Mastering Physics - Mek2 - Assignment 1

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Assignment 1 Due: 11:59pm on Friday, November 25, 2011 Note: To understand how points are awarded, read your instructor's Grading Policy. [Switch to Standard Assignment View]

Block on an Incline A block lies on a plane raised an angle , the force of gravity;

from the horizontal. Three forces act upon the block:

, the normal force; and

, the force of friction. The coefficient of friction

is large enough to prevent the block from sliding .

Part A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? ANSWER:

only only only and and and and

and

Correct

Part B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical?

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ANSWER:

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only only only and and and and

and

Correct

Now you are going to ignore the general rule (actually, a strong suggestion) that you should pick the coordinate system with the most vectors, especially unknown ones, along the coordinate axes. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude

appearing in both the x and y equations, each multiplied by a

trigonometric function. Part C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Hint C.1 Find the y component of

Hint not displayed

Hint C.2 Find the y component of

Hint not displayed

Express your answer in terms of some or all of the variables

,

,

, and .

ANSWER:

Correct

Part D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b.

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Hint D.1 Find the x component of

Hint not displayed

Express your answer in terms of some or all of the variables

,

,

, and .

ANSWER:

Correct

Part E To find the magnitude of the normal force, you must express

in terms of

since

is an

unknown. Using the equations you found in the two previous parts, find an expression for involving

and

but not

.

Hint E.1 How to approach the problem Hint not displayed

ANSWER: =

Correct

Congratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the y-coordinate equation is , which leads immediately to the result obtained here for

.

CONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra.

The Center of Mass of the Earth-Moon-Sun System A common, though incorrect, statement is, "The Moon orbits the Earth." That creates an image of the Moon’s orbit that looks like that shown in the figure. The Earth's gravity pulls on the Moon, causing it to orbit. However, by Newton’s third law, it is known that the Moon exerts a force back on the Earth. Therefore, the Earth should move in response to the Moon. Thus a more accurate statement is, "The Moon and the Earth both orbit the center of mass of the Earth-Moon system."

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In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35×1022 , the mass of the Earth is 6.00×1024 , and the mass of the sun is 2.00×1030 . The distance between the Moon and the Earth is 3.80×105 1.50×108

. The distance between the Earth and the Sun is

.

Part A Calculate the location

of the center of mass of the Earth-Moon system. Use a coordinate

system in which the center of the Earth is at

and the Moon is located in the positive x

direction. Hint A.1 Calculating the center of mass Hint not displayed

Hint A.2 Find the coordinates of the Earth and Moon Hint not displayed

Express your answer in kilometers to three significant figures. ANSWER:

4600 Correct

Part B Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378

and the radius of the Moon is 1737

. Select one of the answers

below:

Choose the correct description of the location of the center of mass of the Earth-Moon system. ANSWER:

The center of mass is exactly in the center between the Earth and the Moon. The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon. The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth. The center of mass is inside the Earth. The center of mass is inside the Moon.

Correct

As you can see, the center of mass for the Earth-Moon system actually lies within the radius of the Earth. For this reason, saying that the Moon orbits the Earth is often a good approximation, though in fact, both the Earth and the Moon orbit that point with a period of

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28 days. The Moon makes large orbits around the center of mass of the Earth-Moon system, whereas the center of the Earth makes small orbits.

Part C Calculate the location of the center of mass of the Earth-Moon-Sun system during a full Moon. A full Moon occurs when the Earth, Moon, and Sun are lined up as shown in the figure. Use a coordinate system in which the center of the sun is at and the Earth and Moon both lie along the positive x direction.

Hint C.1 Calculating the center of mass Hint not displayed

Express your answer in kilometers to three significant figures. ANSWER:

=

456 Correct

The equatorial radius of the Sun is 695,000

. As you can see, the center of mass for

the Sun-Earth-Moon system is well within the Sun. However, if you were to find the center of mass of the Jupiter-Sun system, you would find that it is slightly above the surface of the Sun at 780,000 from the center of the Sun. A distant alien civilization would not be able to see Jupiter directly, because it is far too faint, but they would be able to see the Sun move back and forth as it orbited the center of mass with Jupiter. Because the sun is "wobbling," alien scientists would be able to infer that there was a planet around the Sun.

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This is one of the methods that human scientists are using to identify planets around other stars.

Surprising Exploding Firework A mortar fires a shell of mass

at speed

. The shell explodes at the top of its trajectory

(shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass

and a larger piece of mass

. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance mortar. If there had been no explosion, the shell would have landed a distance

from the from the mortar.

Assume that air resistance and the mass of the shell's explosive charge are negligible.

Part A Find the distance

from the mortar at which the larger piece of the shell lands.

