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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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Samantha Rogers
PHY 201
Seed 18.1
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 forces that act on the ball between the time the child releases it and the time it is caught will be that of gravity as well as the muscle force exerted by the child. Such muscle force will cause the ball to stop and resists the ball’s path of movement. Thus, the muscle acts as a preventive force (or a force that acts opposite of that of the direction of motion).
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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: ->->->->->->->->->->->-> :
peed is similar to velocity, but velocity is a vector. If we look at the fact that vf= 0 m/s when the ball is at its highest point in the air, we can describe the speed of the ball as being the highest when it leaves the child’s hand and when it return to the child’s hand, whereas the speed is at its lowest when the ball is at its highest point (when it stands as still for a moment). Thus, a graph of speed vs. clock time would look very much so like the shape of a V.The ball’s speed would start off high, then get lower, then get higher 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: ->->->->->->->->->->->-> :
Someone alongside the road would see the path of the ball similar to the way the graph looks, as the car is in motion. IF the car was standing still, the ball would appear to be going up and coming down in a fairly straight shaped sort of line. However, since the car is moving, the ball will move slightly with it in an arc sort of shape.
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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: ->->->->->->->->->->->-> :
If the child was coasting along on a bicycle, he or she would have had a very similar path to that of the path from the car. This is because the ball was still thrown from a moving object/vehicle. If the child did not catch the ball, the path of the ball (displacement) would have gone down farther but would have followed the same downward motion. However, the difference would be that the ball would pick up 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: ->->->->->->->->->->->-> :
IF the child drops the ball, then the v0 starts out at 0 m/s and the vf would then have the highest velocity/speed. The direction of motion then changes as well, as when it was thrown the initial direction was up and then the downward motion set in. Here, the initial direction of the motion is downward. The acceleration of the ball will be the force of gravity, as there are no other forces acting on the ball. Thus, the acceleration will be 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 the child holds the ball out of an open window, then the drop of the ball will follow a motion that appears to be a direct, downward, vertical drop from the view of the child. To be more specific, the child’s view will not really see the motion from the building where the ball was dropped if looking downward. From the view of the side street stander, it would seem that the ball follows a straight path, from top to bottom of the window to the ground, vertically.
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30 minutes
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6/26/2012 7:00
Samantha Rogers
PHY 201
Seed 18.2
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: ->->->->->->->->->->->-> :
The ball continues to travel at 10m/s in the horizontal direction.
In .5 second the ball moves 10m/s * .5s = 5 m in that direction.
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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 travels at a constant horizontal velocity but it starts rising more and more slowly, then falls more and more quickly. This results with a curved path. It was rising at a decreasing rate and falling at an increasing rate.
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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 ball travels along the parabolic path.
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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: ->->->->->->->->->->->-> :
vertical direction: `ds = 0. accelerates downward at 9.8m/s^2 for .5 second.
Upward direction is positive.
v0 = (`ds - .5 a `dt^2) / `dt = ( 0 - .5 * (-9.8 m/s^2) * (.5 s)^2) / (.5 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: ->->->->->->->->->->->-> :
vAve = (2.45m/s + 0) /2 = 1.23m/s
`dv = -2.45m/s
`dt = `dv / a = -2.45m/s / (9.8m/s^2)= .25s
vAve * `dt = 1.23m/s * .25s = .31m
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