#$&*
Phy 201
Your 'cq_1_18.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
cq_1_181
#$&*
Phy 201
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.
** **
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : Gravitational. Because the car is not speeding up or slowing down or changing direction, it would be as though
he were doing this inside
his home.
#$&*
What happens to the speed of the ball between release and catch? Describe in some detail; a graph of speed vs. clock time would also be appropriate.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : It's going faster when first release but decelerates (slows down) until it stops mid air then speeds up as it's
going back down.
#$&*
Describe the path of the ball as it would be observed by someone standing along the side of the road.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : They would see this scenario play out as it gets closer to them then farther away.
#$&*
@&
From their perspective the ball would be moving forward at the same time it rises and falls.
What would its path look like?
*@
How would the path differ if the child was coasting along on a bicycle? What if the kid didn't bother to catch the ball? (You know nothing about what happens after
the ball makes contact with the ground, so there's no point in addressing anything that might happen after that point).
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : It would be the same as in the car as long is there was no change in velocity or direction of the bicyle.
#$&*
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 acceleration would be 9.8m/s^2, or it would increase in speed until it hits the floor.
#$&*
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : It would continue to travel in the horizontal direction in the opposite the direction of the car. It the car was
movig forward at 15mph,
then the ball would move backward or -15mph while accelerating vertically at 9.8m/s^2 toward the ground. To the child, the ball would appear to be moving away from him
@&
In the absence of air resistance, what force would cause the ball to slow down from its 15 mph speed? If there is no such force, then what would the child see?
*@
&&&& The ball would go in the same direction as the car, but it would would lose speed on each subsequent bounce until in stopped. I just tried this running with a
basketball in the house. However, I remember as a child seeing things go backward if someone threw something out the car window, and that case maybe they accelerated
after that point.&&&&
@&
The object doesn't really go backward, it slows down rapidly due to air restance while the car keeps moving at the same speed. The object falls behind the car because it moves progressively slower than the car. However it does continue to move forward, and if you watch the object after it hits the pavement you will see it bounding toward you (though falling further and further behind due to your continued velocity).
The is certainly an acceleration involved. The force of air resistance results in a horizontal acceleration equal to that force divided by the mass of the object. Had the object been a massive and dense rock it wouldn't have slowed much, and would not have appeared to move backwards as quickly. as, say, a wad of paper.
So:
Relative to the car the object moves away in the direction opposite the car's motion. This is what you see if you are riding in the car.
Relative to the road the object continues moving forward, but more and more slowly. This is what the observer by the side of the road sees.
*@
&&&&For the question regarding the observer, the observer would see the ball go in the direction of the movement of the car while moving up and down. Could you give me
an example of how I would describe a path?&&&&
@&
I'll give you the shape of the ideal path, but first you need to give me your description.
The object moves forward, ideally (in the absence of air resistance) at a constant horizontal velocity.
It moves upward but more and more slowly before coming to rest (in the upward direction only) for an instant, before falling back down, gaining speed as it falls.
Does this result in straight-lne or curved motion? If the motion is curved, how is it curved?
*@
@&
Suppose the observer on the side of the road was close to the car, so that he was in the path of the ball as the child held it before dropping it. Suppose the child
dropped the ball when only a few feet from the observer. Would the ball suddenly reverse its direction of motion and move away from the observer? What would actually
happen?
*@
&&&&
or behind him while descending toward the ground at the same time.
#$&*
** **
10-15 min
** **
1:57 pm EST 7/6/14
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*0.5s = 5 meters.
#$&*
As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : It simply went out of the child's hand into the air and back down into their hand.
#$&*
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : I don't really understand this question. They would see it as the passenger would see it but from different
angle or perspective. It would
happen so fast, so I don't think the ball would appear to have a drastically different path. Not sure what path is here: horizontal down the road, vertically in the
car or both. I can picture the event in my mind but not sure what is being asked.
@&
The observers don't necessarily even know there's a child and a car. They are concentrating on the ball, which as you have said moves forward during the toss. They
would not see the ball just moving up and down.
*@
#$&*
How fast was the ball moving in the vertical direction at the instant of release? At that instant, what is its velocity as observed by a line of people standing along
the side of the road?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : I started out by decreasing the time interval by half, because I'm only looking at the time from the ball's
release to its highest point
(vf = 0m/s). So solving for v0, I get vf - at or 0m/s - (-9.8m/s^2)(0.25s) = 2.5m/s.
#$&*
How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : 'ds = 2.45m/s*0.25s + 1/2(-9.8m/s^2)(0.25s) = 0.306 m
#$&*
*#&!
@&
Check my notes.
I still want to see your description of the motion of the ball as seen by the observer, particularly the shape of its path, before I give you a completely accurate description of the path.
Use #### to mark insertions.
&#Please see my notes and submit a copy of this document with revisions, comments 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. &#
*@
*#&!*#&!