 600 W device is 3.0 A.What is the rms voltage across it?
27. What is the maximum value of the powerdissipated in a 100 W light bulb?
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28. A 10 Ω heater coil is connected to a 240V ac line. What is the average powerused? What are the maximum and mini-mum values of the instantaneous power?
29. Calculate the resistance and the peak cur-rent in a 1000 W hair dryer connected toa 120 V line.
(a) What is the maximum power whichis dissipated in this hair dryer?
(b) What happens if it is connected to a240 V line in Britain?
30. A magnetic circuit breaker will openits circuit if the instantaneous currentreaches 21.25 A. What is the largest ef-fective current the circuit will carry?
31. You wish to design a fuse which will justallow two 100 W light bulbs, a 700 W hairdryer, and a 150 W stereo to operate on a120 V line. At what instantaneous currentshould the fuse be designed to melt?
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Chapter 6
Waves and Modern Physics
6.1 Quantum Theory
Quantum Theory took almost three decades tocome about, and cannot be credited to any onescientist. It is now the basis for explaining thestructure of matter. The topics in the followingsections involve discussions about things thatwe cannot see and may possibly be beyond ourcomprehension using our present set of rulesand understandings; as with all physics, theyare an attempt to explain and predict what weobserve in a way that we can understand. Theyare models and theories that support one an-other and have been supported experimentally,but they may not actually represent what is re-ally happening. Remember that we cannot seewhat electrons and photons actually are! Thisaspect will be discussed further in section 6.2.
6.1.1 Planck’s Quantum Hypothesis
When an object is heated, it absorbs energy;this energy is then given off in other formsof electromagnetic radiation. This electromag-netic radiation is usually of a frequency belowthe visible spectrum (for low temperatures). Ifan object becomes hot enough, however, it isobserved to emit electromagnetic radiation inthe visible range (light), as shown in the dia-gram below. At the “lower” range (1000 K)of these temperatures, red light begins to beemitted; as an object is heated more and more,higher frequency colors of light (the blue endof the spectrum) are also emitted so that an
extremely hot object (2000 K) will begin toappear white (all of the colors are now beingemitted).
When discussing the spectrum of light emit-ted by an object, we usually discuss blackbod-ies. A blackbody is one that absorbs all radia-tion falling on it, so that any light that is ob-served is light that is being emitted. In otherwords, no light is being reflected from it.
Maxwell’s electromagnetic wave theory doesgive a reason for this electromagnetic radi-ation. It predicts that oscillating electriccharges would produce electromagnetic waves,and objects would emit radiation because ofthis; however, his theory did not accuratelypredict the observed spectrum of light, partic-ularly for the higher frequencies. This is some-times referred to as the ultraviolet catastrophe.
As way of explanation for the observed spec-
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trum, Max Planck suggested in 1900 that theenergy of vibration of the atoms in a solid isnot continuous. In other words, the energyemitted by an atom cannot be just any valuebut can only have discrete values which aremultiples of a minimum value given by
Emin = hf (6.1)
where h is Planck’s Constant, and f is the fre-quency of the oscillation. Plank found h byfitting his formula for the blackbody radia-tion curve to the experiment. Planck’s con-stant has been found experimentally to beh = 6.626× 10−34J · s.
The idea that energy exists only in discreteamounts was a revolutionary idea. The small-est amount of energy possible (hf) is calleda quantum of energy. This is an extremelysmall quantity, as can be seen by the size ofPlanck’s constant; therefore, it would not besignificant in everyday situations. The energyof any molecular vibration could only be somewhole number multiple of this quantum
E = nhf (6.2)
where n is a whole number.Another way of expressing this quantum hy-
pothesis is that not just any amplitude of vi-bration is possible. The possible values for theamplitude are related to the frequency f .
Planck, however, was not entirely happywith this idea. He thought of it as more ofa mathematical device to get the right answerthan an important discovery. He had no ba-sis for suggesting this concept of a quantumof energy other than the fact that it worked— it could be used to accurately predict thespectra of blackbody radiation. Five years af-ter Plank’s hypothesis, Einstein would give itmore credibility in his studies of the photoelec-tric effect.
6.1.2 Photoelectric Effect
When light shines on a metal surface, electronscan be emitted from the surface generating an
electric current. This is known as the photo-electric effect. One of the things that puz-zled scientists about this observed effect wasthat only light above a certain frequency willcause this affect to happen; for example, onlyultraviolet light (even if it is very dim) willcause electrons to be ejected from zinc. If, forexample, red or yellow light is used it cannotcause electrons to be emitted no matter howbright the light is.
Wave theory does not accurately explainphotoelectric effect. Electromagnetic waveshave an energy density associated with them.Based on this theory, any light (regardless offrequency or intensity) would eventually pro-vide enough energy to release electrons; how-ever, if any release occurs, it is always observedto be within one nanosecond.
Although the electromagnetic wave theoryof light does predict that electrons will be re-leased when light shines on a metal (since aforce is exerted on them), it also makes someinaccurate predictions.
• If light intensity is increased, the numberof electrons ejected and their maximumkinetic energy should increase.
• The frequency of the light should not af-fect the kinetic energy of the ejected elec-trons. Only the intensity should affect thekinetic energy of the electrons.
Einstein extended Planck’s quantum theoryto light in 1905. Planck had not suggestedthat light consisted of quanta, only that the en-ergy of the molecular oscillators was quantized;however, since all light ultimately comes from aradiating source, Einstein suggested that lightmay be transmitted as tiny packets called pho-tons. Each photon would have an energy ofhf .
According to Einstein’s photon theory oflight, if a monochromatic light source weremade more intense (brighter), this would im-ply more photons were being transmitted. The
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energy of each of the photons, however, woulddepend only on the frequency (color) of thelight.
Einstein’s Photoelectric Theory consisted ofthree postulates:
• one electron can be ejected upon collisionwith one photon, with the photon losingall of its energy
• some minimum energy Wo (called thework function) is required to release theelectron
• if the energy of the photon is greater thanthe work function (hf > Wo), the electronwill be released. The maximum energy ofthe electron will be the difference betweenthe energy of the photon (hf) and the en-ergy required to release the electron (Wo).
KEmax = hf −Wo (6.3)
Many electrons will require more than thebare minimum (Wo) to escape the metal, andthus the kinetic energy of the electrons may bebelow the maximum.
Einstein’s Photoelectric Theory (if his abovepostulates are accepted) makes certain predic-tions about what should happen in the photo-electric effect:
• an increase in intensity of the light meansmore photons hitting the metal, whichshould mean more electrons being re-leased; the kinetic energy of each electronshould not be changed since the energyof each photon is unchanged (this is onlydetermined by the frequency of the light)
• if the energy of the photon is less than thework function, than no electrons will bereleased. In other words, if f < fo (wheref is the frequency of the incident pho-ton and fo is the threshold frequency(hfo = Wo)), no electrons will be released
• if the frequency of the photon f is in-creased, then KEmax increases linearly
Einstein’s predictions were all verified byMillikan experimentally in 1914.
The diagrams below show how different vari-ables affect the electrons released during thephotoelectric effect.
The quantities of energy calculated at theatomic level are very small. Energy is oftenexpressed in electron volts instead of joules.An electron volt is the amount of energy gainedwhen an electron is accelerated through onevolt. The electron volt is a much smaller unitof energy than a joule
1eV = 1.6× 10−19J
6.1.3 Compton Effect
In 1922, Arthur Compton directed X-rays ofknown wavelength at a graphite target. Alongwith electrons being released from the target(as with the photoelectric effect), X-rays werebeing scattered. Some of the scattered X-raysnow had a lower energy, and thus a lower fre-quency (as indicated by larger wavelength).
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Since
E = hf =hc
λ(6.4)
a larger wavelength λ implies a loss of energyfor the X-ray photons. This shift in energy isknown as the Compton Effect.
Compton proposed that the incident X-rayphoton was acting like a particle that collideswith the electron in the metal; after the colli-sion, the electron gains energy from the X-rayphoton and the X-ray photon now has less en-ergy. The photon does not actually slow down;only its frequency is lowered. If he was correct,the photon and the electron would be experi-encing an elastic collision.
