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course Phy 201
1:43 am , Sept 7
`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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delta position / delta time
10 m / 4 s = 5 m / 2 s = 2.5 m/s
@& The change in the velocity is 10 m/s.
If you divide this by 4 s you get 2.5 m/s^2, not 2.5 m/s.
That is a correct sequence of calculations, if you correct the units, and it's an answer to one of these questions. Not this question, but one that's coming up soon.
The problem here is that 10 m is not a change in position associated with this interval. 10 m/s is the change in velocity, but the change in position is not 10 m.
This comes down to a careful and methodical application of the definition of rate, carefully identifying all the quantities.
This won't give you much trouble, once your see what's needed.*@
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What is the change in the automobile's velocity?
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the first instant: velocity=1.875m/s
the midpoint: velocity=2.5m/s (average)
the second instant: velocity=3.125m/s
velocity is changing by +1.25m/4s or +0.3125m/s
@& Given your assumptions, that's a good breakdown. You have the right idea, and an excellent understanding of the numbers and the linearity of the graph..
However 2.5 m/s is not the average velocity.*@
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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 (comparing the ball's velocity increasing by 2m/s to the automobile's velocity increasing by 0.3125m/s)
@& Given your assumptions that would be a valid and insightful comparison.
However you'll want to find the average rate of change of velocity for the two.*@
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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 automobile during it's 4-second interval (the automobile traveled an average of 20m/s for the 4 seconds equaling a total of 80 meters and the ball traveled a total of1 meter in 1/4 of a second)
@& Good, but the ball would have averaged 2 m/s and traveled 1/2 meter.*@
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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 automobile (covering 20 meters compaired to the 16 that they ball would travel)
@& The ball would speed up to 32 m/s, averaging 16 m/s. In 4 s it would travel 64 meters. The automobile would travel 80 m, as you noted earlier.
As is natural at this point, I believe you are occasionally confusing the various types of rates. With a little more practice you'll sort it all out.*@
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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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KE = 1/2 m v^2
KE = 1/2 (1000 kg) (1.875m/s)^2
KE = 1757.8125 (kg * m^2) / s^2
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What is its KE at the end of the 4-second interval?
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KE = 1/2 m * v^2
KE = 1/2 (1000 kg) * (3.125m/s)^2
KE = 4882.8125 (kg * m^2) / s^2
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What is the change in its KE?
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[4882.8125 (kg * m^2) / s^2] - [1757.8125 (kg * m^2) / s^2] = 3125 (kg * m^2) / s^2
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@& You're using the formula correctly and getting the right units. However the initial speed is 15 m/s and the final is 25 m/s, as given initially.*@
What is the net force acting on this object?
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Note: is the change in KE the same thing as ""a""
F_net = 1000 kg * 3125 (kg * m^2) / s^2
F_net = 3125000 kg^2 * m^2 / s^2
Note: or is the change of velocity the same thing as ""a""
F_net = 1000 kg * 1.25 m/s
F_net = 1250 kg * m / s
@& a is the acceleration, which is the rate of change of velocity with respect to clock time. Its units are m/s^2, not m/s.
You actually calculated a in the very first question, though you misidentified it and didn't quite get the units right.
The correct result here is 1000 kg * 2.5 m/s^2 = 2500 kg m/s^2.*@
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How much work does this net force do?
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'dw = F * 'ds
assuming the second of the two above is correct.
'dw = 1250 kg * m /s * 'ds
@& `ds is the displacement of the automobile, which you previously calculated as 80 meters.*@
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What do you get when you multiply the net force by the time interval?
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F_net * time interval
1250 kg * m / s * 4 s = 5000 kg * m
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`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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Rotated for approximately 1.5 rotations lasting for about 6 seconds
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Answer the same for the dominoes halfway to the center.
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made it 1 full rotation in 5 seconds
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Answer once more for the strap without the dominoes.
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only made it 1/2 of a rotation and lasted 2 seconds
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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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Dominoes on the end: 1.5 rotations / 6 sec = .25 rotations / 1 sec
Dominoes in the middle: 1 rotation / 5 seconds = .2 rotations / 1 sec
No dominoes: 1/2 rotation / 2 sec = .25 rotations / 1 sec
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For each system, how quickly did the rotational velocity change?
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not enough data was taken for these calculations but there was always a trend:
from rest, they would always speed up, and then quickly slow back down
@& You can answer this question by finding the average rate at which the rotational velocity of the strap changed with respect to clock time.*@
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`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 and 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 force exerted by gravity on that car
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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
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Is the net force on the more massive car in the upward or downward direction?
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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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F_net_2, wt_2, T_1, F_net_1, wt_1, T_2
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`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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'dW = F * `ds
'dw = 2000 dynes * 30 cm
'dw = 60000 dynes*cm
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By how much does its KE change? ********
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KE = 1/2 m * v^2
KE = 1/2 *
well what's the mass????
@& You don't know the mass or the velocity, so this formula won't help. But the KE change is related to the work done by the net force (check out the given information once more).*@
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At what rate a is its velocity changing? *******
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0.015 cm / dyne
@& In this case you can't answer the question; I failed to give you enough information. If you knew the mass then, given the net force, you would be able to figure out the acceleration.*@
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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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there was an inconsistency of tension in the rubberband
@& Just the presence of tension on the rubber band would provide more force than gravity. I'm attuned to the acceleration of gravity; when an additional force changes that acceleration I either adjust or miss. Having not anticipated the change I didn't even attempt to adjust, with the result you witnessed.*@
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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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5ft 11in (71 inches)
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What therefore would be its average velocity, assuming it was dropped from rest?
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Velocity = 71 in / .5 sec = 142 in / sec
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At what average rate is its velocity therefore changing?
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4 in / sec
@& Here's where you need to carefully apply the definition of average rate.
This average rate won't have units of in / sec. That would be correct for an average rate of change of position with respect to clock time, but that's not the rate requested here.*@
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`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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Altitude: position
@& The graph shows velocity vs. clock time, not position vs. clock time.*@
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What is the run of the trapezoid and what does it mean?
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Width: time
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What is the slope of the trapezoid and what does it mean?
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rate of change of in y with respect to x, acceleration
@& The y axis represents the velocity, the x axis the clock time. So it's rate of change of velocity with respect to clock time.
That is acceleration, but it's important to reason that out in detail.*@
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What is the average altitude of the trapezoid and what does it mean?
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an average of the y's
@& That's right, but interpretation requires an additional step.
What do the y's represent, and what therefore does the average of the y's represent?*@
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What is the area of the trapezoid and what does it mean?
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average altitude * width, velocity
@& What does ave altitude represent, what does width represent, and what does the product of these quantities represent?*@
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`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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They're really complicated yet simple at the same time. I think with more practice I will of course get better but for now I'm struggling slightly with those 6 ""definitions""
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@& You're not really struggling, you're into the sorting process. You've got all the ideas, and just need to practice applying them carfefully and systematically.*@
@& You're on track to do very well.*@