course Phy 121 èátžø¹îÐÞÎÜŸ¸–ª®úxáÒsÜ£½„Øassignment #021 021. projectiles 2 Physics II 04-07-2008 ¸ž[¯ÈŒs̼àßåЪôRö÷°ÙÐ͉§€¤ assignment #022 022. Motion in force field Physics II 04-07-2008
......!!!!!!!!...................................
06:57:28 `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. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
......!!!!!!!!...................................
RESPONSE --> It appears the object in question is going to experience a bit of drag as it enters this region...since we know it's mass (4 kg) and the force that it experiences (5 N) we would solve for acceleration, a = F/N . The acceleration, then, would be 1.25 m/s^2. (I had to remember not to factor in the 10 m/s velocity since we are just specifically looking at the region where the object experiences some resistance). The net force, directed at 210 degrees, would occur, of course, along both x and y axes, so: acceleration * sin(210) (sin for x) 1.25 m/s^2 * -0.5 = -0.625 m/s^2 (acceleration along x-axis) acceleration * cos(210) (cos for y) 1.25 m/s^2 * -0.87 = -1.088 m/s^2 (acceleration along y-axis) Remembering, of course, that our object was coasting along with velocity of 10 m/s before it entered this area--and that the incremental change in velocity would be constant-- we solve as: 10 m/s - 0.625 m/s = 9.375 m/s for the object to decelerate to zero along the x axis. And, 10 m/s - 1.088 m/s^2 = 8.912 m/s for the velocity to drop to zero along the y axis. confidence assessment: 2
.................................................
......!!!!!!!!...................................
06:57:53 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.
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 2
.................................................
......!!!!!!!!...................................
06:59:00 `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?
......!!!!!!!!...................................
RESPONSE --> Hmmm, not sure....would this involve the use of vectors, Pythagorean formulae, etc.? confidence assessment: 1
.................................................
......!!!!!!!!...................................
07:00:17 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.
......!!!!!!!!...................................
RESPONSE --> I'll have to reread this some more
.................................................