Phy 201
Your 'cq_1_18.1' 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 child in a slowly moving car tosses a ball upward. It rises to a point below the roof of the car and falls back down, at which point the child catches it. During this time the car neither speeds up nor slows down, and does not change direction.
• What force(s) act on the ball between the instant of its release and the instant at which it is caught? You can ignore air resistance.
answer/question/discussion: ->->->->->->->->->->->-> : The force exerted by the child, gravity.
This child has already let go.
• What happens to the speed of the ball between release and catch? Describe in some detail; a graph of speed vs. clock time would also be appropriate.
answer/question/discussion: ->->->->->->->->->->->-> : The speed of the ball will decrease as it moves up, and then increase again as it falls back down. A graph of this behavior would look likea cos graph in which it would leave the hang with a high velocity, dip in the middle and then rise back up to a max velocity.
• Describe the path of the ball as it would be observed by someone standing along the side of the road.
answer/question/discussion: ->->->->->->->->->->->-> : The ball would be moving forward if someone was watching from the side of the road. It would move with the car, as opposed to in the car where it seems like it isn’t moving.
• How would the path differ if the child was coasting along on a bicycle? What if the kid didn't bother to catch the ball? (You know nothing about what happens after the ball makes contact with the ground, so there's no point in addressing anything that might happen after that point).
answer/question/discussion: ->->->->->->->->->->->-> : The path wouldn’t differ except that it would be slower. If the child didn’t catch the ball, it would rise up and move forward, and then land on the ground next to the child’s bike, even if the child was moving. There would be wind resistance however.
• What if the child drops the ball from the (inside) roof of the car to the floor? For the interval between roof and floor, how will the speed of the ball change? What will be the acceleration of the ball? (You know nothing about what happens after the ball makes contact with the floor, so there's no point in addressing anything that might happen after that point).
answer/question/discussion: ->->->->->->->->->->->-> : The speed of the ball will increase as the ball falls to the floor. The acceleration will be 9.8m/s/s.
• What if the child holds the ball out of an open window and drops it. If the ball is dense (e.g., a steel ball) and the car isn't moving very fast, air resistance will have little effect. Describe the motion of the ball as seen by the child. Describe the motion of the ball as seen by an observer by the side of the road. (You know nothing about what happens after the ball makes contact with the ground, so there's no point in addressing anything that might happen after that point).
answer/question/discussion: ->->->->->->->->->->->-> :
To the child, the ball will seem to fall backwards because the car is moving forward, but to someone on the side of the road, the car will seem to fall vertically.
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20 mins
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Because I didn't calculate the forces earlier, I don't have the necessary values to do this part. However, I do understand how to calculate magnitudes with vectors and using the pythagorean theorem method.
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The sum of the x and y components should give you a resultant vector. This resultant vector's magnitude should equal the magnitude of the smaller vectors.
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Since the calculations were not done I do not have the uncertainty values.
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1 hour
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Good, but check the link for further discussion and clarification.
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.