linear motion

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Apr 3, 2024

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Linear Motion in One Dimension Physics 24 Name: Ahmed Elkartoum Date: 02/19/2024 Group Members: __Anwith Telluri, John Xereas, Maddie Gebremichael, Ahmed Elkartoum 1. Horizontal Track Position Graph : To bring the position graph to the screen, click on Graph, then on the appropriate “Run #”. Highlight the linear section of the graph using the highlight icon for increasing x values (motion before the bounce). Use the “curve fit” (linear) tool to determine the best-fit line. m (slope): ______0.433____ (m/s) b (intercept): ______-0.437______ (m) r: ____0.991________ The slope of the position/timeline is the velocity. The coefficient r characterizes the “fit” of the line, the closer r is to one, the higher the confidence that the line represents the experimental points. Equation of the line: _____y=0.433x-0.437___________________________________ Highlight the linear section of the graph for decreasing x values (motion after the bounce). Use the “curve fit” (linear) tool to determine the best-fit line. m (slope): __________-0.327____________ (m/s) b (intercept): _________0.605_____________ (m) r: _____-1__________________ Are you confident that x increases at a constant rate with time? Is equation (1) verified? Justify your answer. Assuming there is an infinite amount of track, this equation should hold up. However, with our graph, we hit the end of the track, causing the position to go negative. Figure 5

Velocity Graph Highlight the velocity graph for the section before the bounce. Using the (mean tool), determine the average (mean) value of the velocity. v (mean) = ______0.276____ m/s Highlight the velocity graph for the section after the bounce. Using the mean tool, determine the average (mean) value of the velocity. v (mean) = _______0.049_______ m/s Does the average value of v match the values of the slope of the position graph? Is equation (2) valid? (Justify your answer.) Yes, it vaguely matches the slope of the position graph. Both equations are valid. Are the velocities before and after the bounce the same? If not, can you propose an explanation for this? No, because there was an inelastic collision with the end of the track. This causes a loss of kinetic energy and thus a loss of velocity. What is the distance traveled by the cart from the beginning to the end of the experiment? _____0.92_____ m What is the displacement of the cart from the beginning to the end of the experiment? ____0.62______ m

Print a copy of both the position and the velocity graphs on one page to be attached to your report. First show the position graph on the screen. Then drag the corresponding velocity run to the page. Both graphs will appear above and below of each other.

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Acceleration during the bounce Using the coordinates tool determine the velocity and time for the cart before and after the bounce. Click on the coordinates tool and drag it to the last velocity point before the bounce starts (last point just before the velocity line starts bending because of the bounce). The coordinates tool shows the time first and then the velocity. Repeat for the first point of the velocity line after the bounce. Velocity before: v 1 = ____0.475______ m/s Time before: t 1 = ___1.15_______ s Velocity after: v 2 = _____0.479_____ m/s Time after: t 2 = ___1.25_______ s Calculate the average acceleration during the bounce (equation (3)): a = ___0.4_____m/s 2 In view of the change in the direction of travel, does your value for the acceleration appear reasonable? Justify your answer. Yes, because the graph shows the acceleration before and after the bounce are similar. Acceleration Graph Highlight the data before and after the bounce and determine the average value of a in each case using the mean tool. a (average): before: __1.339______ m/s 2 after: ___-1.818_______ m/s 2 Is a = 0 (or close to 0) as expected? Yes, because it changes direction at 0

2. Inclined Track (2-inches) Position Graph Equation (1) let us expect that distance is proportional to the square of time. Highlight the position/time cure before the bounce (going down the track.) Use the “curve fit” tool (quadratic) to fit the experimental points. The software will provide 3 numerical values labeled A, B, and C. These values are the parameters appearing in the equation x = At 2 + Bt + C. Comparing this equation to equation (1) for motion, it is apparent that A = (½) a in equation (1). Repeat for the return trip (going up the track.) Down the track: a = __0.234________ m/s 2 Up the track: a = ___-0.314_______ m/s 2 Does equation (1) for distance appear to be verified? Are the values of the acceleration before and after the bounce close to equal? Explain your answers. Yes because they moved the same direction Velocity Graph Following equation (2), the velocity should be a linear function of time. Highlight the velocity line going down the track and use the “curve fit” tool (linear) to determine the best fit line. Repeat for the velocity line going up the track (return trip). Down the track Up the track m (slope) -0.219m/s 2 -0.290m/s 2 r -0,972 -0,973 Do the slopes of the velocity lines before and after the bounce match the values of acceleration calculated before? Yes, they are proportional to the values of acceleration

