SAT II Physics (Gary Graff) Episode 1 Part 3 doc

Chapter 1 PRELIMINPRELIMIN PRELIMINPRELIMIN PRELIMIN ARAR ARAR AR Y CONCEPTSY CONCEPTS Y CONCEPTSY CONCEPTS Y CONCEPTS Peterson’s: www.petersons.com 47 CHAPTER 1 PRELIMINPRELIMIN PRELIMINPRELIMIN PRELIMIN ARAR ARAR AR Y CONCEPTSY CONCEPTS Y CONCEPTSY CONCEPTS Y CONCEPTS SIMPLE EQUASIMPLE EQUA SIMPLE EQUASIMPLE EQUA SIMPLE EQUA TIONS AND ALGEBRATIONS AND ALGEBRA TIONS AND ALGEBRATIONS AND ALGEBRA TIONS AND ALGEBRA Physics is the branch of science that studies physical phenomena. Why does the sound of a starter’s pistol reach an observer standing at the race finish line after the puff of smoke is seen when the gun is fired? Why do some objects float in water, while other objects sink? What causes a ball to roll downhill? These questions and more are the stuff of physics. Sometimes the study of physical phenomena involves observation, experimentation, and calculation. The calculations used in this review are accomplished by using a little algebra, a little geometry, and a little trigonometry. Many relationships in physics can be expressed as equations. For example: F = ma s = vt Sometimes the value to be found is not by itself (isolated) in the equation. Take, for example, the linear motion equation: V f 2 = V 0 2 + 2as Let’s suppose the acceleration a is the only unknown quantity in the equation. We must perform a little algebraic manipulation to isolate a so that it is the only quantity on its side of the equal sign. Before starting the process, remember one simple rule: whatever operation you perform on one side of the equal sign, you must also perform on the other side of the equal sign. Let’s begin by isolating a in the equation below: V f 2 = V 0 2 + 2as We then subtract V o 2 from both sides: V f 2 –V o 2 = V o 2 –V o 2 + 2as Peterson’s SAT II Success: Physics 48 This leads to: V f 2 – V o 2 = 2as Then we divide both sides by 2s: VV s as s f 2 0 2 2 2 2 − = Clearing fractions gives: VV s a f 2 0 2 2 − = If you find this algebra to be difficult, you might need to stop here and review your math skills before proceeding. GRAPHS Some relationships studied in physics exist between experimental values. The relationships between these quantities can be shown using a technique called gg gg g rr rr r aphingaphing aphingaphing aphing. With this technique, one quantity (the independent variable) is carefully controlled, while the other quantity (the dependent variable) is measured at each value of the independent variable. The independent variable is plotted along the x axis of the graph, and the dependent variable is plotted along the y axis of the graph. The slope for the graph is calculated by dividing the ∆Y by the ∆X . Slope y x = ∆ ∆ A graph is a picture of the relationship between two or more quantities. A direct relationship is one in which both quantities increase (or decrease) in the same manner. In an inverse relationship, one quantity increases and the other decreases. A parabolic relation- ship exists when one quantity varies as the square of the other. CHAPTER 1 Peterson’s: www.petersons.com 49 DIRECT RELATIONSHIP GRAPH A Graph A shows a linear relationship between x and y. For every 2 units’ increase in variable x, variable y increases by 1 unit. The slope of the graph is calculated as rise over run. GRAPHS Peterson’s SAT II Success: Physics 50 INVERSE RELATIONSHIP GRAPH B Graph B shows an indirect linear relationship between x and y. CHAPTER 1 Peterson’s: www.petersons.com 51 PARABOLIC RELATIONSHIP GRAPH C Graph C shows a parabolic relationship between x and y. The ability to read a graph is crucial. They are used to display and compare physical concepts. Be sure you can read and evaluate graphs. GRAPHS Peterson’s SAT II Success: Physics 52 RIGHT TRIANGLES Any triangle with a 90° internal