course PHY231 “´¡öƒÚˆˆ™eÄÿÏC¦À¯šœ³T…‘®âaÑ‘Îassignment #019
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18:08:52 `q001. Note that this assignment contains 5 questions. . If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
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RESPONSE --> Distance traveled will be 5 miles: 3^2 + 4^2 = 5^2 And the angle is arctan(4/3) = 53.13 degrees confidence assessment: 3
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18:09:00 If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
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RESPONSE --> self critique assessment: 3
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18:16:50 `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
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RESPONSE --> The x-component is 1307 N: -15 000 cos 265 The y-component is 14 943: -15 000 cos 265 confidence assessment: 2
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18:21:44 The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline.
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RESPONSE --> self critique assessment: 3
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18:24:10 `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
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RESPONSE --> We exert 500 Newtons at -53 degrees based on the calculations we did earlier for the 3 mile, 4 mile travel. confidence assessment: 3
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18:24:22 Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
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RESPONSE --> self critique assessment: 3
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18:31:40 `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction?
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RESPONSE --> The total force will be 234.5 Newtons, directed at 109 degrees from the positive x-axis. Adding the x-components: 200 cos (30) + 300 cos(150) = -86.6 Adding the y-components: 200 sin(30) + 300 sin(150) = 250 Finding the magnitude: sqrt(86.6^2 + 250^2) = 234.5 Finding the angle arctan(-86.6/250) = -70.89 The x-component is negative, so correct for the quadrant: -70.89 + 180 = approximately 109 degrees. confidence assessment: 3
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18:33:44 My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.
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RESPONSE --> self critique assessment: 3
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18:41:49 `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision?
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RESPONSE --> Total momentum before the collision = 145 kg * m/sec: Sum of the x-components of the momenta: 120 cos 60 + 80 cos 330 = 130 kg * m/sec Sum of the y-components of the momenta: 120 sin 60 + 80 sin 330 = 130 kg * m/sec Magnitude of resultant: sqrt( 130^2 = 64^2) = 144.89 confidence assessment: 2
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18:42:03 The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
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RESPONSE --> self critique assessment: 3
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ãž””¸ç¥„j†°Í„½|îêÊ©–ÓjØÍÈî¿ä assignment #019 019. `query 19 Physics I 07-02-2007
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22:23:46 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
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RESPONSE --> To calculate the component parallel to the x-axis, the magnitude of the vector is multiplied by the cosine of the angle the vector forms with the positive x-axis. To calculate the component parallel to the y-axis, the magnitude of the vector is multiplied by the sine of the angle the vector forms with the positive x-axis confidence assessment: 3
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22:24:04 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
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RESPONSE --> self critique assessment: 3
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22:29:39 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
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RESPONSE --> An object sitting at the origin of coordinate axes will be affected in the same way by either: a pair of forces, one parallel to the x axis and one parallel to the y axis a single force whose change in x is equal to the force parallel to the x axis above and whose change in y is equal to the force parallel to the y axis. confidence assessment: 2
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22:29:56 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
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RESPONSE --> self critique assessment: 3
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22:35:06 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> The magnitude of the velocity of a projectile at any given point has two components. They are calculated separately. The horizontal velocity is constant, and usually described as the x-component. A projectile has constant vertical acceleration. The instantaneous velocity can be calculated at a point, and is usually described as the y-component. The magnitude of the velocity of the projectile can be calculated sqrt(x^2 + y^2), and its direction can be found arctan(y/x). confidence assessment: 2
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22:35:17 ** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. **
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RESPONSE --> self critique assessment: 3
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22:40:19 Explain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity.
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RESPONSE --> Assuming air resistance is negligible, the initial horizontal velocity given will be constant. The vertical velocity will change as a result of gravity. The velocity at a specific time is equal to the sum of the initial velocity and the product of the acceleration due to gravity and the time interval. The velocity at a specific distance: sqrt (initial velocity + 2 * acceleraton * distance) confidence assessment: 3
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22:43:53 ** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). **
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RESPONSE --> self critique assessment: 3
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22:48:13 Univ. 8.58 (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?
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RESPONSE --> The impulse is .47 kg * m/sec: Impulse = m(v1 + v2) v = sqrt (2gy) Impulse = m (sqrt (2g * 2) + sqrt(2g * 1.6)) The average force is 237 N: Fave = Impulse / time = .47 N*s / .002s confidence assessment: 3
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22:48:38 ** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.002 s) = -240 Newtons, approx. **
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RESPONSE --> self critique assessment: 3
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22:52:11 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> With more practice (and algebra) I'm able to combine equations better. This makes some of my solutions more efficient, but more importantly shows me relationships that are not intuitive to me. The solutions that don't depend on the mass of the object are surprising. self critique assessment: 3
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