Acceleration during the bounce Using the coordinates tool, determine the velocity and time for the cart before and after the bounce. Click on the coordinates tool and drag it to the last belocity point before the bounce starts (last point just before the velocity line starts bending because of the bounce.) The coordinates tool shows the time first and then the velocity. Repeat for the first point of the velocity line after the bounce. Velocity before: v 1 = _____0.483_____ m/s Time before: t 1 = ___2.8_______ s Velocity after: v 2 = ___0.473_______ m/s Time after: t 2 = ____2.9______ s Calculate the average acceleration during the bounce (equation (3)): a = _____-0.1_____ m/s 2 Is the acceleration during the bounce larger, smaller or the same as the accelerations before and after the bounce (magnitude and direction)? Justify your answer. It should be higher because acceleration is change in velocity. Print a copy of both the position and the velocity graphs on one page to be attached to your report. First show the position graph on the screen. Then drag the corresponding velocity run to the page. Both graphs will appear above and below of each other.

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1. Inclined Track (4-inches) The situation is similar to that of experiment 2 with a stronger acceleration. Position Graph Equation (1) let us expect that distance is proportional to the square of time. Highlight the position/time curve before the bounce (going down the track.) Use the “curve fit” tool (quadratic) to fit the experimental points. The software will provide 3 numerical values labeled A, B, and C. These values are the parameters appearing in the equation x = At 2 + Bt + C. Comparing this equation to equation (1) for motion, it is apparent that A = (½) a in equation (1). Repeat for the return trip (going up the track.) Down the track: a = ____-0.746______ m/s 2 Up the track: a = ___-0.806_______ m/s 2 Does equation (1) for distance appear to be verified? Are the values of the acceleration before and after the bounce close to equal? Explain your answers. Yes, they should be equal

Velocity Graph Following equation (2), the velocity should be a linear function of time. Highlight the velocity line going down the track and use the “curve fit” tool (linear) to determine the best fit line. Repeat for the velocity line going up the track (return trip). Down the track Up the track m (slope) -0.749m/s 2 -0.802m/s 2 r -1 -1 Do the slopes of the velocity lines before and after the bounce match the values of acceleration calculated before? Yes Acceleration during the bounce Using the coordinates tool determine the velocity and time for the cart before and after the bounce. Click on the coordinates tool and drag it to the last velocity point before the bounce starts (last point just before the velocity line starts bending because of the bounce). The coordinates tool shows the time first and then the velocity. Repeat for the first point of the velocity line after the bounce. Velocity before: v 1 = __-0.883________ m/s Time before: t 1 = ____-1.550 s Velocity after: v 2 = ___0.574_______ m/s Time after: t 2 = ____1.850______ s Calculate the average acceleration during the bounce (equation (3)): a = ____0.43______ m/s 2 Is the acceleration during the bounce larger, smaller or the same as the accelerations before and after the bounce (magnitude and direction)? Justify your answer. Larger because acceleration is change in velocity Print a copy of both the position and the velocity graphs on one page to be attached to your report. First show the position graph on the screen. Then drag the corresponding velocity run to the page. Both graphs will appear above and below of each other.

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1. Inclined Track – No Bounce After launch, the cart is subjected to a negative acceleration which will show it down and bring it back toward its starting position. The cart follows a path similar to that of an object thrown upward, which, under the down pull of gravity, returns to its starting point (free fall). The object is subjected to a constant negative acceleration during its whole path and the paths up and down are symmetrical. Position Graph Highlight data points. Using the “curve fit” tool (quadratic) draw a best fit line through the points. The equation has the form x = At2 + Bt + C. The coefficient A should be equal to ½ a, the acceleration applied to the cart. a = ___0.858____ m/s 2 Using the “coordinates tool” determine at what time the maximum height is reached. Choose an interval of time, say 0.2 seconds. On the graph, determine the value of x for this interval of time before and after the maximum: if the free fall path is symmetrical going up and going down, the positions for the times before and after the maximum should be the same. Is the path symmetrical? Justify your answer. Yes because they should be the same Velocity Graph Highlight the data, determine the slope. m (slope) = ___1.06____ m/s2 r: ____0.93______ The slope of the velocity line should be equal to the constant acceleration. Is it? Justify your answer.) Yes, because the slope of the velocity graph is the change in velocity which is the same thing as acceleration. Acceleration Graph Highlight the data. Using the mean tool, determine the average value of the acceleration: a = ___0.989_______ m/s 2 The acceleration should be constant and downward throughout the up and down motion of the cart. Note, particularly that at the moment the cart stops (peak of the path), the negative acceleration is still acting on the cart. What would happen if the acceleration went to zero at the moment the cart reaches its peak and stops briefly? Justify your answer. The cart will remain at rest.

linear motion

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