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Phy 231
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.
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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?
a = 9.8 m/s^2 or 980 cm/s^2
`ds = 120 cm
v0 = 20 cm/s
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
`ds = 120 cm
a = vf – v0 / `ds ; vAve = (vf + v0) / 2 = `ds / `dt
`dt = (2*`ds) / (vf + v0)
a = (vf^2 – v0^2) / (2*`ds)
vf = sqrt((2*`ds*a) + v0^2)
vf = sqrt((2*120 cm*980 cm/s^2) + (20 cm/s)^2)
vf = 485.386 cm/s
`dv = change in velocity = vf – v0 = 485.386 cm/s – 20 cm/s = 465.386 cm/s
Average velocity = vAve = (vf + v0) / 2 = (485.386 cm/s + 20cm/s) / 2 = 252.693 cm/s
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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 = 980 cm/s^2
v0 = 0 cm/s
vf = 80 cm/s
a = vf – v0 / `dt
`dt = (vf – v0) / a
`dt = (80 cm/s – 0 cm/s) / 980 cm/s^2
`dt = 0.0816 s
v0 isn't zero for the uniform-acceleration interval, and 980 cm/s^2 is not a horizontal quantity.
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
vf = 80 cm/s^2
vAve = vf + v0 / 2 = (80 cm/s + 0 cm/s) / 2 = 40 cm/s
`dv = change in velocity = vf – v0 = 80 cm/s – 0 cm/s = 80 cm/s
vAve = `ds / `dt
`ds = (vAve)*(`dt)
`ds = (40 cm/s)*(0.0816 s) = 3.264 cm
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
After the instant of impact with the floor, we can expect that the ball will not be uniformly accelerated.
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• Why does this analysis stop at the instant of impact with the floor?
At the instant of impact wit the floor the acceleration is 0.
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about 30 minutes
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This will require some modification. However you've included most of the necessary steps in the analysis, and it should be easy for you to correct the details.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
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