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assignment #019㣾{WƁyࢌ

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Physics I

11-14-2006

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21:11:44

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 -->

The magnitude is the square root of x^2 + y^2. The x component is the magnitude * cos (theta). The y component is the magnitude*sin (theta). Theta is the arctan (y/x).

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21:11:51

** 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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21:12:41

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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Any force can be split into two equal forces that combine to exert that original vector. The combined component will create that original vector.

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21:12:46

** 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 Fx and Fy.

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21:14:08

Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.

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RESPONSE -->

We must first know the initial velocity of the projectile in horizontal and vertical directions. Then the magnitude is found by the square root of vx^2 and vy^2 summed. The direction is the angle of the vector, which is the arctan of vy/vx.

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21:14:11

** 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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ok

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21:14:53

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 -->

The x velocity is vcos(theta) and the velocity in the y direction is vsin(theta) where v is the magnitude and theta is the angle of the vector.

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21:15:33

** 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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21:15:34

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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21:15:36

** 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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21:15:37

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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You're in good shape here. Let me know if you have questions.