course PHY 121
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20:58:35 `q001. Experiment 6. The horizontal range of a projectile with initial velocity in the horizontal direction is an accurate indicator of its initial velocity. See the corresponding video files on CD #0 (see note in red at the top of this document) entitled Introductor Video Experiments, video clips #11. View also #10. In this experiment we measure the horizontal range of a ball projected horizontally from a ramp. Using the time required to fall to the floor we then determine the average horizontal velocity of the ball during its fall. We test the hypothesis that the horizontal velocity of the falling ball remains the same as at the instant the ball left the ramp. Describe what you saw in the video clip.
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RESPONSE --> In the video clip we saw the instructor drop the ball from various heights after rolling down an inclined ramp. These horizontal distances were measured by allowing the ball to hit carbon paper which made a mark on to white paper after the ball rolled off the table. The instructor then laid out the paper with the ball markings on the ground with a meter stick for us to measure the distances. The 13.3cm trail had to be ran again becuase of an error. On the second trial the results were more consistent.
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21:01:39 As well as other aspects of the setup, you should have described three essential things: You should have described how the ramp was supported so that the end would be horizontal, with the result that the ball leaves the ramp with its velocity solely in the horizontal direction. You should have described how the horizontal distance traveled by the ball is measured And you should have noted that releasing the ball from the same position every time is expected to give the ball the same end-of-the-ramp velocity in every trial.
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RESPONSE --> The ramp was supported by a table that he used a level to determine stability and the drop-off measurements he used a stool, stacks of books and a board to aid in these measurements. And of course it is expected that the instructor released the ball from the same position on every trial.
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21:10:24 `q002. Because of the horizontal orientation of the ramp at the edge of the table, the ball leaves the end of the ramp with a velocity which is in the horizontal direction, with no vertical component. The time required for the ball to reach the floor is independent of its horizontal velocity, as we see from dropping a coin from the edge of a table with one hand as with the other hand we flick another coin off in the horizontal direction (the coins hit the floor simultaneously). Since the initial velocity of the ball in the vertical direction is zero, and since we know the acceleration of gravity, we can determine from the vertical distance how long it takes for the ball to fall. We can therefore use the distance of fall and the horizontal distance traveled to obtain the average horizontal velocity of the ball. Explain in your own words how the vertical and horizontal distances will be used to determine the average horizontal velocity of the ball. For the vertical motion which of the quantities v0, vf, a, `dt and `ds will we know? What relationships and/or equations will we therefore use to calculate the time required for the vertical fall? For the horizontal motion which of the quantities v0, vf, a, `dt and `ds will we know? What relationships or equations will we then use the calculate the average horizontal velocity?
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RESPONSE --> We determine how long it takes the ball to fall by using the equation given to us. `ds = .5 g `dt^2. We solve this equation for dt to get: `dt = `sqrt (2 * `ds / g) The vertical distances are used as the displacement. Now in order to find the average horizontal velocities we will divide the vertical distances which will be used as `ds by the time `dt that we got from the horizontal distances. vAve (horizontal velocity) = `ds / `dt.
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21:13:02 For the vertical motion we will know the initial velocity v0, which is zero; the acceleration a, which is 9.8 meters/second^2; and the vertical displacement, which we will measure in the experiment. Knowing v0, a and `dt we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find the final velocity. We can then average the final velocity with the initial velocity to find the average velocity, which we can then divide into the vertical displacement to determine the time of fall. For the horizontal motion we know only the displacement `ds in the horizontal direction, and the time interval `dt, which is the same is that calculated for the vertical motion. Since wish only to find the average velocity, we will then have what we need, and since vAve = `ds / `dt we can simply divide the horizontal `ds by `dt.
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RESPONSE --> okay, I understand the second part in finding the avg. horizontal velocity. However on the first part I was under the assumption that we needed to use the formula given to fin the time.
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21:13:06 *
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RESPONSE --> ok
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21:13:53 `q003. By intercepting the ball at various heights we can determine the horizontal range and therefore the horizontal velocity of the ball as it leaves the end of the ramp. First, using washers or coins as on the video clip convince yourself that the time required for an object to reach the floor when it initially travels in a purely horizontal direction is independent of its initial horizontal velocity. Explain how you have done this and how it convinced you that the time required is independent of horizontal motion.
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RESPONSE --> I used the cions and they appear to hit at the same place.
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21:14:13 You should have dropped one object while flicking the other off the edge of horizontal table, starting both objects as close to the same instant as possible. You should have noted that if the objects are started at the same instant a hit the floor at the same instant. Since one has zero horizontal velocity while the other has a significant horizontal velocity, this effectively demonstrates the independence of the time of fall from the horizontal velocity.
