assignment #019 Æç÷†¶…ãÑèóàqù®×}‡¢³‚„ Physics I 08-14-2006
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18:06:01 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 --> sin(theta)=y component/magnitude cos(theta)=x component/magnitude
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18:06:15 ** 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 --> ok
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18:07:37 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 --> We can find the force using components one in the x direction and one in the y direction. The magnitude can by found by the pythogorean thereom. We add x components and y components to get a total force if working with a variety of forces.
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18:07:45 ** 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 --> ok
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18:11:06 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> Depends on what information we have concerning the motion. How high the ramp is, how far the projectile traveled
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18:11:51 ** 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 --> ok yes we can find vx and vy with the information i mentioned earlier and then use the pythogorean thereom.
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18:12:42 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 --> we can work backwards with the equations just given to find vx and vy
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18:14:25 ** 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 --> yes you can use cosine=opposite(x component) hypotenuse (magnitude) and also similar using sin for the y component
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18:17:16 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 --> Impulse is m`dv. The average velocity before collision is 3.14 found with `ds=.5a`dt^2. After collisoin I am not sure how to calculate this. But if it is the same then we can just multiply 3.14 by .04 to get an impulse of .13.
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18:19:24 ** 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 --> ok we use the final velocity as it hits the floor because we need to know information about the moment before it hits the floor so that will be twice my number of 3.14 or about 6.3 m/s. I am not sure why rebound velocity can be found with an acceleration of 9.8.
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18:19:30 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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