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course Phy 241
9/06 11
`q001. An automobile is traveling at 15 m/s at one instant, and 4 seconds later it is traveling at 25 m/s, then:What is the average velocity of the automobile, assuming that its velocity changes at a constant rate?
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25 m/s - 15 m/s = 10 m/s divided by 4 s = 2.5 m/s^2
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What is the change in the automobile's velocity?
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it changes 2.5 m/s every 4 seconds
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@& It changes by 2.5 m/s every second. That's what 2.5 m/s^2 means.*@
A ball is dropped in the automobile, and its velocity it observed to change by 2 meters / second in 1/4 of a second.
Which is speeding up more quickly, the ball or the automobile?
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the ball
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@& You need to explain at least briefly why you think it is so.*@
If the ball's speed was 1 meter / second at the beginning of its 1/4-second interval, which traveled further, the automobile during its 4-second interval or the ball during its 1/4-second interval?
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the car which traveled 10 m
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If the ball kept speeding up at the same rate for 4 seconds, which would travel further during the 4-second interval?
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the ball which traveled 16 m
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`q002. When an object of mass m is moving with velocity v, it has the following properties
its kinetic energy is KE = 1/2 m v^2
its momentum is p = m v
Forces acting on objects can change their velocity, momentum and kinetic energy.
When an object of mass m changes its velocity, with respect to clock time, at rate a, then the net force acting on it (i.e., the sum of all the forces acting on it) is F_net = m * a.
If a force F acts through a displacement `ds along the line of the force, then the force does work `dW = F * `ds.
If F happens to be the net force acting on an object, then the KE of that object changes by an amount equal to `dW.
If a net force F_net acts on an object for time interval `dt, then the momentum of that object changes by `dp = F_net * `dt.
We will see later where these definitions come from and what they are good for.
Now, the automobile in the preceding has a mass of 1000 kg.
At the beginning of the 4-second interval, what is its kinetic energy (hereafter abbreviated KE)?
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v = 2.5 m/s so KE = (1/2)1,000 kg * (2.5 m/s)^2 =
500 kg * 6.25 m/s = 3125 kg*m/s
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@& (2.5 m/s)^2 = 6.25 m^2 / s^2, not 6.25 m/s.
In any case v isn't 2.5 m/s.
2.5 m/s^2 is the acceleration. It's not a velocity.*@
What is its KE at the end of the 4-second interval?
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3125 kg*m/s * 4 = 12500 kg*m/s
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What is the change in its KE?
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12,500 kg * m/s - 3125 kg*m/s = 9375 kg*m/s
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@& You need to rethink your KE values for this situaiton.*@
@& I don't think you'll have any trouble, once you identify the two correct velocities.*@
What is the net force acting on this object?
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F_net = m * a
F_net = 9375 kg*m/s * 4s = 37,500 kg*m
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@& 9375 kg m/s is not a mass, and 4 s is not an acceleration.*@
How much work does this net force do?
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work = force * distance traveled so:
37,500kg*m (10 m ) = 375,000 kg*m^2
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@& If you correct the force you'll get the right answer.*@
What do you get when you multiply the net force by the time interval?
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momentum
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@& You can get a quantitative answer here.*@
`q003. Give your results for the experiment with the rotating strap and the dominoes, as indicated below.
When the dominoes were on the ends of the strap, how long did it take the system to come to rest and how far did it rotate?
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took about 3 seconds and 2 1/2 rotations
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Answer the same for the dominoes halfway to the center.
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2 1/2 seconds and 2 rotations
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Answer once more for the strap without the dominoes.
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1 second, 1/2 rotation
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For each system, what was the average rotational velocity (i.e., the average amount of rotation per unit of time)?
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ominoes at end: 2.5 rot / 3s = .833 rotations per second
dominoes middle: 2 rot / 2.5s = .8 rotations per second
no dominoes: .5 rot / 1s = .5 rotations per second
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For each system, how quickly did the rotational velocity change?
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further out the dominoes, the faster the rotational velcity was.
??????????????? not really sure what you want here, change between the systems? or change in general? because i dont know if there was enough information to find the average change in rotational velcity for our given information
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@& How quickly a quantity changes can be answered by calculating the rate of change of that quantity with respect to clock time.*@
`q004. For the cars suspended on opposite sides of the pulley (we call this sort of system an Atwood Machine), four different forces are involved. Gravity pulls down on the more massive car, gravity pulls down on the less massive car, the tension on one end of the string pulls up on the more massive car, and the tension on the other end of the string pulls up on the less massive car. If the pulley is light an frictionless, which is the case here, the tension in the string is the same throughout.
