course Phy 121 sCowStudent Name:
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23:49:30 `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?
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RESPONSE --> Mass = 4kg v0 = x component = 10m/s Force = 5 N a = Fnet/m = 5N / 4kg = 1.25m/s^2 theta = 210 degrees To find the vector components of acceleration we use the magnitude of acceleration, 1.25m/s^2, and the direction of movement, 210 degrees. x component = (1.25m/s^2) * cos(210) = -1.083 m/s^2 y component = (1/25m/s^2) * sin(210) = -0.625 m/s^2 Horizontal direction: In order to find the time duration before the object stops, we use the formula `dt = `dv/a. Since we know the acceleration is -1.08m/s^2, the initial velocity is 10m/s, and the final velocity is 0m/s. `dt = (0-10m/s)/(-1.08m/s^2) = 9.26 seconds Vertical direction: The time duration in this direction is the same, 9.26 seconds. The y velocity while the object moves horizontally is 0m/s. We know that the acceleration is -0.625m/s^2. However, we do not know the final velocity in the y direction. In order to find it, we use the formula vf = vo + a*t. vf = 0m/s + (-0.625m/s^2)(9.26s) = -5.79m/s.
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23:50:48 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.
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RESPONSE --> I understand this problem.
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23:56:03 `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?
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RESPONSE --> `ds = 30m x motion: a = -1.08m/s^2, vo = 10m/s We can find the final velocity since we know `ds, vo, and a. vf^2 = vo^2 + 2a`ds = (10m/s)^2 + 2(-1.08m/s^2)(30m) = 35.2m^2/s^2 vf = 5.93m/s Now that we know initial and final velocity and the acceleration. We can use the formula `dt = `dv/a to find the time duration. `dt = (5.93m/s - 10m/s)/(-1.108m/s^2) = 3.76 seconds. y motion: a = -0.625m/s^2, `dt = 3.76s, vo = 0m/s vf = vo + a * `dt = 0m/s + (-0.625m/s^2)(3.76s) = -2.35m/s The magnitude of the velocity vector is the resultant of the velocity vector components in the horizontal and vertical directions. To find the magnitude, we can consider the x and y components the legs of a right triangle and the magnitude is the hypontenuse. So we use the Pythagorean Theorem to find the magnitude of the velocity vector. magnitude = sqrt (x velocity^2 + y velocity^2) = sqrt [(5.93m/s)^2 + (-2.35m/s)^2] = 6.38 In order to find the direction of the mass' velocity as it leaves the region we calculate, tan-1(y / x). Then we add 360 degrees to the answer because the y component is negative. direction = tan-1(-2.35m/s / 5.93m/s) + 360 degrees = 338.4 degrees
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23:56:55 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.
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RESPONSE --> I understand.
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SľzrҶ|m֞V؞ assignment #022 022. `query 22 Physics I 07-05-2006
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22:45:09 Query gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west
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RESPONSE --> I am not in general physics. Therefore, this problem does not apply to me.
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22:46:05 ** We'll take East to be the positive direction. The origional 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 iimpulse on the tackler is to the East. **
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RESPONSE --> Do I need to know how to do problems like this one for the test?
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