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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: ->->->->->->->->->->->-> scussion:
Once the ball has been released, only gravity is acting on the ball.
The forward motion of the car has set the ball moving at a certain velocity forward, and the ball will still be moving at that velocity, but the force causing that motion is no longer acting on the ball.
The upward motion of the toss has set the ball moving at a certain velocity upward, but that force is no longer acting on the ball once it leaves the child's hand. Instead, gravity causes the ball to accelerate downward, so the initial velocity will decrease as the ball travels upward, slowing down to a moment of rest, and then continue to decrease as the ball falls (speeding up in the negative direction).
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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: ->->->->->->->->->->->-> scussion:
Whatever initial velocity the force of the toss causes, that velocity will only be accurate for an instant as it begins to travel upward. Gravity will immediately begin to accelerate the ball in the downward direction, so the the velocity will decrease until it reaches zero, at which point the ball will come to rest for an instant, and then the velocity will continue to decrease as the ball begins to fall. (The speed will be increasing, in the negative direction, downward.)
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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: ->->->->->->->->->->->-> scussion:
The initial horizontal velocity, established just by moving along at that velocity with the car, will not change if no force such as air resistance causing it to change. So the ball will continue to move forward at the same velocity as the car. Meanwhile, it will rise, come to rest, and then fall back to its starting point. So its path will resemble a concave-downward arc or curve, beginning and ending at the same height. It should a symmetrical curve, as the rate at which the ball slows down on its way up is the same as the rate at which is speeds up on its way down.
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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: ->->->->->->->->->->->-> scussion:
I'm not sure the path would differ if the child was coasting on a bike. The forward velocity should still be consistent unless some other force interferes and causes it to change, so the same conditions would apply: consistent horizontal velocity, initial vertical force of throw (upward) before the interval we are considering, consistent vertical acceleration caused by gravity (downward).
If the child doesn't catch the ball, it would continue to arc downward at the same rate, but experience a negative total vertical displacement.
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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: ->->->->->->->->->->->-> scussion:
When dropped from the inside roof of the car, the ball is subject to acceleration downward due to gravity, as well as the same horizontal constant velocity (established by being in a car moving at that velocity). So the speed will increase due to gravity. The acceleraiton will be constant 980cm/s.
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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: ->->->->->->->->->->->-> scussion:
This should be the same as the previous (dropped from ceiling inside car). The horizontal velocity should be constant, and the vertical motion will be subject to gravity. We will again see a concave downward arc toward the ground, with equal horizontal change per second and increasing vertical change per second.
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20 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: ->->->->->->->->->->->-> scussion:
10m/s * .5s = 5m
The ball travels 5m horizontally.
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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: ->->->->->->->->->->->-> scussion:
From inside the car, the ball would appear to travel straight up and straight 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: ->->->->->->->->->->->-> scussion:
From outside the car, the ball would appear to travel in a concave downward arc (with a very wise base-- it's traveling a much greater distance horizontally than it is vertically).
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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: ->->->->->->->->->->->-> scussion:
I believe that we can assume the upward path takes .25s and the downward path takes .25s. I am pretty sure that since acceleration is constant, we can assume the vertical motion is symmetrical; that each component will take equivalent time and velocity will change in equal magnitude.
So, if that's true, then for the path upward:
a=`dv/`dt
-9.8m/s^2=`dv/.25s
-2.45m/s=`dv
vf=0, so v0=2.45m/s
Observed along the side of the road, the ball would simultaneously have a velocity of 2.45m/s in the vertical direction and 10m/s in the horizontal direction. Using the pythagorean theorem that gives us:
2.45^2+10^=v^2
6.0025+100=v^2
v=about 10.3m/s at an angle
To find the angle:
theta=arctan(2.45/10)
theta= about 13.8deg above the horizontal axis
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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: ->->->->->->->->->->->-> scussion:
vAve=`ds/`dt
(2.45/2)=`ds/.25
`ds(vertical)=about .3m
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*#&!
&#Very good responses. Let me know if you have questions. &#