AP Physics B Summer Assignment Packet 3

Transcription

1 AP Physics B Summer Assignment Packet 3 The assignments included here are to be brought to the first day of class to be submitted. They are: Problems from Conceptual Physics Find the Mistake Straightening Curves Other materials due on the first day of class: Topics 4 and 5 from The Physics Classroom Your list of Big Ideas Read the instructions packet for all the details about these assignments.

2 Name: AP Physics B Summer Assignment Problems From Conceptual Physics All problems are taken directly from Conceptual Physics, 3 rd edition by Paul G. Hewitt Use a separate page to show solutions. For all problems that need it, use g = 9.8 m/s 2 for the acceleration due to gravity. 1) Why is it that an object can accelerate while traveling at constant speed, but not at constant velocity? 2) A ball is thrown straight up. What will be the instantaneous velocity at the top of its path? What will be its acceleration at the top? Why are your answers different? 3) a) Find the speed required to throw a ball straight up and have it return 6.00 seconds later. b) How high does the ball go? 4) If a salmon swims straight upward in the water fast enough to break through the surface at a speed of 5.00 meters per second, how high can it jump above water? 5) In the absence of air resistance, why does the horizontal component of velocity for a projectile remain constant while the vertical component changes? 6) At the instant a ball is thrown horizontally over a level range, a ball held at the side of the first is released and drops to the ground. If air resistance is neglected, which ball strikes the ground first? 7) A projectile is fired straight up at 141 m/s. How fast is it moving at the top of its trajectory? Suppose it is fired upward at 45.0 o above the horizontal plane. How fast is it moving at the top of its curved trajectory? 8) The force of gravity is twice as great on a 2 kg rock as on a 1 kg rock. Why does the 2 kg rock not fall with twice the acceleration? 9) a) Calculate the acceleration if you push with a 20.0 N horizontal force on a 2.0 kg block on a horizontal friction-free air table. b) What acceleration occurs if the friction force is 4.0 N? 10) A horizontal force of 100 N is required to push a crate across a factory floor at a constant speed. What is the net force acting on the crate? What is the force of friction acting on the crate?

3 11) A 10.0 kg mass on a horizontal friction-free air track is accelerated by a string attached to another 10.0 kg mass hanging vertically from a pulley as shown. What is the force due to gravity of the hanging mass? What is the acceleration of the system of both masses? 10.0 kg 10.0 kg 12) Suppose the masses described in the preceding problem are 1.00 kg and kg, respectively. Compare the accelerations when they are interchanged, that is, for the case where the 1.00 kg mass dangles over the pulley, and then for the case where the kg mass dangles over the pulley. What does this indicated about the maximum acceleration of such a system of masses? 13) When you walk along the floor, what pushes you along? 14) When you jump up, the world really does recoil downward. Why can t this motion of the world be noticed? 15) When a rifle is fired, how does the size of the force of the rifle on the bullet compare with the force of the bullet on the rifle? How does the acceleration of the rifle compare with that of the bullet? Defend your answer. 16) If a bicycle and a massive truck have a head-on collision, upon which vehicle is the impact force greater? Which vehicle undergoes the greater change in its motion? Defend your answers. 17) Since the force that acts on a bullet when a gun is fired is equal and opposite to the force that acts on the gun, does this imply a zero net force and therefore the impossibility of an accelerating bullet? Explain. 18) A bug and the windshield of a fast-moving car collide. Indicate whether each of the following statements is true or false. a) The forces of impact on the bug and on the car are the same size. b) The impulses on the bug and on the car are the same size. c) The change in speed of the bug and the car is the same. d) The changes in momentum of the bug and of the car are the same size. 19) If a mouse and an elephant both run with the same kinetic energy, can you say which is running faster? Explain in terms of the equation for KE. 20) A hammer falls off a rooftop and strikes the ground with a certain KE. If it fell from a roof that was four times higher, how would its KE of impact compare? Its speed of impact? (Neglect air resistance.)

