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course Phy 241
9/6 10:00pm
`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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20m/s
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What is the change in the automobile's velocity?
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10m/s
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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, 8m/s^2 compared with 2.5m/s^2
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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.
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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
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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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112,500 Joules
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What is its KE at the end of the 4-second interval?
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312,500 Joules
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What is the change in its KE?
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200,000 Joules
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What is the net force acting on this object?
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2,500 Newtons
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How much work does this net force do?
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200,000 Joules
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What do you get when you multiply the net force by the time interval?
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The change in momentum
A specific quantitative answer could also be given here; clearly you know how to do that calculation and the units, so don't worry about it.
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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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5 seconds, 1 rotation
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Answer the same for the dominoes halfway to the center.
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3 seconds, .75 rotations
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Answer once more for the strap without the dominoes.
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3 seconds, .625 rotations
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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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.2 rotations per second at ends.
.25 rotations per second in middle.
.208 rotations per second without.
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For each system, how quickly did the rotational velocity change?
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-.08 rotations per second^2
-.17 rotations per second^2
-.67 rotations per second^2
respectively
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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 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, because the car moves in the direction of the force of gravity.
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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, the car moves in the direction of the force of 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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Down
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Is the net force on the less massive car in the upward or downward direction?
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Up
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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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I think I am confusing myself on this one. My intuition is that the order of the forces depends on the ratio between the masses of the two cars, but I really can't make heads or tails of this one. I might have to revisit it.
The magnitudes of accelerations are the same for both cars, so the magnitude of the net force is greater on the more massive car.
The tensions T_1 an T_2 are equal. The more massive car accelerates downward so this tension is less than the gravitational force, and similarly is seen to be greater than the gravitational force on the lesser mass.
Simple enough, I suspect, when you see it, but not easy to come up with this early in the course.
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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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60,000 ergs
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By how much does its KE change?
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by 60,000 ergs
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At what rate a is its velocity changing?
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Lets say the car is 5 grams. 400 cm/s^2
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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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The bottom car exerted a force on the top car.
The rubber band was already under tension, which was due to the weight of the lower car, but it's the tension and not the weight of that car that adds to the acceleration of the top car. That tension is equal to the weight of the lower car only at the initial instant. The rubber band immediately begins to shorten, which decreases that tension.
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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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About 1.75 meters.
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What therefore would be its average velocity, assuming it was dropped from rest?
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3.5 meters/second.
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At what aveage rate is its velocity therefore changing?
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14 meters/second^2
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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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The change in velocity.
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What is the run of the trapezoid and what does it mean?
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The change in time.
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What is the slope of the trapezoid and what does it mean?
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Average acceleration
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What is the average altitude of the trapezoid and what does it mean?
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The average velocity.
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What is the area of the trapezoid and what does it mean?
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The displacement.
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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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I am somewhat familiar with them. More understanding is needed though. For now I am fine.
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