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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:
• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
Its horizontal velocity is unchanging; it is constant. Thus, it is simply 40 cm divided by the change in time, .499 s, which equals about 80 cm/s. The final velocity in the vertical direction, as evidenced in the calculations below, is 489 cm/s.
vf^2 = v0^2 + 2 * a * 'ds
vf^2 = 0 cm^2/s^2 + 2 * -980 cm/s/s * -122 cm
vf^2 = 239,120 cm^2/s^2
vf = +- sqrt. ( 239,120 cm^2/s^2 ) ( motion, velocity, and displacement negative )
vf = - 489 cm/s
'ds = ( v0 + vf )/2 * 'dt
-122 cm = ( 0 cm/s + -489 cm/s )/2 * 'dt
-122 cm = -244.5 cm/s * 'dt
.499 s = 'dt
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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: ->->->->->->->->->->->-> :
It's x-component is 20 cm/s ( if right is assumed positive ) and its y-component is -489 cm/s ( if upwards is the positive direction ).
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• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Using Pythagorean theorem and SOHCAHTOA, the speed is (( -489 cm/s )^2 ) + (( 20 cm/s )^2 ) = c^2. C, or the magnitude of the direction of motion is, -489.4 cm/s, 357 degrees of the positive x-axis.
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• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE ( vertical ) = .5 * m * v^2
KE ( v ) = .5 * .07 kg * ( -4.89 m/s )^2 )
KE ( v ) = .035 kg * 23.9121 m^2/s^2
KE ( v ) = .837 J ( est. )
KE ( horizontal ) = .5 * .07 kg * (( .8 m/s )^2 )
KE ( h ) = .0224 J
KE ( total ) = .837 J + .0224 J
KE ( t ) = .8594 J
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• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
The kinetic energy as it left the tabletop was .0224 Joules.
KE ( horizontally only ) = .5 * m * v^2
KE = .5 * .07 kg * ( .8 m/s )^2 )
KE = .035 kg * .64 m^2/s^2
KE = .0224 J
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• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
GPE = m * g * h
GPE = .07 kg * 9.80 m/s/s * 1.22 m
GPE = .837 J
GPE ( floor ) = 0 J ( because its height is assumed 0 m )...
Thus, the change is .837 Joules.
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• How are the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
The initial kinetic energy is the energy of motion, a scalar quantity, that the object posses as it is moving horizontally; it is proportional to the speed. The final kinetic energy is that of the object as it is moving both vertically and horizontally, after it leaves the ledge of the table. And the gravitational potential energy is that energy which is stored, the potential/measure of the ability for work to be done because of its position. It is a shared property between that of the object and the earth. The initial and potential energy are converted into the final kinetic energy plus heat.
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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 ( total ) = .837 J ( vertical ) + .0224 J ( horizontal )
KE ( t ) = .8594 J
K( v ) / K ( total ) * 100 = 97.4% ( est. ) Vertical
K ( h ) / K ( total ) * 100 = 2.6% ( est. ) Horizontal
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25 Minutes
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&#Very good responses. Let me know if you have questions. &#