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Phys310 Wk2

In: Computers and Technology

Submitted By jaamal78
Words 8307
Pages 34
Assignment Week 2
Due: 11:59pm on Sunday, January 18, 2015
You will receive no credit for items you complete after the assignment is due. Grading Policy

A message from your instructor...
The next seven problems are for practice.

Item 1
Learning Goal:
To understand that adding vectors by using geometry and by using components gives the same result, and that manipulating vectors with components is much easier.
Vectors may be manipulated either geometrically or using components. In this problem we consider the addition of two vectors using both of these two methods.
The vectors and have lengths and , respectively, and makes an angle from the direction of .








Vector addition using geometry
Vector addition using geometry is accomplished by putting the tail of one vector (in this case ) on the tip of the other ( ) and using the laws of plane geometry to find








:

,



1.



the length , and angle , of the resultant (or sum) vector,

2.

Vector addition using components
Vector addition using components requires the choice of a coordinate system. In this problem, the x axis is chosen along the direction of . Then the x and y


components of are

and

respectively. This means that the x and y components of are given by



3.
4.

,
.

Part A
Which of the following sets of conditions, if true, would show that the expressions 1 and 2 above define the same vector as expressions 3 and 4?


Check all that apply.
ANSWER:
The two pairs of expressions give the same length and direction for .


The two pairs of expressions give the same length and x component for .


The two pairs of expressions give the same direction and x component for .


The two pairs of expressions give the same length and y component for .


The two pairs of expressions give the same direction and y component for .


The two pairs of expressions give the same x and y components for .


All attempts used; correct answer displayed and

) define the same vector you can show that any of the sets of conditions listed


To show that the two pairs of expressions (for and and for above are met except
They give the same length and x component for .


They give the same length and y component for .


If you consider just the first set of conditions, showing that the two sets of expressions have the same length and x component will imply that the y component has the correct magnitude. However, there is no way to know the sign (i.e., direction) of the y component. To show that the pairs of expressions given above define the same vector , we would need to show that they give the same length and the same x and y components. We will do this in the questions that follow.


Part B
We begin by investigating whether the lengths are the same.
Find the length of the vector starting from the components given in Equations 3 and 4.


Express in terms of , , and .

Hint 1. Apply the Pythagorean theorem
You are given the x and y components in Equations 3 and 4. Simply square these, add them, and take the square root (i.e., apply the Pythagorean theorem) to find the length of vector .
ANSWER:



=

Answer Requested

Part C
The following reasons might explain why the equation for the length just obtained using components is the same as the answer obtained using geometry (Equation 1 above): 1. and are supplementary angles, that is,
.
2. The cosine function satisfies
.
3. Cosine is an even function of its argument, so the extra negative sign in one expression does not matter.

?





Which of these reasons is/are necessary to show that
ANSWER:
1 only
2 only
3 only
1 and 2
2 and 3
1 and 3
1 and 2 and 3

Correct

Having shown that the length of obtained using either geometrical addition OR componentwise addition are the same, all that remains is to show that any one of the


other conditions from part A is satisfied. In the last two parts of the problem, we'll show that the y component of determined geometrically is equal to that given above.


Part D
We begin by finding the y component of from its length and the angle it makes with the x axis, that is, from geometry.


Express the y component of in terms of and .


ANSWER: =

Answer Requested

Part E
Now express the y component of just found by using the geometrical approach in terms of rather than .


Express

in terms of , and .

Hint 1. Law of sines
The law of sines can be used to relate
Express

in terms of

and

.

, , and .

ANSWER: =

Hint 2. Supplementary angles
Angles and are supplementary, which means that

. What is the relationship of

and

?

ANSWER: =

Answer Requested
This is the same as Equation 4 above, obtained using components. Thinking in terms of components, the result is fairly obvious: The y component of is equal


to the the y component of since has no y component in the chosen coordinate system.




At this point you have actually shown that the two vectors are equal by showing that the overall lengths are equal, and also that the y components are equal. You would only have to argue that the x component is positive to complete the proof of equality.
Nevertheless, we are asking you to complete the algebra to show that the x components are equal as well.

