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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: ->->->->->->->->->->->-> :
Well we know in the vertical direction a = 980 cm/s^2, `ds = 122 cm, and v0 = 0
vf = sqrt(0 + 2 * 980 cm/s * 122 cm) = sqrt(239,120 cm^2/s^2) = 489 cm/s
Therefore the `dt to do this would be `dt = (vf - v0) / a = (489 cm/s - 0 cm/s) / 980 cm/s^2 = 0.50 s
Now we can move on to the horizontal direction. a = 0, `ds = 40 cm, `dt = 0.50 s
Therefore vAve = 40 cm / 0.50 s = 80 cm/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: ->->->->->->->->->->->-> :
I think the x component would be 80 cm/s and the y component would be -489 cm/s, since we are looking on the graph. I am not sure if this is correct or not.
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•What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
I suppose we find the magnitude first. magnitude = sqrt(80^2 + -489^2) = 495.5 cm/s To find the vector angle, we take arctan (y component / x component) = -80.7 degrees, which converts to 279.3 degrees in the counterclockwise direction from the x axis.
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•What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
I suppose we use the magnitude value here, but we must convert everything over to m/s, and kg, this is 0.07 kg and 4.96 m/s
KE = .5 * 0.07 kg * (4.96 m/s)^2 = 0.5 * 0.07 kg * 24.6 m^2/s^2 = 0.86 Joules
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•What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
Well it was moving 80 cm/s in the horizontal direction, which is 0.8 m/s
Therefore KE = .5 * .07 kg * (0.8 m/s)^2 = 0.028 Joules
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•What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
I am not sure how to find this still. This concept is very confusing to me. I would assume it would equal -0.832 since that is equal and opposite to the difference of the KE0 and KEf.
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•How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
The difference of the initial KE and final KE is equal and opposite to the PE.
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•How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
I am not sure how to find this either.
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Good.
Check out the discussion below to complete the missing steps.
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