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course phy 121

9/6 10:50

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?

40m/s)/2= 20m/s

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What is the change in the automobile's velocity?

(25-15)/4=10/4=2.5m/s

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@& 2.5 m/s doesn't have the correct units.

Also this calculation doesn't apply to the question, which is the change in the velocity of the car. Your caclulation would correspond to average rate of change of velocity with respect to clock time.

If you use the right words and do the unit calculations rigorously, you'll quickly sort this out. You're definitely on the right track.*@

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?

automobile

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@& You need to explain your reasoning on questions of this nature.*@

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?

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?

automobile

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@& On the last few questions, you need to provide explanations of your reasoning. I'm not going to say yet whether your conclusions were correct or incorrect.*@

`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)?

KE=(1/2)*1000*15m/s^2

=112,500(m^2/s^2)

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• What is its KE at the end of the 4-second interval?

312,500(m^2/s^2)

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• What is the change in its KE?

200,000(m^2/s^2)

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• What is the net force acting on this object?

2500kg(m/s)=2500N

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@& The units are not kg m / s; the units do come out to be Newtons but the Newton is not the same as the kg m/s.*@

• How much work does this net force do?

10000N

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@& It's not clear how you arrived at that quantity.

However work is not measured in Newtons. You need to show the calculation you made, with units throughout, so we both know what you were thinking. With that information I can provide guidance.*@

• What do you get when you multiply the net force by the time interval?

10000N

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@& Again you haven't explained your calculation. However this time the number is correct, though the unit is not Newtons.*@

`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?

8 sec for 1 5/8 rotations

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• Answer the same for the dominoes halfway to the center.

4 sec for ľ rotation

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• Answer once more for the strap without the dominoes.

7 sec for 2 1/8 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)?

Ends=.2 rot/s

Mid=.19 rot/s

w/o=.3 rot/s

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• For each system, how quickly did the rotational velocity change?

****

Ends=.2 rot/s

Mid=.19 rot/s

w/o=.3 rot/s

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@& Those are the average velocities, not how quickly the velocity changes.

How quickly the velocity changes could be indicated by the rate of change of velocity 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?

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?

tension

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• Is the net force on the more massive car in the upward or downward direction?

downward

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• Is the net force on the less massive car in the upward or downward direction?

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).

F_net_1

T_1

Wt_1

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@& The two tensions are equal. They act at opposite ends of the same stretched string, and no other force acts on any part of the string.

*@

`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?

2000 dynes

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@& That's the net force, not the work.

*@

• By how much does its KE change?

2000 dynes

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@& The result should be equal to your result for work, as it is. However your result for work is incorrect; so this answer would change as well.

*@

• At what rate a is its velocity changing?

•

66.66 dynes/cm

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@& You don't say how you got this, but neither the number nor the unit is correct.
*@

`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 speed accelerated at a greater rate than anticipated

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@& That's a correct and true observation, but an explanation would explain why this occurs.*@

... what if given init vel in opp dir ... ?

It would probobly accelerate slower than anticipated

`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.

4.5 ft

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What therefore would be its average velocity, assuming it was dropped from rest?

9ft/s

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At what aveage rate is its velocity therefore changing?

9ft/s

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@& That's average velocity, not average rate of change of velocity with respect to clock time.

You need to systematically apply the definitions.*@

`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?

V_o to v_f

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What is the run of the trapezoid and what does it mean?

Dt= the total clock time covered

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What is the slope of the trapezoid and what does it mean?

Rise/run=(v_f-v_0)/(t_f-t_0)

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What is the average altitude of the trapezoid and what does it mean?

V_f-v_0)/2= average

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@& When you average two quantities you don't subtract them.

The average of 40 and 60 is 50, not the result 10 that would follow from your expression.*@

What is the area of the trapezoid and what does it mean?

V_f-v_0)/2*dt=the change in velocity for the given time span

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@& vf - v0 is the change in velocity.

(vf - v0) / 2 is not an important quantity; it doesn't give you any significant information about the motion. It's just half the change in velocity, which isn't particularly useful. Multiplied by `dt, the result has no meaning.

*@

`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?

Ok but I think that I need more work on this

****

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@& You need to explain your reasoning on questions of this nature.*@

@& On the last few questions, you need to provide explanations of your reasoning. I'm not going to say yet whether your conclusions were correct or incorrect.*@

@& The units are not kg m / s; the units do come out to be Newtons but the Newton is not the same as the kg m/s.*@

@& It's not clear how you arrived at that quantity. However work is not measured in Newtons. You need to show the calculation you made, with units throughout, so we both know what you were thinking. With that information I can provide guidance.*@

@& Again you haven't explained your calculation. However this time the number is correct, though the unit is not Newtons.*@

@& Those are the average velocities, not how quickly the velocity changes. How quickly the velocity changes could be indicated by the rate of change of velocity with respect to clock time.*@

@& The two tensions are equal. They act at opposite ends of the same stretched string, and no other force acts on any part of the string. *@

@& That's the net force, not the work. *@

@& The result should be equal to your result for work, as it is. However your result for work is incorrect; so this answer would change as well. *@

@& You don't say how you got this, but neither the number nor the unit is correct. *@

@& That's a correct and true observation, but an explanation would explain why this occurs.*@

@& That's average velocity, not average rate of change of velocity with respect to clock time. You need to systematically apply the definitions.*@

@& When you average two quantities you don't subtract them. The average of 40 and 60 is 50, not the result 10 that would follow from your expression.*@

@& vf - v0 is the change in velocity. (vf - v0) / 2 is not an important quantity; it doesn't give you any significant information about the motion. It's just half the change in velocity, which isn't particularly useful. Multiplied by `dt, the result has no meaning. *@

@& Like everyone else in both classes, you do need more work on this. But you're off to a good start. Check out my notes.*@

@& It's not clear how you arrived at that quantity.

However work is not measured in Newtons. You need to show the calculation you made, with units throughout, so we both know what you were thinking. With that information I can provide guidance.*@

@& Those are the average velocities, not how quickly the velocity changes.

How quickly the velocity changes could be indicated by the rate of change of velocity with respect to clock time.*@

@& The two tensions are equal. They act at opposite ends of the same stretched string, and no other force acts on any part of the string.

*@

@& That's the net force, not the work.

*@

@& The result should be equal to your result for work, as it is. However your result for work is incorrect; so this answer would change as well.

*@

@& You don't say how you got this, but neither the number nor the unit is correct.
*@

@& That's average velocity, not average rate of change of velocity with respect to clock time.

You need to systematically apply the definitions.*@

@& When you average two quantities you don't subtract them.

The average of 40 and 60 is 50, not the result 10 that would follow from your expression.*@

@& vf - v0 is the change in velocity.

(vf - v0) / 2 is not an important quantity; it doesn't give you any significant information about the motion. It's just half the change in velocity, which isn't particularly useful. Multiplied by `dt, the result has no meaning.

*@

@& Like everyone else in both classes, you do need more work on this. But you're off to a good start.

Check out my notes.*@