Part F
Now find the x component of .


Express your answer in terms of and only.
ANSWER:
=

Answer Requested

Item 2
Learning Goal:
To practice Tactics Box 3.1 Determining the Components of a Vector. and




When a vector is decomposed into component vectors

parallel to the coordinate axes, we can describe each component vector with a single number (a



scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted


TACTICS BOX 3.1 Determining the components of a vector

is the magnitude of the component vector

2. The sign of is positive if points in the positive x direction; it is negative if
3. The y component is determined similarly.


Part A


What is the magnitude of the component vector

shown in the figure?

Express your answer in meters to one significant figure.

ANSWER: = 5



of the x component



1. The absolute value

.

points in the negative x direction.

and

.

Answer Requested

Part B of vector shown in the figure?


What is the sign of the y component

ANSWER: positive negative

Correct

Part C

Express your answers, separated by a comma, in meters to one significant figure.

and

, of vector shown in the figure.


Now, combine the information given in the tactics box above to find the x and y components,

ANSWER:
,

= ­2,­5

,

Answer Requested

Item 3
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.

Part A

Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win?
Ignore air resistance.
Select all that apply.

Hint 1. How to approach the problem
Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track.
ANSWER:
A. Increase the tilt of the yellow track.
B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same.
C. Increase the mass of the ball on the yellow track.
D. Decrease the mass of the ball on the red track.

Correct
If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt?

Item 4
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components.


,


Part A
ANSWER:
= 3,­5,­4




Answer Requested

, and


Let vectors

. Calculate the following, and express your answers as ordered triplets of values separated by commas.

Part B
ANSWER:
= ­5,4,0




Correct

Part C
ANSWER:
= ­6,4,3






Correct

Part D
ANSWER:
= ­3,­2,­11




Correct

Part E
ANSWER:






Correct

= ­11,14,8

Part F
ANSWER:
= 17,­12,­6






Correct

Item 5
Learning Goal:
To practice Problem­Solving Strategy 4.1 for projectile motion problems.
A rock thrown with speed 12.0 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 19.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free­fall acceleration.
PROBLEM­SOLVING STRATEGY 4.1 Projectile motion problems
MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance?
VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch.

Define symbols, and identify what you are trying to find.
SOLVE: The acceleration is known:

and

. Thus, the problem becomes one of two­dimensional kinematics. The kinematic equations are
,

is the same for the horizontal and vertical components of the motion. Find

. from one component, and then use that value for the other component.

ASSESS: Check that your result has the correct units, is reasonable, and answers the question.

Model
Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly.

Visualize

Part A
Which diagram represents an accurate sketch of the rock's trajectory?

Hint 1. The launch angle

ANSWER:



In a projectile's motion, the angle of the initial velocity

above the horizontal is called the launch angle.

Correct

Part B

As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile's acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right.
Where you choose your origin doesn't change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest?
ANSWER:

Solve

At ground level below the point where the rock is launched
At the point where the rock is released
At ground level below the peak of the trajectory
At the peak of the trajectory
At the point where the rock strikes the ground

Correct
It's best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it's generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system.
Now, define symbols representing initial and final position, velocity, and time. Your target variable is representation should be complete now, and similar to the picture below:

, the initial y coordinate of the rock. Your pictorial

Part C
Find the height

from which the rock was launched.

Express your answer in meters to three significant figures.

Hint 1. How to approach the problem
The time needed to move horizontally to the final position = 19.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for .

Hint 2. Find the time spent in the air
How long (

) is the rock in the air?

Express your answer in seconds to three significant figures.

Hint 1. Determine which equation to use
Which of the equations given in the strategy and shown below is the most appropriate to calculate the time
ANSWER:

Hint 2. Find the x component of the initial velocity
What is the x component of the rock's initial velocity?
Express your answer in meters per second to three significant figures.
ANSWER:
= 10.4

ANSWER:

the rock spent in the air?

