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.
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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: ->->->->->->->->->->->-> :
a^2 + b^2 = c^2
(40 cm)^2 + (122 cm)^2 = c^2
1600 cm^2 + 14884 cm^2 = c^2
c^2 = 16484 cm^2
c = 128.39 cm
Vertical:
vf^2 = v0^2 + 2 * a * `ds
vf^2 = (0 m/s)^2 + 2 * 9.8 m/s/s * 1.22 m
vf^2 = 23.912 m^2/s^2
vf = 4.89 m/s
vf = v0 + a * `dt
4.89 m/s = 0 m/s + 9.8 m/s/s * `dt
`dt = .50 sec
Horizontal:
vAve = `ds / `dt
vAve = .4 m / .5 s
vAve = .8 m/s
#$&*
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
Vertical:
vf = a * `dt
vf = 9.8 m/s/s * .49 sec
vf = 4.802 m/s
Horizontal:
vf = .8 m/s
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• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Magnitude = 1.2839 m
Direction = tan^-1 (1.22m/.4m)
= 71.85 degrees
You have calculated the angle of the displacement vector from the end of the ramp to the landing position. This isn't the angle of the velocity vector, but for the moment let's consider a little more closely the angle you have calculated.
Using a standard x-y coordinate system, the y displacement would be negative. The x might or might not also be negative, depending on whether your initial velocity was to the right or left (could be drawn either way).
The angle would then come out either -72 deg (same as 288 deg), or 252 deg (72 deg + 180 deg).
Now let's consider the question of the velocity vector, which is the vector that indicates speed and direction of motion. I believe you figures came out vf_y = 4.8 m/s and vf_x = .8 m/s. As mentioned previously, the y component will be negative, the x component could be either positive or negative. It is these components, not the components of the displacement vector, that should be used to calculate the direction of motion.
#$&*
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5 * m * v^2
KE = .5 * .07 kg * (4.89 m/s)^2
KE = .84 Joules
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• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5 * .07 kg * (0m/s)^2
KE = 0 Joules
The ball was moving as it left the tabletop, presumably at the .8 m/s you found earlier.
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• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
PEgrav = m * g * h
PEgrav = .07 kg(9.8 m/s/s)(1.22 m)
PEgrav = .84 Joules
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• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
KEf – KE0 = `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: ->->->->->->->->->->->-> :
All of the final KE is in the vertical direction.
#$&*
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35 minutes
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You have a few errors, though your work as a whole is very good. It won't take you long to correct them, and it will be worth your time, so do submit the requested revision.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#