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phy121
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: ->->->->->->->->->->->-> :
gravity would be acting against the ball when it is tossed up and would be working on the ball as it falls back down.
Between release and catch, the child’s hand is not touching the ball and does not exert a force on it and the car is also not in contact with the ball and does not exert a force on it.
I would guess the only real force acting on the ball is gravity in the form of the balls weight.
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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 balls velocity slows at it rises stops for an instant and then speeds as it falls. In terms of a graph, the slope of the velocity vs time graph would be constant and would equal the acceleration of gravity. The point where the ball briefly comes to rest would have negative velocity of one side of it and positive velocity on the other side.
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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: ->->->->->->->->->->->-> :
I would think that an observer on the side of the road would see the ball speed up in the vertical direction but not in the horizontal direction making a parabolic path
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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 of the ball would be the same as seen by the child on the bike or someone standing on the road as long as air resistance wasn’t too great. If the speed is greater, the air resistance would cause the ball to accelerate backward in relation to the child.
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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: ->->->->->->->->->->->-> :
I think the ball would of course start off as stationary and then increase its speed as its approaches the ground. The ball would speed up at a constant rate.
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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 there isn’t much air resistance to slow the ball, then it would continue horizontal motion beside the car. The child would see the ball drop straight down but someone looking on from the sidewalk would see a curved path until the ball reaches the road
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25 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 = vAve *`dt
= 10m/sec * .5 seconds
= 5 meters
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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 observer would see the ball travel at a constant horizontal velocity while rising more and more slowly and falling more and more quicker. It would have a parabolic path.
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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 path is a parabola that increases at a decreasing rate and decreases at an increasing rate
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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: ->->->->->->->->->->->-> :
`ds = 0, a = -9.8m/sec^2, `dt = .5 seconds
Ds = v0dt + .5adt^2
0 = v0*.5 seconds + .5*-9.8m/sec^2 *.5seconds^2
V0 = 2.45 m/sec
=
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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: ->->->->->->->->->->->-> :
v0 = 2.45 m/sec vf = 0 and a=-9.8 m/sec^2
vAve = 1.225 m/sec
`dv = -2.45 m/sec
A = dv/`dt
-9.8m/sec^2 = -2.45m/sec/dt
Dt = .25 seconds
ds = vAve *`dt
= 1.225 m/sec*.25 sec =.306 meters
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30 minutes
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&#Very good responses. Let me know if you have questions. &#