assignment97

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course Phy 201

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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(15m/s+25m/s)/2=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 automobile

@& You need to show your reasoning on a problem of this nature, on which there are only a limited number of possible answers.

The same is the case for the next couple of questions.*@

@& You would answer this question most easily by finding the average rate of change of the velocity of each object, with respect to clock time.*@

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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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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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ball

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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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˝*1000kg*15m/s=7500kg m/s

@& The formula is 1/2 m v^2. You have to square that velocity. Likewise on the next question.*@

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

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˝*1000kg*25m/s=12,500kg m/s

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• What is the change in its KE?

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5000kg m/s

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• What is the net force acting on this object?

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Fnet=1,000kg*20m/s Fnet=20,000kg m/s

@& Right idea, but 20 m/s is the average velocity, not the average acceleration. The average acceleration is the average rate of change of velocity with respect to clock time.*@

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• How much work does this net force do?

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12,500kg m/s

@& You don't show how you got this. You need to show the quantities you used in this calculation and how you used them.*@

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

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20,000kg m/s*4s= 80,000kg m

@& If the force was 20 000 kg m/s then this result would be correct. In that sense you're doing the right thing here.

However the net force has different units, and a different number. See my earlier notes.*@

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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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7 seconds rotated 1 and ľ turns (630 degrees)

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

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5 seconds rotated ľ turn (270 degrees)

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• Answer once more for the strap without the dominoes.

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8 seconds 2 and ľ turns (990 degrees)

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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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On ends- 90degrees/second Middle- 54degrees/second None- 123.75degrees/second

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

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Changed most quickly when there were no dominos and least quickly with the dominos in the center.

@& You need to show how you got this result from your data.

The easiest way is to calculate the average rate of change of 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 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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Force exerted by 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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Force exerted by gravity on that car

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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

@& If the downward force of gravity is greater in magnitude than the upward tension force on this car, the net force would be downward. So this answer isn't consistent with your previous answer.

I'll let you decide which answer is correct.*@

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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, F_net_1, Wt_2, wt_1, T2, T1

@& The net force on a car is what you get when you combine a downward gravitataional force (i.e., a weight) with an upward tension. Since the two forces are in opposite directions, the net force will be less in magnitude than the greater of the two forces.

So for example F_net_2 cannot be greater in magnitude than both wt_2 and 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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2000 dynes

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

*@

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• By how much does its KE change?

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It doesn’t change. It would be equal to the net force.

@& Right idea, but you're leaving out some words. The change in KE is not equal to the net force, but to the work done by the net force.*@

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• At what rate a is its velocity changing?

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We are not given the values needed for calculating velocity

@& You aren't asked for velocity, but for the rate at which the velocity is changing.

However I neglected to give you the mass of the car. If you had that you would be able to use it with the net force to find the acceleration, or rate of change of velocity with respect to clock time.*@

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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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Gravitational force pulling it downward

@& If the only force had been the gravitational force on the first car, the instructor would almost certainly have caught it, being used to that acceleration.

The car accelerated at a rate greater than gravitational acceleration because of the tension in the rubber band, which combined with the gravitational force to comprise the net force.*@

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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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5.8 feet

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

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5.8feet/0.5seconds

@& Right. You can finish the calculation, which will give you 10.6 feet / second.*@

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At what aveage rate is its velocity therefore changing?

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5.8 feet/0.5 seconds

@& This is the average velcoity, not the average rate at which the velocity changes.

You need to apply the definition of rate very carefully to this situation.*@

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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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velocity

@& The vertical coordinate of each point is a velocity.

The rise involves both velocities.

What is the expression for the rise, in terms of v0, vf and `dt?*@

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

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Clock time

@& The run would be the change in clock time, not the clock time. No information is given about when the clock started, so you don't know either clock time. But you know how much time elapsed between the two.*@

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

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Slope represents change in velocty

@& You need to give the expression for this quantity in terms of v0, vf and `dt.