angle is defined as a right triangle. In such a triangle, the remaining two internal angles of the right triangle sum to an additional 90°. If you are sure you are dealing with a right triangle, the Pythagorean theoremPythagorean theorem Pythagorean theoremPythagorean theorem Pythagorean theorem becomes a useful tool when trying to find one of the sides. The right triangle above is labeled according to convention. The side opposite the 90° angle is called the hypotenuse and is labeled c. The side beside the smaller of the remaining two angles is called the adjacent side b, and the side across from the smaller angle is called the opposite side a. The Pythagorean theorem is stated mathematically as c 2 = a 2 + b 2 . Thus, should we know the size of any two sides, we can find the missing side. Side Side Side cab acb bca =+ =− =− 22 22 22 Right triangles have special relationships not only with their sides but also with their internal angles. For every right triangle, the sides of that triangle and the internal angles of that triangle form a ratio. CHAPTER 1 Peterson’s: www.petersons.com 53 These are given as: sine or sin cosine θθ θ == = opposite hypotenuse a c adjacent hypot eenuse b c opposite adjacent a b or cos tangent or tan θ θθ = == Further, no matter how long the sides of the triangle may be, the ratio between the three sides remains the same as long as the internal angle remains the same. Triangles 1, 2, 3, and 4 above all have an internal angle of 30°. Thus the sine, cosine, and tangent values are the same in all four triangles. CARTESIAN COORDINATES The construction and use of right triangles is a valuable tool in solving physics problems. Unfortunately, not all measures and quanti- ties begin neatly at zero. Rather than allow this to complicate mat- ters, we simplify by using the Cartesian Coordinate SystemCartesian Coordinate System Cartesian Coordinate SystemCartesian Coordinate System Cartesian Coordinate System in conjunction with triangles. You may also recognize this as the four quadrant x and y system. RIGHT TRIANGLES Peterson’s SAT II Success: Physics 54 • The region between the x and y coordinates on the upper right is called the first quadrant. All x and y quantities in the first quadrant are positive. • The region between the x and y coordinates on the upper left is called the second quadrant. All x quantities in the second quadrant are negative, and all y quantities in the second quadrant are posi- tive. • The region between the x and y coordinates on the lower left is called the third quadrant. All x and y quantities in the third quad- rant are negative. • The region between the x and y coordinates on the lower right is called the fourth quadrant. All x quantities in the fourth quadrant are positive and all y quantities in the fourth quadrant are negative. Quadrant I II III IV X coordinate +x x x +x Y coordinate +y +y y y CHAPTER 1 [...]... 61 CHAPTER 1 Rope a extends 15 ° into the first quadrant Side y = Side r (sin 15 °) = (30 N) (.26) = 7.8N Side x = Side r (cos 15 °) = (30 N) (.97) = 29.1N Both components of rope a are located in the first quadrant y = positive ∴+7.8N x = positive ∴+29.1N 62 Peterson’s SAT II Success: Physics SCALARS AND VECTORS Rope b extends 35 ° into the third quadrant Side y = (side r) (sin 35 °) = (16 N) (.57) = 9.1N... is exactly 18 0° opposite the direction the bird must fly Take the resultant vector and add (or subtract) 18 0° to or from the vector’s direction 18 0° + 31 = 211 ° The bird must fly 5 83 m @ 211 ° from its present position to reach its nest When this is done, the bird will effectively cancel out the resultant vector That’s why its flight will be the equilibrant 60 Peterson’s SAT II Success: Physics SCALARS... (cos 35 °) = (16 N) (.82) = 13 .1N Both components of rope b are located in the third quadrant y = negative ∴−9.1N x = negative ∴ 13 .1N Peterson’s: www.petersons.com 63 CHAPTER 1 Rope c extends 75° into the first quadrant Side y = (side r) (sin 75° ) = (20N) (.97) = 19 .4N Side x = (side r) (cos 75° ) = (20N) (.26) = 5.2N Both components of rope c are located in the first quadrant y = positive ∴ +19 .4 N... we begin: 50 km/hr = ? m/sec There are 10 00 meters in a kilometer, thus we multiply 50 km by 10 00 m/km, yielding 50,000 meters Now we have: 50,000 m/hr = ? m/sec There are 36 00 seconds per hour, thus we multiply one hour by 36 00 sec/hr, yielding 36 00 seconds Divide distance by time for the final value 50, 000m = 13 .88m / sec 36 00 sec 56 Peterson’s SAT II Success: Physics SCALARS AND VECTORS The whole... x = positive ∴+5.2 N 64 Peterson’s SAT II Success: Physics SCALARS AND VECTORS Rope d extends 30 ° into the third quadrant Side y = (side r) (sin 60°) = (30 N) (.87) = 26 .1 Side x = (side r) (cos 60°) = (30 N) (.5) = 15 .0N Both components of rope d are located in the fourth quadrant y = negative ∴−26.1N x = positive ∴ +15 .0 N Peterson’s: www.petersons.com 65 CHAPTER 1 All four ropes have been broken into... the x components Y axis + 7.8N 9.1N +19 .4N 26.1N ROPE Rope a Rope b Rope c Rope d X axis +29.1N 13 .1N + 5.2N +15 .0N –8N in x direction +36 .2N in y direction After combining the x components and the y components, the values are the two components of the new resultant vector Apply the Pythagorean Theorem to find its magnitude r = x 2 + y2 r = (−8N)2 + ( +36 .2 N)2 r = 37 N Now that we know the magnitude... vector r = (30 0m )2 + (500m ) r = 583m This is the magnitude value of the resultant vector The direction for the resultant can be found with a little right triangle trigonometry y x 30 0m tan θ = 500m tan θ = 6 θ = 31 tan θ = Thus the resultant vector (where the bird is located in respect to its starting point) is 583m @ 31 Recall that the bird wanted to fly back to its nest It can’t fly at 31 from its... instantly that there are 60 seconds in one minute and 36 00 seconds in one hour The metric units you should know are: 10 00 meters = 1 kilometer 10 0 centimeters = 1 meter 10 00 millimeters = 1 meter Then there are the equivalent values for volume (liters) and mass (grams) as well There is a list of constants, equivalents, and physics formulas on page 15 Be sure you are very familiar with this information... y = = −.22 = 1 x +36 .2 N This number tells us that the direction of the resultant vector is 1 That means that to find the value of the resultant vector, we simply subtract 1 from 36 0° to find the direction a single rope must pull to equal the pull from the original four ropes The resultant vector is 37 N @ 35 9° Simply stated: One rope pulling with a force of 37 N in a direction of 35 9° accomplishes... value 50, 000m = 13 .88m / sec 36 00 sec 56 Peterson’s SAT II Success: Physics SCALARS AND VECTORS The whole conversion can be made into a one-step calculation by writing each part and solving 1   (50 km/hr) (1, 000 m/km)  36 00sec  = 13 .88m/sec   hr   During the conversion process it is absolutely essential that you state and complete your dimensional analysis SCALARS AND VECTORS SCALARS The treatment . exactly 18 0° opposite the direction the bird must fly. Take the resultant vector and add (or subtract) 18 0° to or from the vector’s direction. 18 0° + 31 = 211 ° The bird must fly 5 83 m @ 211 ° from. (16 N) (.82) = 13 .1N Both components of rope b are located in the third quadrant. y x negative N negative N =∴− =∴− 91 13 1 . . SCALARS AND VECTORS Peterson’s SAT II Success: Physics 64 Rope. 29.1N =∴+ =∴+ CHAPTER 1 Peterson’s: www.petersons.com 63 Rope b extends 35 ° into the third quadrant . Side y = (side r) (sin 35 °) = (16 N) (.57) = 9.1N Side x = (side r) (cos 35 °) = (16 N)