Phy 231
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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:
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What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Okay, so we need dt, so ds = v0(dt) + .5(a)(dt^2), and 1.22m = (.5)(9.8m/s^2)(dt)^2, and dt =~ .5s,
and vf in the vertical direction would be vf^2 = v0^2 + 2(a)(ds), or vf^2 = 2(9.8m/s^2)(1.22m), and vf =~ 4.9m/s
and vAve in the horizontal direction would be .4m/.5s = .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: ->->->->->->->->->->->-> scussion:
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Welp, it's just the two vfs, so for the horizontal it'd be -.8m/s on the coordinate system, and for the vertical, it'd be -4.9m/s.
What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Using p-theorem, it'd be c^2 = (-.8m/s)^2 + (-4.9m/s)^2, and c would be a magnitude of 4.96m/s, but we know it would be negative since it would be in the third quadrant, so it'd be -4.96m/s. Speed can't be negative; speed is the magnitude of the acceleration. Both components of the velocity are negative.
Then for direction, it'd be tan^-1(-4.9m/s/-.8m/s) = 81 deg, but since it'd in the third quadrant, it'd be 261 deg. However, if you'd conducted the experiment from the left instead of the right, it'd be .8m/s and -4.9m/s, and the resultant would be in the fourth quadrant, so it'd be tan^-1(-4.9m/s/.8m/s) = -81, or 279 deg.
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
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KE = .5mv^2, so KE = .5(.07kg)(-4.96m/s)^2 = .861J
What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
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it'd be KE = .5(.07kg)(.8m/s)^2 = .0224J
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> scussion: Ignoring air resistance, it'd be the opposite of dKE, which is .861J - .0224 = .8386J, so dPE = -.8386J
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
finalKE - initialKE + dPE = 0 (ignoring air resistance, etc.)
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How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
well, the horizontal is still going to be .0224J, so if the total is 0.861J, then the vertical accounts for .861-.0224J = .8386J.
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20 min
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&#Good responses. See my notes and let me know if you have questions. &#