Phy 231
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.
** **
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 = 0cm/s ^2 + 2(980cm/s^2)(122cm)
vf^2 = 239120
vf = 489 cm/s (vertical)
489cm/s = 0cm/s + 980cm/s^2 (dt)
Dt = .499s
Horizontal average velocity = 40cm / .499s = 80.1 cm/s
#$&*
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
vertical = -489 cm/s
horizontal = 80.1 cm/s
#$&*
• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
speed = sqrt( 489 cm/s ^2 + 80.1 cm/s ^2) = sqrt(245537.01 cm^2/s^2) = 495.5 cm/s
direction = arctan (-489 / 80.1) = -80.697 + 360 = 279 degrees
#$&*
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KEfinal = (1/2)(70g)(495.5 cm/s)^2 = 8593000 g cm^2/s^2
#$&*
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KEinitial = (1/2)(70g)(80.1cm/s)^2 = 224600 g cm^2/s^2
#$&*
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
PE = 70g * 980cm/s^2 * 122cm = 8369000 g cm^2/s^2
#$&*
• How are the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
The final KE minus initial KE is equal to the change in PE, so change in KE = change in PE.
#$&*
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
final = 8593000 g cm^2/s^2
You have to take the amount of final KE and divide it based on the horizontal and vertical movement. I assume this has something to do with the components???????
The only force is the gravitational force, which is conservative. The horizontal KE is the same as when the ball left the table; the vertical KE is equal to the PE loss. If you figured 1/2 m vx^2 and 1/2 m vy^2 the results would be as indicated.
#$&*
** **
30 minutes
** **
&#This looks good. See my notes. Let me know if you have any questions. &#