We also know that momentum is conservedin any collision, so it would be expected thatthis may be the case here as well. The dif-ficulty here, however, is that a photon has nomass (and p = mv for particles). If we use Ein-stein’s E = mc2 relationship for mass-energyequivalence, we can define a mass equivalenceof m = E/c2. Substituting this into our mo-mentum equation gives
p =E
c2v
but since the speed of a photon is the speed oflight c this simplifies to
p =E
c=
hf
c
or
p =h
λ(6.5)
It is clear that the larger wavelengths ob-served by Compton also indicate a loss of mo-mentum in addition to the loss of energy (fromequation 6.4).
By making careful measurements, Comptonwas able to show that both the energy and mo-mentum gained by these electrons was foundto equal the energy and momentum lost bythe photons (given by equations 6.4 and 6.5).
Both energy and momentum were con-served!
This provided further evidence for the pho-ton theory of light. A photon is a particle thathas energy and momentum, but has no massand travels at the speed of light
6.1.4 de Broglie Hypothesis
Louis de Broglie felt that there was a symme-try in nature. He suggested in 1923 that, sinceelectromagnetic waves had particle properties,then perhaps things thought to be particles(such as electrons) have wave properties.
Equating the momentum of a particle withmass with the momentum of a photon (whichdoes not have mass), he obtained
mv =h
λ
Rearranging this gives an expression for thewavelength of a particle
λ =h
mv(6.6)
which is called the de Broglie wavelength.De Broglie’s work was doubted since par-
ticles had never been observed to have wave-like properties, such as diffraction and inter-ference;1 however, properties of waves such asdiffraction and interference are only observ-able when the size the slits is not much largerthan the wavelength. The wave nature of or-dinary objects is not noticeable because thewavelengths are so small; this is why parti-cles are not generally observed to have waveproperties. The slits required for diffraction orinterference would be much smaller than theobjects themselves.
Objects such as electrons, however, are smallenough that wave properties can be observed.
1In fact, his graduation was held up for one yearuntil Einstein supported the hypothesis and de Brogliegraduated in 1924. He subsequently won the NobelPrize in 1929.
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In 1927, experiments actually showed thatelectrons actually do diffract. The wavelengthassociated with this diffraction was measuredand found to be just what de Broglie had pre-dicted. De Broglie waves are known as matterwaves.
6.1.5 Problems
1. If energy is radiated by all objects, whycan’t we see them in the dark?
2. An HCl molecule vibrates with a naturalfrequency of 8.1 × 1013 Hz. What is thedifference in energy (in joules and electronvolts) between possible values of the oscil-lation energy?
3. A child’s swing has a natural frequency of0.40 Hz.
(a) What is the separation between pos-sible energy values (in joules)?
(b) If the swing reaches a height of 30cm above its lowest point and has amass of 20 kg, what is the value ofthe quantum number n?
(c) Would quantization be measurable inthis case?
4. What is the energy (in joules and electronvolts) of a photon of wavelength
(a) 400 nm
(b) 700 nm
5. If the threshold wavelength in the photo-electric effect increases when the emittingmetal is changed, what can you say aboutthe work functions of the two metals?
6. Explain why the existence of a cutoff fre-quency in the photoelectric effect morestrongly favors a particle theory ratherthan a wave theory of light.
7. Calculate the energy of a photon of bluelight, λ = 450 nm.
8. What is the maximum kinetic energy andspeed of an electron ejected from a sodiumsurface whose work function is 2.28 eVwhen illuminated by light of wavelength
(a) 410 nm
(b) 550 nm
9. Certain types of black-and-white film arenot sensitive to red light. They can be de-veloped with a red “safelight” on. Explainthis on the basis of the photon theory oflight.
10. If an X-ray photon is scattered by an elec-tron, does its wavelength change? If so,does it increase or decrease?
11. Calculate the momentum of a photonwhose wavelength is 500 nm.
12. Calculate the wavelength of a photon hav-ing the same momentum as an electronmoving at 1.0× 106 m/s.
13. Find the speed of an electron having thesame momentum as a photon having awavelength of 0.80 nm.
14. Determine the wavelength of a 0.35 kgbaseball with a speed of 90.0 km/h.
15. Determine the wavelength of an electronthat has been accelerated through a po-tential difference of 100 V .
16. If an electron and a proton travel at thesame speed, which has a shorter wave-length?
17. What are the wavelengths, in meters, of a3.0 eV photon and a 5.0 eV electron?
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6.2 Wave-Particle Duality
Modern physics has required a drastic shift inthe way that we view the world around us. Inthis section we will look at some of the resultsof so called “modern physics” and how theyintegrate and compare to more classical views.
6.2.1 Historical Models of Light
In this section we will discuss and review someof the historical models of light that weretouched upon in your physics 11 course. Wewill start with two models that were proposedaround the same time in the latter part of theseventeenth century.
Newton Particle Model In the latter partof the seventeenth century, a group of scientistsproposed a particle model of light. The mostprominent of these scientists was Isaac New-ton. This model proposed that light was madeup of extremely small particles that travelledextremely fast. It was reasoned that the parti-cles must be extremely small, since two beamsof light could be observed to pass through oneanother without any interference; the particlesmust be moving very fast, since beams of lightappear to travel in straight lines (just as thecurvature of a projectile’s path is reduced asthe particle’s speed is increased).
This model gained acceptance because itcould be used to explain various properties oflight (Newton’s reputation didn’t hurt either).
• Reflection – Light was observed to be re-flected at the same angle as the angle ofincidence; this was also observed when aparticle collided with a surface (for exam-ple, a ball thrown against a wall).
• Refraction – Light appeared to bendwhen going from one medium to another;for example, going from air to water thelight was observed to bend toward the nor-mal. Newton theorized that the light par-ticles are attracted to the the individual
molecules of the medium in which it istravelling. In a uniform medium, the pullwould be the same in all directions and thelight would travel in a straight line. As thelight gets closer to the water, the watermolecules attract the light particles withmore force than the air molecules. Thiscauses the light to change direction as itspeeds up toward the water. It also im-plies that the light would be going fasterin water than in air.
• Dispersion – Newton proposed that dif-ferent colors of light were actually dif-ferent sized particles. As these particlespassed through a prism, the smaller par-ticles were deflected more than the largerparticles which resulted in the white lightbeing split up into the entire spectrum ofcolors. Each color consisted of similarlysized particles that had been lined up.
This particle model of light was the domi-nant model of light for almost two centuries.
Huygens Wave Model Around the sametime as Newton and others were proposing theparticle model of light, another group of sci-entists, led by Christian Huygens, was puttingforward a wave model of light. They proposedthat light actually consists of waves; since allwaves at this time required a medium, thesescientists also proposed that all of space wasfilled with an ether that provided the mediumfor these light waves.
As with Newton’s particle model, Huygen’swave model could be used to explain variousproperties of light.
• Reflection – By observing water waves,it can be observed that they follow thesame law of reflection as light – the angleof incidence is the same as the angle ofreflection.
• Refraction – Again by observing waterwaves, it could be seen that waves bend
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toward the normal when going from deepwater to shallow water, just as light bendstoward the normal going from air to wa-ter; however, waves travel slower in shal-low water than deep water. This wouldimply that light travels slower in waterthan in air, which contradicts Newton’stheory.
• Diffraction – When light goes througha very small pinhole or slit, the result-ing image is slightly blurred, indicating aspreading out of the light. Similarly, wa-ter waves exhibit this effect of bending andspreading out when going through a smallopening.
Huygen’s wave model was not as well ac-cepted as Newton’s particle model, mainly dueto Newton’s reputation; however, by the earlyto mid 1800’s it began to gain more accep-tance for the following reasons. Around the be-ginning of the nineteenth century, Young per-formed his double slit experiment to show thatlight passing through two slits demonstratedthe same interference pattern as two sourcesof water waves; a wave theory of light began tomake more sense now as this alone could ex-plain the interference pattern. Also, in 1850,the speed of light was shown to be lower in wa-ter than in air; this supported Huygen’s theoryof refraction and contradicted Newton’s theoryof refraction.
