course phy 121
cq_1_222phy 121
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:
M = 70g = .07kg
Vert ds = 122cm = 1.22m
Horz ds = 40cm =.4m
F = m*a = .07 * 9.8 = .686N
DW = F * ds = .686N * 1.22m = .83692J
KE = 1/2m*v^2
.83692 = .5 (.07kg) * v^2
v^2 = 23.912
V = 4.88998 = 4.8m/s
&&&& I rounded this wrong. Final vertical velocity = 4.9m/s
Dt = ds/vAve
Dt = 1.22/ (4.9/2) = .49795 s = .5s
Horiz v = ds/dt
V = .4m / .5s = .8 m/s
&&&& I still get horizontal velocity of .8m/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 component = 4.8m/s
&&&&Vertical component = 4.9m/s
Horizontal component = .8m/s
What are its speed and direction of motion at this instant?
answer/question/discussion:
Speed = sqrt 4.8^2 + .8^2 = 23.04 + 0.64 = 4.866 = 4.9 m/s
&&&& Speed = sq rt (4.9^2 + .8^2) = sq rt (24.01 + .064) = 4.964 = 5m/s
Direction = arctan (4.8/.8) = 80.53767 = 80.5 degrees
&&&& Direction = arctan (4.9/.8) = 80.727 = 80.8 degrees
What is its kinetic energy at this instant?
answer/question/discussion:
KE = 1/2m*v^2
KE = .5 (.07kg)* 4.9^2
&&&& KE = .5 (.07kg) * 5^2
&&&& KE = .035 * 25 = .875 J
v is the speed of the ball, not just the vertical component of the velocity. The velocity of the ball has two components; the magnitude of velocity is the speed. Is the magnitude of the velocity only 4.9 m/s? Hint: it isn't much greater than 4.9 m/s, but I believe you will find that it is at least a couple of tenths of a m/s greater.
KE = .035 * 24.01
KE = .84035J
What was its kinetic energy as it left the tabletop?
answer/question/discussion:
KE = 1/2m*v^2
KE = .5 (.07kg) * .8^2
KE = .035 * .64
KE = .0224J
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion:
PE gravity= m*g*h
PE = .07kg * 9.8m/s^2 * 1.22m
PE = .83692
How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion:
My guess would have been that the initial KE + the change in PE would equal the final KE. But it doesn’t. I may have miscalculated somewhere though.
&&&& Init KE + change in PE = final KE
&&&& 0.0224J + 0.83692 = 0.02 + 0.84 = 0.86 which does not = 0.875
You rounded your velocities to two significant figures, which for a velocity of 4.9 or 5.0 m/s would be +-1%. When you square the result the square will be good to within +-2%. .875 is within +-2% of .862.
That is, the difference between the two results is in your roundoff. Had you used three significant figures the two results would have agreed much more closely.
&&&& Looking at my previous answer the change in PE is actually on .003 different from the final KE I calculated previously.
See my previous note, which should help been easily reconcile the difference.
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion
The final KE is .84035J.
The initial KE is .0224J, this is the horizontal portion.
.84035 - .0224 should be the vertical portion (= .81795), which is also the part that I thought would be similar to the gravitational change in PE.
** **
30 mins
&&&& I worked on this for well over an hour and I still don’t know if or where I really went wrong.
** **
Very good thinking, with only one that are, which you should be able to correct without much trouble.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&. &#
The only real error on your first attempt was that you didn't use both components in calculating the speed on which the kinetic energy was based.
The apparent discrepancy in your results this time was simply the result of roundoff error.