What is greater in magnitude, the tension acting on the more massive car or the force exerted by gravity on that car?
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the gravity pulling on the more massive car will ""outweigh"" the gravity pulling on the less massive car, causing the more massive car to fall to the floor or table first, suspending the lighter car in mid air
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@& Neither car is in contact with the other. Both are in contact with the string.
So the net force on each is the resultant of the string tension and the gravitational force on that car.*@
What is greater in magnitude, the tension acting on the less massive car or the force exerted by gravity on that car?
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the tension is greater acting on the less massive car because the gravitational force is greater on the opposite side of the atwood machine because the other car has more mass.
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@& Not bad. Technically the tension is less in magnitude than the weight because the acceleration of the car is in the upward direction.*@
Is the net force on the more massive car in the upward or downward direction?
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net force is more massive downward
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Is the net force on the less massive car in the upward or downward direction?
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upward
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Place in order the magnitudes of the following forces: the net force F_net_1 on the less massive car, the net force F_net_2 on the more massive car, the tension T_1 acting on the less massive car, the tension T_2 acting on the more massive car, the force wt_2 exerted by gravity on the more massive car and the force wt_2 exerted by gravity on the more massive car (wt stands for weight).
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smallest first: wt_1 less massive: F_net_1 less massive car; T_1 less massive car: T_2 more massive; F_net_2 more massive: wt_2 more massive
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@& Except for the net forces, these are in the right order.
If the acceleration is less than half that of gravity (less than about 5 m/s^2), the net forces will both be less in magnitude than any of the other forces.*@
`q005. If a net force of 2000 dynes acts on a toy car through a distance of 30 cm in the direction of the force, then
How much work is done on the car?
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2,000 dynecs*30 cm so 60,000 dynecs*cm
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By how much does its KE change?
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If a force F acts through a displacement `ds along the line of the force, then the force does work `dW = F * `ds.
If F happens to be the net force acting on an object, then the KE of that object changes by an amount equal to `dW.
this being said:
2,000dynecs / 30cm = 66.7 dynecs/cm
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@& F * `ds specifies multiplication, not division.*@
At what rate a is its velocity changing?
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66.7 dynecs/cm
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`q006. Explain why, when the two cars connected by the rubber band chain were dropped, the instructor failed to catch the car as intended. Avoid any reference to the instructor's coordination, reflexes or mental state.
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because the rubber band was pulling down on the higher car, and pulling up on the lower car, causing the higher car to fall at a rate of ""gravity * mass + (force of rubber band chain)"" which was much faster than gravity itself
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... what if given init vel in opp dir ... ?
`q007. It's fairly easy to establish that an object dropped from the instructor's chest height will fall freely to the floor in about 1/2 second.
Estimate how far the object would fall.
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being that the instructor is about 6' tall, the object should fall from chest height which should be roughly 18"" shorter than height which is 4' 6'' per half second
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What therefore would be its average velocity, assuming it was dropped from rest?
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4.5 ft / 1/2 second or 9 ft / second
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At what average rate is its velocity therefore changing?
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9 ft/sec
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@& That's the average velocity, which is the average rate at which position is changing.
You need to apply the definition of average rate of change.*@
`q008. A trapezoid on a graph of velocity v vs. clock time t has altitudes v_0 and v_f. Its width is `dt.
What is the rise of the trapezoid and what does it mean?
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rise is v_f - v_0, this means total distance traveled
@& The rise is vf - v0, as you say.
However vf and v0 are not distances, so vf - v0 isn't a distance.*@
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What is the run of the trapezoid and what does it mean?
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the run is the time, how long it took to get there, or lenght
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What is the slope of the trapezoid and what does it mean?
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rate of change, or velocity
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@& You need to reason this out from the meanings of the two quantities. You do get a rate of change. You have to specify this as a rate of change of something with respect to something else, using the defintion of rate of change.*@
What is the average altitude of the trapezoid and what does it mean?
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displacement. how far it went up
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@& The vertical axis does not measure displacement.
*@
What is the area of the trapezoid and what does it mean?
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area determines the average velocity
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@& This must be reasoned out in terms of the definitions of the quantities used to get the area, and the definition of average rate.
*@
`q009. At the beginning of the second question you were given six bits of information. You are going to need to use this information over and over. You would do well to memorize those six things, though a word-for-word repetition is not necessary. You will probably do so spontaneously as you use them over and over again to understand the behaviors of different systems.
How are you doing with these ideas?
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these past few ideas are a bit sketchy. determining the average rate in change in velocity for some reason is hard for me to get. also the 'dt for work part,
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*#&!*#&!
@& I've inserted a number of notes. Be sure to consider them carefully before reworking this document.*@