Phy 201
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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: ->->->->->->->->->->->-> :
#$&* I find the final vertical velocity by using vf^2 = v0^2 + 2(a * ‘ds) = 0^2 + 2(980cm/s * 122cm) = 489.0 cm/s. And to get the average horizontal velocity, I first use ‘ds = v0 * ‘dt + (.5a * ‘dt^2) to get the ‘dt = 0.50 seconds. And since I’m given the ‘ds_horiz as 40 cm, v_horiz = 40cm / 0.5 sec = 80 cm/s.
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* It’s vertical and horizontal component of its velocity should be just what I answered above, right? But I think they would be twisted a bit. Since the vector triangle’s origin is right at the table’s end, then the vertical component will be the triangle’s x-axis, which would be 489.0 cm/s….. and the horizontal component would be the triangle’s y-axis, which was 80 cm/s.
• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* Using the Pythagorean Theorem, I get sqrt(489cm/s ^2 + 80cm/s ^2) = 495.5 cm/s as the speed. And using arctan(80cm/s / 489cm/s) = 9.3 degrees
should be arcTan(vy / vx)
note that vy is negative
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* KE = 1/2(m * v^2). So 1/2(.07kg * 4.955cm/s) = 0.86 Joules
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* As the ball rolls off the table top, it has no vertical velocity…. only the horizontal. So I think the KE = 1/2(.07kg * 0.80m/s) = .0224 Joules
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* PE_grav = (m * g * y). So I think it’s PE_grav on the tabletop = (.07kg * 9.8m/s^2 * 1.22m) = 0.84 Joules. And once it hits the floor, it should have NO PE, since (.07kg * 9.8m/s^2 * 0m) = 0 Joules. So the ‘dPE = 0 J – 0.84 J = - 0.84 Joules.
• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* Well, ‘dKE + ‘dPE = 0, so ‘dKE = -‘dPE. So the difference in the KE_final and KE_initial would be equal and opposite to the ‘dPE.
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* Um, shouldn’t (80cm/s / 495.5cm/s) * 100% = 16.1% of the magnitude be in the horizontal direction….. and (489cm/s / 495.5cm/s) * 100% = 98.7% be in the vertical direction. I know that these percentages don’t add up to 100%, but I don’t know how else to answer this question….
right idea but KE is not the same as velocity; in fact KE is proportional to the squared velocity
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about 40 minutes....
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I think I'm going to need some detailed explanations of this material. I see how these questions were trying to combine the old and new material, but I simply didn't know or was confused about some of it....
I believe you'll understand the discussion at the link. If you aren't sure, you can submit a revision, though none is required if you don't have questions.
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Solution
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