#$&*
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: ->->->->->->->->->->->-> :
v0 = om/s
m = 70 g
a = 9.8m/s/s
ds=-122 cm = 1.22 m
Vertical: vf2 = v02 + 2a * ds
vf2 = 2(9.8 m.s.s) * -1.22 m = 4.9 = vf
Find dt = ds = (v0 +vf) / 2 * dt
dt = .49
Horizontal: ds = vAve * dt
.40 = vAve * .49
vAve = .82 m/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: ->->->->->->->->->->->-> :
vertical comp. = y comp. = 4.9 m/s
horizontal comp. = x comp. = .82 m/s
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What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
direction: arctan(-122 / 40) + 360 = 288.15deg.
velocity: arctan(4.9 / .82) = 80.50 m/s ???
@& The velocity is a vector. Its magnitude is the speed. The magnitude of a vector is found from its components using the Pythagorean Theorem.*@
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What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
dKE = -dPE = .07 kg * 9.8 m/s * -1.22 = -.837 = 8.37
@& The calculation of the KE change during the fall would read
dKE = -dPE = - .07 kg * 9.8 m/s * -1.22 = 8.37 Joules
This is a good calculation but doesn't take into account all of the KE present at this point.
KE = 1/2 m v^2. Having found v (using the Pythagorean Theorem) you could then easily find the KE.*@
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What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5(.07 kg)(v)2 = I am confused and dont evenknow where I am in the problem???
@& At the ball left the tabletop, what was its speed v? Just plug this value into your equation and you'll get the KE.*@
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What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
NO idea
@& You pretty much calculated that already. See the note on your calculation of `dPE.*@
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How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
The are equalandopposite of each other
@& Once you've done a couple of corrections, you'll know the initial and final KE, as well as the change in PE. Then you can see how they are related.*@
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How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5mv2
Horizontal = .5(.07)(.82)2 = .023
Vertical = .5(.07)(4.9)2 = .84
@& Right.*@
@& You've just about got it. You just need to put it all together.
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