Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

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Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

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Name ____________________________________ Motion in 2D Simulation Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now. 1) Once the simulation opens, click on ‘Show Both’ for Velocity and Acceleration at the top of the page. Now click and drag the red ball around the screen. Make 3 observations about the blue and green arrows (also called vectors) as you drag the ball around. 2) Which color vector (arrow) represents velocity and which one represents acceleration? How can you tell? 3) Try dragging the ball around and around in a circular path. What do you notice about the lengths and directions of the blue and green vectors? Describe their behavior in detail below. 4) Now move the ball at a slow constant speed across the screen. What do you notice now about the vectors? Explain why this happens. 5) What happens to the vectors when you jerk the ball rapidly back and forth across the screen? Explain why this happens. 6) Now click on ‘Circular’ on the bottom. Describe the motion of the ball and the behavior of the two vectors. Is there a force on the ball? How can you tell? Be detailed in your explanations. 7) Click on ‘Simple Harmonic’ on the bottom. Based on the behavior of the ball and the vectors, write a definition of Simple Harmonic Motion.

Name ____________________________________ Motion in 2D Simulation Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D and click on Run Now. 1) Once the simulation opens, click on ‘Show Both’ for Velocity and Acceleration at the top of the page. Now click and drag the red ball around the screen. Make 3 observations about the blue and green arrows (also called vectors) as you drag the ball around. 2) Which color vector (arrow) represents velocity and which one represents acceleration? How can you tell? 3) Try dragging the ball around and around in a circular path. What do you notice about the lengths and directions of the blue and green vectors? Describe their behavior in detail below. 4) Now move the ball at a slow constant speed across the screen. What do you notice now about the vectors? Explain why this happens. 5) What happens to the vectors when you jerk the ball rapidly back and forth across the screen? Explain why this happens. 6) Now click on ‘Circular’ on the bottom. Describe the motion of the ball and the behavior of the two vectors. Is there a force on the ball? How can you tell? Be detailed in your explanations. 7) Click on ‘Simple Harmonic’ on the bottom. Based on the behavior of the ball and the vectors, write a definition of Simple Harmonic Motion.

Name ____________________________________                                      Motion in 2D Simulation   Go to http://phet.colorado.edu/simulations/sims.php?sim=Motion_in_2D … Read More...
Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push Draw the vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Tactics Box 5.2 Identifying Forces Learning Goal: To practice Tactics Box 5.2 Identifying Forces. The first basic step in solving force and motion problems generally involves identifying all of the forces acting on an object. This tactics box provides a step-by-step method for identifying each force in a problem. TACTICS BOX 5.2 Identifying forces Identify the object of interest. This is the object whose motion 1. you wish to study. 2. Draw a picture of the situation. Show the object of interest and all other objects—such as ropes, springs, or surfaces—that touch it. 3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything else is outside. 4. Locate every point on the boundary of this curve where other objects touch the object of interest. These are the points where contact forces are exerted on the object. Name and label each contact force acting on the object. There is at least one force at each point of contact; there may be more than one. When necessary, use subscripts to distinguish forces of the same type. 5. 6. Name and label each long-range force acting on the object. For now, the only long-range force is the gravitational force. Apply these steps to the following problem: A crate is pulled up a rough inclined wood board by a tow rope. Identify the forces on the crate. Part A Which of the following objects are of interest? Check all that apply. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Conceptual Questions on Newton’s 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton’s 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an object–a motion diagram. The dots represent the object’s position at moments separated by equal intervals of time. The dots are connected by arrows representing the object’s average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion. crate earth rope wood board Part A What is the direction of the net force acting on the object at position A? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D upward downward to the left to the right The net force is zero. This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Understanding Newton’s Laws Part A An object cannot remain at rest unless which of the following holds? You did not open hints for this part. ANSWER: Part B If a block is moving to the left at a constant velocity, what can one conclude? You did not open hints for this part. ANSWER: The net force acting on it is zero. The net force acting on it is constant and nonzero. There are no forces at all acting on it. There is only one force acting on it. Part C A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What can you say about the block’s motion? You did not open hints for this part. ANSWER: Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. You did not open hints for this part. ANSWER: There is exactly one force applied to the block. The net force applied to the block is directed to the left. The net force applied to the block is zero. There must be no forces at all applied to the block. 2 kg 3 N 4 N It must be moving to the left. It must be moving to the right. It must be at rest. It could be moving to the left, moving to the right, or be instantaneously at rest. Part E Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. The net force acting on the object __________. Check all that apply. You did not open hints for this part. ANSWER: Tactics Box 5.3 Drawing a Free-Body Diagram Learning Goal: To practice Tactics Box 5.3 Drawing a Free-Body Diagram. A free-body diagram is a diagram that represents the object as a particle and shows all of the forces acting on the object. Learning how to draw such a diagram is a very important skill in solving physics problems. This tactics box explains the essential steps to construct a correct free-body diagram. TACTICS BOX 5.3 Drawing a free-body diagram Identify all forces acting on the object. This step was described 1. in Tactics Box 5.2. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration 4 N 10 N cannot have a magnitude equal to cannot have a magnitude equal to cannot have the same direction as the force with magnitude must have a magnitude greater than 5 N 10 N 10 N 10 N Draw a coordinate system. Use the axes defined in your pictorial representation. If those axes are tilted, for motion along an incline, then the axes of the free-body diagram should be similarly tilted. 2. Represent the object as a dot at the origin of the coordinate axes. This is 3. the particle model. 4. Draw vectors representing each of the identified forces. This was described in Tactics Box 5.1. Be sure to label each force vector. Draw and label the net force vector . Draw this vector beside the diagram, not on the particle. Or, if appropriate, write . Then, check that points in the same direction as the acceleration vector on your motion diagram. 5. Apply these steps to the following problem: Your physics book is sliding on the carpet. Draw a free-body diagram. Part A Which forces are acting on the book? Check all that apply. You did not open hints for this part. ANSWER: F  net F =  net 0 F  net a Part B Draw the most appropriate set of coordinate axes for this problem. The orientation of your vectors will be graded. ANSWER: gravity normal force drag static friction tension kinetic friction spring force Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points.

Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push Draw the vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Tactics Box 5.2 Identifying Forces Learning Goal: To practice Tactics Box 5.2 Identifying Forces. The first basic step in solving force and motion problems generally involves identifying all of the forces acting on an object. This tactics box provides a step-by-step method for identifying each force in a problem. TACTICS BOX 5.2 Identifying forces Identify the object of interest. This is the object whose motion 1. you wish to study. 2. Draw a picture of the situation. Show the object of interest and all other objects—such as ropes, springs, or surfaces—that touch it. 3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything else is outside. 4. Locate every point on the boundary of this curve where other objects touch the object of interest. These are the points where contact forces are exerted on the object. Name and label each contact force acting on the object. There is at least one force at each point of contact; there may be more than one. When necessary, use subscripts to distinguish forces of the same type. 5. 6. Name and label each long-range force acting on the object. For now, the only long-range force is the gravitational force. Apply these steps to the following problem: A crate is pulled up a rough inclined wood board by a tow rope. Identify the forces on the crate. Part A Which of the following objects are of interest? Check all that apply. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Conceptual Questions on Newton’s 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton’s 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an object–a motion diagram. The dots represent the object’s position at moments separated by equal intervals of time. The dots are connected by arrows representing the object’s average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion. crate earth rope wood board Part A What is the direction of the net force acting on the object at position A? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D upward downward to the left to the right The net force is zero. This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Understanding Newton’s Laws Part A An object cannot remain at rest unless which of the following holds? You did not open hints for this part. ANSWER: Part B If a block is moving to the left at a constant velocity, what can one conclude? You did not open hints for this part. ANSWER: The net force acting on it is zero. The net force acting on it is constant and nonzero. There are no forces at all acting on it. There is only one force acting on it. Part C A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What can you say about the block’s motion? You did not open hints for this part. ANSWER: Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. You did not open hints for this part. ANSWER: There is exactly one force applied to the block. The net force applied to the block is directed to the left. The net force applied to the block is zero. There must be no forces at all applied to the block. 2 kg 3 N 4 N It must be moving to the left. It must be moving to the right. It must be at rest. It could be moving to the left, moving to the right, or be instantaneously at rest. Part E Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. The net force acting on the object __________. Check all that apply. You did not open hints for this part. ANSWER: Tactics Box 5.3 Drawing a Free-Body Diagram Learning Goal: To practice Tactics Box 5.3 Drawing a Free-Body Diagram. A free-body diagram is a diagram that represents the object as a particle and shows all of the forces acting on the object. Learning how to draw such a diagram is a very important skill in solving physics problems. This tactics box explains the essential steps to construct a correct free-body diagram. TACTICS BOX 5.3 Drawing a free-body diagram Identify all forces acting on the object. This step was described 1. in Tactics Box 5.2. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration 4 N 10 N cannot have a magnitude equal to cannot have a magnitude equal to cannot have the same direction as the force with magnitude must have a magnitude greater than 5 N 10 N 10 N 10 N Draw a coordinate system. Use the axes defined in your pictorial representation. If those axes are tilted, for motion along an incline, then the axes of the free-body diagram should be similarly tilted. 2. Represent the object as a dot at the origin of the coordinate axes. This is 3. the particle model. 4. Draw vectors representing each of the identified forces. This was described in Tactics Box 5.1. Be sure to label each force vector. Draw and label the net force vector . Draw this vector beside the diagram, not on the particle. Or, if appropriate, write . Then, check that points in the same direction as the acceleration vector on your motion diagram. 5. Apply these steps to the following problem: Your physics book is sliding on the carpet. Draw a free-body diagram. Part A Which forces are acting on the book? Check all that apply. You did not open hints for this part. ANSWER: F  net F =  net 0 F  net a Part B Draw the most appropriate set of coordinate axes for this problem. The orientation of your vectors will be graded. ANSWER: gravity normal force drag static friction tension kinetic friction spring force Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points.

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Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

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Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

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Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

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