Phy 121
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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=sqrt.(2+(980cm/s^2)(122cm) vF=489cm/s…vAve=244.5cm/s. 122cm/244.5cm/s=.5s. 40cm horizontally * .5sec=80cm/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: ->->->->->->->->->->->-> : Y=489cm/s X=80cm/s
• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> : 495cm/s at 80 degrees.
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> : KE=1/2(.07kg)*4.95m/s = .17325N
good but units don't work out; should show you that there's an error
you need to square the velocity
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> : 1/2*(.07kg)*(9.8m/s)=.343N
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> : It lost .16975N
• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> : The change in PE is equal to the difference in KE0 and KEF.
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> : (80cm/s/489cm/s)*100=16.4% horizontally and 83.6% vertically….I think, sort of going out on a limb on this answer.
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20min
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Good overall, but check the link:
&#Please compare your solutions with the expanded discussion at the link
Solution
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