course Phy 121 ??b???a???{???????assignment #019????n???????????Physics I
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16:25:46 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 --> We mulyiply the magnitude by the cosine of the angle.
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16:28:07 ** 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 --> x component = magnitude * cos(angle) y component = magnitude * sin(angle) Magnitude from angle arctan(y component/x component) We use pythagorean theorem to get magnitude
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16:30:02 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 --> One force can be used to accomplish the same result as two forces that are equal to its components. Remember the guy with the ice who was replaced by two. It would seem to be more efficient to use one in most circumstances.
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16:30:11 ** 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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16:32:44 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> Since velocity is a vector, we can imagine horizontal and vertical lines from that point and calculate magnitude and direction using this information. The angle is calculated in the same manner using the arctan ( y / x) method
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16:32:57 ** 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
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16:35:32 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 calculate the x and y components from the initial velocity and these components will correspond to the horizontal and vertical velocities.
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16:35:41 ** 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
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16:35:44 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 -->
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16:35:46 ** 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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16:35:49 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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