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: ->->->->->->->->->->->-> :
The force used to propel the ball into the air and gravity both work on the ball. Also, the force of the car moving forward will be working on the ball, so that it continues to move forward.
For the given interval no force is exerted on the ball by the child or the car.
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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 ball will begin to accelerate in the negative direction once it is released. Then it will reach 0 m/s when gravity causes the velocity to reach 0. Then after the split second it is not moving, the ball will begin to accelerate, this time in the positive direction. The velocity will continue to increase until the ball is caught. The graph will look like a concave up parabola with the vertex at the time when the velocity equals 0m/s.
The direction of the ball's acceleration never changes. The direction of its velocity changes, in response to the constant downward acceleration.
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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 motion of the ball will look like a parabola. The ball leaves the child’s hand and goes up until it stops. Then the ball will come back down in the child’s hand. If the car were not moving the ball would go straight up and down, but since the vehicle was moving, the force is also transferred to the ball and the ball does not have a straight path.
There is no force on the ball associated with the constant-velocity motion of the vehicle.
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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 should be the same if the child is on a bicycle. If the kid did not catch the ball, it would fall to the ground. The velocity would be greater when it hit because it had a greater distance and time to pick up speed. The ball would hit at the same distance from the starting point as the child at that time because the force from the bike is still working on the ball as it falls.
The bike exerts no force as the ball falls. There is no significant force on the ball in the forward direction so its horizontal acceleration is zero (ignoring air resistance). Not only does it not take a force for the ball to continue moving forward at constant velocity, if there was a net force on the ball in the forward direction, it would not move with constant velocity in that direction.
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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 will fall in a straight line from the top of the roof to the floor. The landing point will be directly below the point from which it was dropped. The speed of the ball will increase in the downward direction. The acceleration is the acceleration due to gravity, 9.8m/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: ->->->->->->->->->->->-> :
The ball will fall straight down from the child’s hand when it is released as seen when an outside observer watches it. When the child watches it, the ball will fall straight down also.
The observer will see the ball drop, but will also see the ball move forward at the speed of the car. If the car and the child were suddenly rendered invisible, the observer would see it move in the same familiar manner as a ball projected forward at a velocity matching that of the car.
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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: ->->->->->->->->->->->-> :
The ball will travel 10m/s * .5s = 5 meters in the horizontal 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: ->->->->->->->->->->->-> :
When observed by a passenger, the ball will seem to rise straight into the air and then fall straight back down into the child’s hand. This is because the observer and the ball are moving forward at the same speed.
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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: ->->->->->->->->->->->-> :
When seen by observers out of the car, the ball will follow a parabolic path. It will rise into the air and curve forward, staying above the child’s hand. The ball will stop for a moment in the air (this is where thr slope of the graph is 0) and then begin accelerating down until it reaches the hand, 5 meters past the starting point.
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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: ->->->->->->->->->->->-> :
V0+0m/s / 2 (.25s) = ds
Ds = v0(.25s)+.5(-9.8m/s^2)(.25s ^2)
V0 / 2 (.25s) = .25s v0 - .30625 m
.125 s v0 = .25s v0 - .30625 m
-.125s v0 = -.30625
V0 = 2.45 m/s
As seen by the observers the velocity is 10m/s in the horizontal direction and 2.45m/s in the vertical direction. The magnitude of that vector is:
Sqrt((10m/s)^2 + (2.45m/s)^2)
Sqrt(100m^2/s^2+6.00 m^2/s^2)
Initial velocity = 10.3m/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: ->->->->->->->->->->->-> :
0m/s ^2 = (2.45m/s)^2 +2(-9.8m/s^2)(ds)
-6.0025 m^2/s^2 = -19.6 m/s^2 ds
Ds = .306 meters into the air
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30 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: ->->->->->->->->->->->-> :
The ball will travel 10m/s * .5s = 5 meters in the horizontal direction
#$&*
• As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
answer/question/discussion: ->->->->->->->->->->->-> :
When observed by a passenger, the ball will seem to rise straight into the air and then fall straight back down into the child’s hand. This is because the observer and the ball are moving forward at the same speed.
#$&*
• 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: ->->->->->->->->->->->-> :
When seen by observers out of the car, the ball will follow a parabolic path. It will rise into the air and curve forward, staying above the child’s hand. The ball will stop for a moment in the air (this is where thr slope of the graph is 0) and then begin accelerating down until it reaches the hand, 5 meters past the starting point.
#$&*
• How fast was the ball moving in the vertical direction at the instant of release? At that instant, what is its velocity as observed by a line of people standing along the side of the road?
answer/question/discussion: ->->->->->->->->->->->-> :
V0+0m/s / 2 (.25s) = ds
Ds = v0(.25s)+.5(-9.8m/s^2)(.25s ^2)
V0 / 2 (.25s) = .25s v0 - .30625 m
.125 s v0 = .25s v0 - .30625 m
-.125s v0 = -.30625
V0 = 2.45 m/s
As seen by the observers the velocity is 10m/s in the horizontal direction and 2.45m/s in the vertical direction. The magnitude of that vector is:
Sqrt((10m/s)^2 + (2.45m/s)^2)
Sqrt(100m^2/s^2+6.00 m^2/s^2)
Initial velocity = 10.3m/s
#$&*
• How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> :
0m/s ^2 = (2.45m/s)^2 +2(-9.8m/s^2)(ds)
-6.0025 m^2/s^2 = -19.6 m/s^2 ds
Ds = .306 meters into the air
#$&*
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30 minutes
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No problem, but in the future it's best to submit only one 'seed' question at a time, using the form for that question.
I've inserted a number of notes on the first question. I'm asking for revisions on that question only. The second was correct throughout, but there are some conceptual issues to straighten out on the first.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#