4 Name: AP Physics B Find the Mistake Each problem has the correct answer in italics, and an incorrect approach to solving is shown. For each, you need to: describe, in words, what is wrong about the given solution (e.g., simply saying the equation is wrong is not enough if that is what is wrong, you must say why that equation is wrong); and then solve it correctly. But, do NOT take a different approach. (e.g., if the original solution uses kinematics, do not solve it using energy) For some, more than one incorrect solution is shown. Address the mistakes in both. Throughout this worksheet, the mistakes are never mathematical, nor are the significant figures incorrect. There is something wrong with the physics in each situation. (For any problems involving the acceleration due to gravity, use g = 9.8 m/s 2.) 1) A car is stopped at a red light. When the light turns green, the driver steps on the gas pedal, causing the car to speed up at a uniform rate. It covers 34 meters in 4.0 seconds. What is the car s speed at the end of the 4.0 seconds? (17 m/s) You solve by doing: v = d / t = 34 m / 4.0 s = 8.5 m/s 2) You drop a rock from rest off a 24.0 m tall cliff. It hits the ground 2.21 s later. What is its speed as it hits the ground? (21.7 m/s) You solve by doing: v = d / t = 24.0 m / 2.21 s = 10.9 m/s 3) Paul is driving his car at 22.4 m/s. In 5.65 seconds, he speeds up to 27.8 m/s. How much distance was covered in that time? (142 m) You solve for the distance by doing: v = d / t d = vt = (27.8 m/s)(5.65 s) = 157 m Then you try this (which is also wrong): v = d / t d = vt = (22.4 m/s)(5.65 s) = 127 m 4) Gina is driving behind Paul, and notices a police car. To make sure she is under the speed limit, she slows down from 22.4 m/s to 19.8 m/s in 5.65 seconds. How much distance does she cover in this time? (119 m) You solve for the acceleration by doing: a = Δv / t = (v f v o ) / t = (22.4 m/s 19.8 m/s) / 5.65 s = m/s 2

5 Then, you solve for the distance by doing: d = v o t + ½ at 2 = (22.4 m/s)(5.65 s) + ½(0.460 m/s 2 )(5.65 s) 2 = 134 m 5) A track runner finishes a race and begins to slow down in a uniform manner. In 3.5 seconds, she comes to a complete stop, covering 12 meters while doing so. What was her speed when she finished the race? (6.9 m/s) You solve by doing: v = d / t = 12 m / 3.5 s = 3.4 m/s 6) Robin Hood shoots an arrow horizontally with an initial speed of 38 m/s from a height of 1.8 m. Assuming it does not hit anything along the way, how much time does it take to land, and how far away does it hit the ground from where Robin is standing? (t = 0.61 s; d = 23 m) You solve for the time by doing: v = d / t t = d / v = 1.8 m / 38 m/s = s You correctly solve for the time, and now to solve for the distance it travels, you do: d = v o t + ½ at 2 = (38.0 m/s)(0.61 s) + ½ (9.8 m/s 2 )(0.61 s) 2 = 25 m Find the Mistake Page 2 7) Your friend stands on the roof of her house, which is 12.2 m off the ground, and kicks a soccer ball to you so that it is initially moving horizontally at 13.9 m/s. What is its speed as it hits the ground? (20.8 m/s)

6 You find the time to fall by doing: d = v o t + ½ at 2 t = 2d / a (since v o is initially zero in the y-direction) so, t = 2(12.2 m) / 9.8 m/s 2 = 1.58 s Then, you find the final speed by doing: v f = v o + at = 0 m/s + (9.8 m/s 2 )(1.58 s) = 15.5 m/s 8) As shown to the right, a 6.00 kg mass is pulled upwards by a string. It is being accelerated upwards at a rate of m/s 2. Find the tension in the string. (T = 61.8 N) v a = m/s kg You solve by assuming the tension must be equal to the weight, so T = 58.8 N.

7 AP Physics B Free-Body Diagrams (F) force any generic push or pull that does not fit into one of the other categories (w) weight the force of gravity; always points down (towards the center of the object that is applying the force) (T) tension the force that results from an object s tendency to keep its shape when stretched; usually in strings (N) normal force the reaction force provided by a surface which supports another object; always points perpendicular to the surface (F f ) friction the force that results when two objects slide past (or try to slide past) each other; points in the direction to oppose motion (ΣF or F net ) net force the vector sum of all concurrent forces acting on an object Free-Body Diagrams (FBDs) To analyze a problem involving forces, a free-body diagram is often drawn. In such a diagram, you isolate one object at a time and only draw the forces that act on that object as vectors originating from the object s center. Other information (displacement, velocity, acceleration, etc.) that may be pertinent to the problem may be drawn off to the side, but it is essential that only forces are actually drawn connected to the object itself. The force vectors are labeled with their type of force. When known, the magnitudes of the forces are included. It is easy to combine forces, using the free-body diagram, to determine the net force on the object. Keep in mind, however, that net force is not A force it is the vector sum of forces. As such, a vector for net force does NOT appear on the free-body diagram because that would imply there is a separate force acting on the object, aside from the real forces you have already identified. Example #1 Example #2 Example #3 A 1.5 kg block is lifted up by a string, accelerating at a rate of 0.25 m/s 2. An object of mass m is dragged to the right at constant velocity along a rough surface. Albert rides his sled up a snowy hill. v F net T N a = 0 m/s 2 F net = 0 N v F net a a = 0.25 m/s kg F f m F N v w = 15 N w w F f Notes: the block does have its mass labeled, even though that is not a force, and that is fine; the velocity, acceleration, and net force vectors are shown to the side, not on the FBD; since the tension force must be winning (the block is accelerating upwards), its vector is drawn longer than the weight vector; since there is no surface, there is no normal force Notes: since the problem says the object is moving at a constant velocity, the information about the acceleration and net force are listed; also, since the net force is zero, the vectors for the applied force and friction force are drawn the same length (as are the normal force and weight vectors); the friction force acts to oppose motion Notes: the weight vector always points straight down, no matter what; according to the definition of the normal force, it points perpendicular to the surface; since the problem does not indicate a lack of friction, a friction force is shown; notice the velocity is up the hill, but the acceleration is down the hill (what does that mean?)

