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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cq_1_131
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?
V0 = 20cm/s
Ds = 120cm
A = 980cm/s^2
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
Ds = 120cm
Vf^2 = (20cm/s)^2+2(9.8m/s^2)(120cm)
Vf^2 = 400cm^/s^2+2352cm^2/s^2
you multiplied m/s^2 by cm and got cm^2/s^2. You would get cm * m / s^2.
To avoid this you would want to change to consistent units (e.g., use 980 cm/s^2).
Vf^2 = 2752cm^2/s^2
Vf = 52.5cm/s
Dv = 52.5cm/s-20cm/s = 32.5cm/s
vAve = 52.5cm/s+20cm/s / 2 = 36.25cm/s
&&&&& vf^2 = (20cm/s)^2 +2(980cm/s^2)(120cm)
Vf^2 = 400cm^2/s^2+235200cm^2/s^2
Vf = 485 cm/s
Dv = 485cm/s – 20cm/s = 465 cm/s
vAve = 485cm/s +20cm/s / 2 = 252.5cm/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=0cm/s^2
V0 = 80cm/s
120cm = 36.25cm/s (dt)
Dt = 120cm/36.25cm/s
Dt = 3.31s
&&&&&& 120cm = 252.5cm/s (dt)
Dt = .475s &&&&&&
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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 = 3.31s * 80cm/s = 264.8cm
Vf = 80cm/s
vAve = 80cm/s
dv =0cm/s this is because the horizontal velocity does not change and is not affected by gravitational acceleration
&&&&&& ds = .475s * 80cm/s = 38cm
All of the other measurements are the same &&&&&&&
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
After the ball hits the floor it is not guaranteed that the acceleration will be uniform. Some of the energy in the ball will be lost in the impact and the forces will change to make the upward acceleration different than before. Before the impact the acceleration was uniform.
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• Why does this analysis stop at the instant of impact with the floor?
When the ball hits the floor then a lot of the energy from the fall is lost and also, the same rules do not apply for the situation. The ball will bounce so the acceleration up is going to change and so will the velocity.
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20 minutes
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You made a units error early on. This will affect just about all your results. Should be easy to fix, though.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
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Revision from before
&#This looks very good. Let me know if you have any questions. &#