= 1.83

Hint 3. Find the y component of the initial velocity
What is the y component of the rock's initial velocity?
Express your answer in meters per second to three significant figures.
ANSWER:
= 6.00

ANSWER: = 5.43

Answer Requested

Assess
Part D
A second rock is thrown straight upward with a speed 6.000

. If this rock takes 1.828 to fall to the ground, from what height

was it released?

Express your answer in meters to three significant figures.

Hint 1. Identify the known variables
What are the values of rock lands.

,

,

, and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the

Express your answers to four significant figures in the units shown to the right, separated by commas.
ANSWER:
,

,

, = 0,6.000,1.828,­9.810

,

, ,

Hint 2. Determine which equation to use to find the height

Which equation should you use to find

? Keep in mind that if the positive y axis is upward and the origin is located on the ground,

.

ANSWER:

ANSWER: = 5.43

Answer Requested
Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free­fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 6.000 , and fell the same vertical distance of 5.43 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly.

Item 6
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.

Part A

Which ball (if either) has the greatest speed at the moment of impact?

Hint 1. How to approach this problem
The balls strike the ground at the same time because the horizontal and vertical components of their velocities are independent, meaning that the two balls fall downward at the same rate even though one ball is also moving horizontally.
However, the speed of a ball at any point is the magnitude of the ball's resultant (total) velocity. How does the ball's resultant velocity relate to the vertical and horizontal velocity components?
If you need to, draw diagrams showing the velocities of the two balls at the moment of impact. For each ball, show the vertical and horizontal components of velocity (the latter being zero for the dropped ball), and show the resultant velocity. Which ball is moving faster?
ANSWER:
The ball thrown horizontally
The dropped ball
Both balls have the same speed.

Correct
The two balls have the same vertical velocity when they land, but the thrown ball has an additional horizontal velocity component. Since speed is defined as the magnitude of the resultant velocity vector, the thrown ball is moving faster when it lands.

Item 7
A cannon is fired from the top of a cliff as shown in the figure. Ignore drag (air friction) for this question. Take

as the height of the cliff.

Part A
Which of the paths would the cannonball most likely follow if the cannon barrel is horizontal?

Hint 1. Find the y position as a function of time
Obtain a function for the y position of this cannonball as a function of the time after it is fired.
Use for the magnitude of the acceleration due to gravity, and take

at the base of the cliff, with positive upward.

ANSWER: =

Hint 2. Interpreting your equation
Look at the formula you obtained for
. At way has it fallen down toward the ground?
ANSWER:

the projectile has been flying for 1/2 of the total flight time. According to your formula, what fraction of the

hardly at all
1/4 of the way down
1/2 of the way down
3/4 of the way down.

ANSWER:
A
B
C
D

Correct

Part B
Now the cannon is pointed straight up and fired. (This procedure is not recommended!) Under the conditions already stated (drag is to be ignored) which of the following correctly describes the acceleration of the ball?
ANSWER:
A steadily increasing downward acceleration from the moment the cannonball leaves the cannon barrel until it reaches its highest point
A steadily decreasing upward acceleration from the moment the cannonball leaves the cannon barrel until it reaches its highest point
A constant upward acceleration
A constant downward acceleration

Correct
The acceleration of the cannonball after it is fired is the constant acceleration due to gravity.

A message from your instructor...
The five problems below are for credit and will count towards your homework grade

Item 8

Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system.


Part A
Find the components of the vector with length = 1.00 and angle =20.0 with respect to the x axis as shown.


Enter the x component followed by the y component, separated by a comma.

Hint 1. What is the x component?
Look at the figure shown. points in the positive x direction, so

is positive. Also, the magnitude

is just the length

.



ANSWER: = 0.940,0.342


Correct

Part B
Find the components of the vector with length = 1.00 and angle =20.0 with respect to the x axis as shown.


Enter the x component followed by the y component, separated by a comma.

Hint 1. What is the x component?
The x component is still of the same form, that is,
ANSWER:
= 0.940,0.342

.