8 Name: AP Physics B Summer Assignment Straightening Curves Graphical analysis is a huge part of your studies in physics. When data is plotted, relationships are seen in ways that numbers alone cannot express. The shape of the graph gives more information about the situation being studied, and calculations can be made from the curve that represent other physical quantities. The easiest curve to analyze is a straight line. By definition, a straight line is a curve with constant slope. From algebra, we know that the general equation of a straight line is y = mx + b, where m is the slope and b is the y-intercept. If the data we plot gives a straight line, it is easy for us to calculate the slope of the line or to read the value of the y- or x-intercept all of which may be pertinent to analyzing the problem. We ll start with some easy examples. 1) v f = v o + at This gives the final speed (or velocity) after an object has undergone constant acceleration for some time. You can see that the equation maps to the equation for a line very clearly: v f = v o + a t y = b + mx It is obvious that, when v f is plotted on the vertical axis and t is plotted on the horizontal axis, the slope will represent the acceleration and the y-intercept will represent v o. While this is an easy example, there are some things to note. First, when a straight line is plotted, the slope must be constant and so must the y-intercept. (If the slope were not constant, the line would not be straight, and if the y- intercept were not constant well, what would that even look like?) So, when looking at the equation and choosing what to plot, you must arrange it so that whatever lines up with m and b are constant values. For this particular situation where an object starts at some initial speed, v o, and undergoes acceleration everything is fine because v o cannot change and because we assume the acceleration is uniform. 2) F = qvb It is fine if you do not know what all these terms mean what is important is if you can map it to the straight line equation. So, what to plot? Well, it depends on what is trying to be measured in the experiment. Let s say that the goal is to find q, the charge in coulombs (C) on a small particle, that always travels at the same speed v and feels a force F in a magnetic field B, which is measured in Tesla (T) and can be varied. A good idea is to pull the desired quantity out in front so it physically appears where m, the slope, appears. Whatever is left is then plotted on the horizontal, but you have some choices: F = q(vb) plot F vs. vb slope = q F = (qv)b plot F vs. B slope = qv Either choice is perfectly fine. In the first, you have to calculate the product of v and B, but the slope is what you want. In the second, you have to do a calculation in the end to get q out of the slope. What you cannot plot is F vs. v since v was a constant. What would that look like? It would be a meaningless vertical line. That tells you nothing, except that the two terms are unrelated to each other.

9 Here is sample data for this experiment. The first graph is done for you to start you off. Plot F vs. vb, draw the trend line, and calculate the slope, which should be very close to my value of q. Data: v = 1.6 x 10 4 m/s B = Magnetic Field F = Force (N) (T) x x x x x x 10-3 vb (T. m/s) VERY IMPORTANT Sidebar About Finding Slope: Real data does not form a perfectly straight line. When you draw the best-fit trend line, your aim is to go through the pattern of points. DO NOT pick two points and connect them. This approach may work some (or maybe even most) of the time, but it is a dangerous habit to adopt. When finding slope, you must pick two points that are ON THE LINE. These may be points that are not data points. They will very likely not be data points. They should be two points that are ON THE LINE and are easy to read from the graph. IF the line happens to go through a data point, you may use it, but you should not force the line to do so just to make your life easier. VERY IMPORTANT Sidebar About Setting a Scale: Each scale must start at zero and must be consistent. The two scales do not have to match in intervals. You do not have to label every line. NEVER USE A BREAK IN THE SCALE. It can seriously mess with your calculation of the slope, and sometimes, it will give you a false sense of the pattern. F vs. B F vs. vb

