course Phy 121 ” ¤ùÅô‹‡»Þ¯ú¢ãñ³îЊ´D}dù—Ž¹ôüassignment #019
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22:09:40 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 --> I know that the magnitude of the force is the hypotenuese of the 2 vectors and that the angle is arctan of the 2 components, but I am not sure how to do this in the opposite direction. confidence assessment: 0
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22:11:29 ** 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 --> I guess I was reading the problem wrong, but I did know how to do these parts. I guess I just wasn't sure what the actual answer was called. self critique assessment: 2
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22:14:22 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 --> I think that this is saying that the effects of the components make up the effects of the vector. confidence assessment: 3
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22:16:15 ** 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 --> I see how the two components in different directions make up the vector in between these two components. self critique assessment: 2
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22:18:31 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> If we have the two components we would first calculate the x and y components. Then to find the magnitude, we would solve for the hypotenuese of these two compenents. We find the direction of the velocity by taking arctan (y/x). confidence assessment: 3
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22:18:47 ** 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. self critique assessment: 3
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22:24:36 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 --> For the vertical direction, we would do magnitude sin (theta). For the horizontal direction, we would do magnitued cos (theta). confidence assessment: 3
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22:24:46 ** 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 --> ok. self critique assessment: 3
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22:24:52 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 --> confidence assessment:
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22:24:56 ** 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:
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22:26:12 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I understand vectors a lot better now. I think if I had this for the vector experiment, then I would have been able to understand that a lot more. self critique assessment: 3
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