Correct
The components of still have the same form, that is,

, despite 's placement with respect to the y axis on the drawing.




Part C
Find the components of the vector with length = 1.00 and angle

25.0 as shown.



Enter the x component followed by the y component, separated by a comma.

Hint 1. Method 1: Find the angle that makes with the positive x axis


Angle = 0.436 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis?


Express your answer numerically in degrees.
ANSWER:
115

Hint 2. Method 2: Use vector addition
Look at the figure shown.
.
.




3.
, the x component of is negative, since
Use this information to find
. Similarly, find .


ANSWER:





2.



1.

points in the negative x direction.

= ­0.423,0.906


Correct

Item 9
Learning Goal:
To understand the basic concepts of projectile motion.
Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one­dimensional motions that you have already studied.
One of the most often used techniques in physics is to divide two­ and three­dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component .


Consider a particle with initial velocity that has magnitude 12.0


Part A of ?


What is the x component

Express your answer in meters per second.
ANSWER:
= ­6.00

Correct

Part B of ?


What is the y component

Express your answer in meters per second.
ANSWER:

and is directed 60.0

above the negative x axis.

= 10.4

Correct

Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not.

Part C
Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x­component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle's shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components?
ANSWER:
Both the vertical and horizontal components exhibit motion with constant nonzero acceleration.
The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant­velocity motion.
The vertical component exhibits constant­velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration.
Both the vertical and horizontal components exhibit motion with constant velocity.

Correct
As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude
. Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation
. Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation
.

Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0

.

Part D
How long

does it take for the balls to reach the ground? Use 10

Express your answer in seconds to two significant figures.

for the magnitude of the acceleration due to gravity.

Hint 1. How to approach the problem
The balls are released from rest at a height of 5.0 takes for the balls to reach the ground.

at time

. Using these numbers and basic kinematics, you can determine the amount of time it

ANSWER: = 1.0

Correct
This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero.

Part E
Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart.
Express your answer in meters per second to two significant figures.

Hint 1. How to approach the problem
Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time
, the ball will move horizontally 3.0 . You can assume that its initial x coordinate is
.
ANSWER: = 3.0

Correct
You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time).

Item 10

Learning Goal:
Understand that the acceleration vector is in the direction of the change of the velocity vector.
In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity.
When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x­y plane), the definition of acceleration is in the limit

.







In picturing this vector derivative you can think of the derivative of a vector as an instantaneous quantity by thinking of the velocity of the tip of the arrow as the vector changes in time. Alternatively, you can (for small ) approximate the acceleration as


.







Obviously the difference between and is another vector that can lie in any direction. If it is longer but in the same direction, will be parallel to
.
On the other hand, if has the same magnitude as but is in a slightly different direction, then will be perpendicular to . In general, can differ from in both magnitude and direction, hence can have any direction relative to
.
























This problem contains several examples of this.Consider an object sliding on a frictionless ramp as depicted here. The object is already moving along the ramp toward position 2 when it is at position 1. The following questions concern the direction of the object's acceleration vector, . In this problem, you should find the direction of the acceleration vector by drawing the velocity vector at two points near to the position you are asked about. Note that since the object moves along the track, its velocity vector at a point will be tangent to the track at that point. The acceleration vector will point in the same direction as the vector difference of the two velocities. (This is a result of the equation given above.)










Part A
Which direction best approximates the direction of when the object is at position 1?


Hint 1. Consider the change in velocity
At this point, the object's velocity vector is not changing direction; rather, it is increasing in magnitude. Therefore, the object's acceleration is nearly parallel to its velocity.

ANSWER: straight up downward to the left downward to the right straight down

Correct

Part B
Which direction best approximates the direction of when the object is at position 2?


Hint 1. Consider the change in velocity
At this point, the speed has a local maximum; thus the magnitude of is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of is changing there is an acceleration.




ANSWER: straight up upward to the right straight down downward to the left

Correct
Even though the acceleration is directed straight up, this does not mean that the object is moving straight up.

Part C
Which direction best approximates the direction of when the object is at position 3?


