#$&* course PHY 201 10:08 PM on 4/2/13 022. Motion in force field
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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. STUDENT COMMENT It seems very simple doing all the arithmetic, but getting the right variables in the right spot seems to be my problem. This is still confusing to me? How did you know to find the acceleration out of the question? Why are the accelerations the vectors of the lines? INSTRUCTOR RESPONSE The sequence of ideas goes something like this? From the given information we see that the velocity is changing as a result of a force. Since we are given the net force and the mass we know we can find acceleration. Acceleration is related to the change in velocity. The other thing we need to understand is that the motion of the object in the x direction is independent of the motion in the y direction. Just as with projectiles, the only thing the x and y motions have in common is the time interval. We are given enough information to analyze the motion in the x direction. So we analyze the x components of force, acceleration, velocity and position (the last not necessary in this problem). In the process we find the time required for x velocity to reach zero. We are not given enough information to analyze the y motion, but having found the time interval from the analysis of the x motion, we now have sufficient information. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: know horizontal values are v0, a, and 'dx. Solving for vf = sqrt(v0^2 + 2(a)('dx)) = sqrt((10 m/s)^2 + 2(-1.08 m/s^2)(30 m)) = 5.93 m/s vAve = 'dx / 'dt => 'dt = 'dx / vAve = 30 m / ((5.93 m/s + 10 m/s) / 2 ) = 3.77 s The vertical final velocity would vf = v0 + at = -0.625 m/s^2 * 3.77 s = -2.36 m/s velocity magnitude for the final velocity = sqrt((5.93)^2 + (-2.36)^2) = 6.38 m/s the direction of the final velocity = tan^-1 (-2.36 / 6.38) + 360 = 339.7 degrees confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT COMMENT It makes sense when it explained, but I cannot seem to get the right answers. I guess it will take some time to figure everything out. INSTRUCTOR RESPONSE For the x direction you have acceleration, initial velocity and displacement, from which you can find time interval and final velocity. Having found the time interval you can use the given information to find the final y velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q003. A hockey puck of mass 0.16 kg is sliding across the ice, and is moving at 5 meters / second in the x direction when a sudden gust of wind exerts a force of 0.12 Newtons directed at angle +330 degrees relative to the positive x axis. What are the x and y velocities of the puck after traveling 10 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 0.16 kg, v0 = 5 m/s, a = F / m = 0.12 kg*m/s^2 / 0.16 kg = 0.75 m/s^2 acceleration x component = 0.75 m/s^2 cos 330 = 0.650 m/s^2 acceleration y component = 0.75 m/s^2 sin 330 = -0.375 m/s^2 For horizontal direction, known values are v0, a, and 'dx. Solve for vf. vf = sqrt(v0^2 + 2(a)('dx)) = sqrt((5 m/s)^2 + 2(0.650 m/s^2)(10 m)) = 6.16 m/s vAve = 'dx / 'dt => 'dt = 'dx / vAve = 10 m / ((6.16 m/s + 5 m/s) / 2 ) = 1.79 s for vertical, vf = v0 + at = 0 m/s + (-0.375 m/s^2)(1.79 s) = -0.671 m/s magnitude = sqrt(x^2 + y^2) = sqrt(6.16 m/s)^2 + (-0.671 m/s)^2) = 6.20 m/s angle = tan^-1 (y / x) + 360 = tan^-1 (-0.671 m/s / 6.16 m/s) + 360 = 353.8 deg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: