#$&*
phy231
Your 'cq_1_22.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At
the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so
its vertical velocity is zero. For the interval of free fall:
What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
vo=0cm/s a=980cm/s^2 hds'=40cm vds'=122cm
vf=+-'sqrt(vo^2+2a'ds)=+-'sqrt(2*980cm/s^2*122cm)=+-'sqrt(239120cm^2/s^2)=489cm/s 'dt='dv/a=489m/s/980cm/s^2=.5s
horizonalvAve=40cm/.5s=80cm/s
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Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the
floor?
answer/question/discussion: ->->->->->->->->->->->-> :
v velocity is 489cm/s
h velocity is 80cm/s
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
'sqrt((489m/s)^2+(80cm/s)^2)=495.5cm/s
arctan((80m/s)/(489m/s))=9.29deg
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE=1/2(70g)*(495.5m/s)^2=8.59321*10^6gcm^2/S^4=85.93N I'm not sure about this conversion?????
That would be g cm^2 / s^2, which is (.001 kg) * (.01 m)^2 / s^2 = .001 kg * .0001 m^2 / s^2 = .0000001 kg * m^2 / s^2, or 10^-7 kg m^2 / s^2 = 10^-7 Joules.
So you would have about 8.6 * 10^6 ( 10^-7 Joules) = .86 Joules.
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KE=1/2mv^2=1/2(70g)8(0cm/s)=0N
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
It changes from 85.93N to 0N.
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
They should always be opposites.
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How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
83.4N is in the vertical direction and 6.02N is in the horizontal direction.
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40 min
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11/7/10 1pm
Good, but see my note on the conversion then check the discussion at the link for the last few results. Revision is necessary only if you have questions.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#