Hint 1. Consider the change in velocity
At this point, the speed has a local minimum; thus the magnitude of is not changing. Therefore, no component of the acceleration vector is parallel to the


velocity vector. However, since the direction of is changing there is an acceleration.


ANSWER: upward to the right to the right straight down downward to the right

Correct

Item 11
A man out walking his dog makes one complete pass around a perfectly square city block. He starts at point A and walks clockwise around the block.

be the displacement vector from A to B,




Let

Part A


Which of the following vectors is equal to

?

be the displacement vector from B to C, etc.

Hint 1. Determining a vector
Recall that is a vector representing the displacement of the man and his dog as they walk from point A to point B. This vector has a magnitude equal to one block and a direction along the positive x axis.


Hint 2. Equal vectors
Two vectors are equal if they have the same magnitude and the same direction.
ANSWER:



only


only


only

All of the above
None of the above

Correct
Recall that, for vectors to be equal, they must have the same magnitude and direction.

Part B

ANSWER:



only


only


only

All of the above
None of the above

Correct



Which of the following vectors is equal to

?

Part C




Which of the following vectors is equal to

?

Hint 1. Determining the difference of two vectors
. Thus




to the vector pointing opposite to





can be determined by adding the vector

looks like this:





Carefully perform the vector addition in each of the options and compare the resultant vectors to the one shown above.
ANSWER:
only






only only




All of the above
None of the above

Correct

Item 12
The treasure map in the figure gives the following directions to the buried treasure: "Start at the old oak tree, walk due north for 500 paces, then due east for 110 paces.

Dig." But when you arrive, you find an angry dragon just north of the tree. To avoid the dragon, you set off along the yellow brick road at an angle walking 370 paces you see an opening through the woods.
You may want to review (

pages 70 ­ 74) .

For help with math skills, you may want to review:
Vector Addition
For general problem­solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Vectors.

Part A
Which direction should you go to reach the treasure?
ANSWER:
= 33.7 west of north

Correct

Part B
How far should you go to reach the treasure?
ANSWER:
379 paces

Correct

Item 13

east of north. After

A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the other hits ship B, which is farther away.
The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is .
Note that after Part B the question setup changes slightly.

Part A
What shape is the trajectory (graph of y vs. x) of the shells?
ANSWER:
straight line parabola hyperbola
The shape cannot be determined.

Correct

Part B
For two shells fired at the same speed which statement about the horizontal distance traveled is correct?

Hint 1. Two things to consider
The distance traveled is the product of the x component of the velocity and the time in the air. How does the y component of the velocity affect the "air time"?
What angle would give the longest range?
ANSWER:
The shell fired at a larger angle with respect to the horizontal lands farther away.
The shell fired at an angle closest to 45 degrees lands farther away.
The shell fired at a smaller angle with respect to the horizontal lands farther away.
The lighter shell lands farther away.

Correct

Now, consider for the remaining parts of the question below that both shells are fired at an angle greater than 45 degrees with respect to the horizontal. Remember that enemy ship A is closer than enemy ship B.

Part C
Which shell is fired at the larger angle?

Hint 1. Consider the limiting case
Consider the case in which a shell is fired at 90 degrees above the horizontal (i.e., straight up). What distance will the shell travel? Now lower the angle at which the shell is fired. What happens to the distance ?
ANSWER:
A
B
Both shells are fired at the same angle.

Correct

Part D
Which shell is launched with a greater vertical velocity,

?

ANSWER:
A
B
Both shells are launched with the same vertical velocity.

Correct

Part E
Which shell is launched with a greater horizontal velocity,
ANSWER:

?

A
B
Both shells are launched with the same horizontal velocity.

Correct

Part F
Which shell reaches the greater maximum height?

Hint 1. What determines maximum height?
What determines the maximum height reached by the shell?
ANSWER:
horizontal velocity vertical velocity mass of the shell

ANSWER:
A
B
Both shells reach the same maximum height.

Correct

Part G
Which shell has the longest travel time (time elapsed between being fired and hitting the enemy ship)?

