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PHY 231
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: ->->->->->->->->->->->-> :
Bar any air resistance or abnormal conditions within the car's interior, the gravitational pull on the ball is the main force, causing it to fall quickly after rising. The car's motion inadvertently affects the ball's horizontal motion and direction as well. The ball goes up and down, moving with the momentum of the vehicle, which is partially compensated for by a child most likely throwing the ball in front
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• 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 ball's speed/velocity increases initially from the child's upward thrust/toss ( force ) at an unspecified rate, related to the hand's upward motion. After its brief increase, the speed then decreases, when the downward, negative gravitational pull ( air friction, etc. ) becomes the dominant force in the scenario. The ball reaches a speed of zero meters per second, stops abruptly, and begins its descent with a negative velocity, powered by gravity, until the child catches the ball again.
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• 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's position can be modeled by the equation, y = the initial height + initial velocity ( m/s or equivalent ) multiplied by t - half of the gravitational pull multiplied by t squared, where the position is a function of time. The ball initial increases in height, stops briefly, and descends. The line would not be straight line up and down, as seen from the side; it would be like the hypotenuse of a right triangle, moving with momentum of the vehicle.
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• 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: ->->->->->->->->->->->-> :
All conditions equal, the child was coasting on a bicycle, similar to the way the car moved, the ball's path would be similar, going up and straight with the momentum of the bicycle in front of the child, moving with the same horizontal acceleration as the bicycle. If the kid let the ball drop, the ball's acceleration and velocity would be greater, given its displacement is further, giving it more air time. The acceleration would still be constant, a negative 9.8 m/s^2.
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• 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 acceleration will still be constant at 9.8 m/s^2. But, the position of the ball will not simply be straight down ( not accounting for air resistance ), decreasing from its initial every second. It moves with the same momentum of the vehicle, causing it still to have a horizontal projectile with its drop, similar to a hypotenuse of a right triangle. The velocity will be increasing, if the downward direction is down. The velocity will initial be 0 m/s ( at rest in the child's hand below he/she drops it ).
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• 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: ->->->->->->->->->->->-> :
The ball's initial velocity will still be 0 m/s. It's acceleration will be constant at 9.8 m/s^2. It will go down, with an increasing velocity at a constant rate. If there is no air resistance/friction, the ball will drop straight down. Otherwise, it will keep moving in the same forward momentum as the car, causing the ball to fall in the motion of a hypotenuse of a right triangle.
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25 Minutes
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The 'hypotenuse of a right triangle' motion isn't appropriate; you have constant speed in one direction with increasing speed in the other.
Check this idea, and others, at the link.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then 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.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
&#No revision is necessary unless you have a question, or something you want me to review. &#
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