Change in velocity is involved in the slope, but there's more to the slope than just change in velocity.*@

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

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Average velocity

@& You need to give the expression for this quantity in terms of v0, vf and `dt.*@

@& You have named is correctly.*@

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

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Average rate of change of velocity with respect to time

@& You need to give the expression for this quantity in terms of v0, vf and `dt.*@

@& If you apply the definition of rate to your response, you'll find that it doesn't correspond to the area.*@

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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 ned to practice. I think I generally understand most of it though.

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assignment97

@& You need to show your reasoning on a problem of this nature, on which there are only a limited number of possible answers. The same is the case for the next couple of questions.*@

@& You would answer this question most easily by finding the average rate of change of the velocity of each object, with respect to clock time.*@

@& The formula is 1/2 m v^2. You have to square that velocity. Likewise on the next question.*@

@& Right idea, but 20 m/s is the average velocity, not the average acceleration. The average acceleration is the average rate of change of velocity with respect to clock time.*@

@& You don't show how you got this. You need to show the quantities you used in this calculation and how you used them.*@

@& If the force was 20 000 kg m/s then this result would be correct. In that sense you're doing the right thing here. However the net force has different units, and a different number. See my earlier notes.*@

@& You need to show how you got this result from your data. The easiest way is to calculate the average rate of change of velocity with respect to clock time.*@

@& If the downward force of gravity is greater in magnitude than the upward tension force on this car, the net force would be downward. So this answer isn't consistent with your previous answer. I'll let you decide which answer is correct.*@

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

@& Right idea, but you're leaving out some words. The change in KE is not equal to the net force, but to the work done by the net force.*@

@& You aren't asked for velocity, but for the rate at which the velocity is changing. However I neglected to give you the mass of the car. If you had that you would be able to use it with the net force to find the acceleration, or rate of change of velocity with respect to clock time.*@

@& Right. You can finish the calculation, which will give you 10.6 feet / second.*@

@& This is the average velcoity, not the average rate at which the velocity changes. You need to apply the definition of rate very carefully to this situation.*@

@& The vertical coordinate of each point is a velocity. The rise involves both velocities. What is the expression for the rise, in terms of v0, vf and `dt?*@

@& The run would be the change in clock time, not the clock time. No information is given about when the clock started, so you don't know either clock time. But you know how much time elapsed between the two.*@

@& You need to give the expression for this quantity in terms of v0, vf and `dt. Change in velocity is involved in the slope, but there's more to the slope than just change in velocity.*@

@& You need to give the expression for this quantity in terms of v0, vf and `dt.*@

@& You have named is correctly.*@

@& You need to give the expression for this quantity in terms of v0, vf and `dt.*@

@& If you apply the definition of rate to your response, you'll find that it doesn't correspond to the area.*@

@& As is the case with everyone else in the class, you're getting some and missing some, are on the right track but need more practice. *@

@& You need to show your reasoning on a problem of this nature, on which there are only a limited number of possible answers.

The same is the case for the next couple of questions.*@

@& If the force was 20 000 kg m/s then this result would be correct. In that sense you're doing the right thing here.

However the net force has different units, and a different number. See my earlier notes.*@

@& You need to show how you got this result from your data.

The easiest way is to calculate the average rate of change of velocity with respect to clock time.*@

@& If the downward force of gravity is greater in magnitude than the upward tension force on this car, the net force would be downward. So this answer isn't consistent with your previous answer.

I'll let you decide which answer is correct.*@

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

*@

@& You aren't asked for velocity, but for the rate at which the velocity is changing.

However I neglected to give you the mass of the car. If you had that you would be able to use it with the net force to find the acceleration, or rate of change of velocity with respect to clock time.*@

@& This is the average velcoity, not the average rate at which the velocity changes.

You need to apply the definition of rate very carefully to this situation.*@

@& The vertical coordinate of each point is a velocity.

The rise involves both velocities.

What is the expression for the rise, in terms of v0, vf and `dt?*@

@& You need to give the expression for this quantity in terms of v0, vf and `dt.

Change in velocity is involved in the slope, but there's more to the slope than just change in velocity.*@

@& As is the case with everyone else in the class, you're getting some and missing some, are on the right track but need more practice.

*@