Hint 1. Consider the limiting case

If a shell is fired exactly horizontally (0 degrees) the shell hits the ground right away. As the angle above the horizontal increases, what happens to the time of travel? Does this change as the angle becomes greater than 45 degrees?
ANSWER:
A
B
Both shells have the same travel time.

Correct

Item 14
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 200 above the glacier at a speed of 160
You may want to review (

pages 91 ­ 95) .

For help with math skills, you may want to review:
Mathematical Expressions Involving Squares
For general problem­solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Dock jumping.

Part A
How far short of the target should it drop the package?
Express your answer using three significant figures with the appropriate units.
ANSWER:
1020

Correct

Item 15

.

An old­fashioned single­play vinyl record rotates on a turntable at 45.0

. What are (a) the angular velocity in rad/s and (b) the period of the motion?

Part A
Express your answer with the appropriate units.
ANSWER:
4.71

Correct

Part B
Express your answer with the appropriate units.
ANSWER:
1.33

Correct

Item 16
Your roommate is working on his bicycle and has the bike upside down. He spins the 68.0 three times every second.

Part A
What is the pebble's speed?
Express your answer with the appropriate units.
ANSWER:
6.41

­diameter wheel, and you notice that a pebble stuck in the tread goes by

Correct

Part B
What is the pebble's acceleration?
Express your answer with the appropriate units.
ANSWER:
121

Correct

Item 17
Learning Goal:
To understand how the trajectory of an object depends on its initial velocity, and to understand how air resistance affects the trajectory.
For this problem, use the PhET simulation Projectile Motion. This simulation allows you to fire an object from a cannon, see its trajectory, and measure its range and hang time (the amount of time in the air).

Start the simulation. Press Fire to launch an object. You can choose the object by clicking on one of the objects in the scroll­down menu at top right (a cannonball is not

among the choices). To adjust the cannon barrel’s angle, click and drag on it or type in a numerical value (in degrees). You can also adjust the speed, mass, and diameter of the object by typing in values. Clicking Air Resistance displays settings for (1) the drag coefficient and (2) the altitude (which controls the air density). For this tutorial, we will use an altitude of zero (sea level) and let the drag coefficient be automatically set when the object is chosen.
Play around with the simulation. When you are done, click Erase and select a baseball prior to beginning Part A. Leave Air Resistance unchecked.

Part A
First, you will investigate purely vertical motion. The kinematics equation for vertical motion (ignoring air resistance) is given by
,
where is the initial position (which is 1.2 above the ground due to the wheels of the cannon), is the initial speed, and is the acceleration due to gravity.
Shoot the baseball straight upward (at an angle of

) with an initial speed of 20

.

How long does it take for the baseball to hit the ground?
Express your answer with the appropriate units.
ANSWER:
4.10

Correct (keep in mind that the location where
. This calculation is interesting because

Notice that this value could be determined from the kinematics equation: Approximating the final height to be is inside the cannon, not the ground), the kinematics equation becomes
, or it shows that, for vertical motion, the time the ball is in the air is proportional to its initial speed.

Part B
When the baseball is shot straight upward with an initial speed of 20
, what is the maximum height above its initial location? (Note that the ball’s initial height is denoted by the horizontal white line. It is initially 1.2 above the ground. The yellow box that is below the target on the grass is measuring tape that should be used for this part.)
Express your answer to three significant figures and include the appropriate units.

Hint 1. How to approach the problem
Use the measuring tape to determine the height. Align the plus sign at the beginning of the spool with the horizontal white line, and drag the end of the tape to the maximum height of the ball’s trajectory. You can zoom in or out using the magnification buttons above the Fire and Erase buttons.
ANSWER:

20.4

Correct
Notice that this value could be determined from the kinematics equation: Since you found it takes 4.1 for the ball to reach the ground, it must take roughly
2.1 to reach the maximum height, which is given by (Since 20.4 is considerably longer than the initial height above the ground, it is a good approximation to pretend that the initial position of the baseball was the ground.)

