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Phy 201
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_13.1_labelMessages **
A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
v0 = 20 cm/s, a = 980 cm/s^2, `ds = 120 cm
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf = sqrt(v0^2 + 2 a 'ds).
vf = sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm)
vf = 480 cm/s
vAve = (v0 + vf) / 2
vAve= (20 cm/s + 480 cm/s) / 2
vAve= 250 cm/s
`dt = `ds / vAve
`dt= 120 cm / (250 cm/s)
`dt= .48 sec
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What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
a=0cm/s^2 and v0= 80cm/s
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
`ds = vAve * `dt
= 80 cm/s * .48 sec
= 48.4 cm
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No because the floor will affect the force on the ball.
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Why does this analysis stop at the instant of impact with the floor?
Because analysis is no longer uniform.
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&#Very good work. Let me know if you have questions. &#