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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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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: ->->->->->->->->->->->-> :
Forces = Force throw up (NC), Force gravity down, Force car horizontal.
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The force that threw the ball up happened before the ball was in flight. It had no effect after the instant of release, so should not be listed here.
The car is not in contact with the ball. Earlier the car presumably accelerated, which accelerated the occupant and the ball, but that has already happened and should not be listed.
This leaves only the gravitational force.
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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: ->->->->->->->->->->->-> :
In the 'dy direction, the graph of v vs 'dt starts at a max, decreases at an increasing rate, stops then increases at an increasing rate then a straight line back down to 0 as soon as the child catches the ball.
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Terminology: It's the y direction. `dy stands for the change in the position as measured in the y direction. You would graph velocity vs. t, not vs. a change `dt in clock time.
Acceleration is uniform, so v changes at a constant rate (equal to the acceleration of gravity) throughout.
The analysis ends when the ball touches the child's hand, so whatever happens after that is not part of the interval we are considering.
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The vx vs dt graph is just a horizontal straight line.
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v_x vs t, not v_x vs. `dt.
Using the symbols correctly clarifies our thinking and reduces misconceptions.
You are correct about the nature of the graph.
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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 path of the ball would be a forward moving parabola going up and down in the vertical direction, and moving at a constant rate in the forward direction.
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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 on a bicycle, the ball would decelerate with the child at the same rate.
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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 ball to the child would fall straight down with the motion of the car, it would fall behind the car after it hit the ground. So it would have the same horizontal velocity as the car until it hit the ground.
An observer would see the ball travelling with the car at the same rate of vx until it hit the ground.
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The ball would also travel in an arc, as you described earlier.
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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: ->->->->->->->->->->->-> :
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15 mins
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If the brakes were hit suddenly, the ball would continue to travel forward with the same vox until it hit the back of the front seat or an object in front of the child. Is this a correct statement?
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: ->->->->->->->->->->->-> :
'dsx = vox * 'dt = 10 m/s * 1s = 10 m
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1/2 second covers both the rise and the fall, so the distance is only 5 m.
If the rise alone was 1/2 second the ball would have to go through the roof of the car.
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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 path is a forward moving parabola.
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That is not what a passenger would observe. That's what would be observed from the side of the road.
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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 shape of the path is a concave down parabola.
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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: ->->->->->->->->->->->-> :
vfy = voy*'dt + (1/2*a*'dt^2)
voy = (1/2*a*'dt^2) /'dt = -(1/2 * - 9.8 m/s^2 * 0.5s) = 2.45 m/s
observed by people = v = 'sqrt (vx^2 + vy^2) = 10.3 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: ->->->->->->->->->->->-> :
vavey = (vo + vf) / 2 = 2.45 m/s /2 = 1.225 m/s
'ds = vavey * 'dt = 1.225 m/s * .5 s = 0.61 m
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In some parts of your analysis you are implicitly assuming the (correct) 1/4 second for the rise and 1/4 second for the fall. In some parts you have implicitly assumed 1/2 second for each phase.
Otherwise your analysis is good.
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15 mins
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&#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
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