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course PHY 201
10/15 11 amMore will follow, but this should get you started.
Text and queries
You should submit Query 13 and Query 14 by Monday.
You should work assigned problems from Chapter 6.
Reminder: The Submit Question Form is at http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm.
Deflection of ball down ramp from path
`qx001. How far did the ball travel after leaving the end of the ramp, in the trial without the magnet, and what therefore was its horizontal velocity during the fall, assuming a fall time of .4 second?
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The ball traveled 14.5 cm after leaving the end of the ramp. Given the change in time of .4 s, we can solve for average velocity=14.5 cm/.4s=36.25cm. Vf=2*36.25 cm=72.5.
good, but horizontal velocity is constant during the fall; it doesn't start at 0, so doubling the average velocity is not called for
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`qx002. What is the maximum deflection of the ball, due to the presence of the magnet, from its original straight-line path?
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4 cm
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`qx003. What velocity does the ball therefore attain, in the direction perpendicular to its original straight-line path, as a result of the magnet?
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4 cm/.4 s=10 cm/s= vAve vf=20cm/s
this is more subtle, but the uniform-acceleration phase doesn't start until the ball has passed the magnet, by which time it already has achieved its constant horizontal velocity
there are other assumptions implicit in this approximation, but the bottom line is that the first-order approximation is that the horizontal velocity is your 10 cm/s result
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`qx004. Assume the ball was close enough to the magnet to have its motion significantly influenced for a distance of 5 centimeters. How long did it take for the ball to travel this distance?
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.4 seconds
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`qx005. Assume that the ball has a mass of 60 grams. How much momentum did it gain, in the direction perpendicular to its original line of motion, as a result of the magnet?
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Momentum=m*v 5 cm/.4 s=12.5cm/s=vAve, so Vf=2(12.5 cm/s)=25 cm/s
that would be correct for an acceleration thru 5 cm in .4 s, starting with 0 velocity, but the ball doesn't take .4 seconds to move past the magnet; the .4 seconds is the total time of fall
P=60 g *25 cm/s=1500 g*cm/s
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`qx006. What therefore was the average rate of change of its momentum with respect to clock time, for the 5-cm interval during which the magnet significantly affected its motion?
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P with respect to clock time is (1500 g*cm/s)/.4 s= 3750 g*cm/s^2
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`qx007. What were the measurements by which you can calculate the slope of your ramp? What was the slope of the ramp?
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My ramp was 2 cm tall and 30 cm long. The slope is rise/run so 2 cm/15 cm=0.13
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Acceleration of gravity
`qx008. What was your count for the object dropped in the stairwell? To what time interval in seconds does your count correspond?
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Plastic bead—6 counts. One count is approximately .2 s, so 6 counts *.2 s=1.2 s.
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`qx009. What was your count for the object dropped in the stairwell? To what time interval in seconds does your count correspond?
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Washer—5 counts. One count is approximately .2 s so 5 counts*.2s=1s
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`qx010. What was the distance the object fell (1 block = 8 inches or about 20 centimeters)?
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27 blocks at 20 cm each = 540 cm or 5.4 m
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`qx011. Using the distance of fall and the time interval in seconds, find the acceleration of the falling object.
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Plastic bead -- Vave=5.4 m/1.2 s= 4.5 m/s vf=2* 4.5 m/s=9 m/s a=9 m/s/ 1.2 s=7.5 m/s^2
Washer – Vave= 5.4 m/ 1s= 5.4 m/s. vf= 2* 5.4m/s= 10.8 m/s a=10.8 m/s/1 s= 10.8 m/s^2
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Very good, though you still have a couple of points of confusion on the experiment with the magnet. In any case you're doing almost everything right. We'll go over that in class Monday, and I'm confident you'll understand how it all goes together.
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