By the middle of the nineteenth century, thewave model of light became the more widelyaccepted model of light. This model was not,however, without its problems. For example,there was no evidence of the ether that wassupposedly required for the transmission ofwaves.
Electromagnetic Theory In the latterpart of the nineteenth century, JamesMaxwell improved upon Huygen’s wave model.Maxwell predicted that an accelerating electric
charge will emit interacting electric and mag-netic waves (electromagnetic waves) that re-quire no medium (just as electric and magneticfields require no medium). He further calcu-lated that in order for these waves to continueto travel and interact together, they must betravelling at a speed of 3.0 × 108 m/s — thesame speed as the speed of light!! The logicalconclusion was that light is a type of electro-magnetic wave.
The existence of electromagnetic waves wasdemonstrated a few years later by Hertz. Ac-cording to Maxwell’s theory, light waves arejust a very narrow band of frequencies of thiselectromagnetic wave spectrum.
6.2.2 Modern Theory of Light
Experiments demonstrating the photoelectriceffect and the Compton effect have broughtcredibility back to Newton’s particle model ofthe seventeenth century; however, the wavetheory of light can also explain some aspectsof light such as diffraction, refraction, and in-terference where the particle theory fails.
The two theories, which appear to be incom-patible, each explain certain aspects of the be-havior of light. Neither theory by itself canbe used to explain light. Scientists have cometo accept this and have called it the wave-particle duality of light.
Neils Bohr has proposed the principle ofcomplementarity to summarize this situa-tion. It states that to understand any givenexperiment, we must use either the wave orparticle theory of light; but to understand lightfully, we must refer to both theories. Thetwo aspects of light complement one another.The equation for the energy of a photon itself(E = hf) demonstrates the integration of thetwo theories. The equation represents the en-ergy of a particle on the left side, but on theright side is the frequency of the correspondingwave.
We cannot try to visualize this duality as
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a particle vibrating, or as a wave that has amass; we cannot picture a combination waveand particle. The two aspects of light are dif-ferent “faces” that light shows, depending onwhich property of light is being measured.
In general, when light passes through spaceor a medium, its behavior imitates that of awave; when light interacts with matter, its be-havior is more like that of a particle.
When we try to visualize light, we try tothink of it in terms of what we observe inthe everyday, macroscopic world. We think ofwaves as the water waves that we can easilysee, or a particle as a baseball moving throughthe air, because these are things that we haveobserved to transfer energy from one point toanother. We instinctively want to describelight in these terms; however, there is no rea-son that light should fit our narrow view2 ofthe world around us.
Science simply uses abstractions of the hu-man mind to try to explain and predict theworld around us. In terms of everyday lan-guage and images, light reveals both wave andparticle properties. This does not mean thatlight is either a wave or a particle, or even acombination of the two. It simply means thatin different situations, light behaves similarlyto things (particles and waves) that we haveexperience with.
6.2.3 Modern Theory of Particles
As was shown by de Broglie, this duality ex-tends to particles as well. We must have anunderstanding of both the particle and waveaspects of matter to understand it, but a vi-sual picture is again not possible.
Electrons have traditionally been thought ofas tiny, negatively charged particles. But it hasbeen shown that electrons also exhibit wave
2Our picture of the world around us consists onlyof things large enough to see and that reflect or emitelectromagnetic waves within the range of frequenciesof visible light.
properties. Nobody has ever actually seen anelectron – we have no idea what it “looks” like.For convenience (and to try and preserve oursanity!), we use images and constructs fromour macroscopic world to try and explain themicroscopic world. An electron, like light, isthe set of its properties that we can measure.
It has been said that an electron is a “log-ical construction”. We have grouped the setof properties that we can measure and giventhem the name electron.
6.2.4 Implications
We have referred to the idea that things likelight and electrons are just the sum of theirproperties. We cannot picture what they are,we can only discuss these things in terms oftheir properties. This has some major impli-cations.
Uncertainty Most scientists believe thatthe properties of an object can only be definedby thinking of an experiment that can measurethem. One cannot say that a particle is at acertain location unless it is possible to describean experiment to locate the particle; one can-not say that light diffracts unless it is possibleto describe an experiment to show and mea-sure this diffraction.
This raises another problem: in order tomeasure something, you must interact withit. Consider yourself in a dark room with aping pong ball. In order to locate the ball,you would have to feel your way around. Youwould probably only locate the ball by acci-dentally hitting it with you hand. This wouldtell you where it is, but in the process wouldmove it from that position. You wouldn’t knowwhere it is going.
Applied to a smaller scale, imagine trying tolocate an object such as an electron. To locatethis, suppose we use light (or some other formof electromagnetic radiation). When this radi-ation interacts with the electron, it will actu-
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CHAPTER 6. WAVES AND MODERN PHYSICS 6.2. WAVE-PARTICLE DUALITY
ally transfer its momentum and move the elec-tron.
In addition to the uncertainty associatedwith this interaction, the wave-particle dual-ity contributes even more uncertainty. Objectscan be seen to an accuracy no greater than thewavelength of the radiation used. If we wantan accurate position of a tiny object, we mustuse a small wavelength; but according to equa-tions 6.4 and 6.5, this means that we would beincreasing the energy and momentum of thephoton which would disturb the object evenmore. If, on the other hand, photons of largerwavelength are used then they would have lessof an effect on the object but its position willbe less accurately known.
Thus, the act of measuring actually intro-duces significant uncertainty to either the po-sition or the momentum of the particle. Theposition and momentum of a particle cannotboth be precisely known. This is known as theHeisenberg Uncertainty Principle.
Probability The classical Newtonian viewof the world is that it is deterministic – if weknow the position and velocity of an object atsome point in time, then we can predict its fu-ture position if we know the forces acting onthe object.
Modern physics has seriously questioned thisdeterministic view. We have seen that an elec-tron cannot even be considered to be solelya particle, but has wave properties. Alongwith the Heisenberg uncertainty principle, thismeans that we cannot pinpoint the location ofan electron; we can only calculate probabilitiesthat an electron will be observed at differentplaces. If we cannot say with certainty wherean electron is, than it follows that we cannotpredict with certainty where it will go next.
Since matter is made up of these small par-ticles for which the wave-particle duality isso important, it stands to reason that evenordinary sized particles will be governed byprobability, and not determinism. For exam-
ple, there is a finite probability (although ex-tremely small) that when you through a stonehorizontally it will curve upward!
Granted, the probability that the stone willfollow the expected parabolic path is extremelyhigh; however, it is still a probability and nota certainty. This probability is so high that itgives rise to the appearance of determinism.
In summary, we describe experimental ob-servations on electrons and atoms (and light)using concepts that are familiar to us, suchas waves and particles that exist in space andtime; however, we cannot let ourselves thinkthat electrons and atoms are particles or wavesthat exist in space and time. This distinctionbetween our interpretation of experimental ob-servations and what is really happening is veryimportant.
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CHAPTER 6. WAVES AND MODERN PHYSICS 6.3. MODELS OF THE ATOM
6.3 Models of the Atom
The existence of atoms, and the fact that elec-trons were a part of this structure, was ac-cepted by scientists by 1900. The first modelof the atom visualized the atom as a homo-geneous positive sphere inside of which therewere negative electrons. This was sometimesreferred to as the plum pudding model.
Around 1911, Ernest Rutherford performedan experiment in which he directed positivelycharged alpha particles (helium nuclei) at athin sheet of metal foil. He found that mostof the alpha particles passed through the foilunaffected, but a few were bounced almost di-rectly back. He concluded that the atom ismostly empty space with all of the positivecharge concentrated in a tiny massive centralcore (this is what caused the few alpha parti-cles to bounce away). This became known asthe Rutherford Model. He also suggesteda planetary model where electrons orbit thenucleus. If they were at rest, he argued thatthey would simply be attracted to the positivenucleus. Although a major step forward, thismodel was flawed (as will be seen in the nextsection).