Hint A.1 Find the position of the center of mass in terms of Hint not displayed

Hint A.2 Find the position of the center of mass in terms of Hint not displayed

Express ANSWER:

in terms of

.

=

Correct

Rocket Car

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A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be considered, so the acceleration of the car must not exceed (or five times the acceleration of gravity) during the test. Using the latest materials and technology, the total mass of the car (including the fuel) is 6000 kilograms, and the mass of the fuel is one-third of the total mass of the car (i.e., 2000 killograms). The car is moved to the starting line (and left at rest), at which time the rocket is ignited. The rocket fuel is expelled at a constant speed of 900 meters per second relative to the car, and is burned at a constant rate until used up, which takes only 15 seconds. Ignore all effects of friction in this problem. Part A Find the acceleration

of the car just after the rocket is ignited.

Hint A.1 How to approach the problem Hint not displayed

Hint A.2 find the rate of mass change Hint not displayed

Express your answer to two significant figures. ANSWER:

=

20 Correct

The driver of this car is experiencing just over

, or two times the acceleration one

normally feels due to gravity, at the start of the trip. This is not much different from the acceleration typically experienced by thrill seekers on a roller coaster, so the driver is in no danger on this score.

Part B Find the final acceleration

of the car as the rocket is just about to use up its fuel supply.

Hint B.1 What has changed? Hint not displayed

Hint B.2 Find the final mass Hint not displayed

Express your answer to two significant figures. ANSWER:

=

30 Correct

The driver of this car is experiencing just over

, or three times the acceleration one

normally feels due to gravity, by the end of the trip. This is the maximum acceleration achieved during the trip, and it is still very safe for the driver, who can easily withstand

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over

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with training.

Part C Find the final velocity

of the car just as the rocket is about to use up its fuel supply.

Hint C.1 Find the change in speed Hint not displayed

Express your answer to two significant figures. ANSWER:

= 360 Correct

At the end of the trip, the driver is going a bit over Mach 1, or one times the speed of sound. This problem was based loosely on the breaking of the sound barrier by the ThrustSSC team in October 1997.

Problem 8.111: A Multistage Rocket Suppose the first stage of a two-stage rocket has total mass 1.20×104 is fuel. The total mass of the second stage is 900 relative speed

, of which 700

, of which 1.00×104 is fuel. Assume that the

of ejected material is constant, and ignore any effect of gravity. (The effect of

gravity is small during the firing period if the rate of fuel consumption is large.) Part A Suppose the entire fuel supply carried by the two-stage rocket is utilized in a single-stage rocket with the same total mass of 1.29×104 . In terms of , what is the speed of the rocket, starting from rest, when its fuel is exhausted? ANSWER:

=

1.77 Correct

Part B For the two-stage rocket, what is the speed when the fuel of the first stage is exhausted if the first stage carries the second stage with it to this point? This speed then becomes the initial speed of the second stage. At this point, the second stage separates from the first stage. ANSWER:

=

1.49 Correct

Part C What is the final speed of the second stage?

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ANSWER:

=

3.00 Correct

Part D What value of

is required to give the second stage of the rocket a speed of 8.00

ANSWER:

?

= 2.67 Correct

Pushing a Lawnmower

Consider a lawnmower of weight

which can slide across a horizontal surface with a coefficient of friction

. In this problem the lawnmower is pushed using a massless handle, which makes an angle horizontal. Assume that

with the

, the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward.

Part A Find the magnitude,

, of the force required to slide the lawnmower over the ground at

constant speed by pushing the handle. Hint A.1 How to approach this problem Hint not displayed

Hint A.2 Compute the sum of vertical forces Hint not displayed

Hint A.3 Compute the normal force

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Hint not displayed

Hint A.4 Compute the sum of horizontal forces Hint not displayed

Express the required force in terms of given quantities. ANSWER: =

Correct

Part B The solution for

has a singularity (that is, becomes infinitely large) at a certain angle

angle

, the expression for

will be negative. However, a negative applied force

. For any would

reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all. Find an expression for

.

Hint B.1 How to approach the problem Hint not displayed

ANSWER: =

Correct

You should have found that

, the force required to push the lawnmower at constant speed, was .

Note that this expression becomes infinite when the denominator equals zero: ,

or .

(The phrase " angle

has a singularity at angle

" means that "

goes to infinity at a certain

.")

It's not too hard to understand what this means. Suppose you were pushing straight down on the lawnmower ( degrees). It obviously wouldn't move. But, according to the equation for ,

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when you plug in

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degrees, you get a negative force (which doesn't make sense).

The more vertical you push, the harder it gets to move the lawnmower. At

, it gets

impossible to move it. The force required to move it goes to infinity; you have to push infinitely hard.

Score Summary: Your score on this assignment is 109.2%. You received 15.49 out of a possible total of 16 points, plus 1.98 points of extra credit.

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