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phy 231
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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: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2a`ds
vf = sqrt(v0^2 + 2a`ds) = sqrt(0 + 2(980 cm/s/s)(122cm)) = 489 cm/s
`dt = `dv/`a = (489 cm/s - 0 cm/s)/980 cm/s/s = .5 sec
Horizontal vAve = `ds/`dt = 40cm/.5s = 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: ->->->->->->->->->->->-> :
x and y will be 80 cm/s and 489 cm/s
magnitude = sqrt[(80^2) + (489^2)] = 495.5 cm/s
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
-489 m/s in the vertical direction and +80 cm/s in the horizontal
The resultant vector is arctan (-489 cm/s/80 cm/s)= -80.7 degrees which is below horizontal at a speed of 495.5 cm/s.
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Changing 495.5 cm/s to 4.955 m/s:
KE = 1/2mv^2 = ½ * .07kg * (4.955m/s)^2 = .86 J
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
Changing 80 cm/s to .8 m/s:
KE = 1/2mv^2 = ½ * .07kg * (.8m/s)^2 = .02 J
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
Work done on the ball by g is + and work done on g by ball is (-).
PE = mgh = .7kg * 9.8 m/s/s * .122 m = .84 Joules. The change will be negative since PEg is (-) at -.84 J
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
KE will increase and PE will decrease so that `dKE + `dPE = 0.
`dKE = .86 - .02 = .84 J
`dKE + `dPE = .84 - .84 = 0 J
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&#Very good responses. Let me know if you have questions. &#