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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: ->->->->->->->->->->->-> :
Vertical:
vf^2 = v0^2 + 2a`ds
vf^2 = 2 (9.8 m/s^2) (1.22 m)
vf^2 = 23.914 m^2/s^2
vf = 4.9 m/s
vAve = (0 m/s + 4.9 m/s)/2
vAve = 2.45 m/s
`dt = `ds/vAve
`dt = 1.22 m / 2.45 m/s
`dt = .5 s
Horizontal average velocity:
vAve = `ds/`dt
vAve = .4 m / .5 s
vAve = .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: ->->->->->->->->->->->-> :
4.9 m/s vertical
.8 m/s horizontal (this is because there is no acceleration, so the average velocity is
the initial and final velocities)
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Magnitude:
(.8 m/s)^2 + (4.9 m/s)^2
.64 m^2/s^2 + 24.01 m^2/s^2
24.65 m^2/s^2
approx 5 m/s
Direction of motion:
Because the direction of y is going down to the right, which would land it in the
negative y , positive x quadrant, I used tan-1 (-4.9 m/s / .8 m/s), which gave me -81
degrees. I added 360 degrees to this and got 279 degrees for the angle.
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
.5mv^2
.5 (.07 kg)(5 m/s)^2
.5 (.07 kg)(25 m^2/s^2)
.875 Joules
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
.5 (.07 kg)(.8 m/s)^2
.5 (.07 kg)(.64 m^2/s^2)
.0224 Joules
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
.0224 J - .875 J = -.8526 Joules
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
It is the pull of gravity on the ball downward. The initial KE was from the ball
leaving the table in a horizontal direction.
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How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
Most of the KE is in the vertical direction because the vertical average velocity was
2.45 and the horizontal average velocity was less than .8.
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Excellent.
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