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PHYS 102 Midterm: Midterm 1 Study Guide
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Physics
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PHYS 102
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John Milsom

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U of Arizona -PHYS 102- Midterm 1 Study Guide Midterm Exam Guide 0 Problem Solving Tips 0.1 You can’t memorize your way through a physics class The ▯rst instinct of many life science students is to study for exams by memorizing as much information as possible. This is not a bad instinct | it is generally a successful strategy in biology courses. However, this will not work in physics classeIntroductory physics courses teach problem solving. The only way to get better at problem solving is to practice | this means that you should not only do all of the homework that your professor assigns, you should also work together with a study group to solve or review di▯cult problems. The best way to study for a physics test is to review the solutions and revise your answers to your corrected home- work problems (but don’t try to simply memorize them!). Physics professors are known to put homework problems, either verbatim or slightly modi▯ed on exams. 0.2 How novices and experts solve problems Education research shows that novices and experts approach solving physics problems very di▯erently. If you would like a high grade on your physics exam, you should emulate expert problem solvers, rather than follow your instincts as a novice. How novices (students in physics 101) solve problems: 1. Panic! 1 2. Hunt through the textbook (or equation sheet) until they ▯nd an equa- tion that looks promising or has (some of) the right variables 3. Immediately plug in the numbers given in the problem 4. Plug all of those numbers into their calculator 5. Get a result 6. Relax! The problem has been solved. How experts (physics professors) solve problems: 1. Describe the physical system (objects, forces, interactions) 2. Clarify the geometrical relationships of the system (for example by drawing a picture or adding a coordinate system) 3. Identify the relevant physical principle (this is a Newton’s 3rd law prob- lem, this is a 1-D kinematics problem) 4. Make simplifying assumptions (can we represent this as a particle?, is air resistance important?) 5. Find relevant equations 6. Solve the equations algebraically ▯rst, then plug in numbers 7. Get a result 8. Check the units of the result (to verify the answer is correct) and in- terpret the result (is this reasonable?) 0.3 The mandatory steps for solving a physics problem You may not be an expert problem solver yet, but you can avoid some novice mistakes by following these steps: 1. Examine the situation to determine which physical principles are in- volved 2. Draw a picture 2 3. Write down the things that we know (knowns) and convert units if necessary 4. Write down what we want to ▯gure out (unknowns) 5. Write down an appropriate equation(s) 6. Solve that equation using algebra so that the variable we are interested in is by itself on one side of the equals sign 7. Plug in numbers 8. Use a calculator (or our heads) to do the arithmetic 9. Check to see if our answer is reasonable Most di▯culties that students have when they are doing physics homework assignments come from skipping steps or doing the steps out of order. You will be very tempted to plug in numbers to your equation as soon as possible. Do not fall into this trap! Wait until the end. I promise this will lead to less mistakes and higher exam grades. Drawing a picture will help, even if you are not very artistic, I promise, really. 0.4 Units Another di▯erence between experts and novices at physics problem solving is dealing with units. Novices tend to ignore them, experts are obsessed with them. There is a big di▯erence between 10 km and 10cm (or 10 miles and 10 inches if you are not metrically inclined)! If your units do not cancel when you are solving a problem (10km=10cm) you will get the wrong answer. Professors like to see if you are paying attention on exams by giving you quantities in weird units. One way to make sure that you are solving a problem correctly is to plug in numbers and units into your equation and then cancel the units. You should get an answer with the correct units. If not, you probably did something wrong. 3 0.4.1 Example of our problem solving technique Calculate how long it takes the radio waves to arrive at your current location from a radio station 10 km away. 1. This problem is a straight-forward 1-D kinematics problem, which also requires us to know a little bit about the properties of light. Since we know the displacement of the radio waves and can look up the velocity of light, we should be able to solve for the elapsed time. 2. A picture of our situation: You Tower 10 km 3. We know that we are 10 km away from the tower. Let’s convert our distance in kilometres into a distance in meters, so that we can plug it into an equation in later steps. d = 10 km 1 km = 1000m 1000 m d = 10▯km ▯ 1 ▯m d = 10000 m We also know the speed of the wave, since all light waves, including radio waves, travel at the speed of light. v = c = 3 ▯ 10 m=s 4 4. We want to ▯nd the time (in seconds) that it takes the radio waves to travel 10km. t =? 5. The equation that shows the relationship between speed (v), distance traveled (d) and time it takes to travel (t) for anything, including light waves is : d v = t In other words, our speed is the distance (meters) that we cover per unit time (seconds). 