course PHY231 12/5 245I realize there is a bit of work I have not submitted. with my work and military schedule it has just been easier to print out and work through these as I can. I am attemting to type out and submit some as I can to help with my grade coming to the end of semester. Thank you Question: `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts.
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Given Solution: A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ent: ok ********************************************* Question: `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the change in velocity of the distance in which the object travels through the force field you must look at the acceleration from the last problem. This data can be plugged into one of the equations of motion vf^2 = v0^2 + 2a’ds vf^2 = (10m/s)^2 + 2(-1.0825m/s^2) * 30m vf^2 = 35.05m/s vf = 5.92 m/s With this the average velocity over the 30m range can be found 10m/s + 5.92m/s = 15.92 m/s 15.92m/s / 2 = 7.96m/s = vAve At an average of 7.96m/s it will take 30m/7.96m/s = 3.77s to travel through the area The change in magnitude and direction of the velocity of the mass can be found as such. The x component starts at 10m/s with a decelerating force of -1.0825m/s^2 after 3.77s it will have decelerated 4.08m/s. 10m/s – 4.08m/s = 5.92m/s The y component started with an initial velocity of 0 and was decelerated at -.625m/s^2 for 3.77s it will have changed -.625 * 3.77 = -2.35m/s The magnitude of the velocity of the object exiting the area will be: Sqrt((5.92m/s)^2 + (-2.35m/s)^2) = 6.369m/s Still traveling in the direction of the positive x axis confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.