RRHS Physics 12 Course Notes

J. Burke

2009-2010

c2001-2010

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 Newtons Law of Universal Gravitation . . . . . . . . . . . . . . . . . . . . . 303.2.2 Acceleration Due to Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.3 Satellite Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.4 Keplers 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 Coulombs 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 Ohms 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 *Kirchhoffs 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 Plancks 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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Textbook Correlations

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 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 xy 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 (thelength of the vector) and an angle (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 5m can be used; a 30m displacement

1Note that if we know the magnitude d and the angle, we can use sin and cos identities to solve for dxand 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 = 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 get

a 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 =(x)2 + (y)2

to find the magnitude of c and the angle canbe found using

tan =yx

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, x = ax bx and y =ay by.

1.1.2 Relative Velocity

We saw in section 1.1 that an objects 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 no

absolute 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 2m/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 boats 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 Slimys 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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CHAPTER 1. DYNAMICS EXTENSION 1.1. INTRODUCTION TO VECTORS

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: 10.0m, 30o northof east; 6.0m, 37o east of north; and 12m,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 110m upstream fromwhere it starts. To do so, the pilot musthead the boat at a 45o upstream angle.What is the speed of the rivers 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 planes 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 cars 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 planes 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 240m 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 Newtons 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 Newtons 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 a

box with a rope that makes an angle withthe ground.

Note that the expected acceleration (hori-zontal) for this box and the applied force areneither parallel nor perpendicular, so Newtons2nd 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 usingNewtons 2nd Law.First, the vertical forces. I will take up as

the positive direction; therefore, FN and Fpywill both be positive and Fg will be negative.

may = Fy

may = FN + Fpy Fgand

0 = FN + Fpy Fgsince 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 = FN , where is the coefficient of friction).Now for the horizontal forces:

max = Fx

max = Fpx FfThis can then be used with the horizontal

acceleration. 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 andNewtons 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 Newtons 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 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 (1.2)

andFgy = mg cos (1.3)

We see now by analyzing the perpendicu-lar forces

may = Fy

may = FN Fgym(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 used

to obtain an expression for the parallel accel-eration on the inclined plane

max = Fx

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 0.300m/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 persons 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)

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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 of

the net force on a body in equilibrium musteach be zero, so

Fx = 0

andFy = 0

Looking at the components in the x and ydirection separately, this tells us that in the xdirection

F2x F1x = 0and in the y direction

F1y + F2y Fg = 0The requirement that the net force be zero

is 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 is the product of a force multi-

plied by a distance from the pivot.

= Fd (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.

= 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 conditions

for 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 youdont 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 lifters 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 100N 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 of15.0m. 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 diagram

of 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 = Fx, 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 v2yi

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 use

the 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 the

air as shown below:

Extremely Important!! The arrow in thediagram above represents the velocity vector

for the soccer ball, not the balls actual path!The direction of the arrow indicates the ballsinitial 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 object

launched at an angle should be to resolve theobjects 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, since

there 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 leaves

the 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 =bb2 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-ters 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! Its 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 1500m whena wheel falls off. What horizontal distancewill the wheel travel before it strikes theground and what will the wheels 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 2

g

where v is the initial velocity of theprojectile and is the angle with thehorizontal. (Hint: use the trigono-metric identity sin 2 = 2 sin cos )

(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-nals 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 players 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 with the horizontal. Derive aformula for 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 batters 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 thepucks 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 asHookes Law.The spring constant k is constant for any

given 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 hang

vertically 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 Hookes 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 = 2pim

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

soE = Fd

But F is not constant; it increases linearly aswe move away from equilibrium (Eq 2.6). Sothe average force exerted will be F = 12kx 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 where

the 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 = 12kA

2.

2.2.2 Pendulum Motion

For small displacements ( 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 = 2pi

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 cars 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 spiders 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 0.100m 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 geologists 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 = pa + pb (2.11)

or, since p = mv,

mava +mbvb = mava +mbvb (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 the

right 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 (pa and pb) is equal to the total of themomentum vectors before the collision (pa).

pa = pa + pb (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 be

found 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 = pax + pbx

where the momentum components can befound using the appropriate velocity compo-nents (pax = mavax and pbx = 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 = pay pby

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 = v2a + 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 va and vbwould 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, va and vb (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 of8.61023 kgm/s and 6.21023 kgm/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 105m/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 Newtons 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 of

a 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 (2pir), we get

v =2pirT

(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 =2pivT

(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 NewtonsSecond 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 Fcis 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 the

vectors 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-tons Second Law to this situation, we get

mac = Fc

mac = T Fgwhere 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-tons 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 pushed

outward; it is, in fact, being pulled inward bythe string. Newtons 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 to

explain 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 moons 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 =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 3.0m 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 balls 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 persons weight changed be-cause of the earths 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 60m 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 Newtons Law of UniversalGravitation

In the 1600s, Newton discovered that thisforce depends on the two masses involved andthe distance separating them.

Fg m1m2r2

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 the

constant 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.Newtons Law of Universal Gravitation can

be expressed as

Fg =Gm1m2

r2(3.7)

where G is the proportionality constant and isequal to 6.67 1011 Nm2/kg2. It should benoted that this law allows us to accurately pre-dict results, but not to understand why theyare so. We dont 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 earths curvature matches thecurvature of the falling object. In this way,a satellite can be launched so that it actuallyfalls 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 Newtons Second Law we get

30 RRHS Physics

CHAPTER 3. PLANETARY MOTION 3.2. UNIVERSAL GRAVITATION

F = mac (3.9)

What