6.3.1 Atomic Spectra
As we saw in section 6.1.1, heated solids, liq-uids and dense gases emit light with a con-tinuous spectrum of wavelengths. The con-tinuous nature of this spectrum is due to theinteraction of each atom or molecule with itsneighbor. Less dense gases, where the atomsor molecules are much further away from theirneighbors, emit a discrete spectrum. The emit-ted light is due to individual atoms, not inter-actions between atoms. When viewing thesespectra, individual lines are seen rather thana range of colors. The fact that these spec-tra come from individual atoms and not inter-actions between the atoms means that thesespectra can be used as a fingerprint for identi-
fication.When energy is transferred to atoms, the
atoms absorb this energy and then emit it inthe form of light. The spectrum of a gas isa series of lines of different colors, each linecorresponding to a specific wavelength of lightemitted from the atoms of the gas. This isknown as an emission spectrum. The di-agram below shows an emission spectrum forhydrogen.
A gas that is cool will absorb certain wave-lengths of light that is shone on it. A spectrumwill show dark lines where wavelengths havebeen absorbed. This is known as an absorp-tion spectrum. The picture below shows anabsorption spectrum of sunlight.
The spectrum of sunlight is observed to havesome dark lines. It was deducted that coolgases surrounding the sun absorbed some ofthe wavelengths of sunlight. By analyzingthese wavelengths, the composition of the at-mosphere of the sun was determined. This ishow helium was discovered.
It was observed that cool gaseous elementsabsorb the same wavelengths that they emitwhen excited. These spectra serve as a keyto the structure of the atom, since they areunique to each atom.
The study of spectra is known as spec-troscopy and is an extremely important branchof science. Using spectroscopy, scientists cananalyze unknown materials; in industry, com-position of various products can be verified orused to categorize the products.
Since the spectra resulting from these lowdensity gases is due only to the individual
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6.3. MODELS OF THE ATOM CHAPTER 6. WAVES AND MODERN PHYSICS
atoms (and not the interactions between theatoms, as in solids), any model of the atomshould be able to explain why light is emittedat discrete wavelengths and should be able topredict what these wavelengths will be.
The Rutherford model had two main flaws.
1. Since electrons are orbiting in circularpaths, they are accelerating. Any accel-erating electric charge will give off light(as was seen in Maxwell’s electromagnetictheory in section 6.2.1); as it loses energy,it should slow down and spiral towardsthe nucleus. Thus, the atom would notbe very stable.
2. As the electrons spiraled inward, their fre-quency would increase gradually and sowould the frequency of the light emitted.A continuous range of frequencies wouldtherefore be emitted; this model could notexplain why atoms emit line spectra.
It became clear that Rutherford’s model wasnot sufficient. A student of Rutherford, NeilsBohr, modified Rutherford’s model by inte-grating Planck’s quantum hypothesis.
6.3.2 Bohr Theory
The visible spectrum of hydrogen consists offour lines, as shown in the diagram in sec-tion 6.3.1 - red, green, blue, and violet. TheRutherford model could not explain this, andalso predicted an unstable atom. In 1911, NeilsBohr attempted to unite Rutherford’s nuclearmodel with Einstein and Planck’s quantumtheory. While Rutherford focused on the nu-cleus and the fact that it occupied only a smallpart of the atom, Bohr focused on the electronssurrounding the nucleus. Using quantum the-ory, he suggested that the energy of an electron(and its radius) is quantized.
Bohr postulated that the electron can ex-ist in different energy levels. The small-est energy level is referred to as the groundstate. If an electron absorbs energy, it makes
a transition from the ground state to an ex-cited state; however, it usually remains inthis state for only a fraction of a second. Theelectron then drops back down to the groundstate. Bohr’s theory was that light is onlyemitted when an electrons drops to a lower en-ergy state.
The change in energy of an electron when aphoton is absorbed or emitted is equal to theenergy of the photon; in other words, the dif-ference in energy between the two energy levels(upper and lower) is equal to the energy of thephoton absorbed (in the case of an electronraising energy levels) or emitted (in the caseof an electron dropping energy levels).
The energy of the photon emitted (hf) istherefore given by
hf = Eu − El (6.7)
where Eu is the energy of the electron in thehigher level and El is the energy of the electronin the lower level.
Bohr derived an equation for the energy ofan electron in a specific energy level n in anatom to be
En =−13.6
n2eV (6.8)
where n is called the principal quantumnumber and En is the energy of the electronin electron volts. The energy is negative be-cause energy has to be added to the electronto free it from the force of the nucleus. Thehigher the energy level, the less negative theenergy is (a free electron is defined as havingzero energy).
The number n determines both the radius3
and the energy; both are therefore quantized.The radius increases with n2, while the energydepends on 1/n2 (as can be seen in equation6.8).
When changing energy levels, electrons canjump directly or in steps; for example, going
3These well-defined orbits do not actually exist inthe sense of a planet orbiting the sun.
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CHAPTER 6. WAVES AND MODERN PHYSICS 6.3. MODELS OF THE ATOM
from n=3 to n=1 state, electron can go from 3to 1, or from 3 to 2 and then from 2 to 1. As aresult, three different photons could be emittedin this example. The ground state (the lowestenergy level) exists when n=1. Notice in equa-tion 6.8 that when n=1, the magnitude ofthe energy is the largest; however, the energyis actually at a minimum. This is because forn=1 the electron is closest to the nucleus so itrequires the most energy to be released. As en-ergy is added and the electron goes up levels,the energy En goes up (it gets closer to zero).En represents the amount of energy requiredto free the electron.
The Bohr model works very well for hydro-gen; however, it does not predict the correctspectra for any of the other elements. This wasa major problem with the model. Although itwas the first model to actually explain the dis-crete line spectra, it was obviously not com-plete since it could not be extended to theother elements.
One of the problems with Rutherford’smodel was that it was unstable; an accelerat-ing electron will lose energy and therefore spi-ral into the nucleus. This remained a problemwith Bohr’s model. Bohr did not know how toexplain this, so he simply said that that thelaws of electromagnetism do not hold insidethe atom! This was not generally acceptedvery well by other scientists and remained aproblem with the model of the atom.
6.3.3 Quantum Model
The Bohr model calculated the emission spec-trum and ionization energy of the hydrogenatom, determined energy levels of the ele-ments, and explained some of the chemicalproperties of the elements; however, his pos-tulates could not be explained on the basis ofknown physics and he could not predict thecorrect spectra for any other elements. Hismodel also could not explain why some spec-tral lines were brighter than others and it could
not explain bonding of atoms in molecules.Louis de Broglie, applying his theory of mat-
ter waves, suggested that each electron in theatom is actually a standing wave. Since itwas theorized that electrons move in circles, deBroglie argued that the electron wave must bea circular standing wave. The only waves thatcould exist are waves for which the circum-ference of the circular orbit contains a wholenumber of wavelengths. This provided an ex-planation of the quantized orbits proposed byBohr. This implies that the wave-particle du-ality we discussed earlier is at the root of theatomic structure.
Erwin Schrodinger and Werner Heisenberg,each independently, used de Broglie’s wavemodel to begin a quantum theory of the atom.This theory is known as quantum mechanicsand has been extremely successful in modellingthe microscopic world.
In quantum mechanics, the radius of the or-bit of the electron is not the same as the radiusof planet around the sun, but is actually muchharder to visualize. The electron, since it hasa wave nature, is actually spread out in spacein a cloud of negative charge. This electroncloud can be interpreted as a probability dis-tribution for the electron. If we consider theelectron to be a particle, the density of theelectron cloud predicts the probability that wewill find an electron in a certain area.
There is no defined path that the electronfollows — it is meaningless to even ask howan electron gets from one energy level to an-other. The quantum model of the atom onlypredicts the probability that an electron is ina specific location. The region in which thereis a high probability of finding the electron isreferred to as the electron cloud.
The quantum model predicts the same en-ergy levels for the hydrogen atom as the Bohrmodel does; however, the greater complexityof the quantum model allows it to model theother elements more accurately. The Bohrmodel only had one quantum number (the
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6.3. MODELS OF THE ATOM CHAPTER 6. WAVES AND MODERN PHYSICS
principal quantum number n); the quantummodel uses 3 additional quantum numbers (or-bital (l), magnetic (ml), spin (ms)).