Part C
If the initial speed of the ball is doubled, how does the maximum height change?

Hint 1. How to approach the problem
You can use the simulation to help with this question. Set the initial speed to 40 will likely have to zoom out to see the top of the ball’s trajectory.

, and again measure the maximum height with the measuring tape. You

ANSWER:
The maximum height increases by a factor of two.
The maximum height increases by a factor of 1.4 (square root of 2).
The maximum height increases by a factor of four.

Correct
Since the amount of time it takes to reach the maximum height doubles, and since its average velocity in going upward also doubles (the average velocity is equal to half the initial velocity), the height it reaches before stopping increases by a factor of four (distance is equal to the average velocity multiplied by the time duration). Part D
Erase all the trajectories, and fire the ball vertically again with an initial speed of 20
. As you found earlier, the maximum height is roughly 20 . If the ball isn’t fired vertically, but at an angle less than
, it can reach the same maximum height if its initial speed is faster. Set the initial speed to 25 , and find the angle such that the maximum height is roughly 20 . Experiment by firing the ball with many different angles. You can use the measuring tape to determine the maximum height of the trajectory and compare it to 20 .
What is this angle?
ANSWER:

Correct
Notice that the initial speed in the vertical direction is given by vertically launched ball because they have the same initial speeds in the vertical direction.

. The ball launched at this angle reaches the same height as the

Part E
In the previous part, you found that a ball fired with an initial speed of 25 speed of 20
. Which ball takes longer to land?

and an angle of

reaches the same height as a ball fired vertically with an initial

ANSWER:
Both balls are in the air the same amount of time.
The ball fired at an angle of

stays in the air longer.

The ball fired vertically stays in the air longer.

Correct
The vertical component of the velocity determines how long the ball will be in the air (and its maximum height). The horizontal component of the ball's velocity does not affect this hang time.

Part F
The figure shows two trajectories, made by two balls launched with different angles and possibly different initial speeds.

Based on the figure, for which trajectory was the ball in the air for the greatest amount of time?

Hint 1. How to approach the problem
Think about the result of Part E (think about the relationship between the maximum height of something thrown upward and the amount of time it is in the air).
Does the time in the air depend on the range of the trajectory?
ANSWER:
Trajectory B
It’s impossible to tell solely based on the figure.
Trajectory A
The balls are in the air for the same amount of time.

Correct

All that matters is the vertical height of the trajectory, which is based on the component of the initial velocity in the vertical direction ( trajectory, the more time the ball will be in the air, regardless of the ball’s range or horizontal velocity.

). The higher the

Part G
The range is the distance from the cannon when the ball hits the ground. This distance is given by the horizontal velocity (which is constant) times the amount of time the ball is in the air (which is determined by the vertical component of the initial velocity, as you just discovered).
Set the initial speed to 20
, and fire the ball several times while varying the angle between the cannon and the horizontal. Notice that the digital display near the top gives the range of the ball.

For which angle is the range a maximum (with the initial speed held constant)?
ANSWER:

Correct
When the ball is launched near a level ground, is the optimum angle. If launched with a greater angle, it stays in the air longer, but its horizontal speed is slower, and it won’t go as far. If launched with a smaller angle, its horizontal speed is faster, but it won’t stay in the air as long and it won’t go as far. The product between the horizontal speed and the amount of time in the air is largest when the angle is
.

Part H
How does the range of the object change if its initial velocity is doubled (keeping the angle fixed and less than

)?

ANSWER:
The ball’s range is four times as far.
The ball’s range is eight times as far.
The ball’s range is twice as far.

Correct
Since the vertical component of the velocity is twice as large, it takes twice as long to hit the ground. The horizontal component of the velocity is also twice as large, and since the range is equal to the horizontal velocity times the amount of time the ball is in the air, the range increases by a factor of four. The results of this question and the previous question can be summarized by the range equation, which is
.

