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course Phy 201
10:15am Sept 12Thanks Mr. Smith!
I think I'm starting to get used to the idea of submitting my work online and going back to learn from my mistakes.
I was wondering if there was any way to check our grade for this class... really, just how we're being graded :) making an A in this class is a must for me.
@& You can check the grading criteria in the course of study. www.vhmthphy.vhcc.edu > Physics I > Info then click on the course of study for your course.
That's the official grading policy. What follows here is not official but is consistent with that policy, and pretty much restates it:
Bascially there are four tests and the labs, which comprise most of your grade. The homework and class participation count as 10% or 15%, and you pretty much get full credit on that if you make the appropriate effort.
You have to have a passing lab average to pass the course.
Tests can be retaken. You can't take the same test, but you can take another random version of the test.*@
" `q001. An automobile is traveling at 15 m/s (V1)at one instant, and 4 seconds later it is traveling at 25 m/s(V2), then:
What is the average velocity of the automobile, assuming that its velocity changes at a constant rate?
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average velocity = (15 m/s + 25 m/s) / 2 = 20 m/s
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What is the change in the automobile's velocity?
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25 m/s - 15 m/s = 10 m/s in 4 seconds
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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, because goes from 1 m/s to 32 m/s in 4s where as the automobile goes from 15 m/s to 25 m/s
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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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1 m/s to 3 m/s in 1/4 s = 2 meters in 1/4 s
15 m/s to 25 m/s = 10 meters in 1/4 s
so the car traveled further
@& 1 m/s to 3 m/s implies aveage velocity 2 m/s. In 1/4 s this implies displacement 1/2 meter.
15 m/s to 25 m/s implies average velocity 20 m/s. In 1/4 s the car would travel 5 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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1 m/s ro 3 m/s in 1/4 s = 32 m in 4 s
15 m/s to 25 m in 4 s = 10 meters in 4 s
so the ball traveled further
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@& 1 m/s to 3 m/s in 1/4 sec implies a pickup of 2 m/s in 1/4 sec, or 8 m/s per second. So in 4 seconds the ball would speed up by 32 m/s, giving it a final velocity of 33 m/s, an average velocity os 17 m/s and a displacement of 68 m.
15 m/s to 25 m/s in 4 s implies average velocity 20 m/s, for a displacement of 80 m.*@
`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) (15m/s)^2
KE = 500 kg (225 m^2/s^2)
KE = 112500 (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) * (25m/s)^2
KE = 500 kg (625 m^2/s^2)
KE = 4882.8125 (kg * m^2) / s^2
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@& The preceding step is right, but that calculation doesn't equal the result you get here. You get something over 300 000 kg m^2 / s^2.*@
What is the change in its KE?
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delta KE = K_f - K_o
delta KE = 312500 kgm^2/s^2 - 112500 kgm^2/s^2
delta KE = 200000 kgm^2/s^2
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@& Right.*@
What is the net force acting on this object?
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F_net = m * a
F_net = 1000 kg * 2.5 m/s^2
F_net = 2500 kgm/s^2
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How much work does this net force do?
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'dw = F * 'ds
'dW = 2500 kgm/s^2 * 10 m
'dW = 25000 kgm^2/s^2
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@& The car moves further than 10 m. Check previous notes.
Otherwise you're doing the right thing here.*@
What do you get when you multiply the net force by the time interval?
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F_net * time interval
25000 kgm^2/s^2 * 4 s = 1000 kgm/s
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@& Almost, but your numbers aren't right and you have a unit error.
You get
2500 kg m^2 / s^2 * 4 s = 10 000 kg m^2 / s.*@
`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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@& You just calculated the rotational velocities.
You need to find the average rate of change of rotational velocity 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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@& Your ordering isn't correct, but this is something we'll be discussing in class.*@
`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 *
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@& Change in KE is equal to the work done by the net force, which you calculated previously.
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At what rate a is its velocity changing? *******
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0.015 cm / dyne
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@& You don't have enough information to answer this. My fault.
*@
`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 causing you to misinterpret the force of the falling cars
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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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??
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@& You need to apply the definition of rate of change. Identify the required A and B quantities, etc..*@
`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: velocity
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@& There are two altitudes, representing two velocities.
The rise is not generally equal to either of the altitudes. What is the rise, how is it related to the altitudes, and what does it mean?*@
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
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y-axis = the velocity
x-axis = clock time
rate of change of velocity with respect to clock time
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an average of the y's
an average of the velocities
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What is the area of the trapezoid and what does it mean?
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average altitude (velocity) * width (time interval) = acceleration
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@& average velocity * time interval is not acceleration.
acceleration is rate of change of velocity with respect to clock time, which involves a division.
What do you get if you multiply average velocity by time interval? How is this related to the definition of velocity as a rate of change?*@
`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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&#Your work looks good. See my notes. Let me know if you have any questions. &#