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PHY 121
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: ->->->->->->->->->->->-> n:
`ds = 122cm
A = 9.8m/sec^2
V0 = 0m/sec
Vf^2 = v0^2 + 2(a*`ds)
Vf^2 = 2(9.8m/sec^2 * 1.22m)
Vf^2 = 23.912m^/sec^2
Vf = +-4.89m/sec
4.89m/sec = 0m/sec + 9.8m/sec^2 * `dt
`dt = .5sec
`ds_h = 40cm
`dt = .5sec
vAve = `ds/`dt
vAve = 40cm/.5sec
vAve = 80cm/sec
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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: ->->->->->->->->->->->-> n:
v_y = 4.89m/sec
v_h = .8m/sec
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> n:
magnitude = sqrt[(4.89m/sec)^2 + (.8m/sec)^2)
=4.95m/sec
Theta = arctan[(4.89m/sec)/(.8m/sec)]
=80.7degrees
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> n:
KE = 1/2*m*v^2
KE = 1/2(.07kg)*(4.95m/sec)^2
=.86J
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> n:
v_y = 0m/sec
v_h = .8m/sec
KE = 1/2*(.07g)*(.8m/sec)^2
=.02J
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> n:
Gravity is acting in the same direction as displacement, so work by gravity is positive and work done by the ball against gravity is negative
x `dgravitational_PE = .02J-.86J
=-.84J
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> n:
The `dKE and `dPE when added together, should equal zero indicating they are equal and opposite.
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How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> n:
KE_h = .02J
KE_v = .86J
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*#&!
&#Very good responses. Let me know if you have questions. &#