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RESPONSE --> yes
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21:28:11 `q004. Take data: Record each vertical distance of fall and determine the horizontal distance for each vertical distance as instructed on the video clip. Give your results in an organized table.
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RESPONSE --> Vertical Distance Horizontal Distance 6 cm 9 cm 13.3 cm 13.5 cm 29.7 cm 20.25 cm
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21:28:19 Your table should include for each trial the vertical distance of fall and horizontal distance traveled by the ball after leaving the ramp.
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RESPONSE --> ok
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21:44:59 `q005. Analyze Data: Determine the time required to fall each distance, and use this time with the horizontal distance to find the average horizontal velocity for each vertical distance. Using the distance fallen by the ball and the formula vf^2 = v0^2 + 2 a `ds determine the time required to fall each distance. Note that a is the acceleration of gravity, which is close to 980 cm/s^2 . Using the horizontal distance traveled by the ball during this time interval and the time required to fall, determine for each vertical distance the average horizontal velocity of the ball as it falls. Give your calculations and results.
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RESPONSE --> I used the `ds = .5 g `dt^2 and solved for the `dt to get: `dt = sqrt (2 `ds / g). So for the 1st 6cm vertical drop I got: `dt = sqrt (2 * 9cm / 980 cm/s^2) `dt = sqrt (18 cm 980 cm/s^2) `dt = sqrt (.018s^2) `dt = .134s The second 13.3 cm vertical drop I got: `dt = sqrt ( 2 * 13.5cm / 980 cm/s^2) `dt = sqrt (27 cm / 980 cm/s^2) `dt = sqrt (.028s^2) `dt = .168s The third 29.7 cm vertical drop I got: `dt = `sqrt (2 * 20.25cm / 980 cm/s^2) `dt = `sqrt (40.5cm / 980 cm/s^2) `dt = `sqrt (.041s^2) `dt = .202s Then to get the avg. horizontal velocities: I got for the 6cm vertical drop: 9cm/.134s = 67.164 cm/s = horizontal vAve I got for the 13.3 cm vertical drop: 13.5 cm / .168s = 80.357 cm/s = horizontal vAve I got for the 29.7cm vertical drop: 20.25cm / .202s = 100.248 cm/s = horizontal vAve
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22:47:04 For example, if the vertical distance fallen by the ball was 65 cm and horizontal distance traveled by the ball was 80 cm, you would first have calculated the time required for the ball to fall, using only the vertical motion of the object. The vertical motion starts with initial vertical velocity zero, accelerates at 980 cm/second^2 and travels through vertical displacement 65 cm. You would find that the final velocity was vf = `sqrt(v0^2 + 2 a `ds) = `sqrt( 0 + 2 * 980 cm/s^2 * 65 cm) = 360 cm/s, approx.. The average vertical velocity would therefore be (0 cm/s + 360 cm/s) / 2 = 180 cm/s, and to fall the 65 cm would therefore take `dt = 65 cm / (180 cm/s) = .36 sec, approx.. The 80 cm horizontal displacement would therefore occur in approximately 0.36 seconds, which would imply and average horizontal velocity of 80 cm / .36 sec = 220 cm/s approx..
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RESPONSE --> In doing the calculations the other way I get the following: for the 6 cm: `dt = .166s avg. horizontal velocity = 54.2 cm/s for the 13.3 cm: `dt = .167 avg. horizontal velocity = 80.73 cm/s for the 29.7 cm: `dt = .168s avg. horizontal velocity = 120.64 cm/s
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22:49:53 `q006. Plot y = average horizontal velocity vs. x = time of fall and use your plot to determine whether there is any experimentally significant dependence of average horizontal velocity on time or whether average horizontal velocity seems to be independent of time and therefore constant. Describe your graph. Do the differences in the horizontal velocities seem to be the result of random experimental uncertainties and errors in timing, or does there seemed to be a definite progression in the horizontal velocities as time increases?
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RESPONSE --> I put the time on the x axis and the avg. horizontal velocity on the y axis. The times on ly increased by .001 of a second each time, however the velocitites increased by about 30 - 40 cm/s each time. So there seems to be a definite progression in the horizontal velocities as time increases.
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22:50:19 The horizontal velocities will not differ by a great deal--typically all velocities will be within about 10 percent of the average velocity.
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RESPONSE --> I am not sure why I have the difference in this case.
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22:51:16 `q007. Analysis of Errors Estimate the uncertainties in your data, and indicate the results of these estimates on your graph in the usual manner. Is it possible to draw a horizontal line through the error bars on your graph?
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RESPONSE --> No it is not possible to draw a horizontal line through the error bars on my graph I would have to make the error percentage or bars huge.
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