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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: ->->->->->->->->->->->-> :
for vertical final velocity, vf = sqrt(2*9.8 m/s^2 * 1.22 m) = 4.89 m/s
for horizontal vAve = 0.4 m / 0.499 s = 0.801 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 velocity would be the same as above, 4.89 m/s. Zero acceleration in the horizontal direction would still be 0.801 m/s. vf = v0 + at = 0.801 m/s + 0 m/s^2(0.499) = 0.801 m/s
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• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
mag = sqrt((0.801 m/s)^2 + (4.89 m/s)^2) = 4.96 m/s
direction = tan^-1 (4.89 / 0.801) = 80.7 deg
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If the x and y axes are in the standard orientation, x to the right and y up, then the final y velocity will be -4.89 m/s. This won't affect your result for the magnitude of the velocity; since the y velocity is squared the - sign won't affect that part.
However it does affect the angle, which is arcTan(-4.89 / .801) = -80.7 deg, indicating a final velocity directed at 80.7 deg below the positive x axis.
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• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE of the horizontal motion is 1/2mv^2 = 1/2(0.07 kg)(0.801 m/s)^2 = .022 J
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That is correct, but the kinetic energy of the ball is based on the magnitude of its velocity, which as you have calculated is 4.96 m/s.
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• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
The same as above
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• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
PE = mg(h2 - h1) = 0.07 kg * 9.8 m/s^2(1.22 m - 0 m) = 0.837 J
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• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
By the work equation, Wnet = 'dKE + '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: ->->->->->->->->->->->-> :
Is the vertical motion part of the KE?
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Yes. Check also my previous note.
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Very good on most questions. You will want to make a revision to pull all this together.
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