6. We need to solve this equation, so that t is by itself on one side of the equals sign before we can plug in numbers: d v = t t = d v 7. Now, ▯nally we can plug in numbers 10000 m t = 3 ▯ 10 m=s 8. Using a calculator 10000 m t = 8 3 ▯ 10 m=s t = 0:0000333333 s ▯5 t = 3:33 ▯ 10 s = 0:03 milliseconds 9. Is our answer reasonable? This time, 0.03 milliseconds, is much too small for us to notice, which is good. 5 1 Chapter 2: Kinematics 1.1 Important Equations You should be able to explain the following equations, what concepts they represent, and what each variable represents, and situations where you are likely to use them. ▯x = x f x 0 : displacement ▯t = tf▯ t 0 : elapsed time ▯x v = : Average velocity ▯t ▯v a = : Average acceleration ▯t kinematic equations : v0+ v x = x0+ vtv = 2 v = v0+ at 1 x = x0+ v 0 + at 2 2 v = v 0 2a(x ▯ x )0 1.2 Important Concepts 1.2.1 Vectors, Scalars, and Coordinate Systems A vector is a quantity with both a magnitude and direction. Examples include displacement, velocity, acceleration and force. A scalar is a quantity that has a magnitude, but no direction. Examples include time, temperature, mass and speed. Unless otherwise speci▯ed, motion to the right is usually considered pos- itive, and motion to the left is considered negative. Motion up is usually positive and motion down is negative. However, you are free to de▯ne what- ever positive direction you would like for a problem. Just be very careful that you are consistent! When solving a problem, it is usually a good idea to indicate the positive direction with an arrow in your drawing. 6 1.2.2 Displacement Displacement is di▯erence between the starting and ending positions of an object. The units are meters. It is a vector quantity (you can have a displacement of -2m). Distance traveled is the total length of the path taken between two points. It is a scaler quantity (you cannot travel a distance of -2m). 1.2.3 Time, Velocity, and Speed Velocity is displacement (change in position) per unit time traveled. Its units are m/s. Velocity is a vector quantity (you can travel -2 m/s). A negative velocity indicates that you are traveling in the negative direction (to the left or down, unless otherwise noted). Speed is much like velocity, except it is a scalar quantity. This means that your speed is always a positive number. For example, if you have a velocity of -2 m/s, your speed is 2 m/s. if you have a velocity of 2 m/s, your speed is also 2 m/s. Do not confuse velocity and speed. 1.2.4 Acceleration Acceleration is a change in velocity per unit time. When you accelerate, you are speeding up, slowing down, or changing direction. Its units are m/s . Acceleration is a vector quantity. Be very careful with signs when dealing with acceleration! A negative acceleration does not mean that you are traveling in the negative direction, it means that you are slowing down. The acceleration vector is pointing in the opposite direction as your velocity vector. A positive acceleration does not mean that you are traveling in the positive direction, it means that you are spreading up. The acceleration vector is pointing in the same direction as your velocity vector. 1.2.5 Falling Objects If an object is falling under the in uence of gravity, we know its acceleration. g = 9:80m=s . The acceleration vector points down. According to our sign 2 convention (down = negative), a = ▯g = ▯9:80m=s . Be careful with your signs when solving problems with falling objects. Things will speed up if 7 42 Chapter 2 | Kinematics Figure 2.14 (a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. (b) This car is slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the leFigure 1: (a) Positive velocity, positive acceleration. The car is speedingsite to its direction of motion. (c) This car is moving toward the left, but slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is deup. (b) Positive velocity, negative acceleration. The car is slowing down.r is speeding up as it moves toward the left. It has negative acceleration because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up (not de(c) Negative velocity, negative acceleration. The car is slowing down. (d) Negative velocity, positive acceleration. The car is speeding up. Example 2.1 Calculating Acceleration: A Racehorse Leaves the Gate A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration? 8 dimensional kinematics. Slopes and General Relationships First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axiY-axis and the vertical axis tZe-axis, as in Figure 2.46, a straight-line graph has the general form Z = NY + C. (2.89) Here N is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight linC.is used for the y-intercept, which is the point at which the line crosses the vertical axis. Figure 2.46 A straight-line graph. The equation for a sZ = NY + Cne.is Figure 2: Equation for a straight line. Graph of Displacement vs. Time (a = 0, so v is constant) Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, haveY on the vertical axis anU on the horizontal axis. Figure 2.47 is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada. traveling up. 1.2.6 Graphical Analysis of One-Dimensional Motion The slope of a position vs. time graph is velocity. The slope of a velocity vs. time graph is acceleration. 