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.
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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?
#$&* It is inferred that the ball rolls off the ramp with an initial velocity of 0cm/s. The ball’s vertical displacement is 120cm, and its acceleration in the vertical direction is the acceleration of gravity, which is 980cm/s.
• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
#$&* Even though I think this quantity is astronomically too high, using ‘ds = (vf +v0)/2 * ‘dt, I get a final velocity of 484.85cm/s. The displacement in the vertical direction, as I said earlier, is still 120cm. The change in velocity would be, if my vf is correct, 484.85 and the average velocity would then be 242.4cm/s.
• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
#$&* I would think the ball’s acceleration would remain 980cm/s and its initial velocity in the horizontal direction would be 20cm/s. Using ‘ds = (v0 * ‘dt) + (.5a * ‘dt^2), I get the time interval to be 0.495 seconds.
• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
#$&* I am HORRIBLY confusing myself between the horizontal and vertical aspects of this system. It’s hard to imagine to different types of acceleration acting on the same system, even though I know there must be 2 types. I keep getting the same answers for everyone of these questions. But I do know the horizontal displacement to be 39.6cm because the given horizontal velocity * ‘dt = horiz. ‘ds. That is all I can get for this problem….
You are almost completely on the right track.
• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
#$&* After the impact with the floor, the ball will most likely bounce. In that case, the ball will once again be under the influences of the acceleration of gravity, so I would say that it will begin some sort of uniform acceleration.
• Why does this analysis stop at the instant of impact with the floor?
#$&* Because that is when THIS system ends….
It's because the acceleration changes at that instant, so acceleration is no longer uniform.
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about 40 minutes
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I am horribly confused between the horizontal and vertical aspects of this system. I am mixing this up, using the wrong equations, and doing just about everything possible to impede myself from getting these answers figured out. Any suggestions??
You're not very mixed up at all. Check your solution with that at the given link; you'll be gratifies. Revision is requested only if you have questions and/or want me to check something.
&#Please compare your 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. &#