course Phy 121
4/18 2
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:
Gravity will be acting on the ball as it is released. When it is caught, the actual ball itself will be exerting forces.
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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:
I think that the ball would increase its speed when it is released, but as it begins to reach it’s maximum height will slow down and then as it falls back down will once again regain speed. As time passes, a graph that shows speed vs. clock time would start off increases, show a decrease, and then increase before it came to a stop.
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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:
If the car was maintain a relatively constant speed, then the motion of the vehicle should not affect the motion of the ball as it is released. If someone where observing the ball as it was thrown, then it would simply go up, briefly stop(so brief you probably can’t really tell from just watching) and then it would fall straight back 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:
If the child was coasting along on a bicycle and we assume that since the child is coasting the speed would be slowing down that it may be a little bit more difficult to catch the ball since the ball will not move its horizontal direction. It is only being thrown up in the air and will try to fall back down the same path it was thrown. The child may not be able to catch the ball and it would of course fall to the ground.
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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:
I think the ball would of course start off as stationary and then increase its speed as its approaches the ground. I also recall reading in the beginning of the course that if we dropped two objects that were different masses that they would both fall at the same rate, so that makes me wonder if the object maintains a constant speed as it falls.
Good memory. It's the downward acceleration that's constant. The object speeds up, but at a constant rate.
ef
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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:
As the child observes the ball fall, it will look like it is falling down and then as though it moves backwards while the car is moving forwards. I think that an observer watching would see the same kind of motion.
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&#Please compare your 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. &#
course Phy 121
4/18 3
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:
If it requires the ball a ˝ second to rise and fall back to its start point then I would think that it moved a total of 5 meters in the horizontal direction. If the vehicle travels at a constant speed of 10 meters/second then it seems to make sense that the ball would travel 5 meters in half a second.
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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:
I think that if an observer was watching, that they would see the ball move in a simple upward and then immediate downward motion. As the ball was moving upwards though it might slightly move forward in a horizontal manor on then look a little like it is moving in reverse.
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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 path of the ball would be seen moving slightly forward in a horizontal manor and then look as thought it is moving in an arc like manor and then as it comes back down will look as thought it is moving backwards (so it is moving horizontally in reverse) as it falls down. It will not be seen as moving just straight up and down.
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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:
I’m not one hundred percent sure how to determine the speed of the ball as it was moving in a vertical direction. I would assume that we would use the ˝ second as out time interval and that our change in velocity would be 5 m/s. So to find our average acceleration we would take 5 m/s / .5s = 10 m/s^2
Not far off in your general reasoning, but there's no reason to apply a 5 m/s velocity to the vertical motion.
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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:
I wasn’t sure what to use here to find the maximum height that the ball rose to. I had thought we could use ˝ mk but I’m afraid that is not the right equation to use.
You know time interval, final velocity and acceleration for the vertical motion from release to max height. You can either reason or use the equations of uniformly accelerated motion to analyze this interval.
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&#Please compare your 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. &#