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Phy 231
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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Copy the problem below into a text editor or word processor.
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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 of gravity, and the force of the cars slow movement in the horizontal direction
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@& There is no force associated with movement, only with acceleration. Forward inertia takes care of the forward movement.*@
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 of the ball slows as it reaches its highest point, at the highest point the speed is 0, then accelerates as it comes back down.
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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 be traveling in the upward direction predominately, and also the ball would be moving in the horizontal direction since the car is moving in that direction at a slow pace.
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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 still be up, but the horizontal direction would be opposite the direction of the bicycle. The bicycle could out run the ball falling down, so to speak. Depending how fast the kid is going on the bicycle if he didnt bother to catch it the ball could go over his head and land behind the bicycle, or the ball could just hit him in the head or face before the ball hit the ground.
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@& If the bicycle doesn't speed up, and if air resistance isn't significant, the ball would continue moving in the horizontal direction with unchanged speed and would stay in the same horizontal position with respect to the child.*@
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: ->->->->->->->->->->->-> :
As the ball drops from the roof to the floor the speed will increase, the acceleration will be the force of gravity, 9.8m/s^2.
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@& The acceleration of gravity is g = 9.8 m/s^2. The force of gravity on mass m is m g.*@
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: ->->->->->->->->->->->-> :
The ball will travel diagonally downward opposite the direction the vehicle is going. If the ball is dense the ball will not move as far in the horizontal direction, it will drop faster in the vertical direction. The child would see the ball travel horizontally at a diagonal downward. An observer would see the ball drop downward with little movement in the horizontal direction opposite the cars direction.
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20min
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Copy the problem below into a text editor or word processor.
This form accepts only text so a text editor such as Notepad is fine.
You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.
If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.
As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
________________________________________
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: ->->->->->->->->->->->-> :
10m/s * .5s= 5m
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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: ->->->->->->->->->->->-> :
up, and back 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: ->->->->->->->->->->->-> :
The ball is going up and moving in the horizontal direction of the car. A diagonal line
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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: ->->->->->->->->->->->-> :
.5s * 9.8m/s^2= 4.9m/s=dv,
The people along the side of the road would see the velocity as the speed of the car, and the 4.9m/s upward?
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@& 4.9 m/s is the change in the ball's vertical velocity.
The vertical velocity is continuously changing.*@
How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> :
.5s * 4.9m/s= 2.45m, 2.45m/2=1.225m
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@& 4.9 m/s is the change in the ball's vertical velocity, not its average vertical velocity.*@
*#&!
@& I've inserted a few notes. Be sure to check the links. You have a few misconceptions here, and it's best to straighten them out now rather than later.
Let me know if you have questions.*@
@&
&#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. &#
&#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. &#
*@