Your 'cq_1_12.1' 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 is tossed vertically upward and caught at the position from which it was released.
Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?
answer/question/discussion:
It would not be traveling faster, but I can make an argument for both slower and the same speed. It leaves the hand at a given velocity because of the force behind the toss. As it travels upward it loses velocity at the constant rate of 9.8m/s^2 due to gravity. When it stops it returns and begins to increase in velocity at the same rate due to gravity, but it does not have the initial push behind it this time.
The object didn't have the initial push behind it at the instant of release. The push had at that point ceased, and can have no further effect on the motion of the object. The initial push resulted in the initial upward velocity, but as far as this interval is concerned, that's all it did.
This fact could be of consequence and the ball will not be able to increase to the same velocity as when it left, but I tend to think that it is inconsequential and because of the ball traveling the same distance it will increase to the same velocity. The distance for both is the same, the acceleration magnitude is the same. V0 and Vf just switch places so they are essentially the same…So, the ball would return to the original position at the same speed it was when it was released.
Good. This is the correct conclusion, and the argument you gave is excellent.
What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion:
Air resistance would slow the ball down quicker on the way up, so it would reach the stop point a little sooner than expected if you were just calculating based on acceleration. However, I think it would also provide the same resistance on the way down, therefore making the ball speed up slower on the way back down. If the resistance is the same both on the way up and the way down it should affect the acceleration identically for both directions, and therefore have no affect of the end velocity. It should be the same. It would however affect the distance the ball travels, which would be shorter.
Air resistance exerts a downward force on the upward phase of motion, and an upward force on the downward.
The ball doesn't rise as high, and even if it wasn't for air resistance on the fall it wouldn't attain as much speed. The air resistance during the fall further reduces the final speed.
In terms of energy, air resistance does negative work on both the upward and downward motions (force opposite displacement in both cases, hence negative work), and hence reduces `dW_net_ON, which reduces the final KE.
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20 mins
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I have decided to just move on. Through practice and repetition I will either eventually understand the new information presented in assignment 11 really well, or marginally well. Either way, I have to move on.
You're going to see plenty more of it, and you don't need to master it all on the first encounter.
&#This looks good. See my notes. Let me know if you have any questions. &#