Quantum mechanics uses this model to pre-dict many details about the structure of theatom and is very successful; however, it takespowerful computers to calculate accurate de-tails for many atoms.
6.3.4 Fluorescence and Phosphores-cence
When an atom is excited by a photon from oneenergy state to a higher one, we saw that it ispossible for the electron to return to the lowerstate in two or more jumps. The photons emit-ted will therefore have lower frequencies thanthe one absorbed. This is called fluorescence.Fluorescent objects will emit visible light afterabsorbing ultraviolet radiation.
In a fluorescent light bulb, the applied volt-age accelerates electrons; these electrons col-lide with and excite atoms of the gas in thetube and cause them to emit ultraviolet pho-tons. These photons then strike a fluorescentcoating on the inside of the tube which thenfluoresces (emits photons of visible light).
Phosphorescence works in a similar way;the major difference is that with phospho-rescent materials, when electrons are initiallyexcited they are raised to what is called ametastable state. Metastable states last muchlonger than higher energy levels in typicalatoms (seconds, as compared to 10−8 secondsfor most atoms). In a group of these atoms,some electrons may stay in this metastablestate for over an hour. The result is that lightcan be emitted long after the initial excitation.These materials are used, for example, in lu-minous watch dials.
6.3.5 Problems
1. What are some of the problems with aplanetary model of the atom?
2. How much energy is required to ionize ahydrogen atom in the n = 3 state?
3. At low temperatures, nearly all of theatoms in hydrogen gas will be in theground state. What minimum frequencyphoton is needed if the photoelectric effectis to be observed?
4. How many spectral lines can an atom emitwhen an electron goes from the n = 4 en-ergy level to the ground state.
5. How can the spectrum of hydrogen con-tain so many lines when hydrogen con-tains only one electron?
6. Determine the frequency and wavelengthof the photon emitted when an electrondrops
(a) from E3 to E2 in an excited hydrogenatom
(b) from E4 to E3 in an excited hydrogenatom
7. Calculate the wavelength of all of the pos-sible photons released when an electrondrops from the n = 4 to the n = 2 energylevels in a hydrogen atom. Compare thesewavelengths to the visible spectral lines ofhydrogen in the diagram in section 6.3.1.Explain any discrepancies.
8. Certain dyes and other materials fluoresceby emitting visible light when UV lightfalls on them. Can infrared light producefluorescence?
82 RRHS Physics
Chapter 7
Nuclear Physics
7.1 The Nucleus
In the last chapter, we looked at what is be-lieved about the structure of the atom; we willnow look a bit more in-depth at the structureand workings of the nucleus.
7.1.1 Structure
The number of protons in a neutral atom isequal to the number of electrons and is calledthe atomic number, Z. All atoms of a givenelement have the same number of protons —this number of protons actually determineswhat element it is.
Rutherford postulated the existence of aneutral particle with a mass close to that ofa proton. In 1932, James Chadwick demon-strated the existence of this particle, called aneutron. The sum of the number of neutronsand protons in an atom is called the massnumber, A. The notation used to representparticular atoms is
AZX
where X is the symbol for the element, Z isthe atomic number, and A is the mass number.Sometimes, an element is written as AX, sincethe atomic number Z and the element symbolare redundant; for example, Helium (He) willalways have the atomic number 2.
Atoms of the same element (same number ofprotons) that have different numbers of neu-trons are called isotopes. They have the
same number of electrons and behave the samechemically, but they behave differently in nu-clear reactions. The nucleus of an isotope iscalled a nuclide.
The electric force attracts electrons to thepositive nucleus; however, this same forceshould cause protons to repel each other insidethe nucleus. There must be some other forcethat prevents the protons from repelling. Thisforce is called the strong nuclear force1 andit overcomes electrical repulsion to keep pro-tons together; this force is the same betweenprotons and protons, protons and neutrons,and neutrons and neutrons. This force onlyacts over short distances, so as the distance be-comes greater, the electric force becomes moreimportant. Both protons and neutrons are re-ferred to as nucleons.
7.1.2 Mass Defect
Since the nucleons in a nucleus are held to-gether by this strong nuclear force, work mustbe done to overcome this force if we want toremove one or more nucleons from the nucleus(assuming a stable nucleus). This adds energyto the system. Since we are adding energywhen we remove a nucleon, this means thatthe total energy of all of the parts of the nu-cleus will be more than the total energy of theassembled nucleus.
1This is one of the four forces of nature; the othersbeing the gravitational force, the electromagnetic force,and the weak nuclear force.
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7.1. THE NUCLEUS CHAPTER 7. NUCLEAR PHYSICS
The amount of energy that must be put intoa nucleus in order to break it apart into its neu-trons and protons is called the total bindingenergy; the binding energy per nucleon isthe total binding energy of a nucleus dividedby the mass number A, the total number ofnucleons. Binding energy is not something thenucleus has – it is energy that it lacks relativeto its separate constituents. It is expressed asa negative number.
We know that energy can be expressed as anequivalent amount of mass according to Ein-stein’s
E = mc2 (7.1)
where E is the energy in J , m is the equivalentmass in kg, and c is the speed of light in m/s.
This implies that by adding energy to thesystem, we are actually adding mass. This canbe observed if we compare the mass of a nu-cleus with the mass of the individual nucleonsthat make up the nucleus. The assembled massof a stable nucleus is always less than the sumof the masses of the nucleons that composeit, since energy must be added to take a nu-cleus apart. The difference between the massof a nucleus and the mass of its constituentparts (nucleons) is called the mass defect.Using equation 7.1, the binding energy can becalculated from the experimentally determinedmass defect.
To be stable, the mass of a nucleus must beless than that of its constituents. If the mass ofa nucleus were equal to that of its constituents,it could just fall apart.
The unit of mass used in nuclear physics isthe atomic mass unit, u. One u is defined as112 the mass of 12
6 C nucleus (u = 1.66 × 10−27
kg). Some important values that we will beusing are:
mp = 1.007276 umn = 1.008665 u
where mp is the mass of a proton and mn isthe mass of a neutron.
Using E = mc2, the energy equivalent of 1u can be found to be 931.49 MeV.
In general, the binding energy per nucleonincreases as the mass number A approaches56, which is iron; nuclei heavier than iron havesmaller binding energies. Thus, iron-56 (5626Fe)is the most tightly bound nucleus (it has themost negative binding energy).
In a nuclear reaction, energy is released ifthe nucleus that results from the reaction ismore tightly bound than the original nucleus.In other words, if the total mass of the prod-ucts is less than the total mass of the originalnuclei, some of the mass has been converted toenergy and this energy will be released in thereaction. We will look at this more in the nexttwo sections.
7.1.3 Problems
1. What do different isotopes of an elementhave in common? How are they different?
2. For each of the following, identify the el-ement, the number of protons, and thenumber of neutrons:
(a) 23292 X
(b) 187 X
(c) 11X
(d) 8238X
(e) 24797 X
3. A nuclear reaction produces 9.0 × 1011 Jof energy. What mass was converted?
4. The mass of 21H is 2.014102 u. Calculate
the mass defect and total binding energy.
5. Calculate the total binding energy and thebinding energy per nucleon for 6
3Li (themass of the lithium isotope is 6.015123 u).
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CHAPTER 7. NUCLEAR PHYSICS 7.1. THE NUCLEUS
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7.2. RADIOACTIVE DECAY CHAPTER 7. NUCLEAR PHYSICS
7.2 Radioactive Decay
In 1896, Henri Becquerel discovered that ura-nium was found to darken photographic plateswithout any stimulation when placed nearthem (even when the plates were wrapped).It became apparent that radioactivity was theresult of disintegration or decay of an unstablenucleus, and required no external stimulation.
Many unstable isotopes occur in nature;these isotopes will decay spontaneously. Thisis known as (natural radioactivity); otherunstable isotopes can be produced in the lab-oratory by nuclear reactions; this is known as(artificial radioactivity). We will deal withnatural radioactivity in this section; artificialradioactivity will be addressed in section 7.3.
