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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: ->->->->->->->->->->->-> :
Gravitational force and the force by the child throwing the ball up in the air.
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The force exerted by the child ends at the instant of release, so this force does not act during the interval. The initial velocity was the result of the child's action, but for this interval that action is in the past.
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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 speeds up as it is initially released and begins to slow down as it begins to reach equilibrium, then the ball will speed increase as the ball begins to descend downward
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The ball sped up prior to release, but immediately on release it begins to slow.
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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 would go straight up and then come back down.
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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 car would not be protecting the ball from air resistance. I would imagine that if the kid threw the ball straight up and the bicycle kept going straight then the ball will fall to the ground behind him.
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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: ->->->->->->->->->->->-> :
If the ball was dropped without any additional force applied to it, the gravitational force would be acting on it with an acceleration of 9.8 m/s^2.
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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: ->->->->->->->->->->->-> :
If he drops it, the ball would appear to roll back and fall straight down.
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20 mins
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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: ->->->->->->->->->->->-> :
vAve = 'ds / 'dt, 'ds = 0.5 s * 10 m/s = 5 m
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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: ->->->->->->->->->->->-> :
Straight up and down.
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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: ->->->->->->->->->->->-> :
For the people viewing the ball appears to be traveling horizontally, but going up vertically and then coming down. The Path would be a ball traveling at an angle upward, going across a straight line, and then defending at an angle.
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The horizontal motion is at constant velocity.
The velocity of the vertical motion varies.
So the path will never be straight.
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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: ->->->->->->->->->->->-> :
I cannot visualize this. I know that when an object begins to fall its initial velocity is zero, and the acceleration is 9.8 m/s^2.
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You can analyze the vertical motion of the ball. You know its acceleration to have been 9.8 m/s^2 downward and you know that it rose for 1/2 second before coming to rest for an instant before beginning its fall back down.
What must have been its initial vertical velocity, in order to come to rest in 1/2 second while accelerating downward at 9.8 m/s^2?
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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: ->->->->->->->->->->->-> :
I don't know. 'ds = v0'dt + 0.5a'dt = 0.5a'dt = 2.45 m
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This would be good, except that you made an error that you probably would have caught had you used units.
Specifically the units of 1/2 a `dt would be m/s^2 * s = m/s, not m. So something is wrong with that part of your formula.
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35 mins
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Not bad. It should be easy for you to make the necessary corrections.
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