#$&* course phy 201 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
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): I still don’t understand what to do with the degrees? multiple by velocity? what is the formula. when do you use sin or cos? does the degree speed or slow down the system? cosine x component? sine is y component? is this always?
.............................................
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): 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. ------------------------------------------------ Self-critique rating: 1 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 022. `query 22 ********************************************* Question: `qQuery gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** We'll take East to be the positive direction. The original magnitude and direction of the momentum of the fullback is p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East. The magnitude and direction of the impulse exerted on the fullback will therefore be impulse = change in momentum or impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s. Impulse is negative so the direction is in the negative x direction, i.e., West. Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N The direction is in the negative x direction, i.e., West. The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N. The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0). The impulse on the tackler is to the East. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!