9 64 Chapter 2 | Kinematics Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a projectilegraphs for something falling under the in uence of gravity. Note that thectual path of the rock in space is straight up, and straight down. Discussionposition and velocity of the object are positive for some points negative for 2 The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upwaZ1, since graph switches from positive to negative at the point where the slope of the and W1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, bothmZ gandh Wwitare negative, meaning the rock is below its starting point and continuing to move 3 3 downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still 2 2 −9.80 m/s . Its acceleration −9.80 m/s for the whole trip—while it is moving up and while it is moving down. Note that the values for are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free- fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later. Making Connections: Take-Home Experiment—Reaction Time 1.3 Example Exam Question 1.3.1 Ball A and Ball B roll parallel to each other on a level table top. Below is a stroboscopic photograph, showing the position of the balls at 1 second intervals and a length scale for the table top. (20 points) 1. At what time to the balls have the same speed? Explain your answer. (5 points) The magnitude of velocity is speed. Veloctiy is de▯ned as ▯x=▯t. Since the balls are shown at equal time intervals, they will have the same speed when the distance intervals are the same. If we de▯ne t = 0 at x = 0, this happens between t = 3 and t = 4. 2. On the axis provided, sketch a position vs time, a velocity vs time and an acceleration vs time diagram for both balls. Make sure the points and/or lines on your graphs are clearly labeled. (10 points) see below 3. What is the physical signi▯cance of the slopes of your graphs? (5 points) The slope of the x vs. t graph is velocity. The slope of the v vs. t graph is acceleration. The slope of the a vs. t graph is the change in acceleration over time. 11 x v t t a t 12 2 Chapter 3: Two-Dimensional Kinematics 2.1 Important Equations You should be able to explain the following equations, what concepts they represent, and what each variable represents, and situations where you are likely to use them. vector components : A x Acos▯ A = Asin▯ y adding vector components : R x A + x x R y A + y y q R = R + R 2 : magnitude vector sum x y ▯1 ▯ = tan (Ry=R x : direction vector sum Projectile motion : x = x 0 v x vx= v 0x = vx y = y0+ 12(v 0y+ v yt vy = v 0y▯ gt 1 2 y = y 0 v 0y▯ gt 2 v = v 2 ▯ 2g(y ▯ y0) y 0y Quadratic equation : 2 at + pt + c = 0 b ▯ b ▯ 4ac t = ▯ 2a 13 2.2 Important Concepts 2.2.1 Kinematics in Two Dimensions: An Introduction The horizontal (x) and vertical (y) components of motion can be analyzed separately from one another. This greatly simpli▯es problem solving. Figure 4: A red and yellow ball fall under the in uence of gravity. Even- though the yellow ball has an initial velocity in the x direction and the red ball does not, their y motions are the same. They hit the same y positions at the same time. 2.2.2 Vector Addition and Subtraction: Graphical Methods Graphical method for adding vectors: Make sure that you follow these steps exactly! You should have two arrow heads that touch. Many students will be tempted to draw an arrow from the head of the ▯rst vector to the tail of the second vector in the last step. The might look nicer, because no arrow heads touch, but it will get you the wrong answer. To subtract vectors, put the head of the vector on the other end of your arrow. For vector B, this gets you -B. Then do vector addition, as described above (A + (-B)). 14 Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total displacement vectorUsing a protractor, draw a linȆ relative to the east-west axis% Tof the arrow is proportional to the vector’s magnitude and is measured along the line with a ruler. In t%isof the vector is 10.3 units, anȆ ise direction 29.1º north of east. Vector Addition: Head-to-Tail Method The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow. Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. ThFigure 5: (a)Draw an arrow to represent the ▯rst vector (b) Draw an arrowointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum Dr. The length of thD ais proportional to the vector’s magnitude and is to represent the second vector. Place the tail of the second vector at the measured head of the ▯rst vector (c) Find the sum of the two vectors by drawing anmeasured with a protractor to be 29.1º . arrow from the tail of the ▯rst vector to the head of the second vector Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. 2.2.3 Vector Addition and Subtraction: Analytical Methods Usually it is much easier to put away your ruler and protractor, and use a little trigonometry when you have to deal with vector quantities. Always make sure your calculator is in degree (DEG) mode when using sin and cos. You can deal with the x and y components of your motion separately, then combine them back together at the end. 2.2.4 Projectile Motion A projectile is something that is falling only under the in uence of gravity. The path it follows is called a trajectory. We generally assume that air This OpenStax book is available for free at http://cnx.org/content/col11406/1.9 resistance is negligible for these problems (otherwise they would be way too hard!). Projectile motion is made up of two independent motions: Free fall mo- tion in the vertical direction and uniform motion at constant velocity in the horizontal direction. These motions can be analyzed independently and then combined to understand the total motion of an object. To solve a projectile motion problem: 1. Resolve the the motion into horizontal (x) and vertical components (y) 15 " Y " ≠Z" (3.5) If the vectoA is known, then its magnitud" (its length) and its aȆg(its direction) are known. To "iYand " Z, its x- and y-components, we use the following relationships for a right triangle. " = " cos Ȇ (3.6) Y and " = " sin Ȇ. (3.7) Z Figure 3.27 The magnitudes of the vector coApYandtA Z can be related to the resultaAt and the angȆewith trigonometric Figure 6: How to ▯nd the X and Y components of a vector. identities. Here we se" Y " cos Ȇ and "Z= " sin Ȇ . Suppose, for example, tha| A =xAcos▯ and A = Asyn▯. For example, you will often be givenson walk
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