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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: ->->->->->->->->->->->-> :
The force exerted by the child in the upward direction and gravitational force. There is no force in the direction the car is traveling, as there is no acceleration. The ball would move in that direction also though because it has the same velocity as the car.
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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: ->->->->->->->->->->->-> :
Vertically, the ball as it is tossed upward would have a v_0 and then slow to zero at the top and then begin to accelerate back down until it is caught. The speed in the horizontal direction would match the speed of the car since we are neglecting air resistance.
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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: ->->->->->->->->->->->-> :
It would look like a half circle, or maybe a half oval.
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Close enough; actually the shape would be parabolic.
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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: ->->->->->->->->->->->-> :
By saying coasting, I am assuming that the bike would have acceleration. This would mean that once the ball was in the air, it would not have the same forces as the bike and therefore the ball would come back down behind the bike because the bike would be gaining speed and the ball would not.
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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 would be g. The ball would start with v_0 equals 0 in the vertical direction and accelerate at 9.8 m/s^2 to 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: ->->->->->->->->->->->-> :
To the child, it would appear that the ball dropped straight down and then the car pulled away after the ball hit the ground. To an observer, the ball would fall in a curve downward toward the direction of travel.
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10 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: ->->->->->->->->->->->-> :
10 m/s * .5 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: ->->->->->->->->->->->-> :
a half circle or oval
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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: ->->->->->->->->->->->-> :
v_f=v_0 + a*’dt solve for v_0
v_0 = v_f - a*’dt = 0 - (-9.8 m/s^2)*0.25 s = 2.45 m/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: ->->->->->->->->->->->-> :
‘ds=((v_f + v_0)/2)*’dt = ((0+2.45 m/s)/2)*0.25 s = 0.03 m
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That would be about .3 meter, not .03 meter.
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