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PHY 201
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** CQ_1_22.2_labelMessages **
Samantha Rogers
PHY 201
Seed 22.2
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 = 0 cm/s
a = 980 cm/s^2
d = 122 cm
vf^2 = v0^2 + 2ad
vf^2 = 0^2 + 2*980*122
vf = 489 cm/s
The ball's final velocity in the vertical direction is 489 cm/s.
489 = 0 + 980t
t = 0.5 seconds
vAve = D/t
vAve = 40cm/0.5s = 80 cm/s
The ball's average velocity in the horizontal direction is 80 cm/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: ->->->->->->->->->->->-> :
Assuming 0 horizontal acceleration, its vertical and horizontal components would be:
x-comp: 80 cm/s
y-comp: 489 cm/s
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• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Its speed would be sqrt(80^2 + 489^2) = 495.5 cm/s
Its direction would be arctan(-489/80) = -80.7 + 360 = 279.3 degrees
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• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
K = 1/2 mv^2
K = 1/2 .07*4.955 ^2
K = .86 Joules
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• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
As it left the tabletop, its y-comp velocity was 0, but its x-comp velocity was still 80cm/s. Therefore, its magnitude would be sqr root(0^2 + 80^2) = 80.
K = 1/2 mv^2
K = 1/2 .07 * .80^2
K = .0224 Joules
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• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
U = mg(h2 -h1)
U = .07 * 9.8 * (1.22 - 0)
U = 0.84 Joules
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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 PE + the initial KE is equal to the final KE.
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• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
VERTICAL
K = 1/2mv^2
K = 1/2 * .07 * 4.89^2
K = .084 J
HORIZONTAL
K = 1/2 mv^2
K = 1/2 * .07 * .8 ^2
K = .0224 J
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*#&!
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