10 Straightening Curves Page 2 The examples so far already looked like linear functions ones that produce a straight line, no matter what. But not all equations are so simple. Certain types of equations pop up with enough regularity that we can look at the general form and then match future equations to the standard examples. (For all of the equations that follow, k is simply a constant.) Parabola: y = kx 2 plot y vs. x 2 If you simply plot y vs. x, you will get a parabolic curve. But, if you want to get a value for k, you straighten the curve by plotting y vs. x 2. Simply square the x values. By doing so, you are forcing the parabolic equation to look like a linear one: y = k(x 2 ) y = mx + b Root Curve: y = k x plot y vs. x or plot y 2 vs. x (do you see how that works, too?) Hyperbolic Curve: y = k/x plot y vs. (1/x) or plot (1/y) vs. x Inverse-Square Curve: y = k/x 2 plot y vs. (1/x 2 ) Parabola: y = Root Curve: y = kx 2 k x Hyperbolic Curve: y = k/x Inverse-Square Curve: y = k/x 2 (keeps increasing) (asymptotically approaches both axes) (asymptotically approaches both axes; drops off more quickly than the hyperbolic curve)

11 Below are sets of data. Plot each normally (y vs. x). From the shape of the curve, determine what category it belongs to, and then determine what you could plot to straighten the curve. Use the blank column to calculate the manipulated value of y or x. Plot a new graph, calculate the slope of the line, and use the slope to find the constant k in the relationship. (The slope itself may be k, or you may have to manipulate the slope, depending on what you chose to plot.) Acceptable answers for all the graphs that follow appear on the back of Answers to Numerical Problems from Conceptual Physics. Plot 1 Plot 2 Plot 3 y x y x y x Plot 1: y vs. x Plot 1: Straightened

12 Straightening Curves Page 3 Plot 2: y vs. x Plot 2: Straightened Plot 3: y vs. x Plot 3: Straightened

13 3) w = 2md/t 2 This experiment would be timing an object, pulled by a varying weight hanging over a pulley, and sliding across a smooth level surface for a constant distance. It is fine if you do not know what all these terms mean what is important is if you can map it to the straight line equation: w = 2md/t 2 y = mx + b So, what would be plotted? It is obvious that w would be on the vertical, but for the horizontal, you have some choice. Let s say the goal of the experiment was to measure m, the total mass in the system. A good idea is to pull the desired quantity out in front so it physically appears where m, the slope, appears. Whatever is left is then plotted on the horizontal. w = m(2d/t 2 ) so, plot w vs. (2d/t 2 ) and slope = m However, other options exist: (w/d) = 2m(1/t 2 ) w = (2md)(1/t 2 ) and there are more... so, plot (w/d) vs. (1/t 2 ) plot w vs. (1/t 2 ) and slope = 2m slope = 2md What you cannot do is plot w vs. d because the distance was a constant, so this will not show any relationship (of course the weight hanging over the side was not affected by the distance). Also, although it should be obvious, you cannot plot (w/m) vs. (2d/t 2 ) because you do not know what m is yet. (And how would you calculate it, anyway? The point is that you cannot just play any algebra tricks you want.) Sample data for this experiment follows. Blank columns are left for you to manipulate the data. Choose two of the above arrangements, fill in the top of the data column with what you are plotting and its units, plot the data, calculate the slope, and find the mass from each. (Ideally, the two values of mass that you find should match or be very close.) For each graph, label the axes with quantity and units (e.g.: w (N)), and put a title on the top of the graph that states what is being plotted. Titles for graphs are always listed as, (Vertical) vs. (Horizontal). (Notice the titles of the graphs of F vs. B and F vs. vb are both listed that way.)

14 Straightening Curves Page 4 Data: d = 1.7 m w = weight (N) t = time (s)

15 4) I = W / Rt In this experiment, a device with resistance R is turned on for a given amount of time. For different amounts of current I, the energy given off, W, is measured. Sample data for this experiment follows. Your goal is to find the constant resistance in the circuit. Blank columns are left for you to manipulate the data. Fill in the top of the column with what you are plotting and its units, plot the data, calculate the slope, and use it to find an experimental value of the resistance in ohms (Ω). Make sure to label the axes with quantity and units (e.g.: I (A)), and put a title on the top of the graph that states what is being plotted. Data: t = 300 s I = current (A) W = energy (J)