Notice in the above diagram that stable nu-clei tend to have the same number of neutronsas protons up to a mass number A of 30 or40; beyond this, stable nuclei have more neu-trons than protons. An explanation for thisis that as the nucleus gets bigger, there aremore and more protons repelling each other somore neutrons are needed to exert a strong nu-clear force to hold the nucleus together. If theatomic number gets too large, there are notenough neutrons to do this. As a result, thereare no completely stable nuclides above Z=83.
There are three distinct types of radiation,as will be discussed in the following sections.
7.2.1 Alpha Decay
Alpha (α) particles are nuclei of helium atoms,42He. These nuclei are very tightly bound.They are not very energetic; they can barelypenetrate a piece of paper. Alpha decay oc-curs because the strong nuclear force is unableto hold large nuclei together.
An equation representing alpha decay wouldlook like the following:
22688 Ra →222
86 Rn +42 He
where 22286 Rn is called the daughter nucleus
and 22688 Ra is called the parent nucleus. No-
tice that the mass number decreases by 4 andthe atomic number decreases by 2. This is truefor all alpha decays.
The mass of the parent nucleus is greaterthan the mass of the daughter nucleus plusthe alpha particle;2 the extra energy is carriedaway by the alpha particle as kinetic energy.
Changing from one element into another oneis called transmutation.
Alpha decay occurs because the electricforce of repulsion of the protons overcomes thestrong nuclear force between the nucleons. Re-member that the strong nuclear force cannotact over as large distances as the electric force;therefore, for large nuclei the electric force isable to overcome this strong nuclear force andcause this alpha decay.
7.2.2 Beta Decay
Beta (β) particles are electrons that come outof a nucleus — they are not orbital electrons!It is as if a neutron changes to a proton, usu-ally because there are too many neutrons rel-ative to protons (above stability curve in thediagram shown below). Since the charge was
2This is necessary if the reaction is to occur sponta-neously.
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CHAPTER 7. NUCLEAR PHYSICS 7.2. RADIOACTIVE DECAY
originally neutral, an electron must be releasedto balance the charge of the proton.
Beta particles are more energetic than al-pha particles and can pass through as much as3 mm of aluminum. Beta decay is accompa-nied by the release of a neutrino (or antineu-trino), which has no charge and no mass.3
The weak nuclear force is crucial in Betadecay because the neutrino only interacts withmatter via this weak nuclear force. An exam-ple of a beta decay reaction is shown below:
146 C →14
7 N +0−1 e +0
0 ν
where 0−1e is the beta particle (β−) and 0
0ν isthe antineutrino.
In beta decay, notice that the mass num-ber stays the same but the atomic number in-creases by 1 (transmutation occurs).
There is another kind of β decay in whicha positron (β+) is emitted. A positron hasthe same mass as an electron, but the oppositecharge. It is called the antiparticle to the elec-tron. This can occur if there are too few neu-trons as compared to the number of protons(see the diagram above). One of the protons,by emitting a positron, becomes a neutron.
Another possibility in this situation (too fewneutrons as compared to the number of pro-tons) is an electron capture, in which the nu-cleus captures an orbiting electron from the
3Recent studies have indicated that it may have avery tiny rest mass.
shell. This electron disappears into the nu-cleus, allowing a proton to become a neutron.A neutrino is also emitted.
7.2.3 Gamma Decay
Gamma (γ) rays are high energy photons.They can pass through several cm of lead andstill be detected. For this reason, they can bevery dangerous.
Like an atom, a nucleus can be in an ex-cited state (due to a violent collision or a pre-vious nuclear reaction); when it drops down toa lower energy state, it emits a photon. Thisphoton is known as a gamma ray. In somecases, the nucleus may remain in an excitedstate for some time before it emits a γ ray. Itis then said to be in a metastable state and iscalled an isomer.
Neither the mass number nor the atomicnumber is changed during gamma decay (notransmutation occurs). Other than releas-ing energy, the nucleus does not undergo anychange.
Gamma rays are very similar to X-rays; theyare both high energy photons and even overlapin the electromagnetic spectrum. It is basicallytheir production that is different. Gamma raysoriginate in the nucleus, while X-rays generallyrefer to electron-atom interactions.
7.2.4 Half-lives
All of the nuclei of a radioactive sample do notdecay at the same time – they decay one at atime over a period of time. This is a randomprocess.
The half-life is the time it takes for onehalf of the original isotope (parent nucleus) ina given sample to decay into a different ele-ment (daughter nucleus). Different isotopeshave different half-lives, ranging from fractionsof a second to many thousands of years.
Suppose an isotope has a half-life of 10 years.In 10 years, this means that half of the sample
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7.2. RADIOACTIVE DECAY CHAPTER 7. NUCLEAR PHYSICS
of that isotope will have decayed into a differ-ent element. In another 10 years, half of theremaining sample will have decayed (only one-quarter of the original sample remains).
The activity of a sample is the decay rate ofthat sample. It is proportional to the numberof atoms in a sample, so it is closely related tohalf-life. After one half-life, the activity (or de-cay rate) will also be cut in half. The activityis measured in Bequerel (Bq). One Bequerel isone decay per second.
The diagram below show the number of par-ent nuclei remaining and the decay rate as afunction of time. Notice that the half-life is5700 years.
7.2.5 Problems
1. The isotope 6429Cu is unusual in that it can
decay by γ, β−, or β+ emission. What isthe resulting nuclide in each case?
2. Fill in the missing particle or nucleus.
(a) 4520Ca →? + e− + ν
(b) 5829Cu →? + γ
(c) 4624Cr →46
23 V +?
(d) 23494 Pu →? + α
(e) 23993 Np →239
92 U+?
3. When 2310Ne (mass=22.9945 u) decays to
2311Na (mass=22.9898 u), what is the max-imum kinetic energy of the emitted elec-tron? What is its minimum energy?What is the energy of the neutrino in eachcase?
4. A particular radioactive substance has ahalf-life of 3 years. How much of the sam-ple remains after 12 years?
5. Which will give a higher reading on a ra-diation detector: equal amounts of a ra-dioactive substance that has a short half-life or a radioactive substance that has along half-life?
6. A radioactive bismuth isotope, 21483 Bi,
emits a β particle. Write the completenuclear equation, showing the elementformed.
7. A radioactive polonium isotope, 21084 Po,
emits a α particle. Write the completenuclear equation, showing the elementformed.
8. 23892 U decays by α emission and two succes-sive β emissions back into uranium again.Show the three nuclear decay equationsand predict the atomic mass number ofthe uranium formed.
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CHAPTER 7. NUCLEAR PHYSICS 7.3. ARTIFICIAL RADIOACTIVITY
7.3 Artificial Radioactivity
Radioactive isotopes can be formed from stableisotopes by bombarding them with alpha par-ticles, protons, neutrons, electrons, or gammarays. A nuclear reaction is said to occur whena nucleus is bombarded by another particle, re-sulting in a transmutation. Nuclear reactionscan be man-made (in a laboratory), but theycan also occur in nature.
Enrico Fermi discovered in the 1930’s thatneutrons are most effective at causing nu-clear reactions, since they are not repelledby the positively charged nuclei. Fermi be-gan bombarding the heaviest known element(uranium). This led to the discovery of thetransuranic elements.
7.3.1 Nuclear Fission
It was discovered in 1938, following Fermi’swork, that uranium actually splits in tworoughly equal particles when bombarded by aneutron. This was called nuclear fission, be-cause it resembled cell division. A typical fis-sion reaction is given by
10n +235
92 U →14156 Ba +92
36 Kr + 310n (7.2)
although there are many other possibilities.A tremendous amount of energy is released
because the 23592 U nucleus has a much greater
mass than that of the fission fragments(14156 Ba and 92
36Kr). The fission fragments aremuch more tightly bound than the uraniumnucleus.
It was observed that extra neutrons wereproduced in these fission reactions, and a sin-gle neutron was required to start a fission re-action. It was reasoned that these extra neu-trons could be used to start other reactions,resulting in a sustained chain reaction. Thiswould provide enormous amounts of energy.The first nuclear reactor (research) based onthis concept was constructed at the Universityof Chicago in 1942.