Part I
Now, let’s see what happens when the cannon is high above the ground. Click on the wheel of the cannon, and drag it upward as far as it goes (about 21 ground). Set the initial velocity to 20 , and fire several balls while varying the angle.
For what angle is the range the greatest?

above the

ANSWER:

Correct
Since the cannon is very high off the ground, the ball will be in the air for an appreciable amount of time even if the ball is launched nearly horizontally. Thus, the amount of time the ball is in the air isn’t proportional to the vertical component of the initial velocity (as it was when the cannon was on the ground). This means that the initial horizontal velocity is more important, resulting in an optimal angle less than
. You should realize that the range equation given in Part H,
,
is not valid when the initial height is not zero. You can also verify that, if you change the initial velocity, the optimal angle also changes!

Part J
So far in this tutorial, you have been launching a baseball. Let’s see what happens to the trajectory if you launch something bigger and heavier, like a Buick car.
Compare the trajectory and range of the baseball to that of the Buick car, using the same initial speed and angle (e.g.,
). (Be sure that air resistance is still turned off.) Which statement is true?
ANSWER:
The trajectories and thus the range of the Buick and the baseball are identical.
The trajectories differ; the range of the Buick is shorter than that of the baseball.
The trajectories differ; the range of the Buick is longer than that of the baseball.

Correct
Since we are ignoring air resistance, the trajectory of the object does not depend on its mass or size. In the next part, you will turn on air resistance and discover what changes.

Part K
In the previous part, you discovered that the trajectory of an object does not depend on the object’s size or mass. But if you have ever seen a parachutist or a feather falling, you know this isn’t really true. That is because we have been neglecting air resistance, and we will now study its effects here.
Select Air Resistance for the simulation. Fire a baseball with an initial speed of roughly 20 and an angle of
. Compare the trajectory to the case without air

resistance. How do the trajectories differ?
ANSWER:
The trajectory with air resistance has a longer range.
The trajectory with air resistance has a shorter range.
The trajectories are identical.

Correct
Air resistance is a force due to the object ramming through the air molecules, and is always in the opposite direction to the object’s velocity. This means the air resistance force will slow the object down, resulting in a shorter range (the simulation assumes the air is still; there is no strong tailwind).

Part L
Notice that you can adjust the diameter (and mass) of any object (e.g., you can make a really big baseball). What happens to the trajectory (with air resistance on) when you increase the diameter while keeping the mass constant?
ANSWER:
Increasing the size makes the range of the trajectory increase.
Increasing the size makes the range of the trajectory decrease.
The size of the object doesn’t affect the trajectory.

Correct
Since the surface area increases if the diameter increases, the object is sweeping through more air, causing more collisions, and a greater force of air drag (in fact, if the diameter is doubled, for a given speed, the force of air drag is increased by a factor of four). This greater force of air drag causes the object to slow down more quickly, resulting in a slower average speed and a shorter range.

Part M

You might think that it is never a good approximation to ignore air resistance. However, often it is. Fire the baseball without air resistance, and then fire it with air resistance (same angle and initial speed). Then, adjust the mass of the baseball (increase it and decrease it) and see what happens to the trajectory. Don’t change the diameter. When does the range with air resistance approach the range without air resistance?
ANSWER:

The range with air resistance approaches the range without air resistance as the mass of the baseball is decreased.
The range with air resistance approaches the range without air resistance as the mass of the baseball is increased.
It never does. Regardless of the mass, the range with air resistance is always shorter than the range without.

Correct
As the mass is increased, the force of gravity on the baseball becomes larger. The force due to air drag just depends on the speed and the size of the object, so it doesn’t change if the mass changes. As the mass gets large enough, the force of gravity becomes much larger than the air drag force, and so the air drag force becomes negligible. This results in a trajectory nearly the same as when air resistance is turned off. Thus, for small, dense objects (like rocks and bowling balls), air resistance is typically unimportant, but for objects with a low density (like feathers) or a very large surface area (like parachutists), air resistance is very important. Score Summary:
Your score on this assignment is 98.8%.
You received 24.69 out of a possible total of 25 points.

PhET Interactive Simulations
University of Colorado http://phet.colorado.edu…...

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