#$&*
Phy 241
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: ->->->->->->->->->->->-> :
v_f = sqrt(0 + 2*9.81 m/s^2 * .122 m)
= 1.547 m/s
vf(y) = v0 + a * `dt
'dt = (v_f- v_0) / a
= 1.547 m/s - 0 / 9.81 m/s^2
= 0.1577 s
(v_ave)x = 'ds/'dt
= 0.040 m / 0.1577 s
= 0.2536 m/s
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122 meters is 1.22 meters, not .122 meters.
However the method of solution is correct, and you have a correct solution if the fall is 12.2 cm.
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F_net*'ds = 1/2m*v^2
0.070 kg * -9.81 m/s^2 * 0.122 cm = 1/2 * 0.070 kg * v^2
v = sqrt(2 * 0.837774 / 0.07)
= 1.54 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: ->->->->->->->->->->->-> :
v_fx = 0.2536 * 2
= .5073 m/s
v_fy = 1.547 m/s
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
R = sqrt(1.547^2 + 0.5073^2)
= 1.63 m/s @ 71.84 degrees
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = 1/2 m * v ^2
= 1/2 * 0.070 g * (1.63 m/s)^2
= 0.093 N-m
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = 1/2 m * v^2
= 0
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
0.093 N-m
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The change in gravitational PE would be
`dPE = 070 kg * 9.8 m/s^2 * .122 m, which is significantly less than .093 N m.
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
KE_intial = 0
KE_final = -PE
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The ball started with nonzero kinetic energy.
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#$&*
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
KE_x = 1/2 * m * v_x^2
= 1/2 * 0.070 kg * 0.06431 m/s
= 0.0022509 N-m
KE_y = 1/2 * m * v_y^2
= 1/2 * 0.070 * 2.393
= 0.08376 N-m
#$&*
** **
10 minutes
** **
2pm 10/14/2012
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You've got some decimal-point errors (some but not all of which I've pointed out), and some other more important errors in your solution.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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