The first use of nuclear fission was theatomic bomb used in World War II. PresidentRoosevelt authorized the Manhattan Projectto research and attempt to build an atomicbomb. Under the direction of Robert Oppen-heimer, the top scientists in Europe and theU.S. developed the first nuclear bomb.
This bomb consisted of two masses of ura-nium, each less than the critical mass requiredfor the bomb. To detonate the bomb, the twomasses would be brought together quickly. Achain reaction would begin and a tremendousamount of energy would be released. A bombusing uranium was dropped on Hiroshima, andone using plutonium was dropped on Nagasaki.This ended the war.
When a fission bomb explodes, radioactivefission fragments are released into the atmo-sphere; this is known as radioactive fallout.This fallout is a concern with nuclear testing.If these fission fragments enter our food chain,they can be much more dangerous than thefallout itself. Alpha and beta particles can usu-ally be prevented from entering our bodies byclothing and skin; however, if the radioactivesource enters our body through our food, theseparticles are in direct contact with our cells.
7.3.2 Nuclear Reactors
There are some problems associated with thepractical use of fission in nuclear reactors:
1. The neutrons emitted during the reac-tion shown in equation 7.2 are moving toofast; they must be slowed down to be ab-sorbed by 235
92 U . This is accomplished witha moderator, often deuterium4 (21H) orgraphite (which consists of 12
6 C). A mod-erator is most effective if the atoms areclose to the mass of the neutrons.
2. Naturally occurring uranium is 99.3%23892 U and only 0.7 % of the fissionable
4which can be used in the form of heavy water.
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7.3. ARTIFICIAL RADIOACTIVITY CHAPTER 7. NUCLEAR PHYSICS
23592 U ; to sustain a chain reaction, the ura-nium must be enriched5 so that is is 2-5% 235
92 U . Without enough fissionable ura-nium, too many of the neutrons will beabsorbed by the nonfissionable materials.There is also only a limited supply of ura-nium.
3. Some neutrons may escape before havinga chance to cause further fissions; someminimum critical mass is needed (usu-ally a few kg).
Most people are aware of the dangers of nu-clear reactions. The fission fragments fromthese reactions have many more neutrons thanprotons and are unstable (they are radioac-tive). There is a danger associated with thedisposal of these materials, particularly sincethey usually have large half-lives.
In a nuclear reactor that is being used to pro-duce electrical energy, the heat from the fissionreaction is used to boil water; this producessteam which is then used to turn a generator.6
The core of the reactor consists of fuel tosustain the nuclear reaction (sealed in metalrods) and a moderator, which was discussedearlier. Also present are control rods, usuallycontaining cadmium; these control the rate ofthe reaction. To slow the reaction down, thecontrol rods are fully inserted into the reactorso that they can absorb the neutrons. Becauseof the high temperatures reached in the reac-tor, a coolant is also necessary to take awaysome of the excess heat.
Breeder reactors are a particular type ofreactor that actually creates more fissionablefuel than was there originally. One of thebyproducts is 239
94 Pu, which is created when23892 U absorbs neutrons. This 239
94 Pu is fission-able, and can be separated to be used as fuel;however, this plutonium has an extremely longhalf-life of 24000 years and is very toxic. It
5This is not usually necessary if the reactor is usingheavy water as a moderator.
6see the diagram on page 932 of your textbook.
can also easily be used to construct a nuclearbomb.
CANDU Reactor This reactor has beendeveloped for use by Atomic Energy CanadaLimited (AECL). The major difference be-tween the CANDU reactor and other reactorsis that it uses heavy water as a moderator andcoolant. Since heavy water is a better mod-erator than natural water, the reactor can usenatural uranium instead of enriched uranium,which is very expensive. It has a simplified de-sign, so it can be built where technology is lim-ited. Because of its design, it has a higher life-time capacity and has longer operating cyclesthan other types of nuclear reactors. There arepresently CANDU reactors in Ontario, Que-bec, and New Brunswick.
7.3.3 Nuclear Fusion
In nuclear fusion, nuclei with smaller massescombine to give a nucleus with a larger mass(this is the process that occurs in the stars).As long as this larger mass is more tightlybound than the smaller masses, energy willbe released. For example, helium is extremelytightly bound; any reaction resulting in the for-mation of helium will very likely release energy.
The series of reactions that occur in the suninvolves the following steps:
11H +1
1 H →21 H +0
1 e +00 ν
11H +2
1 H →32 He
32He +3
2 He →42 He + 21
1H
The first two reactions would have to occurtwice. The net result is that 4 protons pro-duce one α particle (He), 2 positrons and 2neutrinos.
Nuclear fusion has many features whichmake it more attractive than nuclear fission.Some of the benefits of nuclear fusion include:
1. The energy released is greater (for a givenmass of fuel) than that released in fission.
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CHAPTER 7. NUCLEAR PHYSICS 7.3. ARTIFICIAL RADIOACTIVITY
2. There is less of a radioactive waste prob-lem than there is associated with nuclearfission (the products are mainly hydrogenand helium).
3. The fuel is plentiful (such as deuterium,which is available in the oceans)
We do not presently have any practical nu-clear reactors, so obviously there are someproblems with controlled fusion reactions.Some of the problems associated with nuclearfusion are:
1. Fusion reactions require extremely hightemperatures (108 K). This is higher thanany known material can stand. Thesetemperatures are needed to make positivenuclei travel fast enough to get close toone another;7 for this reason, fusion reac-tions are often referred to as thermonu-clear reactions.
2. Once this high temperature is achieved,it is very difficult to control the reaction(or to even contain it) to obtain usableenergy. Controlled fusion has not yetbeen attained. This is not necessarily aproblem when designing a bomb, but it isa problem with a nuclear reactor.
Attempts have been made to use magneticfields to confine reaction, but as of now thisrequires more energy than is produced in thefusion reaction, and all of the particles can stillnot be contained in the field. A few yearsago, a couple of scientists published a paperin which they believed that they had producedcold fusion, but their claims were soon shownto be wrong. At present, the only way that weknow of to produce fusion is at extremely hightemperatures, and there is no way to control itat these temperatures.
7they must get close enough together for the strongnuclear force to act
7.3.4 Problems
Atomic masses of selected isotopes for use withproblems.
Isotope Atomic Mass11H 1.007825 u21H 2.014102 u31H 3.016049 u32He 3.016029 u42He 4.002603
14156 Ba 140.9141 u9236Kr 91.9250 u8838Sr 87.905625 u
13654 Xe 135.90722 u23592 U 235.043925 u23892 U 238.050786 u
1. Why must the fission process release neu-trons if it is to be useful?
2. Why are neutrons such good projectilesfor producing nuclear reactions?
3. Calculate the energy released in the fissionreaction
10n +235
92 U →8838 Sr +136
54 Xe + 1210n
4. What is the energy released in the fissionreaction that is given in equation 7.2?
5. How many fission reactions take place persecond in a 25 MW reactor? Assume that200 MeV is released per fission.
6. How much energy is released when twodeuterium nuclei fuse to form 3
2He withthe release of a neutron?
7. The reaction in the sun was said to use 4protons to produce a 4
2He nucleus (ignor-ing positrons and neutrinos). How muchenergy would this release?
8. List three medical uses of radioactivity.
9. The fission of a uranium nucleus and thefusion of four hydrogen nuclei both pro-duce energy.
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7.3. ARTIFICIAL RADIOACTIVITY CHAPTER 7. NUCLEAR PHYSICS
(a) Which produces more energy?
(b) Does the fission of 1 kg of uraniumnuclei or the fusion of 1 kg of hydro-gen nuclei produce more energy?
(c) Why are your answers to parts a andb different?
10. The energy released in the fission of oneatom of 235
92 U is 200 MeV.
(a) How many atoms are in 1.00 kg ofuranium-235?
(b) How much energy would be releasedif all of the atoms in this 1.00 kg un-derwent fission?
(c) A typical large nuclear reactor pro-duces fission energy at a rate of3600 MW. How many kilograms ofuranium-235 would be used in oneyear?
