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Phy 121
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?
I would say that the v0 = 20 cm/s, the displacement was 120 cm, and the acceleration in the vertical direction was 980 cm/s^2.
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
I would think that the vf = 0 cm/s, the change in velocity to be vf - v0 = -20 cm/s, and the average velocity to be -10 cm/s.
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vf would be the final velocity attained, just before striking the floor, by the falling ball.
After striking the floor the acceleration changes, and you would no longer be in a uniform-acceleration interval.
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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?
#$&*I would say that the initial velocity = 80 cm/s and the acceleration must be 0 in the horizontal direction. We find the clock time by using the information in the vertical direction. `dt = -20 cm / 980 cm/s^2 = 0.02 s
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If you divide displacement by acceleration you don't get the change in clock time. You would get that by dividing displacement by average velocity.
Note that the units of your calculation would be s^2 and not 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?
vf = 0, vAve = (80 cm/s + 0 cm/s) / 2 = 40 cm/s, therefore `ds = 40 cm/s * 0.02 s = .8 cm
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
I would imagine the acceleration to be equal to zero at the instant of impact with the floor
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Why does this analysis stop at the instant of impact with the floor?
Because the final velocities in both the vertical and horizontal directions will equal 0 cm/s.
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*#&!
&#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
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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Check my notes, then check the discussion at the link given here.
There is no need for a revision if you understand everything.
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