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course Phy 241
10/12 1
`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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19cm, vAve = 47.5cm/s
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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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20.5cm was max and was obtained with magnet 1.5cm from end of the ramp.
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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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vAve 51.25cm/s
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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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Asumming that the vAve was constant where we found it to be 47.5cm/s( which was 19cm/.4s), then it would take approx .1053sec at 47.5cm/s to travel 5cm.
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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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For 1st velocity of 47.5cm/s, when we mult by 60g(mass) we get momentum of 2850(g*cm)/s
For 2nd velocity of 51.25cm/s, when we mult by 60g(mass) we get momentum of 3075(g*cm)/s
So 225(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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It took approx .1053sec to travel the 5cm and the momentum was approx 2850(g*cm)/s at begening of interval and 3075(g*cm)/s at the end so ave rate is [(2850+3075)/2]/.1053 = approx 28133.9031g*cm
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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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Ramp was approx 30cm long and I used 3 dominoes to hold it up which is approx 2.4cm(we will round to 2.5cm because ball was on top of ramp not the dominoes). So rise is -2.5cm and run is 30cm ..
Slope = -2.5cm/30cm =-.0833cm
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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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Bead was 2 sec
Marble was 1.2 sec
Nut was .75 sec
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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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?????
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`qx010. What was the distance the object fell (1 block = 8 inches or about 20 centimeters)?
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Approx 540cm 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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For
Bead, a = 2.7m/s^2
Marble, a = 5.4m/s^2
Nut , a = 19.2m/s^2
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University Physics Students Only
`qx012. Find the initial velocity of the falling ball, based on your measurements of x, y, x0, y0 and your determination of sin(theta) and cos(theta).
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????Having some trouble putting together. I have symbolic solution for v0cos and sin are solved for. I having trouble knowing what values go into for x and x0, y and y0?????
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x0 and y0 would be the coordinates of the end of the ramp
x and y would be the coordinates of the landing point
`qx013. Find the time of fall of the ball.
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Had trouble with previous problem, do not have v0.
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`qx014. Questions about uncertainties:
* What is your estimated uncertainty in the measurements you used to determine sin(theta)?
* What therefore is your uncertainty in sin(theta)?
* What uncertainty does this introduce into your determination of v0?
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We have to use the Pythagorean Theorem so that we get,
(-2.5)^2 + b^2 = (30)^2
b = approx 29.8957cm
So that Sine(theta) = 29.8957cm +- .0001%
Uncertainty for v0 would be sin(.0001%), I believe???????
sin(theta) is the 'rise' of the ramp divided by its length
your 'rise' is 2.5 cm, probably +- at least .1 cm. That's +-4% accuracy. You can't get +-.0001% accuracy for a quantity based on a measurement with 4% error.
`qx015. Based on a single trapezoid of the graph of F vs. x for magnet forces, explain the meaning of the slope and the area of this graph.
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Slope is change in force divided by change in position, which is rate of change of force wrt position.
Area is ave force value of interval*width(which is displacement) and gives you work done
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`qx016. Based on a single trapezoid of the graph of PE vs. x for magnet forces, explain the meaning of the slope and the area of this graph.
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Slope is change in PE divided by change in position, which is force.
Area is ave value of PE over the interval*width(which is displacement) and givs you force needed for displacment
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"
I posted a couple of notes related to questions 1-10 yesterday.
I've inserted some notes for #'s 12-16 in this document. You can resubmit just this last part of the document, with solutions and/or questions.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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