11. The first atomic bomb released 1.0× 1014
J of energy. What was the mass of theuranium-235 that was fissioned to producethis energy?
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Appendix A
Analysis of Data
A.1 Experimental Data
In any scientific experiment, there are errorspresent. Some of these may be due to humanerrors, others may be inherent in the instru-ments that we are using. Because these errorsaffect the accuracy and precision of our results,their analysis is extremely important in any ex-periment. Errors in an experiment can gener-ally be classified as resulting from two sources:
Instrument Error It is safe to say that allof the instruments that we use have some errorbuilt in to them. The instrument may havebeen damaged at some point, or their may be aproblem with the calibration of the instrument.Take a meter stick, for example. The woodmay shrink or warp, the thickness of the linesmay vary, the ends of the stick may be chipped.All of these factors will contribute some errorto the experiment.
In addition to the fact that instruments mayhave ”flaws”, instruments are designed to mea-sure within certain limits. In our meter stickexample, the device is only calibrated in mil-limeters; therefore, use of this meter stick hasan uncertainty associated with it. Supposethat a measurement is between 2.3 cm and 2.4cm. We only know that the correct measure-ment is 2.3*, where the * digit is some num-ber between 0 and 9. If the actual measure-ment appears past the halfway point between2.3 and 2.4, we might estimate it to be 2.37;
however, this 7 is only an estimate. It may besmaller or larger. The uncertainty in this mea-surement is in the second decimal place. Wecould say that the uncertainty is at least 0.1mm, and probably even more.
Human Error The error introduced by theperson using the instrument is often evenlarger than that due to the instrument itself.Errors may come from such things as improperpositioning of the instrument, wrong positionof the eye with respect to the scale and the ob-ject to be measured, and judging the final digit(see above). Practice with any particular in-strument will generally improve one’s accuracywith that instrument.
In writing lab reports, you will be expectedto do an error analysis. You should attemptto be as specific as possibly in this analysis. Inother words, do not write ”human error”or ”instrument error” as your sources oferror. Be Specific! Errors in procedure, er-rors in calculation, errors due to rounding off,and errors due to mismeasurement are not le-gitimate. In addition to estimating the uncer-tainty off specific measurements as describedabove, focus your attention on the discrepan-cies between the assumptions made during theanalysis of your data based on theoretic con-siderations and the actual conditions presentduring the collection of data. For example,was friction considered to be constant, or neg-ligent? Were masses of ropes or strings ac-
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A.2. STATISTICAL ANALYSIS APPENDIX A. ANALYSIS OF DATA
counted for? Were objects that were assumedto be fixed in one place actually allowed tomove? These are the kinds of questions youshould ask yourself.
It is a good idea, either in discussing thesources of error or in the conclusion, to suggestways the experiment might be improved.
A.1.1 Precision and Random Errors
If you repeat an experiment several times, youcannot expect to get the same result everytime. Instruments and human error will causedifferences in your results (errors). As longas these errors are random, you would expectthat about half of your measurements wouldbe too small and half too large. You would as-sume that they will tend to cancel out providedenough measurements are taken. For this rea-son, scientists generally repeat experiments toobtain a large number of estimates that can beaveraged together to obtain a more reliable es-timate. The more random error we have in ourexperiment, the less precise our results are.
In our error analysis, we will deal with an-alyzing results which we assume have randomerror; this type of error is present in all ex-periments. Section A.2 will look at ways toestimate the precision of our results.
Just because it is expected that there will berandom error associated with the lab, this isnot an excuse to be careless. The goal in anyexperiment should be to reduce this randomerror as much as possible in order to increasethe confidence we have in our final result. Thisgoal is achieved by being careful in taking mea-surements and ensuring that the instrumentsare in good working order.
A.1.2 Accuracy and Systematic Er-rors
The other case is if the errors are systematic;that is, the measurements are always too high,or too low. This may result from a mistake in
calibrating instruments, from the person con-ducting the experiment making the same mis-take for each repetition, or from an error inher-ent to the technique for measuring the prop-erty. In this case, no matter how many esti-mates are averaged together, the final resultwill still be different from the true value. Inthis case, we may end up with a very preciseestimate, but it will not be very accurate.
In the case of systematic error, a mistakehas usually been made at some point in theexperiment or there was a problem with theequipment used. This type of error is generallymore serious, as it cannot be eliminated with-out locating the source of the problem. Addingto the difficulty is that there may be manysystematic errors present of which we have noknowledge.
A.2 Statistical Analysis
The precision of the data can be quantitativelyexpressed with a statistical analysis. This typeof analysis will give us some idea of how muchuncertainty can be assigned to our measuredvalue due to random errors only. It doesnot address any possible systematic errors.
A.2.1 Standard Deviation
The standard deviation (σ) of a data set is auseful measure of the uncertainty in any exper-imental result. It is basically a statistical mea-sure of the spread of the data. The smaller thisvalue, the more precise the data is considered(all of the experimental results would probablybe pretty close to the average). A large valuewould mean that the experimental results werenot all close to the average value that was cal-culated. The more data points that we have,the smaller the standard deviation should be.This is why we do many trials when performinga scientific experiment.
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APPENDIX A. ANALYSIS OF DATA A.2. STATISTICAL ANALYSIS
The standard deviation is given by
σ =
√(x1 − x)2 + (x2 − x)2 + · · · (xN − x)2
N − 1(A.1)
where xi are the individual measurements, xis the average of all the values, and N is thenumber of measurements.
When examining the data, you may findthat a few of the values are especially far fromthe rest. It is often reasonable to exclude thesevalues from any analysis since it is likely thatthese values result from some mistake in per-forming or recording that particular measure-ment. The data points that remain after thisanalysis are the ones that would be used forcomputing the mean and the standard devia-tion.
A.2.2 Confidence Intervals
The standard deviation can be used to obtainconfidence limits for our results. A confidencelimit (δ) for an average of a group of measure-ments can be defined as
δ =tσ√N
(A.2)
so that an average x with confidence intervalscan be expressed as x± δ. The relevant valuesfor t are given in the table.
What this means is that if we want a 95%confidence interval, and we took 12 measure-ments, we would use t = 2.20. A 95% confi-dence interval means that there is a 95% prob-ability that the true average 1 lies within theconfidence limits. To be even more sure thatthe true average is within our estimate, wecould use a 99% confidence limit which give awider range of possible values, but 95% confi-dence intervals are the most common measureof confidence in scientific studies.
Consider an example where we took 9 mea-surements, and got an average value x of 4.70
1obtained by repeating the experiment under theexact same conditions an infinite number of times
Table A.1: Values of t for various confidenceintervals
N Confidence interval of(no. of trials) 80% 90% 95% 99%
2 3.08 6.31 12.7 63.73 1.89 2.92 4.30 9.924 1.64 2.35 3.18 5.845 1.53 2.13 2.78 4.606 1.48 2.02 2.57 4.037 1.44 1.94 2.45 3.718 1.42 1.90 2.36 3.509 1.40 1.86 2.31 3.3610 1.38 1.83 2.26 3.2511 1.37 1.81 2.23 3.1712 1.36 1.80 2.20 3.1113 1.36 1.78 2.18 3.0614 1.35 1.77 2.16 3.0115 1.34 1.76 2.14 2.98∞ 1.29 1.64 1.96 2.58
and a standard deviation σ of 0.45. To ob-tain a 95% confidence interval, we would uset = 2.31 to obtain a confidence limit of ±0.35.Our confidence interval (or our best estimate)would then be 4.70±0.35, or in other words wecan say with a 95% degree of confidence thattrue experimental average is in the range of4.35 to 5.05. If we know the theoretical valueto be 4.8, then we can say that the data sup-ports the theory since this is in the range ofour uncertainty. If we have a theoretical valueof 5.4, then our estimate would be statisticallydifferent from this. In this case, the differencemay be due to systematic errors and this wouldhave to be investigated and rectified, if possi-ble.
Note, however, that this type of error anal-ysis does not take into consideration anysystematic errors present in the lab. Itonly addresses the random errors in the databy providing a quantitative measure of the pre-cision of our results.
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