phy121
Your 'cq_1_22.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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?
`dt = sqrt[.122m / (.5* 9.8m/s/s)] = .158sec
yvF = 9.8m/s/s * .158sec = 1.55m/s or 1550cm/s
xvAve = .04m / .158sec = .253m/s or 253cm/s
Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the
instant before striking the floor?
x = 1.55m/s theta (0*)
y = .253m/s theta (270*)
What are its speed and direction of motion at this instant?
1.57m/s theta (9.27*)
What is its kinetic energy at this instant?
v0 is equal to .253m/s theta (0*) due to it only having motion in the x-axis at the moment it leaves the table:
KE = (.5*m*vF^2) - (.5*m*v0) = (.5 * .07Kg * 1.57m/s^2) - (.5 * .07kg * .253m/s^2) = .084031J
What was its kinetic energy as it left the tabletop?
KE = .5 * .07kg * .253m/s^2) = .00224J
What is the change in its gravitational potential energy from the tabletop to the floor?
it decreases by:
.5 * .07kg * 1.55m/s^2 = .084088J
How are the the initial KE, the final KE and the change in PE related?
-PE = initial KE + final KE
How much of the final KE is in the horizontal direction and how much in the vertical?
.00224J is horizontal and .084088J in the vertical direcion.
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Very good, but be careful of conversions. Your work is correct for distances of 122 mm and 40 mm, but 122 cm is 1.22 m and 40 cm is .40 m. No need to correct this.