cq_1_222

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: finding the PE (m*g*y) or (.07 * 9.8 * 1.22) = .837 J

setting that to be the KE of the fall, we find vf = sqrt(.837 / (.5*.07)) = 4.89 m/s

vAve in horizontal is .4/.51 = .787 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: vf = 4.89 m/s in the vertical, and vf = .787 m/s in the horizontal

• What are its speed and direction of motion at this instant?

answer/question/discussion: v = sqrt(.787^2 + 4.89^2) = 4.95 m/s

• What is its kinetic energy at this instant?

answer/question/discussion: KE = .5*.07*(4.95^2) = .86 J

• What was its kinetic energy as it left the tabletop?

answer/question/discussion: 0 J

the ball was moving at .787 m/s, according to your calculation; the KE wasn't zero

• What is the change in its gravitational potential energy from the tabletop to the floor?

answer/question/discussion: .837 J

• How are the the initial KE, the final KE and the change in PE related?

answer/question/discussion: the PE calculated at the beginning is transferred to KE throughout the fall of the ball, therefore they’re all related.

• How much of the final KE is in the horizontal direction and how much in the vertical?

answer/question/discussion: KEh = .5(.07)(.787^2) = .022 J

KEv = .5(.07)(4.89^2) = .8369 J

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20 min

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Everything is right except for two of your answers; however your error in calculating initial KE prevented you from correctly articulating the details of the energy conservation in this situation.

&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.

Solution

This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#