#$&*
phy121
Your 'cq_1_22.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_22.2_labelMessages **
A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:
• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
vertical component
v0 = 0, `ds = 1.22 meters, a = 9.8 m/sec^2
vf^2 = v0^2 + 2*a*`ds
= 0^2 +2*9.8m/sec^2 *1.22 m
Vf = 4.9 m/sec
vAVe = 4.9m/sec/2 = 2.45 m/sec
v=`ds/`dt
2.45 m/sec = 1.22m/`dt
`dt=.50 sec
For the horiztonal component
`ds = .4 m, `dt = .50 sec
V = `ds/`dt
= .4m/.50 sec = .8 m/sec
#$&*
• 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: ->->->->->->->->->->->-> :
.8 m/sec vertical component and -4.9 m/sec horizontal component
#$&*
• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
we would use our vector that shows .8 m/sec and -4.9 m/sec
magnitude = squareroot (.8m/sec ^ 2 + -4.9m/sec^2 )=4.96 m/sec = speeds
the angle is arctan (-4.9m/sec / .8m/sec) + 180 =99.27 degrees
#$&*
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5m*v^2 = .5*.07kg*(4.9 m/sec)^2 = .84 joules
@&
You would want to use the 4.96 m/s magnitude of the velocity.
*@
#$&*
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
when the object left the tabletop its velocity was .8m/sec
KE = .5m*v^2 = .5*.07kg*.8m/sec^2= .02 joules
#$&*
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
the change in PE is equal to the work donedd by the object against forces that are conservative. So the gravitational potential energy would be negative so the change in gravitational potential energy is -.84 joules
#$&*
• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
`dKE = KEf - KE0
`dKE + `dPE =0
.84 joules +-.84 joules =0
#$&*
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
KEf in the horizontal direction =.02 joules
KEf in the vertical direction = .84 joules
@&
The total KE would therefore be .86 Joules, consistent with the result that would have been obtained using 4.96 m/s as the velocity.
*@
#$&*
** **
25 minutes
** **
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
@&
Good, but see my notes. Also check the discussion at the link below (no revision is necessary).
*@