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Phy 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.
** 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: ->->->->->->->->->->->-> :
In the vertical direction the initial velocity is 0, the displacement is -1.22m, and the acceleration is -9.8m/s^2. To determine the final velocity:
Vf = `sqrt 2(-9.8m/s^2)(-1.22m)
Vf = -4.9m/s
vAve = -4.9m/s / 2 = -2.45m/s
`dt = -1.22m / -2.45m/s = 0.5s
If the time interval is 0.5s, and if in this time interval an object experienced a horizontal displacement of 0.4m, the velocity will be 0.4m / 0.5s = 0.8m/s
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• 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: ->->->->->->->->->->->-> :
The x component of the velocity is 0.8m/s
The y component is -4.9m/s
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• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
The speed is the magnitude of the x and y components:
Magnitude(speed) = `sqrt (0.8m/s)^2 + (-4.9m/s)^2 = 4.9m/s
The direction of motion is the angle:
Angle = arctan (-4.9m/s) / (0.8m/s) + 180 degrees = 100 degrees. ???It makes more sense for me to add 180 here but I have also seen circumstances where 360 degrees is added and im unclear as to which is which????
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You add 180 degrees when the x component is negative. In this case the y component is negative.
Your arcTan will give you an angle of about -80 degrees.
The angle is measured from the positive x axis.
If you rotate -80 degrees from the positive x axis, you are at the same position as if you had rotated +280 degrees.
Seeing this another way, the positive x axis is at 0 degrees, or 360 degrees, or 720 degrees, etc.. Every 360 degrees you end up in the same place. So -80 degrees from the x axis is -80 degrees from 360 degrees, or 280 degrees. It would also be the same as 640 degrees, though this isn't relevant here.
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• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = ˝ (0.07kg)(-4.9m/s)^2 = -0.84 J
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• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
???I suppose in light of this question that it would be proper to use the known constant or final velocity rather than the magnitude of the velocity???
The object left the table at 0.8m/s:
KE = ˝ (0.07kg (0.8m/s)^2 = 0.0224 J
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• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
The change in potential energy is equal and opposite the change in kinetic energy, which is final kinetic energy (-0.84J) - initial velocity (0.0224J) = -0.8624J.
KE = -PE
KE = -(-0.8624) = 0.8624 Gravity acts negatively downward so if we acted against gravity to pick the ball up and give it potential energy we would be doing positive work. This means the kinetic energy will have to be negative.
???Im not sure my signs are right on some of these and that may have thrown off my answers????
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• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
The change in potential energy is equal and opposite the change in kinetic energy, which is equal to the final kinetic energy subtracted from the initial kinetic energy.
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&#Good responses. See my notes and let me know if you have questions. &#
Phy 121
Your 'cq_1_08.2' 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_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point? answer/question/discussion:Had to work on this one a while but it was pretty fun(don't judge). Here's what I'm thinking. I'm only given two values.. The 12m above ground would be an initial position but I'm pretty confident that doesn't tell me much at this point. So, what I did was create a `dt since I was fairly certain that the value was attainable in a short number of tries. Here's my reasoning:
⁃ I know: v0 = 15m/s
⁃ a = -10m/s^2-----which is also the slope of the uniform line
⁃ I input: `dt = 1s
⁃ `dv = vf - v0
⁃ a = `dv / `dt, and `dv = a * `dt, vf - v0 = a * `dt, vf = a * `dt + v0
⁃ vf = -10m/s^2 * 1s + 15m/s = 5m/s = vf
⁃ I want to know how far it goes in the + direction AND how long it takes. I can reason that the ball reaches its highest point when it has ran out of velocity, so vf = 0m/s.
⁃ vf = v0 + a * `dt
⁃ (vf - v0) / a = `dt
⁃ (0m/s -15m/s) / -10m/s^2 = 1.5s
⁃ vAve = (v0 + vf) / 2 = (15m/s + 0m/s) / 2 = 7.5m/s
⁃ `ds = vAve * `dt
⁃ `ds = 7.5m/s * (1.5s) = 11.25m above the original 12m so actually 11.25m + 12m = 23.25m
⁃ In hindsight, I also could've used vf = 0m/s. I did a lot of extra work but I guess that's called learning? #$&*
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It's almost always better to figure it out on your own, then if necessary see how it could have been done more simply.
Your reasoning was excellent throughout.
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? answer/question/discussion:I know: v0 = 0(at the top of its travel) and I'm pretty sure that I'm traveling in the negative direction now but I'm not sure how to comp for it. I changed the acceleration to positive since the ball is traveling in the same direction as the acceleration, and I thought that the change in position was negative but I'm not in any way referencing it to the initial throw up, at least not yet.
a = 10m/s^2
`ds = 23.25m
I need to know:
vf = 21.6m/s
`dt =
vf^2 = v0^2 + 2(a)(`ds)
= 0 + 2(10m/s^2)(23.25m)
vf^2 = -465m^2/s^2
+-sqrt vf^2 = +- sqrt 465m^2/s^2
vf = 21.6m/s
`dt = `dv / a
`dt = (21.6m/s - 0m/s) / 10m/s^2 = 2.16s
So the ball is traveling at a rate of 21.6m/s when it hits the ground.
After the initial toss the ball travels in the + direction for 1.5s, then in the - direction from 2.16s, so 1.5s + 2.16s = 3.66s
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Very good.
Note that you could have left the upward direction as positive. The acceleration and the displacement would have both been negative, so you would still have found that vf = +- sqrt(465 m^2 / s^2). In this case you would have selected the negative value, vf = -21.6 m/s.
Also be sure to check the discussion at the link.
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&#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. &#
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