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Phy 241
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: ->->->->->->->->->->->-> :
If there is no air resistance, the ball would be acted on by gravity.
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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 speed would start at 0, rise as time goes on at a constant rate until its upward momentum is overcome by velocity. Then it would start falling at a constant rate until the ball is caught and its speed is 0.
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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: ->->->->->->->->->->->-> :
There would of course be the up and down travel, but the ball would be seen to be moving in the direction of the car at its speed.
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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: ->->->->->->->->->->->-> :
The path would be very similar if the child was on a bike instead of a car. If the child didn’t catch it, the ball would keep accelerating due to gravity, past the point of release, until it hits the ground. All the while it would be moving in the direction of the bike at release with the same horizontal speed.
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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 ball will accelerate downwards at 9.8m/s^2 due to gravity until it hits the floor.
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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: ->->->->->->->->->->->-> :
From the child’s perspective the ball will drop straight down until it hits the ground since it will be moving in the same direction\speed of the car. To an observer, the ball will move vertically due to gravity, but it will also move horizontally in the direction of the car until it hits the ground.
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about 5 minutes
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A child in a car tosses a ball upward so that after release it requires 1/2 second to rise and fall back into the child's hand at the same height from which it was released. The car is traveling at a constant speed of 10 meters / second in the horizontal direction.
• Between release and catch, how far did the ball travel in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds=10m/s*1s=10m
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• As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball traveled 10m horizontally and 1.225m up and down from and to the child’s hand.
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• Sketch the path of the ball as observed by a line of people standing along the side of the road. Describe your sketch. What was shape of the path of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
The graph looks like the top of a triangle since vertical acceleration is constant and horizontal speed is constant.
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Speed isn't constant in both directions. It is in the horizontal direction, but in the vertical direction it's the acceleration that's constant.
The result is that the ball is never traveling along a straight line, though when it comes to rest in the vertical direction its velocity is for an instant purely horizontal.
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• How fast was the ball moving in the vertical direction at the instant of release? At that instant, what is its velocity as observed by a line of people standing along the side of the road?
answer/question/discussion: ->->->->->->->->->->->-> :
At the instant of release the ball would have been moving up at 4.8m/s since it takes gravity .5s to overcome its speed. That is its vertical velocity as observed outside and its horizontal velocity is 10m/s.
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• How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> :
`dv=9.8m/s^2*.5s=4.9m/s ; vf=4.9m/s ; v0=0m/s ; vAve=2.45m/s ; `ds=2.45m/s*.5s=1.225m
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Good, but you'll want to think a little more about the shape of the ball's path.
No need for a revision but do check my notes and the discussion at the link.
&#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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