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, what are the ball's initial velocity, displacement and acceleration in the vertical direction?
Initial velocity = 20 cm/s
a = 980 cm/s/s
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^2 = v0^2 + 2 * a * ds
vf^2 = (20 cm/s)^2 + 2 * 980 cm/s/s * 120 cm
vf^2 = 400 cm/s^2 + 235200 cm^2/s^2
vf = +-485.4 cm/s
final velocity is +485.4 cm/s because it is downward.
vAve = (20 cm/s + 485.4 cm/s)/2
vAve = 252.7 cm/s
485.4 cm/s – 20 cm/s = 465.4 cm/s
dt = 120 cm/(252.7 cm/s)
dt = 0.475 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 = 0 cm/s/s
v0 = 80 cm/s
dt = 0.475 sec
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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
ds = 80 cm/s * 0.475 sec
ds = 38 cm
Vf = 80 cm/s
vAve = (80 cm/s + 80 cm/s)/2 = 80 cm/s
ds = 38 cm
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No, it will no longer be constant because the net force will then change.
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• Why does this analysis stop at the instant of impact with the floor?
The acceleration is not in uniform acceleration anymore so the equations would not be applicable.
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40 minutes
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&#This looks very good. Let me know if you have any questions. &#