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PHY 201
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.
** CQ_1_22.2_labelMessages **
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: ->->->->->->->->->->->-> :
a^2 + b^2 = c^2
(40 cm)^2 + (122 cm)^2 = c^2
1600 cm^2 + 14884 cm^2 = c^2
c^2 = 16484 cm^2
c = 128.39 cm
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This is the correct length of the straight line between the initial and final points.
However the ball speeds up in the vertical direction while it remains at constant velocity in the horizontal. So it doesn't follow a straight line.
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Vertical:
vf^2 = v0^2 + 2 * a * `ds
vf^2 = (0 m/s)^2 + 2 * 9.8 m/s/s * 1.22 m
vf^2 = 23.912 m^2/s^2
vf = 4.89 m/s
vf = v0 + a * `dt
4.89 m/s = 0 m/s + 9.8 m/s/s * `dt
`dt = .50 sec
Horizontal:
vAve = `ds / `dt
vAve = .4 m / .5 s
vAve = .8 m/s
Vertical:
vf = a * `dt
vf = 9.8 m/s/s * .49 sec
vf = 4.802 m/s
Horizontal:
vf = .8 m/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: ->->->->->->->->->->->-> :
Vertical:
vf = a * `dt
vf = 9.8 m/s/s * .49 sec
vf = 4.802 m/s
Horizontal:
vf = .8 m/s
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•What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
288 deg.
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•What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5 * m * v^2
KE = .5 * .07 kg * (4.89 m/s)^2
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You haven't indicated where the 4.89 m/s comes from. It isn't indicated in any of your previous results.
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KE = .84 Joules
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•What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> : .8 m/s
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•
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
PEgrav = m * g * h
PEgrav = .07 kg(9.8 m/s/s)(1.22 m)
PEgrav = .84 Joules
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•How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
KEf - KE0 = `dPE
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•How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
All of the final KE is in the vertical direction.
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1 hour w/ help
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